geometry
3D geometry
geometric transformation
reflection
videos

[quote="jpark"]I finally got a Rubik's Cube today!!! :D  :D

I never knew this was so addicting..[/quote]

You rock! Have you yet stunned your classmates by solving it with eyes closed?

I just read over an old discussion about what the IMO team should pull off during the three minutes on the ceremony. If you get on the team, this year the trick might as well be all 6 team members standing in a row on stage solving Rubic's cubes, and finishing at the same time. Optional rider: passing the cubes to each other every 10 seconds or so while solving them.

Anyway, have fun with the cube! Once you master it, you should try solving a hypercube sometime.

Wow Vika..

You finally joined!!
Anyways, I can solve a cube every 10 minutes now if I screw up once or less!! I'm getting better.. I'll catch up to David soon..  :D  :D

[quote="jpark"] I'll catch up to David soon..  :D  [/quote]
 :o

[quote="jpark"]
Anyways, I can solve a cube every 10 minutes now if I screw up once or less!! I'm getting better.. I'll catch up to David soon..  :D  :D[/quote]

Lol...Good luck with THAT!

And Vic~good to see you here.

Jenn, you can have my previous avatar(the Rubik's cube one).
You can also have my previous signature too: There are 43252003274489856000 different positions for Rubik's cube! :D
____________________________________________________________________
EDITED: I just found out that the world record for single Rubik's 3x3x3 Cube is broken!!!
It used to be 12.11 seconds, but it's now 11.75 seconds. :o

[quote="YawningLion"][quote="jpark"]Wow Vika..

You finally joined!!
[/quote]
Yep, I finally made it here - it feels like being in Winter camp again! Besides, upon rereading some old posts, I found out that someone in this world - other than me - likes climbing trees (I think it was Sarah). Mooo (se)!

Question: how do you guys get to include your favourite theorem/conjecture with each of your messages?  I would put the Continuum Hypothesis if I could.

Jenn, when do you prognose to be ready for the 4x4x4 Rubic's cube? Praying to the Moose at regular intervals (preferably stopwatch-measured) will guarantee success. Good luck![/quote]

You like climbing treez~!~!~!~!~!

Haha I guess you weren't there that night when ppl were deciding what to do, then i was like *really loud* "LET'S GO CLIMB A TREE!!!!!!"

[quote="YawningLion"]
Yep, I finally made it here - it feels like being in Winter camp again! Besides, upon rereading some old posts, I found out that someone in this world - other than me - likes climbing trees (I think it was Sarah). Mooo (se)! [/quote]

 :D

I climbed that tree too!  It was a nice tree.  That was a fun night too (the night after we gave Alex his midnight concert, eh?)  Fun times.  I'll definitely be showing up this year at seminar (just to see some of you that I haven't seen in ages!)  Good luck this contest season!

[quote="lightrhee"]Jenn, you can have my previous avatar(the Rubik's cube one).
You can also have my previous signature too: There are 43252003274489856000 different positions for Rubik's cube! :D[/quote]

hahaha, somehow, I don't feel that I deserve those yet..

When I get down to 3 minutes-ish, maybe I'll take them from you.. :P
Thanks David!

[quote="YawningLion"][quote="TheDreamer"] You like climbing treez~!~!~!~!~!

Haha I guess you weren't there that night when ppl were deciding what to do, then i was like *really loud* "LET'S GO CLIMB A TREE!!!!!!"[/quote]

I wasn't there on the last night - did you guys actually climb anything (given that it was probably pitch dark)? On the good side, when climbing in the dark there is no height fright, since you don't see how high up you are. We should have a monumental tree-climbing expedition on the next Seminar.[/quote]

Haha, that will give me motivation to try to make seminar!

But lol, I don't think it's a good thing if you don't know how high up you are T.T

[quote="YawningLion"]

I wasn't there on the last night - did you guys actually climb anything (given that it was probably pitch dark)? On the good side, when climbing in the dark there is no height fright, since you don't see how high up you are. We should have a monumental tree-climbing expedition on the next Seminar.[/quote]

Going up was fine.  I think I slipped down. heh, I swore there used to be a branch where I put my foot.  I think yufei has pics of the tree climbing.

[quote="Sunny"]
Going up was fine.  I think I slipped down. heh, I swore there used to be a branch where I put my foot.  I think yufei has pics of the tree climbing.[/quote]

Were you actually climbing in the dark? - if yes, the pictures probably look quite mysterious   :ninja: . The funnest thing is when a crucial branch lets you climb up, but on your way down decides it doesn't like you, and treacherously breaks. Then comes the time to unleash your parachute - or your flying skills. I once had to stretch like a ballerina (never thought I was so flexible ...) to avoid spending the rest of my life on that tree and joining the squirrel community.

[quote="lightrhee"][/quote][quote="lowest_reynolds"][quote="Elyot"][quote="YawningLion"]...that one puzzle that looks like a spiral, but is really just a bunch of circles.[/quote]

Do you, by chance, mean Fraser's spiral?

[img]http://www.psy.ritsumei.ac.jp/~akitaoka/FraSpr01.gif[/img]

:lol:[/quote]

I opened this page for ONE SECOND and then I get blown away by this huge spiral thingy...

These things boggle my mind.  I swear you Canadians are going to give someone a disease or some mental illness or something... :rotfl:[/quote][quote="lightrhee"]
 :rotfl:[/quote]
Whoa, that's a cool pic. :o
I just followed the path of the second largest circle...*headache* :!:

 :idea: BTW, are we QUOTING again?  :rotfl:

[quote="lowest_reynolds"]These things boggle my mind.  I swear you Canadians are going to give someone a disease or some mental illness or something... :rotfl:[/quote]

We try.   :D

There's more illness-bearing images on the same site.

http://www.psy.ritsumei.ac.jp/~akitaoka/

Now they spin, and now they don't!  :S

NOW I'm suffering from a severe case of migraine...(Especially since I'm looking at those now at 11pm...)
Thanks for the link though.
Just a warning: really be careful when you click on that...It is bad for mental health. :!:  :noo:

[quote="YY"]NOW I'm suffering from a severe case of migraine...(Especially since I'm looking at those now at 11pm...)
Just a warning: really be careful when you click on that...It is bad for mental health. :!:  :noo:[/quote]

But these eyeboggling images are good for training your eyes to look at anything unbearable - such as homework at 3 am  ;). I hope you still managed to fall asleep after gazing at them ... Don't worry, soon your eyes will grow some armour and get used to this kind of challenge.

Fraser's spiral is cool - especially if you try to play something like chess on its background. Only it'll be an infinite chessboard inside a cone, and its squares converge to the vertex. Someone should make up the rules for these "Fraser chess".

:arrow: Edit: After looking at the spiral again, I realized it's probably not on a cone but on the inside of a cylindrical 'tunnel'. Actually, it would still be the same game of 'infinite chess' as the game on the inside of a cone.

P.S. Is this the first historical occurrence of the usage of the :arrow: emoticon?

[quote="Elyot"][quote="YawningLion"]
Fraser's spiral is cool - especially if you try to play something like chess on its background. Only it'll be an infinite chessboard inside a cone, and its squares converge to the vertex. Someone should make up the rules for these "Fraser chess".[/quote]

[img]http://i44.photobucket.com/albums/f28/passisasdf/spiralchess.gif[/img]

Couldn't resist a photoshop opportunity.  :lol:   :lol:

By the way... it actually is checkmate![/quote]

Hahahaha.   :rotfl:

What the frick, you even put a caption on the black king!

That pic's the funniest thing I've seen this week.

[quote="british8985"]but I just REMEMBERED days before that
HE IS COLOURBLIND
how can he do that??[/quote]

Who is too colourblind to do what? Fraser or your math teacher? And what could he play well?

Mind-boggled,

[quote="YawningLion"]I wonder whether they sell these ...[/quote]

Oooh you could so go into business making crazy math games.  Make a fortune selling Fraser Chess to math and chess people everywhere.  lol

[quote="YawningLion"][size=150]Fraser Chess Rules![/size]
('rules' can be either a verb or a noun)

Elyot, in your Checkmate situation in Fraser chess, is the white king being checked by the black queen that hangs upside down? I initially mistook the bishops for pawns, and wondered how the checkmate works out ... It's so much harder to catch a king here, on an infinite chessboard with no edges/corners to bully the king in, so congrats to your black figures!

We could make an additional rule regarding the black and white stripes that make up the circles. Besides bishops being able to follow these circles, we could allow any figure to follow a stripe of its colour, which would involve jumping two squares along the circle. This will add even more chaos to the game!

And how will we station the figures at the beginning? Could be in rows along two circles, but that would be equivalent to stationing them along diagonals on a normal board, which changes everything. Or we could place the figures in rows pointing at the vertex on opposite sides of the cone. Or let's just appeal for the services of a random number generator.

How would we define the 'edge of the board' where a pawn can become a queen? Maybe Fraser chess is just way too symmetric to have this rule ...

And it will be possible to build a chunk of an actual Fraser board - a cone with the vertex cut off, to which magnetized figures can stick. Then the edges would be defined, but the board would just be a variation of the normal board curled around and glued to itself. That'll still be so interesting to play, actually, and all the normal rules hold (such as the pawn to queen rule). I wonder whether they sell these ...[/quote]

You know the joke's over when someone's taken the thread seriously.

[quote="Singular"]
You know the joke's over when someone's taken the thread seriously.[/quote]

 :D Well, I'm crazy, so taking non-serious things seriously is my hobby. If given the time and resources, I can go on a whole rant about Fraser chess, make up a bunch of mutually inconsistent rules, and eventually write a novel where the figures in Fraser chess learn to moo. Can someone fund me please?

P.S. I don't take many things in life seriously, actually. Life is too amusing.

[quote="YawningLion"]
Elyot, in your Checkmate situation in Fraser chess, is the white king being checked by the black queen that hangs upside down?
[/quote]

Yes.

[quote="YawningLion"]
And it will be possible to build a chunk of an actual Fraser board - a cone with the vertex cut off, to which magnetized figures can stick. Then the edges would be defined, but the board would just be a variation of the normal board curled around and glued to itself. That'll still be so interesting to play, actually, and all the normal rules hold (such as the pawn to queen rule). I wonder whether they sell these ...[/quote]

erm... have fun with that.  I think go would work better than chess.

[quote="YawningLion"]
- It is technically imaginable to fall from a great height, but survive by grabbing onto branches on the way and slowing yourself down. I haven't tried, though. I should when I start seeking a creative and exciting method of suicide some 100 years from now.  [/quote]

Ah yes, very possible.

http://video.google.com/videoplay?docid=-5704371950004408401&q=parkour

The particular segment starts at about 2:50.

Wow... that's one of the funniest and best edited parkour/traceur videos I've seen in a long time!  Thanks for the share!


If  $x,y,z>0$  prove that  $\sum\sqrt{\frac{xy}{(x+z)(y+z)}}\geq\frac{3}{2}$
Thank  you !

sign reversed!

$\sum\sqrt{\frac{xy}{(x+z)(y+z)}}\leq\frac{3}{2}$

Note that $\sqrt{\frac{xy}{(x+z)(y+z)}}\leq\frac{1}{2}(\frac{x}{x+y}+\frac{y}{y+z})$.........

Do you see that??
 :)


Hi,  I have been struggling with the following problem:

1/(4 x 9) + 1/(9 x 14) + 1/(14 x 19) + ... + 1/(1999 x 2004)

Any ideas?

Cool :).  And thanks!

So it will reduce the problem to 1/5 x (1/4 - 1999/2004) = 25 / 501


modular arithmetic
quadratics
induction
function
number theory unsolved
number theory

Problem: Find all positive integer $n$ and prime number $p$ such that :
Any $a_1,a_2,...,a_n\in\{1,2,...,p-1\}$ we have $\sum_{k=1}^na_k^2\not\equiv0\pmod{p}$

Repaired ....!

I think, something wong here. Since for each $n\geq 2$ we can find $p\mid n$, so $\sum_{k=1}^n1^2\equiv0\pmod{p}$.

I think that $n<3$ and $p=4k+3$.

Let's first show that for $n\ge 3$, all residues $\pmod p$ can be written as $\sum_{i=1}^k a_i^2$. Let $A$ be the set of quadratic residues $\pmod p$. The Cauchy-Davenport Theorem states that (we're working in $\mathbb Z_p$ here) $|A+A|\ge \min(p,2|A|-1)=2|A|-1=p-2$. From here we can show by induction on $n$ that $|A+A+A+\ldots+A|\ge p$ where there are at least $3$ $A$'s in that sum of sets. In particular, $0\in A+A+A+\ldots+A$.

On the other hand, it's well-known that $-1$ is a quadratic residue of $p$ iff $p=4k+1$ (this deals with the case $n=2$). Of course, $n=1$ always works.

Is there something wrong with this?

P.S. Could you please elaborate on that "p-adic analysis needs it thing"? What's the connection with p-adic analysis?

Edit: I only considered $p>2$. For $p=2$ $n$ has to be odd in order to satisfy the requirement $\sum_{i=1}^n a_i^2\ne 0,\ \forall a_i\in\{1\}$ :D.

I forgot The Cauchy-Davenport Theorem :(
Anyway, everyone can look it here http://www.mathlinks.ro/Forum/viewtopic.php?t=19496 .

thanks for very quickly solution, but it seems that there is somthing wrong in here: Take $p=3$ and $(n,3)=1$ then any $a=1$ or $2$ we have $a^2\equiv1\pmod{3}$
so
\[a_1^2+a_2^2+\cdots+a_n^2\equiv n\not\equiv0\pmod{3}\]

Ps: We are reading a paper on Green p_adic functions in $Q_p^2$ ( 2_dimision p_adic space). From his proof we think we could extend his method in the general case , which is very important. That's why i am wondering why he (the author of the paper, he lived in Russian) did not study it? You can imagine as following. In the expression definition of Green function has a fraction which the denometer (a/b ==> b= denometer) has for
$|x|_p^2=\sum_{k=1}^nx_k^2$ so if it is zero then there is sigular at $0$ and the problem becomes very hard!

Yes, that's because of that $p-2$ in the proof. It's too small when $p=3$. I think we should also treat the case $p=3$ manually, and I hope it's Ok for $p\ge 5$.. :) Any more counterexamples? :)

oh thanks so much and  no more  :D , just a question that is where is the hat of my Mr Green :D


function
inequalities
inequalities proposed

Let $ a,b,c$ be nonnegative real numbers. Prove that
\[ \sum \frac{a^2}{5a^2\plus{}(b\plus{}c)^2} \le \frac{1}{3}.\]
:)
Equality holds for $ a\equal{}b\equal{}c$ or $ a\equal{}b,c\equal{}0$.

Hope you like it!

[quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers. Prove that
\[ \sum \frac {a^2}{5a^2 \plus{} (b \plus{} c)^2} \le \frac {1}{3}.
\]
:)
Equality holds for $ a \equal{} b \equal{} c$ or $ a \equal{} b,c \equal{} 0$.

Hope you like it![/quote]
Is there a solution without SOS? :)
VasC, what is your solution?

[quote="Vasc"][quote="can_hang2007"] Is there a solution without SOS? :)[/quote]
There is a simple solution using RCF-Theorem for $ a \plus{} b \plus{} c \equal{} 1$, since the function $ f(x) \equal{} \frac { \minus{} x^2}{5x^2 \plus{} (1 \minus{} x)^2}$ is convex for $ \frac 1{3} \le x < 1$. Therefore, it suffices to prove the original inequality for $ b \equal{} c \equal{} 1$.[/quote]
Thank you, VasC.
The first solution of mine is the following
Rewrite the inequality as
\[ (\sum a^2)(\sum \frac{1}{5a^2\plus{}(b\plus{}c)^2}) \plus{}2\sum \frac{bc}{5a^2\plus{}(b\plus{}c)^2} \ge \frac{5}{3}\]
Now, by the Cauchy Schwarz, we have
\[ \sum \frac{1}{5a^2\plus{}(b\plus{}c)^2} \ge \frac{49(\sum a)^2}{\sum (a\plus{}3b\plus{}3c)^2(5a^2\plus{}(b\plus{}c)^2)}\]
\[ \frac{bc}{5a^2\plus{}(b\plus{}c)^2}  \ge \frac{(\sum bc)^2}{\sum bc(5a^2\plus{}(b\plus{}c)^2)}\equal{}\frac{\sum bc}{(\sum a)^2}\]
And finally, we have to prove
\[ \frac{49(\sum a^2)(\sum a)^2}{\sum (a\plus{}3b\plus{}3c)^2(5a^2\plus{}(b\plus{}c)^2)}\plus{}\frac{2\sum bc}{(\sum a)^2} \ge \frac{5}{3}\]
It is easy by Schur Inequality. :)

[hide]
PS: I named this techiques as CYH, it means "Cẩn Yêu Hằng" (in Vietnamese) and "Can Love Hang" (in English)... I want everybody to know that I love Hang very much. [/hide]

[quote="can_hang2007"][quote="Vasc"][quote="can_hang2007"] Is there a solution without SOS? :)[/quote]
There is a simple solution using RCF-Theorem for $ a \plus{} b \plus{} c \equal{} 1$, since the function $ f(x) \equal{} \frac { \minus{} x^2}{5x^2 \plus{} (1 \minus{} x)^2}$ is convex for $ \frac 1{3} \le x < 1$. Therefore, it suffices to prove the original inequality for $ b \equal{} c \equal{} 1$.[/quote]
Thank you, VasC.
The first solution of mine is the following
Rewrite the inequality as
\[ (\sum a^2)(\sum \frac {1}{5a^2 \plus{} (b \plus{} c)^2}) \plus{} 2\sum \frac {bc}{5a^2 \plus{} (b \plus{} c)^2} \ge \frac {5}{3}
\]
Now, by the Cauchy Schwarz, we have
\[ \sum \frac {1}{5a^2 \plus{} (b \plus{} c)^2} \ge \frac {49(\sum a)^2}{\sum (a \plus{} 3b \plus{} 3c)^2(5a^2 \plus{} (b \plus{} c)^2)}
\]

\[ \frac {bc}{5a^2 \plus{} (b \plus{} c)^2} \ge \frac {(\sum bc)^2}{\sum bc(5a^2 \plus{} (b \plus{} c)^2)} \equal{} \frac {\sum bc}{(\sum a)^2}
\]
And finally, we have to prove
\[ \frac {49(\sum a^2)(\sum a)^2}{\sum (a \plus{} 3b \plus{} 3c)^2(5a^2 \plus{} (b \plus{} c)^2)} \plus{} \frac {2\sum bc}{(\sum a)^2} \ge \frac {5}{3}
\]
It is easy by Schur Inequality. :)

[hide]
PS: I named this techiques as CYH, it means "Cẩn Yêu Hằng" (in Vietnamese) and "Can Love Hang" (in English)... I want everybody to know that I love Hang very much. [/hide][/quote]

Nice proof,VQBC  :)

[quote=luofangxiang][img]http://s2.sinaimg.cn/middle/0018zOAxgy71yk1nIhbb1&690[/img][/quote]

luofangxiang,You are good at proof...,is good!

BQ

[b]Also Make Sense[/b]
Let $a,b,c$ be nonnegative real numbers. Prove that
$$\sum \frac{a}{7a^2+5(b+c)^2} \le \frac{1}{3(a+b+c)}.$$

[quote=Grotex]Let $a,b,c$ be nonnegative real numbers. Prove that
$$\sum \frac{a^2}{2a^2+(b+c)^2} \ge \frac{1}{2}.$$[/quote]
It very easy!
$$\sum \frac{a^2}{2a^2+(b+c)^2} \ge \sum \frac{a^2}{2a^2+2b^2+2c^2}= \frac{1}{2}.$$

[quote=dm_edogawasonichi][quote=Grotex]Let $a,b,c$ be nonnegative real numbers. Prove that
$$\sum \frac{a^2}{a^2+2(b+c)^2} \ge \frac{1}{a+b+c}.$$[/quote]
Try $a=b=c=1/3$  ;)[/quote]

Oh iot is just a typo.
I will edit it.

[quote=Grotex]Let $a,b,c$ be nonnegative real numbers. Prove that
$$\sum \frac{a}{a^2+2(b+c)^2} \ge \frac{1}{a+b+c}.$$[/quote]

Cauchy inequality
 :-D

the original inequality can be improved further, but then counter CS can't work then

@above: Mind post your improved version?


calculus
integration
real analysis
real analysis unsolved

Calculate integral $ \int_{0}^{\plus{}\infty}{\frac{e^{\minus{}x^2}}{\left(x^2\plus{}\frac{1}{2}\right)^2}\; dx}$.

hello, your integral $ \int_{0}^{\plus{}\infty}{\frac{e^{\minus{}x^2}}{\left(x^2\plus{}\frac{1}{2}\right)^2}\; dx}\equal{}\sqrt{\pi}$.
Sonnhard.

[quote="sos440"]Is there an elegant way to obtain this result? My solution seems far from obvious...
[/quote]

Instead of converting to a double integral, we can directly go with

$ I \equal{} \minus{}\int_{0}^{\infty} \frac{e^{\minus{}x^2}}{2x}\frac{d}{dx}\frac{1}{x^2\plus{}1/2}$ and then integrate by parts which gives a similar cancellation to the one you have. But here the boundary terms will diverge and we have to check that they cancel. So I will not call it elegant.

[quote="Dr Sonnhard Graubner"]hello, your integral $ \int_{0}^{ \plus{} \infty}{\frac {e^{ \minus{} x^2}}{\left(x^2 \plus{} \frac {1}{2}\right)^2}\; dx} \equal{} \sqrt {\pi}$.
Sonnhard.[/quote]

Sonnhard, why you don't post solution? Only answer.


superior algebra
superior algebra solved

Show that if n is odd then the set of all n-cycle cosists of two conjugacy classes of equal size in An

Same topic as this one:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=58651[/url]


Let $ a,b,c,d$ be nonnegative real numbers. Prove that
\[ \frac{ab}{a\plus{}b\plus{}c}\plus{}\frac{bc}{b\plus{}c\plus{}d}\plus{}\frac{cd}{c\plus{}d\plus{}a}\plus{}\frac{da}{d\plus{}a\plus{}b} \le \frac{1}{3}(a\plus{}b\plus{}c\plus{}d).\]
Equality holds if $ a\equal{}b\equal{}c\equal{}d$ :)

PS: We can prove it by Cauchy Schwarz Inequality. :)

[quote="can_hang2007"]Let $ a,b,c,d$ be nonnegative real numbers. Prove that
\[ \frac {ab}{a \plus{} b \plus{} c} \plus{} \frac {bc}{b \plus{} c \plus{} d} \plus{} \frac {cd}{c \plus{} d \plus{} a} \plus{} \frac {da}{d \plus{} a \plus{} b} \le \frac {1}{3}(a \plus{} b \plus{} c \plus{} d).
\]
Equality holds if $ a \equal{} b \equal{} c \equal{} d$ :)

PS: We can prove it by Cauchy Schwarz Inequality. :)[/quote]
$ \frac{ab}{a\plus{}b\plus{}c}  \leq \frac{ab}{9a}\plus{}\frac{ab}{9b} \plus{}\frac{ab}{9c}$
hence :  we  need  prove that :
$ \frac{ab}{c} \plus{} \frac{bc}{d} \plus{}\frac{cd}{a} \plus{}\frac{da}{b} \leq a\plus{}b\plus{}c \plus{}d$

[quote="Hong Quy"]
hence :  we  need  prove that :
$ \frac {ab}{c} \plus{} \frac {bc}{d} \plus{} \frac {cd}{a} \plus{} \frac {da}{b} \leq a \plus{} b \plus{} c \plus{} d$[/quote]
Which wrong. :wink:

[quote="can_hang2007"]
PS: We can prove it by Cauchy Schwarz Inequality. :)[/quote]

How to use Cauchy Schwarz Inequality？

But I still don't know how to use Cauchy Schwarz Inequality？ :(

Can you post your solution?
Thank you very much!

[quote="can_hang2007"]Let $ a,b,c,d$ be nonnegative real numbers. Prove that
\[ \frac {ab}{a \plus{} b \plus{} c} \plus{} \frac {bc}{b \plus{} c \plus{} d} \plus{} \frac {cd}{c \plus{} d \plus{} a} \plus{} \frac {da}{d \plus{} a \plus{} b} \le \frac {1}{3}(a \plus{} b \plus{} c \plus{} d).
\]
Equality holds if $ a \equal{} b \equal{} c \equal{} d$ :)

PS: We can prove it by Cauchy Schwarz Inequality. :)[/quote]

Now can we see a proof?  :)

[quote="can_hang2007"]Let $ a,b,c,d$ be nonnegative real numbers. Prove that
\[ \frac {ab}{a \plus{} b \plus{} c} \plus{} \frac {bc}{b \plus{} c \plus{} d} \plus{} \frac {cd}{c \plus{} d \plus{} a} \plus{} \frac {da}{d \plus{} a \plus{} b} \le \frac {1}{3}(a \plus{} b \plus{} c \plus{} d).
\]
Equality holds if $ a \equal{} b \equal{} c \equal{} d$ :)

PS: We can prove it by Cauchy Schwarz Inequality. :)[/quote]

Where can I find the proof?

[quote="can_hang2007"]for mitdac123[/quote]
Where r u mitdac?? We r waiting to see the solution! :P

[quote="Inequalities Master"]OK, can_hang2007.Give us a proof, please.[/quote]
This is my present for mitdac123, my younger brother. So I still wait for his answer.

Hello can_hang2007! I have posted an approach previously. Will u please modify it by giving me some clue!
I failed to reach the $ \frac{1}{3}$ factor. :(

[quote="Sunkern_sunflora"]It doesn't seem a very easy problem. So far I could prove is that $ L.H.S.\leq{a \plus{} b \plus{} c \plus{} d}$, but I didn't obtain the $ \frac {1}{3}$ factor.

Here's my approach;

$ \sum_{cyc} \frac {ab}{a \plus{} b \plus{} c}\leq {\sum_{cyc} \frac {ab}{a \plus{} b}}$

$ \Rightarrow\sum_{cyc} \frac {ab}{a \plus{} b} \equal{} \frac {1}{2}(\sum_{cyc} \frac {2ab}{a \plus{} b})$

$ \Rightarrow\sum_{cyc} \frac {ab}{a \plus{} b}\leq \frac {1}{2}(\sum_{cyc} \frac {a^2 \plus{} b^2}{a \plus{} b})\leq \frac {1}{2}(\sum_{cyc} {(a \plus{} b)}) \equal{} a \plus{} b \plus{} c \plus{} d$.

Still I didn't get where we are to use $ C.S.$.[/quote]
Dear sunkern,
I saw this approach of your, but I think the inequality
\[ \sum \frac{ab}{a\plus{}b\plus{}c} \le a\plus{}b\plus{}c\plus{}d\]
is trivial.
since
\[ \frac{ab}{a\plus{}b\plus{}c} \le \frac{ab}{b} \equal{}a\]

[quote="Sunkern_sunflora"]I know that it was a stupid proof, I am eager to know some hint! Plz, give me very, very little hint so that your brother couldn't guess from it! At least tell how to use the C.S.! PLZ.!!!  :maybe:[/quote]
My hint is taken notice that
\[ \frac{4ab}{a\plus{}b\plus{}c} \equal{} a\plus{}b \minus{}\frac{(a\minus{}b)^2\plus{}ac \plus{}bc}{a\plus{}b\plus{}c}\]
:)

yess I noticed that before :d , we can set :$ a\plus{}b\plus{}c\plus{}d\equal{}1$, use cauchy-shwarz ,and this very nice inequality $ 2\sqrt{ab} \plus{}  \left| a\minus{}b \right|  \geq a\plus{}b$ , am I right ??  :maybe:

[quote="can_hang2007"]Let $ a,b,c,d$ be nonnegative real numbers. Prove that
\[ \frac {ab}{a \plus{} b \plus{} c} \plus{} \frac {bc}{b \plus{} c \plus{} d} \plus{} \frac {cd}{c \plus{} d \plus{} a} \plus{} \frac {da}{d \plus{} a \plus{} b} \le \frac {1}{3}(a \plus{} b \plus{} c \plus{} d).
\]
Equality holds if $ a \equal{} b \equal{} c \equal{} d$ :)

PS: We can prove it by Cauchy Schwarz Inequality. :)[/quote]

We also have:

[quote="anas"]yess I noticed that before :d , we can set :$ a \plus{} b \plus{} c \plus{} d \equal{} 1$, use cauchy-shwarz ,and this very nice inequality $ 2\sqrt {ab} \plus{} \left| a \minus{} b \right| \geq a \plus{} b$ , am I right ??  :maybe:[/quote]
More details?

[quote="can_hang2007"]Let $ a,b,c,d$ be nonnegative real numbers. Prove that
\[ \frac {ab}{a \plus{} b \plus{} c} \plus{} \frac {bc}{b \plus{} c \plus{} d} \plus{} \frac {cd}{c \plus{} d \plus{} a} \plus{} \frac {da}{d \plus{} a \plus{} b} \le \frac {1}{3}(a \plus{} b \plus{} c \plus{} d).\]
Equality holds if $ a \equal{} b \equal{} c \equal{} d$ :)

PS: We can prove it by Cauchy Schwarz Inequality. :)[/quote]
See here: http://can-hang2007.blogspot.com/2010/01/inequality-22-v-q-b-can.html

My dear friend,
your proof is very nice but I don't understand why it must be valid only for $ k \geq \frac{2}{7}$ At first sight your proof seems to me valid for any $ k\geq 0$ but surely I am wrong :blush: but I don't understand why :blush: Very sorry

[quote="manlio"]My dear friend,
your proof is very nice but I don't understand why it must be valid only for $ k \geq \frac {2}{7}$ At first sight your proof seems to me valid for any $ k\geq 0$ but surely I am wrong :blush: but I don't understand why :blush: Very sorry[/quote]
My dear friend,

The inequality
$ \frac{ab}{a\plus{}kb\plus{}c}\plus{}\frac{bc}{b\plus{}kc\plus{}d}\plus{}\frac{cd}{c\plus{}kd\plus{}a}\plus{}\frac{da}{d\plus{}ka\plus{}b} \le \frac{a\plus{}b\plus{}c\plus{}d}{k\plus{}2}$
does not hold for all $ k \ge 0.$

For example, it is false for $ k\equal{}\frac{1}{5}$ at the point $ (a,b,c,d)\equal{}(1/2,1,0,0).$

I guess that this inequality holds for all $ k \ge \frac{1}{4}$ (a little remark to see that for $ k\equal{}1/4,$ we have equality for $ (a,b,c,d) \sim (1/2,1,0,0)$). I am still trying on it.

Thank you very much, my dear friend, now I found my mistake. Very sorry :blush:


geometry
geometric transformation
rotation
reflection
vector
trigonometry
algebra

A circle is rotation with constant angular speed. It is heading to a wall with constant velocity, forming an angle with the normal.
We know the coefficient of friction between the 2 objects, and know too that the collision is an elastic one.

What is:
the angle of reflection of the circle;
the final angular speed;
the final speed;
the energy loss due the friction.

I Was thinking, there is a way to solve this with hamilton? Or what is the simplest way to solve? Because I do not have many ideas, I was trying the hardest way.  :P  But there need to be a easy.

Do you have a diagram?  I can't really picture this scenario so it is hard for me to come up with a potential solution.

I will do better. As a attachement is an animation of the problem.

Note: The initial angular velocity in the animation is zero to be sure you will see the ball rotation after it hit the wall.

Note 2: There is no gravity. $ g\equal{}0$

Wait a second, how can there be energy loss due to friction if the collision is elastic?

I have a solution to the problem, but it assumes that no energy is lost.  It doesn't require Hamiltonian mechanics.

...or maybe the definition of an elastic collision is that only the linear kinetic energy is unchanged?  I don't remember.  But then why would the problem ask for the final speed?

In an elastic collision, kinetic energy is conserved.  Of course, other energy can be lost but it can't leave the system.

OK, so the reason I'm confused is that I'm considering the circular object and the wall to be a closed system; the "ideal" wall wouldn't move after the elastic collision, so I would assume that the SUM of the circular object's linear and rotational kinetic energies before and after the collision is constant.  This assumption allows one to come up with a system of equations involving impulse, angular impulse, etc. that allow you to solve the problem, but then I don't see why the question would ask for the energy lost due to friction.

So in sum, I am confused by the problem statement's usage of the term "elastic collision" - does this mean that ALL kinetic energy is conserved, including both linear and angular kinetic energy (which I assumed), or just linear kinetic energy?

Or on second thought, it could be possible to come up with a system of equations that only involve momentum/impulse and angular momentum/angular impulse without any energy considerations to solve the problem, in which case the "elastic" in the problem statement is unnecessary.  I remember solving a very similar problem before that way.

I just messed up with the definition of Elasticity in the problem. Sorry. I will now make a new statement without those assumptions that lead to a confusion.

There a Ball and a Wall with coefficient of friction know. There is  no gravity.

You hit the ball in the wall, with that, the ball wil have a NEW angular speed, a new vector velocity.

[b]The Energy is not necessary conserved[/b] (if will lose some part of the kinetic energy to the wall when it hit it, the friction with the wall will make the ball rotate, and for that will "consume" energy)

The problem is to find the Final attributes of the ball.

OK, then.  In that case, here's a start.  Call the final variables $ \theta'$, $ v'$, and $ \omega'$.

Let $ J_\perp$ and $ J_\parallel$ be the components of the impulse exerted on the circular object by the wall.  Then considering the change in momentum:
\[ J_\perp \equal{} m(v\cos\theta \plus{} v'\cos\theta')
\]

\[ J_\parallel \equal{} m(v\sin\theta \minus{} v'\sin\theta')
\]
Considering the change in angular momentum:
\[ I(\omega \minus{} \omega') \equal{} J_\parallel r
\]
where $ I$ is the moment of inertia, presumably $ \frac {1}{2}mr^2$.

Considering friction:
\[ J_\parallel \equal{} \mu J_\perp
\]
Unfortunately, that's only four equations in five unknown variables, so there is one more equation that I have yet to think of.  Which is kind of annoying because I feel like I've solved this exact problem before.

Edit: Note that if you assumed no kinetic energy was lost, we'd also have $ \frac {1}{2}mv^2 \plus{} \frac {1}{2}I\omega^2 \equal{} \frac {1}{2}mv'^2 \plus{} \frac {1}{2}I\omega'^2$, which would provide the fifth equation.  But I guess that doesn't make sense physically.

If there isn't sliding, the friction doesn't make any work...

If there is no friction, there is no rolling.  Simple as that.

I didn't say no friction, but no sliding.

No sliding implies $ x\equal{}r\theta$, $ v\equal{}r\omega$, and $ a\equal{}r\alpha$.


trigonometry
geometry
cyclic quadrilateral
trig identities
Law of Cosines

A cyclic quadrilateral is a quadrilateral that can be inscribed in a circle, such as a square or rectangle.

[quote="Yooper"]Given any four line segments that can form a quadrilateral, can they always be arranged to form a cyclic quadrilateral?  If yes, does the order of the four segments matter?

A HS student posed this to me, and none of us have made any progress.  My gut instinct is "Yes", but a proof?[/quote]

Is it just me, or does 4 line segments define a quadrilateral?

Then in that case, the answer would be no.

Another handy term to know is incyclic.

Incyclic quadrilaterals are quadrilaterals are are circumscribable about a circle. I believe a parallelgram is incyclic.

A cyclic quadrilateral has opposite angles adding up to 180 degrees. You should be able to prove it from there.

An incyclic quadrilateral has the sums of opposite sides equal.

4 segment lengths do not determine the quadrilateral (think of rhombi with side length x - they're not all the same).

Does anyone have a geometric answer to Yooper's question?

You can get there with the Law of Cosines.  Let our quad. be ABCD.  Let the side lengths be

a :ge: b :ge: c :ge: d

where sides with length a & b meet at C.  Let diagonal BD have length x.  If the quadrilateral is cyclic, the law of cosines on BCD and BAD gives:

x:^2: = a:^2: + b:^2: - 2ab cos C

and

x:^2: = c:^2: + d:^2: + 2cd cos C

(where cos A = -cos C since we have a cyclic quadrilateral)

If these two equations can be solved for a valid value of cos C, then we can build a cyclic quadrilateral.

Subtract the first from the second, solve for cos C:

cos C = (a:^2: + b:^2: - c:^2: - d:^2:)/(2ab + 2cd)

We wish to show that the mess on the right is always :le: 1.  It is always positive since a :ge: b :ge: c :ge: d.

Since a :le: b+c+d (since this is a quadrilateral, the three smallest sides must add up to a number at least as large as the largest side), we have

a-b :le: c+d

(a-b):^2: :le: (c+d):^2:

A little rearranging and we get to:

(a:^2: + b:^2: - c:^2: - d:^2:)/(2ab + 2cd) :le: 1

Thus, we can always solve for an angle C such that we can construct a quadrilateral with side lengths a,b,c,d that is cyclic.

I'm guessing there's a more elegant geometric approach...

Wait, I'm slow.

COuld you explain why the last step is valid?

And also, how do u prove congruent polygons then?

Given a, b, c, d, if I construct an triangle with side lengths a and b and included angle with measure C, where cos C is the expression I gave, the third side will have length x.  As my algebra showed, the triangle with sides c and d and included angle 180:^o: - C will also have third side with length x.  I glue these triangles together along the common side length, and I get a quadrilateral with sides a,b,c,d and opposite angles that are supplementary.

Not sure what you're asking about congruent polygons - you have to prove that all the sides & all the angles are equal.

[color=cyan]Well, you don't need all the sides and angles.  For example, if you had all the sides in order and one specifc angle, and you know if it's concave or convex, that makes a unique quadrilateral.  Or, 4 angles and 3 sides.  or, for that matter, 3 angles and 3 sides.  (Although you also have to know the order.)  There are lots of different sets of things you can get to know if it makes a unique quadrilateral -- but generally speaking, you need a lot more than for triangles.[/color]

I can only refer you to the excellent book Geometry Revisited (H. S. M. Coxeter and S. L. Greitzer) at this time, as I do not have it with me. There is a passage in that about "degrees of flexibility" in a quadrilateral, given it's four sides. This will probably be both the key to the elegant geometric method mentioned, and I know that it explicitly deals with whether the order of the sides matter (the second question in the first post).

Richard - I found your proof elegant in the design of the proof and in that its only primary tool was the Low of Cosines.

I was expecting the last line of the proof to be "Since angles A and C are supplementary, ABCD is cyclic. "  Therefore you caught me a off guard when you started the proof by stating that A and C are supplementary.  At first glance, it seemed that you were assuming what you were trying to prove.  But as you amplified in your follow-up, the key was that you were able to construct two triangles with supplementary angles A and C whose opposite sides were congruent of length x - hence they could be joined to formed the desired cyclic quadrilateral.  Nice!

Poincare - thanks for the "Geometry Revisited" reference.  Yes, both parts of my original problem are discussed in Section 3.2.  Interestingly, the above result was stated without proof, simply stating:  "The degree of freedom enables us to increase or decrease two opposite angles till they are supplementary..."

The second part of my original question, "Does the order of the four segments matter?", is handled very elegantly by above authors by twice dissecting and rearranging the polygon that Richard proved existed with the order a, b, c, and d.  Give it a try!  Hmmm, that reveals that all three distinct orderings of the four segments are cyclic on the SAME circle - nice!  By the way, it does take a moment to verify that there are only three distinct orderings - that's not 'obvious'.

Yes, I thought it would be in there. I believe they were trying to appeal more to your physical intuition in that section, but they still showed very elegantly the point about the order of the sides.

P.S. My name's not Poincare; that's my title, I think. I'm not quite sure.


So, a lot of us Bay Area people happen to know how to solve a Rubik's Cube, I think. Well I know for a fact that me, Eric, Matt, and Taylor can, and since Lynnelle knows about Macky I think she can to.

You can race on this website. It's a virtual cube applet, and you control the turns with keys. It might take some getting used to, tough.

Tell me if anyone is interested in racing, we can arrange a time :) .

http://www.ryanheise.com/cube/speed/

Looks great. After I'm done with this Calc practice I'll start trying it out. I might be able to speed cube faster this way ;-)

Oh yeah, I forgot to mention there's also a single player version, and they give you key controls there.

One thing that annoys me is the lack of B, B', u, u', f, and f'. It screws me up all over the place.

And on the top, there's a link called the Cube Database. Check it out, it's really awesome. People can enter algorithms for their various methods (They have Fridrich, Heise, Petrus, Roux (Although there's nothing in here, lol), and Zborowski-Bruchem). Each method has its own set of steps (For Fridrich they have F2L, OLL, and PLL) and they have full sets of algorithms. There's an awesome virtual cube that does the algorithm for you. The only problem is that it doesn't give the algorithm in notation.

They also have a blindfold cube version, I haven't tried that yet (I suck at blindfolded, my record is 3:56 set a few months ago, and now I can't even do it).

So yeah, check it out people.

wow
thats hard
i finished the second layer (yeah go bad methods)
then i screwed up

you're right, that thing really needs B, B', u, u', f, and f' :(

Whoa...I'm totally not used to the controls. *whips out real cube*

man these controls suck....
first try i got a time of 15 minutes 58 sec. wtf?

Not bad, I got 14 minutes something on my first try. Then I never got higher than 4 after that, except once when I forgot the algorithm and had to look it up, so it was 5:5x.

I think I average like 45 right now, my record is 23.88.

yay okay this is the first time i've actually done it and not messed up/had to go halfway through
which ive done about 5 times :D
and i got 3:28.81 :)
albert, you are SCARY :ninja:

EDIT:  3:25:41

wow, macky's 17th...
woulda expected higher

Well I don't think he uses this much... anyway, he's not that much into cubing anymore I think, cuz he needs to concentrate on school.

Dude I suck at that applet...

Does anyone want to arrange a time to race?

okay, but you'd pwn me anyways :P

w00t new record:  2:04:06 :)
considering how bad my method is and how much i suck at this applet, that's pretty good :D

ew this thing is hard
i like the non-virtual version


It's very hard i think. Nobody wants to challenge it??


How many ways can you change a $ \$5$ bill into a combination of dimes or quarters?

Dimes are worth $ \$0.10$ and quarters are worth $ \$0.25$. $ \text{LCM}(\$0.10, \$0.25) \equal{} \$0.50$. That means every $ \$0.50$ we have a combination of quarters and dimes. There are $ \$5\div\$0.50 \equal{} 10$. But our answer is not $ 10$! We have to take into account that there may be $ 0$ quarters and $ 50$ dimes (or $ 20$ quarters and $ 0$ dimes). So our final answer is $ 10 \plus{} 1 \equal{} \boxed{11}$.

The number of quarters must be even; otherwise the remaining amount to make $ \$5$ can't be comprised of just dimes. The case with the most quarters is if all of the coins used are quarters, which would be $ \frac {\$5}{\$0.25} \equal{} 20$ quarters. The case with the least quarters would be if all of the coins used were dimes, thus with $ 0$ quarters. Now, we just count cases in between and add them up, finding that there are $ \fbox{11}$ ways to make change for a $ \$5$ bill with only dimes and quarters.

EDIT: BEATEN


induction
search
combinatorics proposed
combinatorics

Let a convex polygonon with $n$ vertices with $n\geq 4$. We divide (at any way)  the polygonon at triangles, whose vertices are vertices of the polygonon,such that the triangles has no two common interior points. We paint with black the triangles, whose $2$  sides are sides of the polygonon and with red the triangles whose only $1$ side is side of the polygonon and with white the trianges whose sides are not sides of the polygonon. Prove that the black triangles are always, the thewhite triangles+2 triangles.

Please I want from the moderators to send all the problems to Olympiad recousrces.
I will put all the for problems from 2004

Consider the dual graph of the triangulation. This graph is a tree, and the problem can be rephrased as follows: given a tree with at least two vertices and with maximal degree $\le 3$, prove that the number of vertices with degree $3$ is equal to the number of triangles with degree $1$ minus $2$.

We can prove the assertion by induction: it's trivial for the smallest such tree, namely the tree with two vertices. Now assume we have shown this for all trees with $\le n-1$ vertices, and we have such a tree with $n\ge 3$ vertices. Eliminate a terminal vertex $x$ of the tree. This terminal vertex must have been connected to a vertex $y$ of degree $2$ or $3$.

If the degree of $y$ was $2$, then in the new tree, $y$ has degree $1$, meaning that the number of vertices of degree $1$ stays the same. Also, the number of vertices of degree $3$ stays the same, and the induction hypothesis gives the result.

If, on the other hand, the degree of $y$ was $3$, now it's $2$. Both the number of vertices of degree $1$ and the number of vertices of degree $3$ have decreased by $1$, and, again, we're done, by the induction hypothesis.

P.S.

You might also want to search for it, since I'm sure it's been posted before. Just search for words like "JBMO 2004", "polygon" (there's no such thing as a "polygonon" in English, by the way :)), "triangles", etc. I remember Pierre posting a solution.

[quote="grobber"]I remember Pierre posting a solution.[/quote]

I remember it too. And the official solution given just after was really nice and easy.

Pierre.

This problem was proposed by Calin Popescu. ;)


MIT college Harvard Princeton Columbia Stanford Duke

Do you need a PhD or Masters from a top univeristiy like MIT or Harvard to get a job as a quant? If yes, then suppose I don't get adimission into MIT or Harvard, then what univerisity should I apply to- what are some of the top universities besides MIT and Harvard?  :)

Harvard's got an awesome Math program, and their other sciences are pretty darn spiffy good too.

Why not try Caltech too, while you're at it.  I think Duke is getting more and more attention these days, aren't they?

p.s.
I was just at the Stanford Weekend, and at the math department thing this super cool freshman dude was doing a putnam problem for all the profro's.   :lol:
Actually, my host was a math major, and I think he said that Stanford was trying to get a bigger name for itself in Math.  Nice.

Harvard certainly has a very strong math program both for undergraduates and for graduate students.

[quote="Izzybee"]I might have the wrong idea, but as far as I know, Harvard is more famous for law and business related subjects. To study science in general (and therefore math) I'd recommed:

US:
Columbia University (New York)
Stanford University

International:
The University of Western Ontario, Canada.
The University of Waterloo, Canada
Max Planck Institute, Germany
The University of Varsovia, Poland
University of Barcelona, Spain[/quote]

UWO is a pretty mediocre school overall, although apparently their math program is decent.  Columbia is a nice school, but there are many, many more prestigious (and probably better) schools for sciences and math in North America.  I can't speak to the rest of your post, but it's fair to say that 'you have the wrong idea'.

[quote="orl"][quote="Izzybee"]

International:
...
Max Planck Institute, Germany
..
[/quote]

Those are different kinds of research institutions and it is not any university.[/quote]
On the contrary, I know that the MPI has at least math programs for MD and phd. And don't universities do research? ;)

[quote="blahblahblah"][quote="Izzybee"]I might have the wrong idea, but as far as I know, Harvard is more famous for law and business related subjects. To study science in general (and therefore math) I'd recommed:

...[/quote]

UWO is a pretty mediocre school overall, although apparently their math program is decent.  Columbia is a nice school, but there are many, many more prestigious (and probably better) schools for sciences and math in North America.  I can't speak to the rest of your post, but it's fair to say that 'you have the wrong idea'.[/quote]

Well, the thing is that the universities (and institute) I named are the ones I've heard are good  from my father, and researchers I've met: I've never seen math papers published by people who works a Harvard, but that may be because I've never met anyone who collaborates with researchers employed there.

Those are all great suggestions. But my original question was: What are the good universities to go to in order to become a [u][b]QUANT[/b][/u]?
I am sure that one can get a great math education from all the universities listed in those suggestions but are they good enough if one is trying to pursue a career as a quant- because you need to graduate from really really prestigious and well known university to become a quant?
You need MAGIC WORDS like 'MIT' or 'Harvard' on your resume, otherwise your employeer will throw your resume in garbage. Any suggestions on what are some other MAGIC WORDS besides MIT and Harvard. Thx.

As was suggested [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=47236]the last time this question was asked[/url], a good place to look is the [url=http://www.janestcapital.com/careers.html#recruiting-schedule]Jane Street recruiting calendar[/url].

Hey, you forgot Polyechnique school (l'X) and "Ecole Normale Supérieure" (E.N.S) in France  :P

From what my college advisor said, I'd think Cambridge is a great international one as well...


calculus
equation
rectangle
algebra
area
IMO Shortlist

Ali Barber, the carpet merchant, has a rectangular piece of carpet whose dimensions are unknown. Unfortunately, his tape measure is broken and he has no other measuring instruments. However, he finds that if he lays it flat on the floor of either of his storerooms, then each corner of the carpet touches a different wall of that room. He knows that the sides of the carpet are integral numbers of feet and that his two storerooms have the same (unknown) length, but widths of 38 feet and 50 feet respectively. What are the carpet dimensions?

[quote="amparvardi"]Sorry,anyone can help me solving this problem?
tanks.[/quote]
1) What are the rules for including a $ (x,y)$ carpet in a $ (L,a)$ room ... ?
==============================================
... such that each corner of the carpet is on a different wall.
Let $ \theta$ the angle between carpet side $ x$ and wall $ L$. We get :
$ L\equal{}x\cos(\theta)\plus{}y\sin(\theta)$
$ a\equal{}x\sin(\theta)\plus{}y\cos(\theta)$

So $ \cos(\theta)\equal{}\frac{xL\minus{}ay}{x^2\minus{}y^2}$ and $ \sin(\theta)\equal{}\frac{yL\minus{}ax}{y^2\minus{}x^2}$

And so the condition is $ (xL\minus{}ay)^2\plus{}(yL\minus{}ax)^2\equal{}(x^2\minus{}y^2)^2$

Which may be written $ \boxed{(x^2\plus{}y^2)(L^2\plus{}a^2)\minus{}4aLxy\equal{}(x^2\minus{}y^2)^2}$

3) resolution of the problem
==================
Let $ (x,y)$ be the dimension of the carpet (integers)
Let $ (L,a\equal{}50)$ and $ (L,b\equal{}38)$ the dimension of the two rooms. We get :

$ (x^2\plus{}y^2)(L^2\plus{}a^2)\minus{}4aLxy\equal{}(x^2\minus{}y^2)^2$
$ (x^2\plus{}y^2)(L^2\plus{}b^2)\minus{}4bLxy\equal{}(x^2\minus{}y^2)^2$

Subtracting, we get $ (x^2\plus{}y^2)(a\plus{}b)\equal{}4Lxy$ and so $ L\equal{}\frac{(x^2\plus{}y^2)(a\plus{}b)}{4xy}$

Plugging this in the first equation,we get $ (x^2\plus{}y^2)(\frac{(x^2\plus{}y^2)^2(a\plus{}b)^2}{16x^2y^2}\plus{}a^2)\minus{}(x^2\plus{}y^2)(a^2\plus{}ab)\equal{}(x^2\minus{}y^2)^2$

So $ (x^2\plus{}y^2)((x^2\plus{}y^2)^2(a\plus{}b)^2\minus{}16abx^2y^2)\equal{}16x^2y^2(x^2\minus{}y^2)^2$

So $ (x^2\plus{}y^2)((x^2\plus{}y^2)^2(a\plus{}b)^2\minus{}16abx^2y^2)\equal{}16x^2y^2((x^2\plus{}y^2)^2\minus{}4x^2y^2)$

Let $ x\equal{}pu$ and $ y\equal{}pv$ with $ \gcd(x,y)\equal{}p$ and so $ \gcd(u,v)\equal{}1$. The equation becomes :

$ (u^2\plus{}v^2)((u^2\plus{}v^2)^2(a\plus{}b)^2\minus{}16abu^2v^2)\equal{}16p^2u^2v^2((u^2\plus{}v^2)^2\minus{}4u^2v^2)$

So $ u^2v^2 | (u^2\plus{}v^2)^3(a\plus{}b)^2$ and so $ uv|a\plus{}b$ (remember $ \gcd(u,v)\equal{}1$)

Wlog say $ u> v$, we get $ uv|88$ and so $ (u,v)\in\{(2,1),(4,1),(8,1),$ $ (11,1),(22,1),(44,1),$ $ (88,1),(11,2),(11,4),(11,8)\}$

The equation $ (u^2\plus{}v^2)((u^2\plus{}v^2)^2(a\plus{}b)^2\minus{}16abu^2v^2)\equal{}16p^2u^2v^2((u^2\plus{}v^2)^2\minus{}4u^2v^2)$ allows to find $ p^2$ from $ (u,v)$ :

Applying  this to the $ 10$ couples we got, only one gives a perfect square : $ (2,1)$ gives $ p^2\equal{}625$

Hence the unique solution $ \boxed{(x,y)\equal{}(50,25)}$ which implies $ L\equal{}55$ (and so is an integer too, which was not a condition of the problem) and it is easy to check back that these numbers match the requirements.


geometry
3D geometry
sphere
algebra
function
domain
search

Admittedly, I am new to this forum and to combinatorics.  I have done some reading of sample combinatoric problems but don’t seem to find one that fits my case.  So, I am asking for some help and direction.

Here is my problem:
Consider a sentence of 10 words.  Of these 10 words, 6 have synonym replacements.  But the number of synonym replacements can vary for each of the 6 words.  Consider this list of variations:

Alpha character = word; Numeric value = synonym variations

A	3
B	4
C	2
D	1
E	2
F	5

The rule is that at least 3 of the 6 words must be different in each sentence combination.  In other words, a pattern of any 3 words (or more) cannot be repeated in any other combination.

For example (number indicates word-synonym variation):
Combination 1) A1, B3, C1, D1, E1, F2 is an acceptable combination of word synonyms
Combination 2) A1, B2, C1, D1, E1, F3 fails the rule because only B2 and F3 (only 2) are different from Combination 1).
Combination 3) A1, B2, C1, D1, E2, F3 passes the rule because B2, E2, F3 are different from Combination 1).

Questions:
1.	In combinatoric terms, what is the proper way to describe this problem? (so I can better understand the appropriate termonology)
2.	What combinatoric theory applies to this problem? (I want to understand the theory so I can apply it to other similar problems).
3.	What is the correct mathematic formula to apply to this problem to determine the number of sentance combinations?  (Again, I would like to better understand how to work out similar problems of this type).

Thank you in advance for your replies and guidance.

30 views in a few hours – that speaks well to the thread title, but no replies.  Well, this is a real business problem that requires a real solution.  It is not an academic exercise.  So any help (even pointing me in the right direction) would be appreciated.  Perhaps a problem like this has been discussed in an earlier thread?

Here is what I know.  The problem appears to be a Permutation problem because order counts.  “B” synonyms cannot appear before “A” synonyms.

And repetition is NOT permitted, because once a set of 3 synonyms is chosen, that set cannot be chosen again (as a set).

If the number of synonyms were the same for each word, the problem would be easier (I think), but they are not.  And that is what I am having trouble getting my head around.

Am I heading in the right direction? Anyone?

Don't be impatient -- if anything, it makes me (and probably others) less inclined to spend my time helping you out.

Am I right that the question is, "What is the most number of different combinations we can have such that any two of them disagree in at least 3 positions?"  If so, your question appears to be related to error-correcting codes (e.g. the Hamming code).

JBL, tks for your reply.  Yes, I was getting frustrated - staring at the white board for hours.  Tks for your encouragement.

I decided to rephrase the problem in the form of a spreadsheet example.  In stead of trying to use a thousand words and use terminology that I am unfamiliar with, this spreadsheet model should (hopefully) describe the problem better.  In addition to providing a graphic, I have attached the spreadsheet to this post (file: synonym_selection_model).

The top portion of the spreadsheet describes a sentence.  6 of 10 words in the sentence can be disambiguated with wordnet.  The remaining 3 can not be disambiguated and are represented with a "----".  Of those words that can be disambiguated, wordnet returns hypernyms, hyponyms, etc (types of synonyms).  They are represented by A1, A2, A3, etc.

The lower part of the spreadsheet attempts to record permutations that satisfy the rule.  The permutations were developed manually.  Obviously, there are many more permutations that could be listed that pass the test.  I have left the spreadsheet in a state that shows that permutation 11 fails as it does not have enough unique synonym changes (with C1, there are only 2 differences compared to the selected row).  If the C1 value were changed to C2, it would pass.

My question, then, is can the number of possible permutations be modeled mathematically?  If YES, then how?

Thank you again for you kind responses.

[img]http://www.truepath.com/synonym_selection_model.gif[/img]

Okay, good.  So, let me rephrase my question:  in your example, the image you've attached shows that we have a "conflict" between permutations 11 and 13, i.e. they agree in too many positions.  Am I correct that you are trying to construct a list as large as possible such that no two elements of the list conflict?  (The other, much easier possibility is that you're trying to construct a list of sentences that don't conflict with a given, fixed sentence (but that are allowed to conflict with each other).)

If so, the problem is related to sphere packing -- you have a set (all the sentences you could possibly make by your available choices) and a distance (the number of positions at which two sentences differ) and you want to know how many points you can choose such that the distance between any two points is at least 3.  Unfortunately, I believe this problem is difficult to solve in such generality.  The "typical" situation for this problem is when we have two choices, 0 or 1, for each position (what you call words, but this conflicts with the usual terminology), and we're looking for a set of strings such that each pair differs in a certain number of positions.  This question lies in the domain of coding theory, and coding theorists are typically interested not in the maximal size of such a set but in a set that is both "large" and efficiently-constructable.  Even in this special context, no one knows what the largest size of a set of strings with a given distance is (as a function of the length of the strings (in your terminology, the number of disambiguable words) and the number of positions at which we want them to differ), and in fact this is a big open question.  (One of my office-mates says that the number is known to within some smallish factor, say 2.)  If you will only be interested in "small" examples (not more than a handful of words in your sentence, and each word with only a handful of synonyms), then it's probably possible to come up with a program that solves the problem approximately (e.g. with some type of greedy or probabilistic approach), but (1) I don't really know what a good way to go about this would be, and (2) even if I did, I lack sufficient programming experience to actually implement an attack.  (Of course, some exhaustive search is possible, but even for your relatively small example it will take a very, very, very long time.)  This is probably not very helpful to you, but I think that's because you've picked a very hard question :)

Wow Joel,

Thank you very much for your insight.  YES, your description of my problem seems to be accurate.

We have already figured out a way to program the solution, although we have not done it yet.  The process is fairly straightforward and similar to the process shown in the spreadsheet.

1) Construct an algorithm that will produce every permutation of every word in the sentence.  This data goes into an array (array “A”).
2) Then choose the first row of the array and start another array (array “B”).
3) Then choose the second permutation from array "A" and compare it with every row of array “B” (similar to my spreadsheet).  If this permutation passes the test, then add it to array “B’.
4) Continue the sequence till all permutations from array “A” are tested.
5) Then count the rows in array “B”.

But we wanted to know before hand how many permutations would be in array "B" so that we could 'tune' other system variables before actually running the program.

In reality, there will be several sentence and several paragraphs.  This will be a CPU intensive task but seems to have some real value.

Even though your reply did not have the answer I was hopeful for, you did give me an answer.  And that, in and of itself, is of value to me.  Thank you again for your insight.


If:
$ ab\equal{}(a\minus{}b\minus{}c)(a\minus{}b\plus{}c)$
$ 7a^2\plus{}2c^2\equal{}(3a\plus{}c)(3a\minus{}c)$
for positive $ (a,b,c)$

Prove that:
$ (a\plus{}b)^2\equal{}2a^2\minus{}b^2$
$ a\plus{}b>c$
$ a\plus{}c>b$
$ b\plus{}c>a$

this is impossible. $ ab > 0$ and $ a\plus{}c > b$ implies $ a\minus{}b\minus{}c >0 \iff a>b\plus{}c$, which contradicts what you are asking us to prove.

If:
$ ab \equal{} (b\plus{}c\minus{}a)(a \minus{} b \plus{} c)$
$ 7a^2 \plus{} 2c^2 \equal{} (3a \plus{} c)(3a \minus{} c)$
for positive $ (a,b,c)$

Prove that:
$ (a \plus{} b)^2 \equal{} 2a^2 \minus{} b^2$
$ a \plus{} b > c$
$ a \plus{} c > b$
$ b \plus{} c > a$

sorry about the typo

So from the first given,
$ ab\equal{}c^2\minus{}(a\minus{}b)^2$
$ c^2\equal{}ab\plus{}(a\minus{}b)^2$
Substituting in I get,
$ a^2\plus{}3b^2\equal{}3ab$
Which is not equivalent to what you wrote.


Write two natural number $ a$ at paper 1 and $ b$ at paper 2. Then write $ \gcd(a,b)+1$ at paper 1 and $ \text{lcm}(a,b)-1$ at paper 2. Blablabla
Find all $ (a,b)$ such that:
(a) Numbers at paper 1 are bounded
(b) Numbers at paper 2 are bounded

Example: (3,5) -> (2,14) -> (1,13) -> (2,12) -> ...

Hmm..I can prove that, except for some finite and clearly determined values for which we have cycles, the sequence of numbers on paper $ 2$ is divergent to $ \plus{} \infty$.
Moreover, if the inital numbers have opposite parities, the sequence on paper $ 1$ is bounded (it is constant after some steps).

I think that the sequence on paper $ 1$ is always bounded, but the general case seems to be difficult to handle, unless there is some trick :wink:

Pierre.


How many even natural-number factors does $ n \equal{} 2^2 \cdot 3^1 \cdot 7^2$ have?

I just counted all the possible factors and subtracted the odd ones.

Or, you could also count the number of factors that include a $ 2$ in their prime factorization.  To do that, we count the number of factors in $ 2^1\cdot3^1\cdot7^2$ (kinda hard to explain why though...).  To count the number of factors a number has, we simply increase each exponent in their prime factorization by $ 1$, and them multiply them together.  We have $ (1\plus{}1)(1\plus{}1)(2\plus{}1)\equal{}2\cdot2\cdot3\equal{}\boxed{12}$

[quote=horsefeedapples]What I did was count the total number of factors which is $3 \times 2 \times 3$ and then subtract the number of factors which are not even which is $2 \times 3$. Therefore $18-6=12$[/quote]

That is exactly what I did, too!!!!!

This is problem is really easy for a level 18
[hide = easy solution]Even means it must have a factor of 2, so we can have 2 options for the number of 2's, 2 options for the number of 3's, and 3 options for the number of 7's. 2 x 2 x 3 = 12.[/hide]

[quote=peelybonehead]This problems requires the knowledge of number theory to solve it efficiently[/quote]

... is it not number theory?

[quote=peelybonehead]This problems requires the knowledge of number theory to solve it efficiently[/quote]

are you ok peelybonehead? is this satire or did someone hack into your account

[quote=peelybonehead]This problems requires the knowledge of number theory to solve it efficiently[/quote]

this isn't a number theory problem ...?


After 2/3 of the boys left a school dance, the ratio of girls to boys was 9:5. Then 21 girls left, and the ratio became 3:4. how many students total remained at the dance?

how would u do this?

 :ninja:  :rotfl:  :huh:

[hide]Since after $\frac{2}{3}$ of the boys were gone, from the first info, we get: $\frac{g}{\frac{1}{3}b}= \frac{9}{5}$. From the second info, we get: $\frac{g-21}{\frac{1}{3}b}= \frac{3}{4}$

From the first equation, $g=\frac{3b}{5}$

From the second equation, $g-21 = \frac{b}{4}$

Subtracting the second equation from the first, $21=\frac{7b}{20}$. It's easy to see that $b=60$. Also, we can substitute this into the first equation to get $g=36$.

So there are $36-21=15$ girls left and $\frac{60}{3}=20$ boys left. :D [/hide]

[quote="mr. math"]
how would u do this?[/quote]
you forgot to answer the question vishalarul. :P

can u give a little more detail on how to solve for g and b. thnx

[hide]Call the number of girls $g$ and the number of boys $b$.  After two-thirds of the boys leave the dance, $\frac13$ of the boys are left.  We also know that for every 5 boys, there are 9 girls.  We only have one third of the original boys, so the equation is $5g=\frac13\times9b\Rightarrow 5g=3b$.

After 21 girls leave, the equation becomes
$4(g-21)=3(\frac13b)$
$4g-84=b$

Now, solve the equations.
$4g-b=84$
$5g-3b=0$

$12g-3b=252$
$5g-3b=0$

$7g=252$
$g=36$
$b=60$

Plugging back into the equations, there will be $\frac13\times60=20$ boys and $36-21=15$ girls, making $45$ total students.[/hide]

This is a typical mathcounts problem; you might want to check in the mathcounts forum for similar problems...


For positive reals $x_{1}, x_{2}, \cdots, x_{n}$. Prove that

\[n+\prod x_{i}\ge 1+\sum x_{i}\]


hmm im interested in bunking with anyone for nats if they would like to on Aops....ill put their name down

btw
im a girl so..

idk anyone else who wants to bunk with someone else...i guess put ur requests down!!!!

Erm, I suppose many people on AoPS have friends here, but haven't necessarily met all of them :roll:, so may not be interested in bunking with people they haven't ever met.

Dont you bunk with people youve never met anyway?

well, they usually try to put teammates together, but if there's an odd number of males or females, then someone is paired with another teammember.  I believe if nothing is indicated, they pair teammates alphabetically, then people not on the same team by their state name (alphabetically).  This is just based on observation (my 6th grade year, we had four guys, and we were paired by last names alphabetically), same in 7th, then in 8th 3 guys and a girl, so one of my teammates roomed with someone from the Virgin Islands.  I don't remember who the girl roomed with though...

resees, haha, um, anyway, i dont think there are much girl AoPSers going to nats. So far ur one of the only girls i know goin to nats. :roll:

well, except this one person

hmm thats why i am asking....in hopes that some girl will se this and will mediate my prayers and i wont be stuck with some constipated idiot....no offense........

Constipated idiot? Just cause you have an AoPS account doesnt mean you wont be a constipated idiot. I speak from experience.... just kidding....

hmm yes but if it someone from AOPS...u can telll whether they sound sort of like a constipated idiot

While we're talking about "constipated idiots"

I just wanted to say that minimal time is actually spent in your room, aside from sleeping and etc, no big chunks of time spend in the room jump out at me from memory.  I would say the largest was the time spent before the banquet getting dressed and ready.  If you don't like your roommate, it shouldn't be a big deal, most of the time spent there is just sleeping...  Also, at least for my team, a lot of down time was spent talking amongst ourselves, playing card games, distributing stuff, etc, which usually happened in one central room.

Look [url=http://mathcounts.org/Document.Doc?id=341]here[/url], choose a random girl and put her name down.

In my experience nat roommate assignment is almost always random and almost always with someone else of your own team.

Roommates are fun and cheifly useful when making prank calls. This not recommended before the competition. Probably not recommended after either ;).

Hmmm Resees I might consider you too? But im not sure cuz i have other ppl on my list:

dragonlaird7
LHM
HolyMushroomspwnthywrld17 (if she makes nats, which I believe is highly likely)

If you're a girl, you can ask me to add you on my rooming consideration list. Please PM me (with what state you're from and what place you got)
I pretty much know anyways that I (as the only girl on my team) isn't gonna be with a person who I met face-to-face before nats. So yeah, it'd be nice if my roommate is a fellow AoPSer :) ....But as long as she isn't annoying/snoring/whatever the crap might be! :wink:

when  i was makin this , i didnt realize how unimportatnt bunking is...i thought like atleast 1/3 of ur time would be with ur bunk partner...whatever...continue this discussion if u want.....but if i guess a mod wants to, delete it

if you make nats, u will get to stay at the hotel for 2 nights rite?

You get to stay for THREE nights. (Thursday, Friday, Saturday) :)

[quote="AIME15"]Hm, if I make nats all I'll put there is

"anyone but Maximilian Schindler, Runpeng Liu, and Victor Wang".[/quote]

[b]or[/b], not and. :P

hmm, I think I'm already forced to room with Runpeng Liu...

What if you requested a roommate from a different state? :P

Do they let siblings bunk together (brother and sister)?

What about..3 in one room, 1 in another?

See..there's this one teammate of ours...who we're SCARED of (for reasons I'm not going to say here..PM if you really want :P). No one wants to be with him.

[quote="solafidefarms"]Can confirm no. Nor bunking with parents - you'll be assigned a room, though if your coach lets you you can probably not actually sleep therein.[/quote]

But my brother is on my team...

I'm sure if you ask, they will let you. But why would you want to?


geometry
incenter
circumcircle
geometric transformation
geometry proposed

[quote="livetolove212"]Let $ I_a, I_b, I_c$ be excenters of triangle $ ABC$. We know that $ OI$ is [b]Euler's line[/b] of triangle $ I_aI_bI_c$
On the other side, $ (A'), (B'), (C')$ concur at G so G is the circumcenter of triangle $ A'B'C'$
Note that $ A'B'//AB, A'C'//AC$ and $ AI$ is the bisector of angle $ BAC$ so $ A'I$ is the bisector of angle $ B'A'C'$
Similarly we get I is the incenter of triangle $ A'B'C'$.
Let $ I'_a, I'_b, I'_c$ is the excenter of triangle $ A'B'C'$. It's easy to see that $ I'_aI'_b//I_aI_b, I'_bI'_c//I_bI_c, I'_cI'a//I_cI_a$ then [b]Euler's line[/b] of two triangles $ I_aI_bI_c$ and $ I'_aI'_bI'_c$ are parallel, but they have common point I so $ G,I,O$ are collinear.
We complete the proof.[/quote]
dear [b]livetolove212[/b]
why G is the circumcenter of triangle $ A'B'C'$?

[quote="livetolove212"]Because $ (A'), (B'), (C')$ are congruent and pass through G.[/quote]
I am sorry ,I did not notice the condition "D, E, and F are congruent circles"!
 :blush:

[quote="77ant"]For a triangle ABC with incenter I and circumcenter O. Circles D, E, and F are congruent circles passing through point G. Circle D is tangent to AB and AC, circle E is tangent to AB and BC, and circle F is tangent to AC and BC. Prove that O, G, and I are collinear.[/quote]
Since $ D,E,F$ are equidistant from the respective pairs of  sides of $ \triangle ABC,$ it follows that $ \triangle DEF$ and $ \triangle ABC$  are homothetic with homothetic center $ I.$ $ X$ is obviously circumcenter of $ \triangle DEF.$ Hence $ O,X$ and the homothetic center $ I$ are collinear. In fact, the concurrency point $ X$ of the circles $ (D),(E),(F)$ is the insimilicenter of $ (I) \sim (O).$


Hello!

find an integer $ n$ such that

\[ \sigma(n)\plus{}500\equal{}\sigma(n\plus{}2)\]
where $ \sigma(n)$ is the sum of divisors of $ n$.

well, $ n\equal{}2005$ is a solution. But, is it the only one?


analytic geometry
function
geometry
probability and stats

Given two independent random variables $ X$ and $ Y$ and a $ \sigma$-algebras $ A$ , are the random variables  $ E(X|A)$ and $ E(Y|A)$ necesserly independent ? by using geometric anologie and projections that donesn't hold ! ( $ L_X$ and $ L_Y$  are ortogonals does'nt mean that the projections of $ X$ and $ Y$ on $ L_A$ are so  :huh: ! )

You are right. They may become dependent. The simplest example is when $ X$ and $ Y$ are the coordinate functions on $ \{\minus{}1,1\}^2$ and $ A$ is the $ \sigma$-algebra generated by $ \{(\minus{}1,\minus{}1)\}$, $ \{(1,\minus{}1),(\minus{}1,1)\}$ and $ \{(1,1)\}$, in which case the conditional expectations are identical and non-constant.

Thank you fedja, I use the similarity with geometry to have an idea but I couldn't see it with random variables, could you please explain what is the conditional expectation in that case ? is it $ \minus{}1$ on $ \{(\minus{}1,\minus{}1)\}$, $ 1$ on $ \{(1,1)\}$ and $ 0$ on $ \{(\minus{}1,1);(1,\minus{}1)\}$ ?

all ot off thanks , your exemple is so helpful because it responds to my original question which is : we dont have always $ E(XY|A) \equal{} E(X|A)E(Y|A)$ even for independent random variables $ X$ and $ Y$   :roll: !


This looks more like a "limits" problem  :D

The key is to observe that $\frac{k}{n^2+n} \leq \frac{k}{n^2+k} \leq \frac{k}{n^2+1}$ and then sum up.

We get $S > \frac12$ and $S < \frac{n(n+1)}{2(n^2+1)}$. The inequality which remains to be proved is $\frac{n(n+1)}{2(n^2+1)} < \frac{n+1}{2n} \Leftrightarrow n^2<n^2+1$, true.

[quote="cezar lupu"]Actually, perfect-radio is right here. From the sandwich theorem we can deduce easily that
$\lim_{n\to\infty} \frac{1}{n^{2}+1}+\frac{2}{n^{2}+2}+\ldots+\frac{n}{n^{2}+n}=\frac{1}{2}.$ ;)[/quote]??
He is right but he doesn't need that at all... we just have $\sum k = n(n+1)/2$

I think a proof using AM-HM inequality works as well (direct application for LHS of the inequality)

(for RHS:)

$\sum {\frac{k}{n^2+k}}= \sum {1-\frac{n^2}{n^2+k}=n-n^2\sum \frac 1{n^2+k}}$

and then use AM-HM for $\sum \frac 1{n^2+k}$

I think we can also use Jensens inequality on the function $\frac{x}{n^2+x}$ and get a solution.

@above, using Jensen's we obtain:

$$f(1) + \cdots + f(n) \leq 1 - \frac{n^2 + 1}{2n^2 + n + 1}$$

Good luck trying to prove that.


geometry
inequalities
analytic geometry
function
triangle inequality
inequalities proposed

For positive $a,b,c$, prove that \[\sqrt{a^2+ab+b^2}+\sqrt{b^2+bc+c^2}>\sqrt{c^2+ca+a^2}\]

I know it can be solved easily by geometry, but, please, give a non-geometric proof.

Are you kidding?

It is clear $\sqrt{a^2+ab+b^2}> a+\frac{b}{2}$ and $\sqrt{b^2+bc+c^2}> \frac{b}{2}+c$, so
\[\sqrt{a^2+ab+b^2}+\sqrt{b^2+bc+c^2}> a+c >\sqrt{a^2+ac+c^2}.\]

That's all.

maybe I'm really crazy :blush: ........I even haven't thought that it can be solved easily without using geometry........ :(

Maybe you meant
\[
\sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2} \geq \sqrt{c^2+ca+a^2}
\]

Yeee! It is much more interesting to find non-geometrical solution for this one, though geometrical solution is evident.

mecrazywong,
were you possibly thinking about this inequality instead?
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=7791[/url]

In Canada, there's a correspondence program for olympiad-level problems called, unsurprisingly, olymon.

[quote="billzhao"]\[
\sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2} \geq \sqrt{c^2+ca+a^2}
\][/quote]

Well, we can algebrize the geometric solution by replacing the points by their respective coordinates: At first, remember the inequality

$\sqrt{x^2+y^2}+\sqrt{X^2+Y^2} \geq \sqrt{\left(x+X\right)^2+\left(y+Y\right)^2}$

for any four real numbers x, y, X, Y. (This is, of course, the triangle inequality, but you can also show it algebraically by squaring and applying Cauchy-Schwarz.) Now apply this inequality to $x=\frac{b}{2}-a$, $y=\frac{\sqrt3 b}{2}$, $X=-\frac{b+c}{2}$, $Y=\frac{\sqrt3 \left(c-b\right)}{2}$. After a quick calculation, you get the inequality in question.

Intuition tells me that a more natural proof could be found on the base of the famous principle that the sum of the minima of several functions is $\leq$ to the minimum of the sum. Any ideas how to do this precisely?

  Darij


To achieve a better end, please use proper grammar and spelling in your posts while making sure that they make sense and are clear.
I cannot understand your above post.

[quote="ackter78692"]Mary,Larry, and Harry are going to Disney world together. Mary and Larry drove halfway together that distance was 580miles they figure the gas to where they picked up Harry was 68.50 dollars. find out what is each persons share of gas.[/quote]

I am guessing Mary and Larry drove 580 miles together and spent 68.50 dollars on gas. So that means they went 1160/137 miles per dollar of gas. If Mary Larry and Harry drive 580 miles together then Mary needs to pay 34.25 dollars for the first half and 22.83 cents for the second half. Larry would pay the same as Mary and Harry would pay 34.25.

Mary=58.08
Larry=58.08
Harry=34.25


which one are you talking abt? the one on aops or the one of mathematics? :)

See [url=http://en.wikipedia.org/wiki/Asymptote]here[/url]  :)

i cud of done it
plzzzzzzzzzzzz tell in simple terms

basically, it's a line which [i]tends to[/i] being a tangent to a curve, as the curve tends to plus or minus $ \infty$
refer  the examples given on wiki n you'll understand better. :)


happy independence day. rather belated   :o
we have it today  :D

well i shud say same
rather i bit tooooooo belated.. :P


a,b,c are positive reals. Prove that
$ \frac {a}{(b\plus{}c)^2} \plus{} \frac {b}{(a\plus{}c)^2} \plus{} \frac {c}{(b\plus{}a)^2} \ge \frac{9}{4(a\plus{}b\plus{}c)}$


[quote="Arvind_sn"]Billy Bob goes to Las Vegas and is at a casino there. He is playing a game where he has a full card deck (52 cards, no jokers) and draws from the deck randomly. What is a probability that Billy Bob either gets a spade or a king?[/quote]

[hide]16 possibilities. $\frac{16}{52}=\boxed{\frac{4}{13}}$[/hide]

Look again SplashD.

[hide]You see, the probability of pulling out a spade is $\frac{13}{52}$. Then, the probability of pulling out a king is $\frac{4}{52}$. Add them together. We get $\frac{17}{52}$. Then, we must subtract $\frac{1}{52}$ (We do not want to count the king of spades twice) therefore yielding $\frac{16}{52}$. That is our answer. You were right when you said 16 possibilities, but I do not know why you did 52 choose 2. The answer that my source had was different.[/hide]

 :wink:

[quote="SplashD"][quote="lotrgreengrapes7926"][quote="nutz_for2.718281828"]He said exactly what you said, arvid sn. You can also get that using PIE.[/quote]
SplashD edited his answer. :wink: :D[/quote]

Yeah, it just doesn't look like it because it's hidden. :P[/quote]

Wow, I'm going to put a hide in my signature now so people can't quote me and see when I edited :O.

[quote="ShoeFactory"][/quote][quote="SplashD"][quote="lotrgreengrapes7926"][quote="nutz_for2.718281828"]He said exactly what you said, arvid sn. You can also get that using PIE.[/quote]
SplashD edited his answer. :wink: :D[/quote]

Yeah, it just doesn't look like it because it's hidden. :P[/quote][quote="ShoeFactory"]

Wow, I'm going to put a hide in my signature now so people can't quote me and see when I edited :O.[/quote]

Yeah, but if make it too 'hidden', mods will be pming you here and there.

chance of getting a spade= 13/52
chance of getting a king= 4/52
13/52+ 4/15 = 17/52
But, you count the king twice so 17/52- 1/52= 16/52= 4/13

[quote="Arvind_sn"]Billy Bob goes to Las Vegas and is at a casino there. He is playing a game where he has a full card deck (52 cards, no jokers) and draws from the deck randomly. What is a probability that Billy Bob either gets a spade or a king?[/quote]

[hide]spade: 1/4
king: 1/13
both: 16/52=8/26=4/13[/hide]

the chance of getting a king is $\frac{4}{52}$ and the chance of getting a spade is $\frac{13}{52}$ and when you combine the two fractions you get $\frac{4}{52}+\frac{13}{52}=\frac{17}{52}$ but you dont double count the king of spades so it is $\frac{17}{52}-\frac{1}{52}=\boxed{\frac{16}{52}}$

[quote="daermon"]u forgot to reduce  :wink:[/quote]

i dont understand...

The answer you got, $\frac{16}{52}$, isn't in simplest form. You can write this in simplest form by reducing. You divide both the numerator and the denominator by four. Effectively, you are dividing the whole fraction by one. You are dividing it by $\frac{4}{4}$. Our final answer should be $\frac{4}{13}$

 :wink:


geometry
symmetry
circumcircle
geometric transformation
reflection

Geometry, and if I find something as interesting as it, I'll put it too ;)

AD is the angle bisector, P a point on it, AM the median. Show that PK is perpendicular to BC.

EF is parallel to AP, B is the midpoint of EF.

A spiral symmetry from O moves a circle to the other one.

To prove: $AB \cdot CD \cdot EF = BC \cdot DE \cdot FA$

Very nice problem.
L, the midpoint of KL, is on the circumcircle of ABC. Prove that I,O,N lie on a line.

AB parallel to CD. M,N are midpoints.
Prove that EN is the bisector of ANB.

HM symmedian of HBC, $A_{1}R$ symmedian of $A_{1}B_{1}C_{1}$.

P' is the inverse of P. A' is symmetric to A. To prove the collinearity of B,P',A'.

BK is parallel to CY, PM to XC. Prove that $\angle BKM = 90$.

T is the reflection of O on BC. The lines AKT and NM are perpendicular.

E,F are the feet of angle bisectors. Prove that $OI_{A}\perp EF$

Just prove it, use transformations.  :wink:

AK:KB = 1:2, ABC is isosceles, F is a midpoint, C' is a midpoint of arc AB. Prove that perpendicularity.

ABC is similar to $A_{1}B_{1}C_{1}$, the vertices of one are on the lines of the other. Prove that the ortocentre of $A_{1}B_{1}C_{1}$ is the circumcentre of ABC.

http://www.mathlinks.ro/Forum/viewtopic.php?t=140631

The messiest problem  :D

Italy, Great Britain and Azerbaijan are going to Balkan 2007!  :lol:

Haha, that's the first time I've seen Azerbaijan in the Balkans.

To be honest, the graphic first reminded me of http://www.albinoblacksheep.com/flash/end.php :D

  dg

To prove: AL + BH + CK = AK + BL + CH

http://www.mathlinks.ro/Forum/posting.php?mode=reply&t=143475

Given the circumference, a point P on it and a secant r, find the circumference that touches P, r, and the circumference.

Start with the main circumference, C and the secant AB. Prove that the red circumferences are coaxial.

http://www.mathlinks.ro/Forum/posting.php?mode=reply&t=143626

Prove that K is the midpoint of MN.

http://www.mathlinks.ro/Forum/viewtopic.php?t=143661

M is the midpoint of PA. PA,PB are tangents. Prove that PN = 2MN.

http://www.mathlinks.ro/Forum/viewtopic.php?t=145001

$A',B'$ are midpoints, O the circumcentre, $OM \perp KL$. Prove that $\frac{KM}{ML}= \frac{BD}{DA}$.

http://www.mathlinks.ro/Forum/viewtopic.php?t=145077

The two incircles are tangent => ABCD circumscribed, MNPQ inscribed

Solution to this problem:
http://www.mathlinks.ro/viewtopic.php?t=153371


trigonometry
limit
algebra unsolved
algebra

Suppose $ x \in (0,\pi)$ .
Let $ a_0\equal{}\sin x$ and $ a_n\equal{}\sin a_{n\minus{}1}$, $ n\equal{}1,2,....$ Prove that :
               $ a_1^3<\frac{1}{n} \sum_{k\equal{}1}^{n} ka_k^2<3$ for all $ n$

We prove that :
$ \frac {1}{n} \sum^{n}_{k \equal{} 1} ka_k^2 < \frac {1}{n \plus{} 1} \sum^{n \plus{} 1}_{k \equal{} 1} ka_k^2$
Which is equivalent to :
$ \sum^{n}_{k \equal{} 1} ka_k^2 < n(n \plus{} 1) a_{n \plus{} 1}^2$ (Trivial because $ a_n \le a_{n \plus{} 1}$)
Therefore,
$ LHS < a_1^2 \le \frac {1}{n} \sum^{n}_{k \equal{} 1} ka_k^2 < \lim_{n \rightarrow \plus{} \infty} \frac {\sum^{n}_{k \equal{} 1}k a_k^2 }{n}$
$ \equal{} \lim_{n \rightarrow \plus{} \infty} (n \plus{} 1) a_{n \plus{} 1}^2 \equal{} \lim_{n \rightarrow \plus{} \infty} \frac {1}{\frac {1}{a_{n \plus{} 1}^2} \minus{} \frac {1}{a_n^2}}$
$ \lim_{x \rightarrow 0^ \plus{} } \frac {1}{\frac {1}{sinx^2} \minus{} \frac {1}{x^2}} \equal{} 3$

[quote="Vincent Gilbert"]We prove that :
$ \frac {1}{n} \sum^{n}_{k \equal{} 1} ka_k^2 < \frac {1}{n \plus{} 1} \sum^{n \plus{} 1}_{k \equal{} 1} ka_k^2$
Which is equivalent to :
$ \sum^{n}_{k \equal{} 1} ka_k^2 < n(n \plus{} 1) a_{n \plus{} 1}^2$ (Trivial because $ a_n \le a_{n \plus{} 1}$)[/quote]
No,We have:$ a_{n\plus{}1}\equal{}\sin a_n <a_n$

Hix :( .You're right . :(( How could I get a  silly mistake  like it =.=


trigonometry
function
calculus
integration
trig identities

I am not really sure if this is the correct board for this question, but i noticed the "trig identities" thread and also realized that it is in high school where graphing calculators become of use.  So.

I received a TI-86 graphing calculator for my birthday last year.  It is an exceptionally useful tool to me, even though i cannot figure out how to use most of its functions and applications.  I have a desire to create some programs, but i am not sure how.

The first program that came to mind was a "trig identities" program.  I am not sure if the programs section of the calculator is a good place for this, though.  Perhaps it is pointless, i am not really sure as i am inexperienced.

Would it be a good idea to have a program such as this?  maybe it is less of a program and more of a displayed list.  I know next year in calculus i can use graphing calculators on tests, but no notes, so even a list might be a good thing to have.

How would i go about making this program?  (you dont have to list all the trig identities, just the commands that make them display).  I already took a shot at making it, but it said error, of course.

Are there any other programs that are practically required for a graphing calculator?

Thanks for your help.  Sorry if this is in the wrong board, and sorry if you guys are offended by my use of the calculator.  Sometimes the calculator is a requirement...though i try to do my work in my head or, at worst, with a pencil and paper.

Personally, I don't find listing trig identities so useful, as I more often than not derive them on the spot.  But in terms of calculator programming in general, you should probably consult your manual or local computer geek.  If you just want to fiddle around with some programs, I suggest learning a little C or something.  I can't stand calculator programming, as I find the typing in stuff way too slow. :D

I agree with probability1.01, it is much better to understand the trig identities or at the very least memorize them.  If you are taking AP calculus, a large portion of the test is non-calculator.  Also, the AP test does not require that you know any really advanced identities $\left(e^{x i}=i\sin(x)+\cos(x)\right)$, just make sure you know the basics.
$\sin^2(x) + \cos^2(x) =1$  (you can derive most from here)
$\cos(2x) = 2\cos^2(x) -1$ (they like this one for integrals)
$\sin(2x)=2\sin(x)\cos(x)$.
However, a calculator can be a great learning tool.  I've figured out a lot with my TI-89, and on tests it solves almost everything. :D


Nobody wants to think about this one. You don't know what you miss: nothing.


trigonometry inequalities function calculus derivative

Wow.... i knew the hint you gave me before i saw the spoiler..... i must be getting better at Trig  :D

[hide="solution"]
set $x=\tan{A}, y=\tan{B}, z=\tan{C}$. The first statement is true but i do not want to prove right now. there should be one by, i think, Pranay somewhere on the forum.

Anyways, the second one transforms to:

$\tan{A}(1-\tan^2{B})(1-\tan^2{C})+\tan{B}(1-\tan^2{C})(1-\tan^2{A})+\tan{C}(1-\tan^2{A})(1-\tan^2{B})=4(\tan{A}\tan{B}\tan{C})$
$\tan{A}\sec{B}\sec{C}+\tan{B}\sec{C}\sec{A}+\tan{C}\sec{A}\sec{B}=4(\tan{A}\tan{B}\tan{C})$
$\frac{\sin{A}+\sin{B}+\sin{C}}{\cos{A}\cos{B}\cos{C}}=4(\tan{A}\tan{B}\tan{C})$
$\sin{A}+\sin{B}+\sin{C} = 4(\sin{A}\sin{B}\sin{C})$

the last one could be proven quite easily, so i am done.
[/hide]

Nice job!

Prove for $0<a,b,c \in \Re$ where $a+b+c=abc$ that \[\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}}+\frac{1}{\sqrt{1+c^2}} \leq \frac{3}{2}\]

[quote="agolsme"]From the Arbelos:

If $x + y + z = xyz$, show that
$x(1 - y^2)(1 - z^2) + y(1-z^2)(1-x^2) + z(1-x^2)(1-y^2)=4xyz$

[hide="the hint they give is"]$\tan{A} + \tan{B} + \tan{C} = \tan{A} \tan{B} \tan{C}$[/hide][/quote]

if you dont see how the hint plays in (but use xxreddevilzxx's method over this if you do), you can expand it all to get:

$x+y+z - (x^2y + xy^2+x^2z+y^2z+x^2z+y^2z) +x^2y^2z+z^2yz^2+xy^2z^2 - 4xyz = 0$

Upon inspection somewhat.. you can factor this into:

$(-1 + x y + x z + y z) (-(x+y+z) + x y z) = 0$

But, since $x+y+z = xyz$, the second part of the parentheses becomes 0, and satisfies the equation. tah dah!

and a proof for $\tan A+\tan B+\tan C = \tan A \tan B \tan C$ (if $A+B+C = 180$, is in the trig identity thread right [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=16103&postorder=asc&highlight=identity+list&start=40]here[/url].)

can someone post the proof that xxreddevilzxx ended with?

[quote="towersfreak2006"][quote="DPopov"]Nice job!

Prove for $a,b,c \in \Re$ where $a+b+c=abc$ that \[\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}}+\frac{1}{\sqrt{1+c^2}} \leq \frac{3}{2}\][/quote]
This is not true for $a=b=c=0$, since we then get $1+1+1\not\leq\frac{3}{2}$.[/quote]

hmm well he probably means where a,b,c are not equal to zero

and im assuming the same hint applies here as well.

$\frac{1}{\sqrt{1+\tan^2 A}} +\frac{1}{\sqrt{1+\tan^2 B}} +\frac{1}{\sqrt{1+\tan^2 B}} \Longleftrightarrow$

$\cos A + \cos B  + \cos C \leq \frac{3}{2}$

that i dont know how to prove but it was the last identity in that same thread.... left unproven :( anyone have any thoughts on it?

[quote="towersfreak2006"][quote="DPopov"]Nice job!

Prove for $a,b,c \in \Re$ where $a+b+c=abc$ that \[\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}}+\frac{1}{\sqrt{1+c^2}} \leq \frac{3}{2}\][/quote]
This is not true for $a=b=c=0$, since we then get $1+1+1\not\leq\frac{3}{2}$.

How did xxreddevilzxx get from
\[\sum_{cyc}\tan{A}(1-\tan^2{B})(1-\tan^2{C})=4(\tan{A}\tan{B}\tan{C})\] to \[\tan{A}\sec{B}\sec{C}+\tan{B}\sec{C}\sec{A}+\tan{C}\sec{A}\sec{B}=4(\tan{A}\tan{B}\tan{C})\]

EDIT: Cleaned up my mess.[/quote]

That's the step I am confused about as well. (and the same step I was stuck on when I was trying it)

sorry i will go back and try again. I forgot the square signs, which could take the problem into a totally different way :blush:

Yes $a,b,c >0$ and PK the proof follows easily from Jensen's.

[quote="calculuslover800"][quote="xxreddevilzxx"]
$\tan{A}(1-\tan^2{B})(1-\tan^2{C})+\tan{B}(1-\tan^2{C})(1-\tan^2{A})+\tan{C}(1-\tan^2{A})(1-\tan^2{B})=4(\tan{A}\tan{B}\tan{C})$
$\tan{A}\sec{B}\sec{C}+\tan{B}\sec{C}\sec{A}+\tan{C}\sec{A}\sec{B}=4(\tan{A}\tan{B}\tan{C})$
$\frac{\sin{A}+\sin{B}+\sin{C}}{\cos{A}\cos{B}\cos{C}}=4(\tan{A}\tan{B}\tan{C})$
$\sin{A}+\sin{B}+\sin{C} = 4(\sin{A}\sin{B}\sin{C})$

the last one could be proven quite easily, so i am done.[/quote]

There is an identity that says $1 + \tan^2{\theta} = \sec^2{\theta}$, however saying that
$1 - \tan^2{\theta} = \sec^2{\theta}$ is [u]wrong[/u].  As a result of this, the last statement,

$\sin{A}+\sin{B}+\sin{C} = 4(\sin{A}\sin{B}\sin{C})$

is incorrect.  As a counter example, when $A=B=C=45^o$,

$\sin{45^o}+\sin{45^o}+\sin{45^o} = 4(\sin{45^o}\sin{45^o}\sin{45^o})$
$3(\frac{\sqrt{2}}2) = 4(\frac{\sqrt{2}}2)^3$
$\sqrt{\frac92} \neq \sqrt{32}$

Which shows that
$\sin{A}+\sin{B}+\sin{C} \neq 4(\sin{A}\sin{B}\sin{C})$[/quote]

i said i made an error. I forgot to square some of the stuff in the middle.

[hide="Complete Solution to My Problem"]

Let $a=\tan A$, $b=\tan B$, and $c=\tan C$ since it is well known that $\tan A \tan B \tan C=\tan A+\tan B+\tan C$. The given inequality then transforms into $\cos A+\cos B+\cos C \leq \frac{3}{2}$. Let $w_1=w_2=w_3=\frac{1}{3}$ and $f(x)=\cos x$. Applying Jensen's on the concave function we obtain$\frac{1}{3}(\cos A+\cos B+\cos C) \leq \cos{\frac{A+B+C}{3}}$. Since $A+B+C=180$ the inequality becomes $\cos A+\cos B+\cos C \leq \frac{3}{2}$ as desired.

[/hide]

Cosine is not concave on $(0,180)$. If we say $f(x) = \cos x \Rightarrow f''(x) = -cos x$, from the second derivative, we see that $f''(x)$ is negative on $(0,90)$ and positive on $(90,180)$. This tells us that $f(x)$ is convex on $(0,90)$ and concave on $(90,180)$. So, you can't apply Jensen's. Sorry. But since this thread is all about SWEET identities. Here is another one:
$\cos A + \cos B + \cos C = 4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2} + 1$
[hide="Spoiler"]Man, Try to do it, but if you can't, I'm kind of proud of my proof :lol: So, go to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=187198&highlight=#187198]it[/url].[/hide]

where did that stickied identity stuff go anyways?

http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=Trig+identity&t=16103

By the way Porkchop, $\cos(x)$ is convex on $(90, 180)$ and concave on $(0, 90)$ and the inequality is true if $A,B,C<90$. In any case at most one of the angles can be $>90$ and it's easy to see that the LHS becomes even smaller in this case so the result does follow from Jensen but it just requires a little casework which I was too lazy to show  :lol:


function
induction
limit
geometry
geometric transformation
algebra
functional equation

Find all function $f(x)$ is continuous on inteval [0,1] anh satifies the condition $f(x)+f(y)=f(x+y), \forall x,y\in [0,1]$

This is a well known problem.

By a simple induction, $f(nk)=nf(k)$ if $n$ is an integer and $k\in[0,\frac{1}{n}]$, and so $f\left(\frac{1}{n}\right)=\frac{1}{n}f(1)$. Now, for positive integers $p$ and $q$ with $\frac{p}{q}\in[0,1]$ we have $f\left(\frac{p}{q}\right)=pf\left(\frac{1}{q}\right)=\frac{p}{q}f(1)$.

So $f(x)=kx$ for all $x\in[0,1]\cap\mathbb{Q}$.

But for any real number $\alpha\in[0,1]$, there exists a sequence of rationals $(a_{n})_{n}$ converging to $\alpha$, and so by continuity $f(\alpha)=\lim_{n\rightarrow\infty}f(a_{n})=\lim_{n\rightarrow\infty}ka_{n}=k\alpha$, and so $f(x)=kx$ for all $x\in[0,1]$, where $k$ is an arbitrary real number.

This condition was weakened in 1875 by Darboux who showed that it was only necessary for the function to be continuous at one point ([url=http://en.wikipedia.org/wiki/Cauchy_functional_equation]Source[/url]).

If an additive function is continuous at one point, then it is easy to show (by translation) that it is continuous at all points.

Oke, Assume $f$ is continuous at point $x_{0}$. For abitrary point $x_{1}\in R$ we have:
$f(x)=f(x-x_{1}+x_{0})+f(x_{1})-f(x_{0})\forall x\in R$
From that derives:
$\lim_{x\to x_{1}}f(x)=\lim_{x\to x_{1}}(f(x-x_{1}+x_{0})+f(x_{1})-f(x_{0}))=f(x_{0})+f(x_{1})-f(x_{0})=f(x_{1})$
So, $f$ is continuous at $x_{1}$, deduce $f$ is continuous on $R$

[quote="hien"]I have a question, and don't know it is true or false:
" If $f$ ia additive, i.e $f(x+y)=f(x)+f(y),\forall x,y\in\mathbb{R}$ then $f$ is continuous on $\mathbb{R}$[/quote]
In 1905, G. Hamel proved that there are non-continuous, non-monotonic, non-bounded solutions.
[url=http://en.wikipedia.org/wiki/Cauchy_functional_equation]Wikipedia[/url]
[url=http://mathworld.wolfram.com/CauchyFunctionalEquation.html]Mathworld[/url]


Let $ a^2+b^2+c^2+d^2=1 $ . Find the maximum value of the expression :
  $ P= a^3(b+c+d)+b^3(a+c+d)+c^3(a+b+d)+d^3(a+b+c) $

I think it's wrong, since 3 is never attained?

The variables don't have to be positive.
We will prove the inequality:
$P=\sum a^3(b+c+d) \leq \frac{3}{4} = \frac{3}{4}(a^2+b^2+c^2+d^2)^2=\frac{3}{4}(a^4 + b^4 + c^4 + d^4 + 2a^2b^2 + 2b^2c^2 + 2c^2a^2 + 2d^2a^2 + 2d^2b^2 + 2d^2c^2)$
which is obviously equivalent to:
$4(\sum a^3(b+c+d)) \leq 3(a^4 + b^4 + c^4 + d^4) + 6(a^2b^2 + b^2c^2 + c^2a^2 + d^2a^2 + d^2b^2)$

Now, we will prove that:
$a^4 + 6a^2b^2 + b^4 \geq 4a^3b + 4ab^3$
holds. In fact, it's equivalent to: $(a-b)^4 \geq 0$ which is clearly non-negative.
Now, we make similar inequalities, then we sum them up and we are done.
So: $P\leq 3/4$ and equality holds for $a=b=c=d=1/2$ or $a=b=c=d=-1/2$.


calculus
integration
number theory
Diophantine equation
modular arithmetic

[hide]Rewrite the equation[/hide]
[hide]modulo, modulo, modulo[/hide]
Here is the solution.
[hide] y=432/19-(23/19)x
y=14mod19-4mod19x
Since x,y are integers,
4mod19x=14mod19
x=13
Substituting into the original equation,
y=7.

(13,7)
[/hide][/hide]

Umm, actually, there are infinitely many solutions to that equation.  However, there is only one POSITIVE INTEGRAL solution.

Other:  (-6, 30)

This is just a Diophantine equation, and there are infinitely many or no solutions to all Diophantine equations.

[quote="jclarkemathL314159"]NO CALCULATORS

I posted a problem similar to this, but I actually figured out how to solve these.
23x+19y=432.
Find the only integer solution set (x,y).[/quote]

[hide="My answer"]

Using modular arithmetic  in 19 we get that

4x is congruent to 14 (mod 19)

It can easily be found out that the smallest value of x that satisfies this is 13 so the value of y is 7.

But there are infinitely many "integer" solutions.

$(13,7)$[/hide]


abstract algebra
complex numbers
absolute value
superior algebra
superior algebra unsolved

How to prove that $ \text{GL}\left(n,\mathbb{C}\right)\slash H$, where $ H \equal{} \left\{A: \det A > 0\right\}$, is isomorphic to group of all complex numbers with module $ 1$?

Thanks.

Equivalently, if $ U$ is the group of complex numbers with absolute value 1, then you need to find a homomorphism $ f: \ GL(n,\mathbb{C})\rightarrow U$ which is onto and satisfies $ f(A)\equal{}1$ if and only if $ \det A$ is a positive real number.

[hide="such as"]
\[ f(A)\equal{}\frac{\det A}{|\det A|}\]
[/hide]

Thank you!

And I already solved this problem myself in same way :)


factorial
logarithms
limit
real analysis
real analysis unsolved

Find the asymptotic behavior of

$f(x) = \sum_{n=0}^\infty \frac{n^{x}}{n!}$

as $x \to \infty$.

I could prove it is at least exponential but that's as far as I got. :| It looks so inviting but in fact it is hard!

A rough estimate: the terms decay (super-)exponentially starting  about at $n=[x/\log x]$ -- the exponential tail can be ignored. For $n=[x/\log x]$ we have $\log(n^{x}/n!)\approx x\log x$. The preceding terms don't contribute very much since there are not so many of them. Hence $\log f(x)\approx x\log x$. Of course, this does not answer the question about $f(x)$.

(I use $F\approx G$ to mean $F=O(G)$ and $G=O(F)$.)

I think in fact

$\lim_{x \to \infty}\frac{f(x)}{e\cdot \Gamma(x+1)}\to 1$

but I don't have proof. :|

I wonder what about for $\lambda \in \mathbb{R}^{+}$ fixed,

$f_\lambda(x) = \sum_{n=0}^\infty \frac{n^{x}\lambda^{n}}{n!}$

I think the $\lambda$ has a pretty significant effect. But I don't even have an educated guess as to what. Maybe it's like $\Gamma(\lambda x)$ ? ...