Here is the official web-site for IMO'2007 in Hanoi:

................     http://www.imo2007.edu.vn/

How would anybody know already?

Also, why do people sometimes write VietNam? I never see it that way in the U.S.

[quote="N.T.TUAN"][quote="rem"]It's not empty, it's under construction. [/quote]
I know  :D  are you go to VietNam in 2007?[/quote]
I don't know yet, we have not written any of the MOs in our country yet, but I am training hard and desperately want to go to IMO 2007.

it is nice to be in the IMO... but it will be more good when you will gain a medal! good luck!

dl

[quote="rem"]It's not empty, it's under construction.  :rotfl:
I like the animation :D

gjw, do you know when it will be updated?[/quote]

The IMO'2007 site actually is alive.
In case ...  http://www.imo2007.edu.vn/   ... does not work, you should
try ...  http://www.imo2007.edu.vn/index.php?module=ViewCat&CatId=1

A very interesting site!

Even the IMO 2007 competition problems seem to be already online:
http://www.imo2007.edu.vn/index.php?module=problem
http://www.imo2007.edu.vn/Problems/ENGDay1.pdf
http://www.imo2007.edu.vn/Problems/ENGDay2.pdf
http://www.imo2007.edu.vn/Problems/VIEDay1.pdf
http://www.imo2007.edu.vn/Problems/VIEDay2.pdf

But the files are all protected, and I do not have permission to access them.

I bet the problems are [u]not[/u] online ;)
That would require them to exist :D

[quote="test20"]But the files are all protected, and I do not have permission to access them.[/quote]

I love these servers that write Forbidden automatically when encountering a 404 error...

  darij

It's a nice website. :)
I think they have just created the place where they will post the problems and have created an authorization thing for the link, and they have not posted the problems yet.

maybe it'll a place for all of you to introduce yourselves :D .
 http://www.imo2007.edu.vn/Problems/ENGDay1.pdf
the link is exist,but it links to nowhere :D it's only available when IMO finished :D wait  :lol:

[quote="test20"][quote="test20"]A very interesting site!

Even the IMO 2007 competition problems seem to be already online:
http://www.imo2007.edu.vn/index.php?module=problem
http://www.imo2007.edu.vn/Problems/ENGDay1.pdf
http://www.imo2007.edu.vn/Problems/ENGDay2.pdf

But the files are all protected, and I do not have permission to access them.[/quote]

Also the solutions to the IMO'2007 competition problems seem to be online:
http://www.imo2007.edu.vn/Solutions/ENGDay1.pdf
http://www.imo2007.edu.vn/Solutions/ENGDay2.pdf

But again, I do not have permission to access them.[/quote]

did you anyway read what darij wrote? :)

btw. i will be really glad if they finally bother to rent professionalists to make site for such "non-important" event as IMO. maybe sense of aesthetics is uncommon among mathmaticians, but yes, i do find this site utterly ugly :)

o and well, mexican IMO site was pretty OK. but, nevermind :)

They say that so far they are just testing the information and how it is displayed. I hope they will make it look nicer.
Because the site:
http://www.imo2007.edu.vn/
has better aesthetics but you cannot click on anything!
NVM they changed it. A few days ago they had a good picture, but now it is gone! :o

haha :D, I think that creators found out that the people are looking into the site!

[quote="test20"][quote="test20"]
Also the solutions to the IMO'2007 competition problems seem to be online:
http://www.imo2007.edu.vn/Solutions/ENGDay1.pdf
http://www.imo2007.edu.vn/Solutions/ENGDay2.pdf

But again, I do not have permission to access them.[/quote]

And even the final report on IMO'2007 in Vietnam is already online:
http://www.imo2007.edu.vn/Reports/FinalReport.pdf

Of course, I do not have permission to access it.
But it should contain plenty of interesting information...[/quote]

$test20$,it was a nice joke  :D ,but it is already not funy :P .

However,it seems to me that you already know all the results of IMO 2007,don't you?Then,could you tell me what I get there? I would be greatful  to you :rotfl:  :P .

[quote="Tiks"]
However,it seems to me that you already know all the results of IMO 2007,
don't you?
Then,could you tell me what I get there? I would be greatful to you.[/quote]

Unfortunately, the system does not give me permission to access these files.
The answer to your question is most probably contained in
http://www.imo2007.edu.vn/Reports/FinalRankingOfParticipants.pdf

If you manage to break into their system, then you may be able to find the
answer to your question there.

The IMO website is no longer under construction:

http://www.imo2007.edu.vn/

[quote="orl"]I guess the web site is not the fastest around.  :wink:

Does anybody know whether Le Hung Viet Bao will be an IMO coordinator, as initially assumed, in Vietnam as he seems to go to the "Summer School and Conference on Automorphic Forms and Shimura Varieties" from 09.07.2007 until 27.07.2007 whereas IMO 2007 takes place from 19th-31st July.

http://cdsagenda5.ictp.trieste.it/listOfParticipants.php?id=a06207

And yeah for towers of Hanoi.  :D[/quote]
He said that those last days et the Summer School arenot important, and he will escape it some days before. :d

Maths in general and IMO in particular themselves are enough good-looking! So the WebSite needn't! :D  :lol:

[quote="rem"]The website is so dead. I wonder when they will post the names and/or photos of participants, I am sure some teams have already been selected...
And I really hope they will make the website look better. :roll:[/quote]

Hey rem, I think it will not take long until they update that names and photos of contestants. Our team,for exapmple, is already selected and we gave our photos to our leader to send them to the organizers. So,I guess we don't have to wait too long for the updates to come out;).

I'm not sure if that will be so quick.

Our team has been selected and we sent our photos about two weeks ago and there are still no updates.


geometry
3D geometry
number theory solved
number theory

Prove, that there are no natural numbers m, n satisfying equation:

[b]m^3 + 13 = 2^n[/b]

It's obvious that n can't be 1. So say $n \geq 1$. Now we get that  $m^3 \equiv -1 (mod \; 4)$. But that's impossible

The problem is quite straightforward - look at the residues of cubes and powers of two modulo 13. You get that $n=3k$ for some natural $k$ then equation can be rewritten as:

$13=(m-2^k)(m^2+m2^k+2^{2k})$ from which it is easy to get contradiction.

R.H.S= $2^n\equiv 1,2,4  \mbox{mod} 7 $.
for n>0 m must be odd , $\Rightarrow m^3 \equiv 0,\pm1   \mbox{mod} 7 $.
L.H.S = $ m^3 + 13 \equiv O,5,-1 \mbox{mod} 7 $. then there are no natural numbers m, n satisfy the equation


Twenty-five different positive integers add to 2008. What is the largest value that the least of them can have?

[hide=" not too hard"]call the smallest number x
the smallest the sum can possibly be is x+x+1+x+2+x+3 ... +x+24 = 25x+300 (it can be anything greater than or equal to this)
so we have $ 25x\plus{}300 \le 2008 \implies x \le 68.32$
but x is an integer so x=68[/hide]

[quote="stevenmeow"]call the smallest number x
the smallest the sum can possibly be is x+x+1+x+2+x+3 ... +x+24 = 25x+300 (it can be anything greater than or equal to this)
so we have $ 25x \plus{} 300 \le 2008 \implies x \le 68.32$
[b]but x is an integer so x=68[/b][/quote]

I used this exact same logic, but i couldn't understand the last part (bold). I understand how its an integer, but how can you do that???

[hide="Hint Chain"]What is the largest integer that is less than or equal to 68.32?
[hide]68.
So what's the maximal value of $ x$?
[hide]=68.  :wink: [/hide][/hide][/hide]


induction
binomial coefficients
Divisibility Theory

Suppose that $m=nq$, where $n$ and $q$ are positive integers. Prove that the sum of binomial coefficients \[\sum_{k=0}^{n-1}{ \gcd(n, k)q \choose \gcd(n, k)}\] is divisible by $m$.

I don't know if I can only give a hint:
[b]Lemma 1[/b]:${p^{\alpha}dq-1 \choose p^{\alpha}d-1} \equiv {p^{\alpha-1}dq-1 \choose p^{\alpha-1}d-1}(\mod p^{\alpha+1}),\alpha>=1 $
[b]Lemma 2[/b]:$f(n,q)=q\sum_{d|n}{{dq-1\choose d}\phi(\frac{n}{d})}$
[b]Lemma 3[/b]:$pf(\frac{n}{p},q)\equiv f(n,q)$(We prove the Lemma 3with Lemma2 and Lemma1)
then by Lemma 3,we can proved by induction $nq|f(n,q)$

Thank you for your advice,I need to be patient and careful in CMO :lol:

My friend,if you still have any patience,please read the following solution :)

[b]Lemma[/b]$ X_{\alpha}\equiv X_{\alpha \plus{} 1},(modp^{\alpha \plus{} 1})$,
where$ p$ is a prime $ X_{\alpha} \equal{} \binom{dp^{\alpha}q \minus{} 1}{dp^{\alpha} \minus{} 1}$

OK.set $ f(x) \equal{} (x \minus{} 1)(x \minus{} 1)...(x \minus{} p \plus{} 1)$ where $ p$ is a prime.

use definition of $ \binom{m}{n}$ to expand  $ X_{\alpha},X_{\alpha \plus{} 1}$

We get $ X_{\alpha \plus{} 1} \minus{} X_{\alpha} \equal{} X_{\alpha}(\frac {f(dp^{\alpha \plus{} 1}q)f(dp^{\alpha \plus{} 1}q \minus{} p)...f(dp^{\alpha \plus{} 1}q \minus{} dp^{\alpha \plus{} 1} \plus{} p)}{f(dp^{\alpha \plus{} 1})f(dp^{\alpha \plus{} 1} \minus{} p)...f(p)} \minus{} 1)$

Note that $ f(kp)$ is not a multiple of $ p$

We're enough to show

$ f(dp^{\alpha \plus{} 1}q)f(dp^{\alpha \plus{} 1}q \minus{} p)...f(dp^{\alpha \plus{} 1}q \minus{} dp^{\alpha \plus{} 1} \plus{} p)$

$ \equiv f(dp^{\alpha \plus{} 1})f(dp^{\alpha \plus{} 1} \minus{} p)...f(p),modp^{\alpha \plus{} 1}$

It is obvious because $ f(dp^{\alpha \plus{} 1}q \minus{} kp)\equiv f(dp^{\alpha \plus{} 1} \minus{} kp),modp^{\alpha \plus{} 1}$ for any integer $ k$

which completes the proof for our lemma

Let's come back to our initial problem:

It is equivalent to $ \leftrightarrow n|\sum_{d|n}\binom{dq \minus{} 1}{d \minus{} 1}\phi(\frac {n}{d})$

If we denote $ n$ by $ n \equal{} p^{\alpha}m$ where $ (m,p) \equal{} 1$ we are sufficient to show

$ \leftrightarrow p^{\alpha}|\sum_{d|m}\sum_{k \equal{} 0}^{\alpha}\binom{dp^{k}q \minus{} 1}{dp^{k} \minus{} 1}\phi(\frac {p^{\alpha \minus{} k}m}{d})$

Moreover we are sufficient to show $ p^{\alpha}|A_{\alpha}$

where$ A_{\alpha} \equal{} \sum_{k \equal{} 0}^{\alpha}\binom{dp^{k}q \minus{} 1}{dp^{k} \minus{} 1}\phi(\frac {p^{\alpha \minus{} k}m}{d})$

Let's induction on $ \alpha$

Note that $ A_{\alpha \plus{} 1} \equal{} \sum_{k \equal{} 0}^{\alpha \plus{} 1}\binom{dp^{k}q \minus{} 1}{dp^{k} \minus{} 1}\phi(\frac {p^{\alpha \plus{} 1 \minus{} k}m}{d})$

and $ \phi(\frac {mp}{d}) \equal{} (p \minus{} 1)\phi(\frac {m}{d})$,$ \phi(\frac {mp^a}{d}) \equal{} p\phi(\frac {m^{a \minus{} 1}}{d})$for $ a > 1$

We get $ A_{\alpha \plus{} 1} \equal{} pA_{\alpha} \plus{} \binom{dqp^{\alpha \plus{} 1} \minus{} 1}{dp^{\alpha \plus{} 1} \minus{} 1}\phi(\frac {m}{d}) \minus{} \binom{dqp^{\alpha} \minus{} 1}{dp^{\alpha} \minus{} 1}\phi(\frac {m}{d})$

Thus $ A_{\alpha \plus{} 1}\equiv \phi(\frac {m}{d}) (\binom{dqp^{\alpha \plus{} 1} \minus{} 1}{dp^{\alpha \plus{} 1} \minus{} 1} \minus{} \binom{dqp^{\alpha} \minus{} 1}{dp^{\alpha} \minus{} 1}) \equiv 0,(mod p^{\alpha \plus{} 1})$ by the lemma

which completes our induction

QED

Is it OK now?

and dear moderator Peter,could you please delete all of my previous posts(including the attachments)Thank you in advance!!! :)

[quote="hxy09"]and dear moderator Peter,could you please delete all of my previous posts(including the attachments)Thank you in advance!!! :)[/quote]Done. I hope this problem has finally found its solution now.

There is a, in my opinion, much nicer solution using Burnside's Lemma.

Consider the set $ X $ of regular $ nq $-gons where exactly $ n $ of the vertices are colored red, and the other $ n(q - 1) $ of the vertices are colored blue.  Let $ G $ be the set of rotations of a regular $ nq $-gon by $ \frac{2k\pi}{nq} $ radians for $ k \in \{0, 1, 2, \dots, nq - 1\}. $  Consider the group action of $ G $ on $ X. $  Because no element of $ X $ is fixed by a rotation of $ \frac{2k\pi}{nq} $ radians if $ q \nmid k $ we have by Burnside's Lemma that the number of orbits of this group action is given by:
\[ \frac{1}{nq}\sum_{k = 0}^{n - 1}\binom{\text{gcd}(n, k)q}{\text{gcd}(n, k)} \]
Which must be an integer so we are done.


We have $ a$ and $ b$ are positive and: $ \sqrt{ab}\plus{} \sqrt{(a\plus{}1)(b\plus{}1)} \equal{} \sqrt[4]{2007}$
Find the value of :
$ S \equal{} \sqrt{a(b\plus{}1)} \plus{} \sqrt{b(a\plus{}1)}$

[hide]$ \sqrt {ab} \plus{} \sqrt {(a \plus{} 1)(b \plus{} 1)} \equal{} \sqrt [4]{2007}$

$ ab \plus{} ab \plus{} a \plus{} b \plus{} 1 \plus{} 2\sqrt {ab(a \plus{} 1)(b \plus{} 1)} \equal{} \sqrt [2]{2007}$

$ S^2 \equal{} 2\sqrt {ab(a \plus{} 1)(b \plus{} 1)} \plus{} 2ab \plus{} a \plus{} b  \equal{} \sqrt [2]{2007} \minus{} 1$[/hide]

EDIT: corrected a sign

[hide="Solution"]Note that $ \sqrt{x}\plus{}\sqrt{y}\equal{}\sqrt{x\plus{}y\plus{}2\sqrt{xy}}$.

Hence, $ \sqrt{ab}\plus{}\sqrt{(a\plus{}1)(b\plus{}1)}\equal{}\sqrt{ab\plus{}ab\plus{}a\plus{}b\plus{}1\plus{}2\sqrt{ab(a\plus{}1)(b\plus{}1)}}$.

Also, $ \sqrt{a(b\plus{}1)}\plus{}\sqrt{b(a\plus{}1)}\equal{}\sqrt{ab\plus{}ab\plus{}a\plus{}b\plus{}2\sqrt{ab(a\plus{}1)(b\plus{}1)}}$.

Since $ ab\plus{}ab\plus{}a\plus{}b\plus{}1\plus{}2\sqrt{ab(a\plus{}1)(b\plus{}1)}\equal{}\sqrt{2007}$, $ \sqrt{a(b\plus{}1)}\plus{}\sqrt{b(a\plus{}1)}\equal{}\sqrt{\sqrt{2007}\minus{}1}$.

Answer: $ \sqrt{\sqrt{2007}\minus{}1}$
[/hide]

[hide="solution"]
Squaring the given equation gives
\[ ab\plus{}ab\plus{}a\plus{}b\plus{}1\plus{}2\sqrt{ab(a\plus{}1)(b\plus{}1)}\equal{}\sqrt{2007}\]

Squaring $ S$ gives
\[ S^2\equal{}ab\plus{}a\plus{}ab\plus{}b\plus{}2\sqrt{ab(a\plus{}1)(b\plus{}1)}\]

Note that $ S^2\equal{}\sqrt{2007}\minus{}1$. Hence $ S\equal{}\sqrt{\sqrt{2007}\minus{}1}$.[/hide]


algorithm
function
number theory open
number theory

Let $ n\equal{}9$,
$ r_1\equal{}\frac{3}{5},r_2\equal{}\frac{67375}{108},r_3\equal{}\frac{175}{18},r_4\equal{}\frac{55}{39},r_5\equal{}\frac 13,r_6\equal{}\frac{26}{77},r_7\equal{}\frac 67,r_8\equal{}\frac{9}{13},$ $ r_n\equal{}189$.
Let sequence $ N_k$ defined by: $ N_0\equal{}2$ and $ N_{k\plus{}1}\equal{}N_kr_i$, were $ r_i$ is first $ r_j$, suth that $ N_kr_j$ is integer.
It give $ N_1\equal{}378,N_2\equal{}3675,...$.
Prove:
1) If $ N_k\equal{}2^m$, then $ m$ is prime.
2) For any prime p exist j, suth that $ N_j\equal{}2^p$.
3) There are not sequence of rationals $ r_1,...,r_n,n\le 8$ suth that 1),2) holds.

[quote="Rust"]Let $ n = 9$,
$ r_1 = \frac {3}{5},r_2 = \frac {67375}{108},r_3 = \frac {175}{18},r_4 = \frac {55}{39},r_5 = \frac 13,r_6 = \frac {26}{77},r_7 = \frac 67,r_8 = \frac {9}{13},$ $ r_n = 189$.
Let sequence $ N_k$ defined by: $ N_0 = 2$ and $ N_{k + 1} = N_kr_i$, were $ r_i$ is first $ r_j$, suth that $ N_kr_j$ is integer.
It give $ N_1 = 378,N_2 = 3675,...$.
Prove:
1) If $ N_k = 2^m$, then $ m$ is prime.
2) For any prime p exist j, suth that $ N_j = 2^p$.
3) There are not sequence of rationals $ r_1,...,r_n,n\le 8$ suth that 1),2) holds.[/quote]
but if $ N_j=2^{13}$ for exemple $ \Longrightarrow\ \frac{2^{13}}{N_{j-1}}=r_i=\frac{a}{b}: \ (gcd(a,b)=1)$ for some $ i\in\{1,2,...,9\}$
that gives $ a=2^{h}$ for some $ h$ but there aren't such $ r_i$   :!:   :!:

[quote="aviateurpilot"]sorry, we can have $ \frac {2^{13}}{N_{j \minus{} 1}} \equal{} r_5 \equal{} \frac {2^{0}}{3}$ :)[/quote]
Yes. $ N_{2560}\equal{}3*2^{13}$ and $ N_{2561}\equal{}r_i*N_{2560}$, were $ r_i$ is first $ r_j$, suth that $ N_{2560}*r_j$ is integer. It is $ r_5\equal{}\frac 13$.  :lol:

[quote="aviateurpilot"]we can work with application $ f: \mathbb{N}\to \mathbb{Z}^{6}: \ f(x) \equal{} (a_1,a_2,a_3,a_4,a_5)\iff x \equal{} \prod_{k \equal{} 1}^{6}p_{k}^{a_k}$ where $ p_k \equal{} k^{th}\ prime$
[/quote]
Yes. See [url]http://mathworld.wolfram.com/FRACTRAN.html[/url]. But Convey's algoritm had 14 rationals.

But $ a_9\equal{}m\minus{}1$ give noting. Why $ m$ is prime?

[quote="Rust"]But $ a_9 \equal{} m \minus{} 1$ give noting. Why $ m$ is prime?[/quote]
(sorry for ma bad english, i can't write what i want clairly !! :oops: )
i didn't say that it gives $ m$ prime.
i say just that if we prove $ 1)$ and $ 2)$ this result ($ a_9\equal{}m\minus{}1$) gives more
gives that $ \exists (x_n)....$ :)

i'am clair now?

For any strongly increase recursive function $ f: N\to N, f(1)>1$ exist suth algorithm and $ a_n\equal{}f(n)\minus{}1$. For another algorithm $ f(n)$ is not prime. For example it is not hard built Convey's algorithm for $ f(n)\equal{}F_{n\plus{}2}$.

[quote="Rust"]For any strongly increase recursive function $ f: N\to N, f(1) > 1$ exist suth algorithm and $ a_n \equal{} f(n) \minus{} 1$. For another algorithm $ f(n)$ is not prime. For example it is not hard built Convey's algorithm for $ f(n) \equal{} F_{n \plus{} 2}$.[/quote]
you want say that ther exists an algorithm such that $ \exists i: \ N_i\equal{}2^{m}\iff m\equal{}f(j)$ :$ f(j)$ not necessary prime.

can you gives me an example?

[quote="aviateurpilot"][quote="Rust"]For any strongly increase recursive function $ f: N\to N, f(1) > 1$ exist suth algorithm and $ a_n \equal{} f(n) \minus{} 1$. For another algorithm $ f(n)$ is not prime. For example it is not hard built Convey's algorithm for $ f(n) \equal{} F_{n \plus{} 2}$.[/quote]
you want say that ther exists an algorithm such that $ \exists i: \ N_i \equal{} 2^{m}\iff m \equal{} f(j)$ :$ f(j)$ not necessary prime.

can you gives me an example?[/quote]
Yes. It is obviosly if you know calculations in nets Petry.


This is equation: $ c{\rm{os}}2t \minus{} \sin 2t \plus{} \cos t \equal{} 0$.

hello, by applying the addition formulas we get
$ 2\cos^2(t)\minus{}1\minus{}2\sin(t)\cos(t)\plus{}\cos(t)\equal{}0$. This equation is equivalent with
$ \frac{2(1\minus{}\tan(\frac{1}{2}t)^2)^2}{(1\plus{}\tan(\frac{1}{2}t)^2)^2}\minus{}1\minus{}\frac{4\tan(\frac{1}{2}t)(1\minus{}\tan(\frac{1}{2}t)^2)}{(1\plus{}\tan(\frac{1}{2}t)^2)^2}\plus{}\frac{1\minus{}\tan(\frac{1}{2}t)^2}{1\plus{}\tan(\frac{1}{2}t)^2}\equal{}0$
Substituting $ \tan(\frac{1}{2}t)\equal{}z$ we get
$ \frac{2(1\minus{}z^2)^2}{(1\plus{}z^2)^2}\minus{}1\minus{}\frac{4z(1\minus{}z^2)}{(1\plus{}z^2)^2}\plus{}\frac{1\minus{}z^2}{1\plus{}z^2}\equal{}0$. By simplifying we get
$ 2z^3\minus{}3z^2\minus{}2z\plus{}1\equal{}0$.
Now you can solve this equation.
Sonnhard.

Can we write $ \left\{ \begin{array}{l}
 \sin t \equal{} x \\
 \cos t \equal{} y \\
 \end{array} \right.$ then to equation system?


geometry
circumcircle
conics
ellipse
hyperbola
analytic geometry
projective geometry

I dunno I came up with this in about a minute.

For $ \triangle ABC$, let the tangent at $ A$ to the circumcircle meet $ BC$ at $ A'$. Define $ B'$ and $ C'$ analogously. Prove that $ A'$, $ B'$, and $ C'$ are collinear.

What's wrong with my diagram?

[img]http://i25.tinypic.com/2v9onma.gif[/img]

You have C' as the intersection of two tangents. It should lie on AB.

an extended form of ab right?

edit: I'm guessing that its an extended form of AB or else you would have pointed that out

bah I solved it but then realized I made a bad assumption -_-

[quote="not_trig"]dude what it is pascal yawn done[/quote]

Eh how do you use pascal's for this...I thought Pascal's applied only to hexagons o.O

do you make a hexagon with A',B',C',A,B,C (which won't work anyway since I don't think A'B'C'ABC is cyclic) and then do like the converse of pascal's (which I'm not even sure holds) o.O

[quote="#H34N1"]I think this works...

[hide]Invert around $ A$ with a radius sufficiently large but smaller than $ AB$. Then $ A$ goes $ A'$ to infinity, $ B$ and $ C$ to $ B'$ and $ C'$ respectively and remain collinear. $ A',B',C'$ are collinear. So, since the tangent lines must intersect with $ A'B'$, $ B'C'$, and $ A'C'$, then it is clear that the intersection points $ X'$, $ Y'$, and $ Z'$ are collinear. Re-inverting, we are done.[/hide][/quote]

The inverses of three collinear points are not necessarily collinear, so I'm not sure how you deduced $ X$, $ Y$, and $ Z$ being collinear from $ X'$, $ Y'$, and $ Z'$ being collinear, and I don't quite see how you got that anyway? Maybe I just misunderstand what you said...

Anyway, about Pascal, you use hexagon $ AABBCC$. Yeah, this may be kind of weird. But basically the line "AA" means the tangent to the circle at point A. Just consider two points A and B on a circle and the limit of line AB as B gets closer to A.

So for Pascal's on AABBCC, you get the three points being collinear

AA and BC = tangent at point A and BC
AB and CC = tangent at point C and AB
BB and CA = tangent at point B and CA

Okay I was totally not thinking when I wrote that :dry:...I'll rethink it
I don't know why I thought that something must be collinear to intersect :maybe:

Wait...what's wrong with my logic again? X is a point on line BC. Y and Z are defined analogously. So since the inverses of A,B,C are collinear, X,Y,Z lie on that line so they're also collinear....unless I messed up again?

Also how is AABBCC concyclic?

Yes, its true that A', B', and C' would be collinear (with A' being at infinity). Now, are you saying that X, Y, and Z should lie on B'C'? For that, you would need to define the radius of the circle of inversion.

If you mean that X', Y', and Z' should like on B'C', that wouldn't be true either; B, C, and Z are collinear, but that doesn't mean B', C', and Z' collinear. (It does mean that AB'C'Z' is cyclic).

AABBCC is concyclic because all of its points lie on one circle (the circumcircle of triangle ABC). Then AA is the tangent line to that circle, and so are BB and CC.

[quote="CatalystOfNostalgia"]what if the stuff was on an ELLIPSE INSTEAD  :o

OR A HYPERBOLA[/hide][/quote]

then we use coordinates :)

Pascal's theorem does not only apply to circles. It works for any conic section. So the solution would be exactly the same.  :D

Oh I actually forgot about that.  :blush:

On the page I read about Pascal's theorem, it said that the projective plane is its home and that in the Euclidean (?) plane, it has exceptions.  I don't actually remember if it was the Euclidean plane or not, but what is the projective plane? And what are the exceptions?

Probably, the exception refers to when two lines would be parallel. Like, in hexagon ABCDEF, if AB is parallel to DE, then they would meet at an infinity point. However, you could then say that GH (where G is the intersection of BC and EF, H the intersection of CD and FA), would have to pass through this same point at infinity, meaning it would be parallel to AB and DE.


function
topology
real analysis
real analysis unsolved

Can any one tell me if there does or does not exist a function that has uncountably many discontinuities on the reals but is continues on the extended reals?

Motivation: Consider the function f(x)={ Tan(x)          -Pi/2<x<Pi/2
                                                       -infinity         x =< -Pi/2
                                                         infinity         x => Pi/2}

what do you mean "continues on the extended reals" ??

When a function isn't defined at a point, analysts don't call that a discontinuity.

Sorry, let me clarify what I was saying. Does there exist a function f:R-->[0,infinity] that has uncountably many discontinuities in R (non extended reals), but is continuous in the extended reals (my definition of continuous in the extended reals means for every open subset of [0,infinity], say (a,infinity] the preimage of that open set is open in R)?

And sorry about my previous example, please consider the function
f(x)={|sec(x)|      (Pi/2 - n*Pi) <x< (Pi/2+n*Pi) where n is an integer
        infinity       x= n*Pi/2     where n is an integer}
This function has countably many "discontinuities" (in the sense like poles) if our range does not include infinity, but it is continuous if our range does include infinity (I believe it is continuous).

Yes. Let $ E$ be the Cantor set, and set $ f(x) \equal{} 1/d(x,E)$ for $ x \notin E$, $ f(x) \equal{} \infty$ for all $ x \in E$.

I don't think this function is continuous in the extended reals.

[quote="byakkoo"]Actually, what do you mean by f(x) = 1/d(x,E), or simply what do you mean by d(x,E)?[/quote]

$ d(x,E)$ is the distance from $ x$ to $ E$, and it is not very good manners to say my example fails before you even understand what it is.


find all positive integers n s.t. d(n) + :phi: (n) = n
where d(n) is the number of positive divisors of n (including 1 and n)
and :phi: (n) is the number of positive integers <= n that are relatively prime  to n.

n can be [color=white]6, 8 or 9.
Look for n such that the set {m:1<m<n, m is not coprime to n and m is not a divisor of n} has only 1 member.[/color]

nice insight, when I first tried the problem i tried using equations for d(n) and  :phi: (n) and got nowhere quickly, until i made the same realization you did.

anyway, here is an extension to this problem:

find all n in positive integers such that  :stddev: (n) +  :phi: (n) = 2n
where  :stddev: (n) = the sum of the positive divisors of n (including 1 and n)
and  :phi: (n) is the number of positive integers <= n that are relatively prime to n.

How about this ...

[hide]If n is a prime, then  :stddev:(n)=n+1 and  :phi:(n)=n-1 so  :stddev:(n)+ :phi:(n)=2n.  So all primes have this property.

We wil prove that if n is composite then  :stddev:(n)+ :phi:(n)>2n, so no composite numbers have the property.

Suppose n has N proper divisors (i.e. divisors that are not equal to n or 1). Let d be a proper divisor of n, and let D be the sum of the proper divisors of n.

Then  :stddev:(n) = D + n + 1

To estimate  :phi:(n), we note that a proper divisor d will divide (n/d)-1 integers less than n. So to get an estimate of the number of integers <=n and *not* co-prime to n, we can sum (n/d)-1 over proper divisors d, and add 1 for n itself. This will over-estimate the number of integers not co-prime to n, because some integers will be counted twice. So

#(integers not co-prime to n) <= sum over d[(n/d)-1] + 1

Noting that the sum over d of n/d is D (because every n/d is another proper divisor of n) and the sum over d of 1 is the number of proper divisors, N, we have

#(integers not co-prime to n) <= D - N + 1

=>  :phi:(n) >= n - D + N - 1

=>  :stddev:(n) +  :phi:(n) >= 2n + N

So if n is composite (i.e. N>0) then  :stddev:(n)+ :phi:(n)>2n.[/hide]

good job, my solution for this problem was a bit overkill (it used Mbius inversion)


function
trigonometry
integration
real analysis
real analysis unsolved

Suppose $k \in \mathbb R$, $k >0$, and that $f$ is a real valued function on an interval $I$ such that $f''(x)$ exists for all $x \in I$, and $f''(x)+k^{2}f(x) = 0$ for all $x \in I$.  Show that there exist constants $A$ and $B$ such that $f(x) = A \cos kx+B \sin kx$.

Question: do we need to assume $f$ is bounded on $I$ to be able to say the power series we construct for $f$ converges to $f$ ?

let $f(x) \vec i$ denote the displacement of a particle inmoving in  $x$ direction then

then $f^{''}(x)$ $=$ $-k^{2}f(x)$

represents a $s.h.m$ in the $x$ direction so if you have studied harmonic motion this is done and will lead to the sol^n by cckek

thanks for these solutions, but i am specifically trying to solve it with power series

Edit: i figured out my question now


Functional Analysis
advanced fields
advanced fields unsolved

Hi everybody ! We got some exam problems to prepare ourselves for the final and I can't understand how to solve this one :(

We define a weighted shift $ W$ on a Hilbert space $ H$. It acts on a basis by $ We_n \equal{} w_n e_{n \plus{} 1}$ where $ \{ w_n \}$ is a bounded sequence of complex numbers.

First we show $ \parallel{} W \parallel{} \equal{} \sup_{n}{|w_n|} < \infty$ so $ W$ is well-defined as a bounded operator.

[u]Now, we have to show that there is a diagonal unitary $ U_{\theta}$ such that $ U _{\theta}WU_{\theta}^{\star} \equal{} e^{i\theta}W$ for all $ \theta \in (0,2\pi)$ and then deduce that $ \sigma(W)$ (the spectrum) has circular symmetry.[/u]

So we have to find a diagonal $ U_{\theta}$ which almost commutes with $ W$,  $ U _{\theta}W \equal{} e^{i\theta}WU_{\theta}$ and $ U_{\theta}$ preserve ALL the structure of the Hilbert space (because unitary)... and don't really know what to do, I tried many $ U_{\theta}$ but each fails at least one of those conditions...

any help/hint :) ??

If $ U_{\theta}$ is diagonal, $ U_{\theta} e_n \equal{} a_n e_n$ and so LHS $ \equal{} w_n a_{n \plus{} 1} e_{n \plus{} 1}$ and the RHS $ \equal{} e^{i\theta}a_n w_n e_{n \plus{} 1}$ so we should have recursively $ a_{n \plus{} 1} \equal{} e^{i (n \plus{} 1) \theta} a_0$, (where $ w_n$ is not 0). Also here $ \{e_n\}$ is not necessarily orthogonal anymore but it doesn't really matter.

And then $ \langle U_{\theta} e_n, e_m \rangle \equal{} e^{i n \theta} a_0 \langle e_n, e_m \rangle \equal{} \langle e_n, e^{ \minus{} i n \theta} \bar{a_0} e_m \rangle \equal{} \langle e_n, U_{\theta}^{ \minus{} 1}e_m \rangle$ so $ e_m \equal{} U_{\theta} (e^{ \minus{} i n \theta} \bar{a_0} e_m) \equal{} e^{i (m \minus{} n) \theta} \bar{a_0} a_0 e_m$ but this is true for all $ m,n$ and $ \theta$ is fixed so it leads to nothing and something went wrong in my development....  :huh:  :huh:

Working on functional assignment huh? ;)
How hard is the course? I am taking it next winter (year and term).
From what I heard at the PMC his exam's are going to be knowledge based but not difficult =)

Okay i'm stupid in fact just work backward as above and find a recursive recursive relation for the coefficients....
It should be something like $ U_{\theta} \equal{} diag(d_1, d_2,...)$ and $ d_{n\plus{}1} \equal{} d_n e^{i \theta}$, begin with $ d_0 \equal{} 1$ so $ |d_i| \equal{} 1$ and the operator is unitary !


What are the common substitution ciphers involved in puzzle-solving?

could someone please answer my question?

OK, It's been almost a month since I asked this question and no one has responded yet.
If someone doesn't answer soon, I'll blow my top.  :bomb:

:huuh:
Erm, I asked for substitution ciphers used in [b]puzzle-solving[/b]. From what I read in Wikipedia, it doesn't look like [b]SHA-512[/b] is used to solve puzzles.
Or maybe you were joking  :?:
In which case,  :)

To make myself clear, I'm just looking for a sort of round-up of common substitution ciphers, like the caesar cipher, which are used in puzzles quite often.

Fbjzj zkiss rwolp skcwt kooki tjlbs djifb sVwos zckdh o.

Not if its any sort of substitution. What sort of sentence ends in a one-letter word?

You better rethink your way you look at block codes  :wink:

Well, seeing as I haven't worked with ciphers much, how should I be looking at them? What are block codes, anyway? :huh:

[b]Regarding post #4 :[/b] C'mon, I was (clearly) just kidding. Why did someone give me such a bad rating?

Anyway, I wasn't able to decode [b]aleph0[/b]'s post. Could someone help me with that?
And I'm still hoping for a list of common ciphers.


Γεια σας παιδια,ειμαι νεο μελος και ελπιζω να βρω καποιες απαντησεις σε σχεση με τα μθηματικα.Μετα απο ενα διαστημα αδρανειας ειμαι πλεον σε θεση ν ασχοληθω και παλι καπως πιο σοβαρα με τα μαθηματικα και συγκεκριμενα με μαθηματικες ολυμπιαδες.Οποτε θελω να μου προτεινετε καποια καλα βιβλια αναλυσης και γραμμικης αλγεβρας στα ελληνικα αλλα και στα αγγλικα.Επισης αλλη μια ερωτηση εκτος απο seemous kai IMC υπαρχουν αλλοι διαγωνισμοι για φοιτητες?
Ευχαριστω

Αυτό που χρειάζεσαι 100% είναι το Putnam and Beyond.
Γενικά καλό θα ήταν να πας και στα μαθήματα της μαθηματικής κάθε Σ/Κ 2-5.

Όσο για υπόλοιπα βιβλία:
1)Το "problems and theorems in Analysis I &II" τoυ Polya είναι πολύ γερό βιβλίο, τίγκα ασκήσεις και όχι μόνο ανάλυσης. Γενικά αξίζει αλλά δεν το έχω ακουμπήσει οσο θα θελα
2)Αν θες για λίγο πιο αφηρημένη γραμμική αλγεβρα - όχι ακριβώς άμεσα προετοιμασία για διαγωνισμούς αλλά πολύ ισχυρό υπόβαθρο θα σου πρότεινα το "Linear Algebra and its applications" του Lax.

Ευχαριστω πολυ.Μηπως υπαρχει καποιο αξιολογο ελληνικο βιβλιο γιατι τ αγγλικα μου δεν νομιζω να με βοηθησουν ?

Χαιρομαι για αυτην την ελληνικη εκδοση κι ευχαριστω και παλι

[quote="lemonidas"]To putnam and beyond είναι μεταφρασμένο στα ελληνικά και πρέπει να το παρεις 100%. Τώρα τα άλλα δύο δεν νομίζω να έχουν μεταφραστεί... Οπότε ας απαντήσει κάποιος άλλος για ελληνικά βιβλία  :P[/quote]
To putnam and beyond στα ελληνικα :!:  :?:
ποιες εκδοσεις το εχουν μεταφρασει????


function
complex analysis
complex analysis unsolved

May someone help me in the following question?

Let
$ f_n(z)\equal{}1\plus{}\dfrac{1}{z}\plus{}\dfrac{1}{2!z^2}\plus{}\ldots\plus{}\dfrac{1}{n!z^n},  \, n\in\mathbb N.$
Prove that, for each $ \rho>0$, there is a natural $ n_0$ such that if $ n\geq n_0$, then all zeros of $ f_n$ belong to the disc $ |z|<\rho$.

Thanks in advance.

Note that $ \sum_{j\equal{}0}^{\plus{}\infty}\dfrac{1}{j!z^{j}} \equal{} \exp (1/z)$ and use Hurwitz's Theorem.

[quote="croiss"]... use Hurwitz's Theorem.[/quote] whose proof can rely on Rouché's theorem. ;)


[quote="2006"]The real number$\ a,b$ satisfy the condition   $\aa^{2}+b^{2}$ $\leq 1$
prove that     $\((ac-bd)^{2}$ $\geq$ $\((a^{2}+b^{2}-1)(a^{2}+b^{2}-1)$
                                   good luck!!       :D[/quote]
Is it "The real number$\ a,b$ satisfy the condition   $\aa^{2}+b^{2}$ $\leq 1$
prove that     $\((ac-bd-1)^{2}$ $\geq$ $\((a^{2}+b^{2}-1)(c^{2}+d^{2}-1)$
?
I think it is a famous inequality called "Azell"(I am not sure about the name)
It is a application of Chauchy
And it holds for all $c,d$

can i ask you : what is c and d
is c.c+d.d<=1 , too

Sorry,I have made a mistake. :oops: I have edited my post.


Let $ a, b, c >0$ intigers such that  $ a^2\plus{}2005^2\equal{}b^2\plus{}c^2$   and    $ \text{max}\{a,b,c\} < 2005$.

Find minmum and maximum $ a\plus{}b\plus{}c.$

If $ b^2\plus{}c^2\equal{}N$ number $ b\plus{}c$ had maximum value, when $ b\equal{}c, b\plus{}c\equal{}\sqrt{2N}$, minimum when one of them minimal, second maximal value.
Therefore, if condition is $ max(a,b,c)\le 2005$, then maximum $ a\plus{}b\plus{}c\equal{}3*2005$, minimum, when $ a\equal{}1,b\equal{}1,c\equal{}2005, a\plus{}b\plus{}c\equal{}2007$.


[quote="Pirkuliyev Rovsen"]1)$ \left\{\begin{array}{l} x^{3} \minus{} y \equal{} 6 \\
y^{3} \minus{} z \equal{} 6 \\
z^{3} \minus{} x \equal{} 6 \end{array}\right.$ [/quote]
[hide]
Let $ x > 2.$ Hence, $ z^3 \equal{} x \plus{} 6 > 8,$ which gives $ z > 2.$
Hence, $ y^3 \equal{} z \plus{} 6 > 8,$ which gives $ y > 2.$
Similarly, if $ x < 2$ we obtain: $ z < 2$ and $ y < 2,$ but it's contradiction
because $ \sum_{cyc}(x^3 \minus{} y \minus{} 6) \equal{} 0\Leftrightarrow\sum_{cyc}(x \minus{} 2)(x^2 \plus{} 2x \plus{} 3) \equal{} 0.$
[/hide]

Were you proving the only solution is (2, 2, 2)?

Something tells me that's the only solution to the other one is (2, 2, 2) as well...

I'll give it a try, I know this is a horrible proof...lol

[hide="not rigourous at all"]If you increase x by a value a, then y will also have to increase, but by a value greater than a.  Then z must also increase by a value bigger then a.  Then x will also increase by a value greater than a, contradiction.  Prove it similarly with the decreasing x.[/hide]


algebra
polynomial
function
LaTeX
algebra proposed

Let $ P\in\mathbb{R}[x]$ such that $ \deg P=n$ and $ P$ has $ n$ real roots in $ (1,\infty)$. Prove that $ Q(x)=(x+1)P(x)P'(x)+P^{2}(x)+xP^{2}(x)$ has at least $ 2n-1$ distinct real roots.


search
geometry
circumcircle
trigonometry
angle bisector
geometry unsolved

In triangle $ABC$, $D$, $E$ and $F$ are points on $BC$ such that $AD$, $AE$ and $AF$ are an altitude, an angle-bisector and a median, respectively. If $AD=12$, $AE=13$ and $AF=m$, determine the values of $m$ for which $\angle BAC$ is acute, right and obtuse.

Does my solution work? It seems correct to me, but the numbers aren't exactly what you'd call pretty.

[hide="Solution"]Because there's only one value of $m$ such that $\angle BAC=90^\circ ,$ we search for that first. Let this distinct value of $m$ be $x$.

First, we prove that when $\angle BAC=90^\circ$, $\angle DAE=\angle EAF.$

[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=812[/img]

In the diagram above, we know that $\angle BAE=\angle CAE.$ Hence, $\angle DAE=\angle EAF$ if and only if $\angle BAD=\angle CAF.$ Notice that triangle $AFC$ is isosceles because $AF=CF=\frac12BC$. Thus, $\angle FAC=\angle FCA$. Additionally, $\angle CBA=90^\circ- \angle BCA.$ However, triangle $ADB$ is right, so $\angle BAD=90^\circ- \angle CBA=\angle FCA$. Therefore, $\angle BAD=\angle FCA=\angle CAF$ and our proof is complete.

[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=813[/img]

Now, we continue with the right triangle case. Let $EF=y$. By the angle bisector theorem, $\frac5{12}=\frac yx \implies y= \frac{5}{12}x.$ Additionally, Stewart's Theorem tells us that $13^2=12x\left( 1-\frac{(5+y)^2}{(12+x)^2}.\right)$

Plugging in $y=\frac5{12}x$, we obtain $13^2=12x\left( 1-\frac{(5+\frac5{12}x)^2}{(12+x)^2}.\right)$ However, notice that $\left(\frac{(5+\frac5{12}x)^2}{(12+x)^2}\right)=\left(\frac{(60+5x)^2}{144(12+x)^2}\right)=\frac{25}{144}!$ Hence, $169=12x-\frac{25}{12}x$ and $x=\frac{2028}{119}.$

Now, note that if $m>x$, $\angle BAC>90^\circ$ and if $m<x$, $\angle BAC<90^\circ$. We now need the smallest and largest possible values of $m$. This means that any value of $m$ greater than the smallest possible value of $m$ and smaller than $x$ makes $\angle BAC$ acute, and any value of $m$ less than the largest possible value of $m$ and greater than $x$ makes $\angle BAC$ obtuse.

[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=814&sid=314cd3dce246fe70f5f5a977bc27f6c0[/img]

We now need to find the extreme values of $m.$ Notice that triangle $ADE$ is always fixed as a $5-12-13$ right triangle, so we construct this triangle.

We can now select points $B$ and $C$ anywhere on line $DE$, so long as $\angle BAE=\angle CAE$. Though this effects the length of the median, it has absolutely no effect on the proportion of triangle $ADE.$

Notice that by selecting $B$ and $C$ such that $\angle BAC$ is infinitely close to $180^\circ,$ the length of median $AF$ approaches infinity. By selecting $B$ and $C$ such that $\angle BAC$ is infinitely close to $0^\circ$, the length of $AF$ approaches $13,$ the length of the angle bisector.

In conclusion, $\angle BAC$ is acute when $13<m<\frac{2028}{119}$, right when $m=\frac{2028}{119}$, and obtuse when $\frac{2028}{119}<m<\infty .$[/hide]

Let $\angle A = 90^\circ$ be right, $a^2 = b^2 + c^2.$ Let $\angle A = 90^\circ$ be right, $a^2 = b^2 + c^2.$ A-symmedian is identical with A-altitude. because A-symmedian connects A with the intersection of the circumcircle tangents at B, C, but these are both perpendicular to the diameter BC, i.e., parallel. Hence A-symmedian is also parallel to them, i.e., perpendicular to BC and $\angle EAD = \angle EAF.$

median: $m = AF = \frac a 2$

altitude: $h = AD = \frac{bc}{a}\ \Longrightarrow\ bc = ah$

bisector: $k = AE = \frac{bc \sqrt 2}{b + c}\ \Longrightarrow\ k^2(b^2 + 2bc + c^2) = 2b^2c^2$

Substituting $bc = ah$ and $a^2 = b^2 + c^2$ into the last equation:

$k^2(a^2 + 2ah) = 2a^2h^2$

Reducing by a:

$k^2(a + 2h) = 2ah^2\ \Longrightarrow\ a = \frac{2hk^2}{2h^2 - k^2}$

$m = \frac a 2 = \frac{hk^2}{2h^2 - k^2} = \frac{12 \cdot 169}{288 - 169} = \frac{2028}{119}$

simpler way[hide]Notice that $\angle DAE=\angle EAF$. Let $\angle DAF= \delta$ :D than $\cos \delta=\frac{12}{13}$ and $\cos 2\delta=2\cos^2 \delta -1= 2*\frac{144}{169} -1=\frac{119}{169}=\frac{12}x$ so $x= \frac{169*12}{119}$[/hide]


Well folks, I was told that many geniuses work at D.E.Shaw, New York ( at other D.E.Shaw offices, but mostly at New York), I was also told that the interview process that seeks prospective candidates for a Quantitative Analyst etc comprises many grueling rounds of interviews which includes proving theorems, etc.

I would like to know if the above information that is accurate?

I would like to know if the geniuses(genius in what sense) working at Shaw are indeed geniuses/ do they have excellent credentials?

I thank you all.

Regards

I don't know if I'd call the interview process "grueling".  It's certainly a lot easier than, say, the USAMO.

As for genius, that's not a label I particularly like.  There are many IMO medalists and Putnam winners at Shaw.  If you want to call them "geniuses", then, yes, I suppose there are plenty of geniuses there.  During the 4 years I was there, my coworkers were an intellectually impressive lot -- probably on average more interesting and impressive than the people I went to MOP with.  Maybe on average not quite as good at math as the MOPpers, but more able to apply their skills to a broader range of problems, and more able to operate in uncertain environments.  MOP had more people who were uncomfortable outside a very narrow range of problems (i.e., theoretical math problems), but these observed differences might just be because the MOPpers were all 15 or 16 years old as opposed to the 25-40 year olds I worked with at Shaw.


Given are $\triangle ABC$ area $S$, with any $M$ on plane prove that
$(MA+MB+MC)(a^{2}\frac{MB}{MC}+b^{2}\frac{MC}{MA}+c^{2}\frac{MA}{MB})\ge 4S(a+b+c)$  :D

I'm not sure yet but it looks like Cauchy and using fact that $S=\frac{r(a+b+c)}{2}$

Are you sure me@home, can you write clearly?

No I'm not sure at all, that's just what I thought of when I first saw the problem. \[(MA+MB+MC)(a^{2}\frac{MB}{MC}+b^{2}\frac{MC}{MA}+c^{2}\frac{MA}{MB})\ge 2(a+b+c)^{2}r\] By Cauchy Schwarz, \[(MA+MB+MC)(a^{2}\frac{MB}{MC}+b^{2}\frac{MC}{MA}+c^{2}\frac{MA}{MB})\ge \left(\sum_{cyc}a\sqrt{MB}\right)^{2}\] That is what I was talking about in the first post. Now I am not very good with geometric inequalities so I am not sure what to do from here, but it seems very plausible that it is an easy one step.

[color=purple]This following problem is true, too  :D and it's my own 100% :D
   $ (MA \plus{} MB \plus{} MC)(\frac {MB}{MC}bc \plus{} \frac {MC}{MA}ca \plus{} \frac {MA}{MB}ab)\ge 4S(a \plus{} b \plus{} c)$[/color]


[quote="manlio"]For $ a,b,c$ positive reals

$ \sum_{cyc} \frac {a^2}{b^2} \plus{} 6 \geq \frac {3}{2}\sum_{cyc} \frac {a^2 \plus{} b^2}{ab}$[/quote]

<=> $ 2(x^2\plus{}y^2\plus{}z^2)\plus{}12 \ge 3(x\plus{}y\plus{}z\plus{}xy\plus{}yz\plus{}zx)$ for $ abc\equal{}1$
using ABC method  :wink:

Have you a proof with classical inequalities?

Thank you very much.

You can use mixing variables Manlio . Let x=a^2 etc and then abc=1 and let a be the min

Then $ f(a,b,c)-f(a,\sqt{bc},\sqrt{bc})=(y-z)^2(2y^2+2z^2+4yz-3-3x^2)\geq 0$

Then you have to prove that $ f(a,\sqt{bc},\sqrt{bc})\geq -12$ which is equivalent to

$ \frac{(x-1)^2(2x^4+4x^3+3x^2-4x+1)}{x^2}\geq 0$ which clearly holds

dear manlio  :) I think you should try use "p,q,r technieque" because it is very powerful for this kind ...
The inequality is equivalent to:
\[ 2p^2\plus{}12\minus{}3p\minus{}7q\ge 0\]
Note that $ p\equal{}a\plus{}b\plus{}c,q\equal{}ab\plus{}bc\plus{}ca,r\equal{}abc\equal{}1$. Using Schur's inequality we have: $ \minus{}q\ge \minus{} \frac{9\plus{}p^3}{9p}$. So we have:

\[ 2p^2\plus{}12\minus{}3p\minus{}7q\ge 2p^2\plus{}12\minus{}3p\minus{} \frac{7(9\plus{}p^3)}{9p}\ge 0\]
\[ \iff \frac{(p\minus{}3)(p^2\minus{}9p\plus{}21)}{4p}\ge 0\]
It is clearly true because $ p\ge 3\sqrt[3]{r}\equal{}3, p^2\minus{}9p\plus{}21\ge 0$. We are done. Manlio, you can read more "p,q,r technique" in my article. Contact me if you want to have one. Good luck to you  :wink:

Many thanks, Zaizai.

P.S. Is "p,q,r tecnique" an article of many pages you have already attached to an old post of yours?


linear algebra
matrix
geometry
geometric transformation
rotation
rotation matrix

(i) Find a real $ 3 \times 3$ matrix with eigenvalues $ 1, i, \minus{}i$.
(ii) Construct a $ 3 \times 3$ non-zero real matrix which has all three eigenvalues zero.

For (ii) just write a triangular matrix with all elements of main diagonal equal to zero (such matrix will then be nilpotent). For (i), I think its time for you to show some work of your own.

For (i) I have just obtained the answer: a rotation matrix $ \left( \begin{array}{ccc} 0 & \minus{}1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{array} \right)$. This was because of a hint that asked me to think geometrically - how else could this problem be approached?

For (ii) I have a question about its phrasing - does non-zero real matrix mean "no zero entries"? Otherwise I have $ \left( \begin{array}{ccc} 0 & 1 & 2 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{array} \right)$

We can probably do that with no entries zero.

Let $ N\equal{}\begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix}.$

$ P\equal{}\begin{bmatrix}1&\minus{}1&2\\ 3&\minus{}4&4\\ 1&\minus{}2&1\end{bmatrix}.$

We can compute that $ P^{\minus{}1}\equal{}\begin{bmatrix}\minus{}4&3&\minus{}4\\1&1&\minus{}2\\2&\minus{}1&1\end{bmatrix}.$

(There's a little bit of rigging there: I've got a list somewhere of $ 3\times 3$ integer matrices with determinant $ \pm 1.$)

Now compute $ PNP^{\minus{}1}\equal{}A\equal{}\begin{bmatrix}\minus{}3&2&\minus{}3\\\minus{}11&7&\minus{}10\\\minus{}5&3&\minus{}4\end{bmatrix}$

If you multiply it out, you'll see that $ A^3\equal{}0.$

[quote="Kent Merryfield"]We can compute that $ P^{ \minus{} 1} \equal{} \begin{bmatrix} \minus{} 4 & 3 & \minus{} 4 \\
1 & 1 & \minus{} 2 \\
2 & \minus{} 1 & 1\end{bmatrix}.$

(There's a little bit of rigging there: I've got a list somewhere of $ 3\times 3$ integer matrices with determinant $ \pm 1.$)

Now compute $ PNP^{ \minus{} 1} \equal{} A \equal{} \begin{bmatrix} \minus{} 3 & 2 & \minus{} 3 \\
\minus{} 11 & 7 & \minus{} 10 \\
\minus{} 5 & 3 & \minus{} 4\end{bmatrix}$[/quote]

Sorry, why would you need the determinants of $ P$ to be $ \pm 1$? That sort of list does come in handy though :P

I don't [i]need[/i] the determinants to be $ \pm1.$ It's just that it's less painful to write out integer matrices and not have fractions everywhere. Just a minor convenience.

Oh, I see. So we can conclude that if $ \mathsf{N}$ is a nilpotent matrix, then $ \exists$ some invertible matrix $ \mathsf{P}$ such that $ \mathsf{PNP}^{\minus{}1}$ is equal to a nilpotent matrix with all entries non-zero?

Observe that $ A \equal{} PNP^{ \minus{} 1} \equal{} > A^k \equal{} PN^kP^{ \minus{} 1}$, $ k \in \mathbb{N}$, so that if N is nilpotent so is any matrix similar to N. In the language of linear transformations there is nothing fancy about that, since similarity is just the representation of the same operator relative to different bases. In fact, we can even find nilpotent matrices $ N \in M_{n \times n}$ such that $ N, N^2, N^3, ..., N^{n \minus{} 1}$ have all elements different from zero and $ N^k \equal{} O_{n \times n}$. Two more remarks:

1) For every nilpotent matrix, the smallest $ k$ for which $ N^k \equal{} O_{n \times n}$ is called the index or degree of N.

2) We can show that, if $ k$ is the index of N, then $ k \leq n$.


LaTeX
logarithms
calculus
calculus computations

Hello again, I need help on another differential equation. I still don't know how to use LaTeX so I just scanned the question:

[img]http://i6.photobucket.com/albums/y208/DemonicImpact/Question2.jpg[/img]

Because the dependent variable is missing, I start out by doing the following substitutions:
p=y'
dp/dx=y''

This gives the following differential equation:

dp/dx = 2x + (x^2 - p)^2

However, I don't know what to do after that since I don't know what kind of an differential equation it is...

Please help.

Thanks.

Let $ z\equal{}x^2\minus{}y'$, you have ... :)

A $ \text{\LaTeX}$ version of the equation:
$ y''=2x+(x^2-y')^2$, with initial conditions $ y'(2)=5,y(2)=\frac{5}{3}$
After the substitution $ y'(x)=p(x)$, this becomes $ p'=2x+(x^2-p)^2$, with initial condition $ p(2)=5$.

Quote me to see how the $ \text{\LaTeX}$ works.

Now, for a hint. Try one more substitution: $ g(x)=p(x)-x^2$. (This is the same as kunny's idea)

Hint

Step 1. Set $ z \equal{} x^2 \minus{} y'$ to get the differential equation with respect to $ z \equal{} z(x).$

Step 2. Solve the differential equation obtained in (1) with the initial condition.

Step 3. Rewrite the reasult of (1) into the differential equation with respect to $ y \equal{} y(x).$

Step 4. Solve the differential equation obtained in (1) with the initial condition.

[hide="Answer"]$ y \equal{} \frac {1}{3}x^3 \minus{} \ln |x \minus{} 3| \minus{} 1.$[/hide]

Thanks for the help guys and thanks for showing me how LaTeX works. I'll try to use it the next time that I have a problem.


search
real analysis
real analysis unsolved

seems that I can never understand the 'density' and 'cardinality' concepts.. can anyone help me with the following questions?

1. show that the real number system is uncountable and, in fact, has the same cardinality as the set of all subsets of the integers.

2. prove that there are an infinte number of rational numbers in between any two distinct real numbers..

3. prove that the irrational numbers are dense in R.

1. Search (here or on google) for Cantor's diagonalization argument.

2. The sum of two rational numbers is rational. You can find a rational number as small as you like, $\frac{1}{n}$. Suppose the distance between the two given numbers is $x$. Then if you can find one rational number between the two given numbers, surely you can find at least $nx-1$ for any $n$.

3. I think you can use density of rationals (#2) and [b]mlok[/b]'s idea here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=109340


limit
integration
logarithms
calculus
calculus computations

Prove that $ \lim_{n\to\infty}{a_n} \equal{} L\implies$ $ \lim_{n\to\infty}\int_{\frac {1}{a_n}}^{a_n}\frac {1}{x(1 \plus{} x^n)}\ \mathrm {dx} \equal{} \ln L\ .$

I get something a little bit different, namely that

$ \int_{\frac1a}^a\frac1{x(1\plus{}x^n)}\,dx\equal{}\ln a$ no matter what $ n$ is.

Hence to have the limit you want, we'd need $ a_n\to L,$ not $ \sqrt[n]{a_n}\to L.$

Let $ I\equal{}\int_{\frac1a}^a\frac1{x(1\plus{}x^n)}\,dx.$ Let $ u\equal{}\frac1x$ and note that $ \frac{du}{u}\equal{}\frac{dx}{x}.$ (That's absorbing the minus sign into the flip of the endpoints.)

$ I\equal{}\int_{\frac1a}^a\frac{1}{u\left(1\plus{}\frac1{u^n}\right)}\,du\equal{}
\int_{\frac1a}^a\frac{x^n}{x(1\plus{}x^n)}\,dx.$ (Turning $ u$ back into $ x.$)

Add to get

$ 2I\equal{}\int_{\frac1a}^a\frac{1\plus{}x^n}{x(1\plus{}x^n)}\,dx\equal{}
\int_{\frac1a}^a\frac{dx}{x}\equal{}2\ln a.$

The rest follows.

[quote="Kent Merryfield"]
$ I \equal{} \int_{\frac1a}^a\frac {1}{u\left(1 \plus{} \frac1{u^n}\right)}\,du \equal{} \int_{\frac1a}^a\frac {x^n}{x(1 \plus{} x^n)}\,dx.$ (Turning $ u$ back into $ x.$)[/quote]
I've seen this several times, what's the name of this property?, can be proven?

Yup, that's what I got as well. Notice that:
$ \frac{1}{x(x^n\plus{}1)}\equal{}\frac{1}{x}\minus{}\frac{x^{n\minus{}1}}{x^n\plus{}1}$. Thus,
$ \int_{\frac{1}{a_n}}^{a_n}\frac{1}{x(1\plus{}x^n)}dx\equal{}
\ln{a_n}^2\minus{}\frac{1}{n}\ln{\frac{{a_n}^n\plus{}1}{{a_n}^{\minus{}n}\plus{}1}}\equal{}
\ln{a_n}.$

[quote="Kent Merryfield"]Hence to have the limit you want, we'd need $ a_n\to L\ .$[/quote]
Thank you, you are right ! I corrected the upper enunciation : $ \boxed {\int_{\frac1a}^a\frac1{x(1 \plus{} x^n)}\ \mathrm {dx} \equal{} \ln a}\ .$

[color=red][b]I.[/b][/color] $ \int_{\frac1a}^a\frac1{x(1 \plus{} x^n)}\ \mathrm {dx} \equal{} \frac 1n\cdot\int_{\frac 1a}^a\frac {\left(x^n\right)'}{x^n\left(1 \plus{} x^n\right)}\ \mathrm {dx} \equal{} \frac 1n\cdot\int_{\frac {1}{a^n}}^{a^n}\frac {1}{x(1 \plus{} x)}\ \mathrm {dx} \equal{} \ln a\ .$

[color=red][b]II.[/b][/color] $ \left\|\begin{array}{c} x\ : \ \equal{} \frac 1x \\
 \\
\mathrm {dx}\ : \ \equal{} \minus{} \frac {1}{x^2}\cdot\mathrm {dx}\end{array}\right\|$ $ \implies$ $ \int_{\frac1a}^a\frac1{x(1 \plus{} x^n)}\ \mathrm {dx} \equal{}$ $ \frac 1n\cdot\int_{\frac 1a}^a\frac {\left(1\plus{}x^n\right)'}{1 \plus{} x^n}\ \mathrm {dx} \equal{} \ln a\ .$


1. How many ways can 72 be written as the sum of one or more consecutive integers with the integers only being positive

2. Same question but now you can use negative integers (such as -71+(-70)+...+71+72)

[quote="alanchou"]the problem never said as a sum of 2 consecutive integers, it just said sum of consecutive integers[/quote]

ya I made it more clearer just in case

[quote="the future"][quote="abacadaea"]i get no solution,
it can be expressed as n+(n+1)=72 where n is an integer. 2n+1 must be odd, but 72 is even...[/quote]

sorry it can be multiple consecutive integers.

like 4+5+6+7+..+12=72[/quote]
You just find the largest possible number of consecutive integers and go from there.
For an odd number of intergers, you just divide 72 by that number and if it is divisible, then there's a way.
If it is an even interger, you add or minus 1 from 72 and divide that number by your interger, but since 72+1 = 73 and 72-1 = 71, no even number of intergers will work.
So yeah,
just go from:
-3,-2,-1,0,1,...12
To...
23,24,25

Get the factors(divisors) and do it.

As ZhangPeijin has mentioned, we can not use the even factors. It's not going to work.

So the only odd factor that I could find was [i]3[/i] and [i]9[/i].
So [b]2[/b] ways of expressing it with only [u]positive [/u]integers.

With negative, you can express the same one that we did on the positive integers. But this time, 1 actually counts, because you can do [i]-71~72[/i]

So [b]3[/b] ways of expressing it with negative the integers.

So I think the answer for #2 is [b]5[/b].

donkey kong yea you basically have the right answer...

I dont know if I was clear enough but you can write it sa 72=72 soits 3 and 6 I think.


geometry
circumcircle
trigonometry
power of a point
radical axis
geometry proposed

Let $ABC$ be triangle with its circumcircle $O$ .now consider the Arc $AC$ on the $O$
(the one not having $B$), we are going to give another translated Arc for it with the name of $A^,C^,$ such that Arcs $AC, A^,C^,$ are symmetric wrt the line $AC$
(actually $A=A^,$ and $C=C^,$ but any way.)
now for any arbitrary point on $A^,C^,$ like $R$,  let the line $AR$ meets $BC$ at
$S$ and $CR$ meets $AB$ at $K$. prove that circumecircles of the triangles $SRK$ are sharing a same radical axis (or they are having  :) ) with movement of $R$.

P.S We found this at our geometry classes but do any one know an officail source?

Easy but  nice  I enjoy  problem like this.
It' easy to see that $BSRK$ is cyclic. let circumcircle of $BSRK$ intersect translated Arc AC(not contain $B$) at $R,T$.
two triangle $ATS$ and $TKC$ are similiar. then $\frac{TS}{TK}=\frac{AS}{CK}=\frac{AT}{TC}$ from sinus theorem in triangles $ASC$ and$ AKC$ we can get:
$\frac{AS}{CK}=\frac{AT}{TC}=\frac{\sin C}{\sin A}$
therfore point $T$ is constant. and we are done.

  A.Mathx


On a 3 hour, 30 minute entrance test for a private high school, there are 150 questions. There are 60 math questions and 90 English questions. Twice as much time is given for each math question as is given for each English question. How much time, in hours, is given for the 60 math questions?

[hide]So totally there are 210 minutes with 60 math and 90 english.

So observing we find that if we give 2 mins forr a math and 1 min for a eglish weget 210

so totally there are 120 mins for math.[/hide]

[hide]3hr, 30 min = 210min.
We solve using two equations.

$ 2M=E$
$ 60M+90E=210$

$ 60M+90(2M)=210$
$ 60M+180M=210$
$ 240M=210$
Time for each Math question is $ M=210/240$ or $ 7/8=87.5%
$
$ .875(210)=183.75min$
[b]3hrs, 3min, 45sec on math problems.[/b][/hide]

This seems wrong... will someone please correct me if I am?

[quote="bluecarneal"]$ 2M\equal{}E$[/quote]
That should be $ M\equal{}2E$.  :wink:


I would like to present a nice problem by a friend of mine [b][size=109]Nikos Kyriazis[/size][/b] from Thessaloniki – Hellas.

$\bullet$ Three similar isosceles triangles have as common the vertex opposite their bases and it is formed a convex hexagon by the other vertices. Prove that the lines connecting the midpoints of the opposite sides of such that hexagon, are concurrent at one point.

Kostas Vittas.

PS. Based on this problem we can have an easy proof of another problem presented by [b][size=109]''gemath''[/size][/b] at http://www.mathlinks.ro/Forum/viewtopic.php?t=131455

Very nice problem Kostas
General problem:Given convex hexagon $ABCDEF$ lines connecting the midpoints of the opposite sides of hexagon are concurrent iff $S_{ACE}=S_{BDF}$
Here in your problem we can prove $\triangle ACE =\triangle BDF$ because they have three equal side. Indeed we can see $AC=R^\alpha(BD)$ with
$\alpha =\angle AOB+\angle BOC$ thus $AC=BD$ similarly for $CE=DF,EA=FB$.Thus $S_{ACE}=S_{BDF}$
For the general problem it is [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=hexagon&t=17902]here[/url] in mathlinks and there is a proof of me in it!
But I think you found this problem base on synthetic geometry, can you post your solution Kostas?  :)

[quote="gemath"] ...
But I think you found this problem base on synthetic geometry, can you post your solution Kostas?  :)[/quote]

Thank you gemath, for your reference. The solution I have found by Nikos Kyriazis, based on the synthetic geometry and with pleasure, I will write it neatly and post here later.

Give me please a little time because this is a difficult solution based on several Lemmas, here and there, in Kyriazis’s textbook.

Kostas Vittas.

ok you are welcome, I will wait your solution  :)

Now, I see that it is not so difficult problem. Based on the below Lemma, we can prove that the triangles $\bigtriangleup K'LM,$ $\bigtriangleup L'KM,$ $\bigtriangleup M'KL,$ are similar isosceles ones.

Hence, the segment lines $KK',$ $LL',$ $MM',$ are concurrent at one point and the proof is completed.

[b][size=109]LEMMA.[/size][/b] – A triangle $ABC,$ is given and let $ABD,$ $ACE$ be two similar right triangles, erected on $AB,$ $AC$ respectively, outwardly to it, such that $\angle ADB = \angle AEC = 90^{o}$ and $\angle BAD = \angle CAE = \angle \phi.$

If $M,$ is the midpoint of $BC,$ prove that $MD = ME$ and $\angle MDE = \angle MED = \angle \phi.$

Kostas Vittas.

PS. I will post here later the synthetic proof I have in mind, of the Lemma.

[quote="vittasko"]

[b][size=109]LEMMA.[/size][/b] – A triangle $ABC,$ is given and let $ABD,$ $ACE$ be two similar right triangles, erected on $AB,$ $AC$ respectively, outwardly to it, such that $\angle ADB = \angle AEC = 90^{o}$ and $\angle BAD = \angle CAE = \angle \phi.$

If $M,$ is the midpoint of $BC,$ prove that $MD = ME$ and $\angle MDE = \angle MED = \angle \phi.$[/quote]

[b][size=117]PROOF OF THE LEMMA[/size][/b]

We denote as $B',$ $C',$ the symmetric points of $B,$ $C$ with respect to $D,$ $E,$ respectively.

It is easy to show that the triangles $\bigtriangleup AB'C,$ $ABC',$ are congruent, because of $AB' = AB,$ $AC = AC'$ and $\angle B'AC = \angle BAC' = \angle A+\angle 2\phi.$

Hence, we have that $B'C = BC'$ and then $MD\parallel = \frac{B'C}{2}= \frac{BC'}{2}\parallel = ME.$ That is the midpoint $M,$ of $BC,$ lies on the midperpendicular of the segment $DE.$

We draw the circumcircle $(O_{1})$ of $\bigtriangleup ADE$ and let be the points $K\equiv AB\cap (O_{1})$ and $L\equiv AC\cap (O_{1}).$ So, because of $\angle DAK = \angle EAL,$ we conclude that the quadrilateral $KDEL,$ is an isosceles trapezium. That is the intersection point $M',$ of $DL,$ $EK,$ lies on the common midperpendicular of $DE,$ $KL$ and we will prove that $M'\equiv M.$ It is enough to prove that the point $M',$ lies on $BC.$

We denote as $N,$ the intersection point of $BD,$ $CE$ and it is easy to show that this point lies on $(O_{1}),$ as the antidiametric point $($ =diametric opposite $)$ of $A,$ because of $AD\perp ND$ and $AE\perp NE.$

We consider the non-convex inscribed hexagon $ALDNEK$ and so, by Pascal’s theorem, we have that the points $B\equiv AK\cap DN,$ $M'\equiv LD\cap EK$ and $C\equiv AL\cap NE,$ are collinear. Hence, we conclude that $M'\equiv M.$

It is easy now, to show that $\angle DEM = \angle DLK = \angle DAK = \angle \phi$ and the proof is completed.

Kostas Vittas.

PS. I have now to check, if the proof of the Kyriazis’s result I presented, is the same or similar to him one.


geometry
parallelogram
geometry proposed

Let $ABCD$ be a paralelogram ($\angle{A}< \frac{\pi}{2}$) and the circle with diameter $AC$, and the intersections $E,F$ of $BC,CD$ respectively with the circle, and $P$ the point where the tangent in $A$ to the circle cuts $BD$. Prove that $P,E,F$ are colinear.

WE CONSIDER {M}=CE$\cap$AP   {N}=CF$\cap$AP
USING MENELAUS IN $\triangle$ CMN with P-F-E=>
$\frac{CE}{EM}$ $\cdot$ $\frac{MP}{PN}$ $\cdot$ $\frac{NF}{FC}$=1<=>

$\frac{NF}{ME}$=$\frac{CF}{CE}$$\frac{PN}{MP}$
USING THE POWER OF THE CIRCLE FORMULA=>

NA$^{2}$=[u]NF[/u] $\cdot$ NC (1)

MA$^{2}$=[u]MC[/u] $\cdot$ ME (2) =>

$\frac{NF}{ME}$=$\frac{NA^{2}}{MA^{2}}$ $\cdot$ $\frac{MC}{NC}$=$\frac{ND^{2}}{DC^{2}}$ $\cdot$ $\frac{MC}{DC}$=$\frac{ND^{2}}{AB^{2}}$ $\cdot$ $\frac{MC}{NC}$=$\frac{PN^{2}}{PA^{2}}$ $\cdot$ $\frac{MC}{NC}$.
We can easily observe that $\frac{PN^{2}}{PA^{2}}$=$\frac{PN}{PM}$ =>

$\frac{NF}{ME}$=$\frac{PN}{PM}$ $\cdot$ $\frac{MC}{NC}$
we must prove that: $\frac{MC}{NC}$ =$\frac{CF}{CE}$
CF=CN-NF
CE=CM-MA =>
$\frac{CF}{CE}$=$\frac{CN-NF}{CM-MA}$ and using (1) and (2)=>

$\frac{CF}{CE}$=$\frac{CN^{2}-NA^{2}}{CM^{2}-MA^{2}}$ $\cdot$ $\frac{MC}{NC}$

but CN$^{2}$-NA$^{2}$=CM$^{2}$-MA$^{2}$=AC$^{2}$ =>
$\frac{CF}{CE}$=$\frac{MC}{NC}$ and the problem is solved


inequalities
algebra
polynomial
n-variable inequality

For a given positive integer $n\ge 2$, suppose positive integers $a_i$ where $1\le i\le n$ satisfy $a_1<a_2<\ldots <a_n$ and $\sum_{i=1}^n \frac{1}{a_i}\le 1$. Prove that, for any real number $x$, the following inequality holds
\[\left(\sum_{i=1}^n\frac{1}{a_i^2+x^2}\right)^2\le\frac{1}{2}\cdot\frac{1}{a_1(a_1-1)+x^2} \]

[i]Li Shenghong[/i]

If $ a_1,a_2,\cdots,a_n(n\geq2)$ be natural numbers such that

$ a_1 < a_2 < \cdots < a_n$ and $ \sum_{k \equal{} 1}^{n}\frac {1}{a_k}\leq1$,

then the following inequlity

$ \left(\sum_{k \equal{} 1}^{n}\frac {1}{a_{k}^2 \plus{} x}\right)^2\leq \frac {m}{a_1(a_1 \minus{} 1) \plus{} x}$

holds for all nonnegative real number $ x$ if $ m \equal{} 0.3229383\cdots$,

which is unique real root of the following irreducible polynomial over $ \mathbb{Q}:$

$ 5180741959680000m^5 \minus{} 10225012469299200m^4 \plus{} 7842687249241912m^3$

$ \minus{} 2829413217657726m^2 \plus{} 654489379537533m \minus{} 87403180032000;$

with equality if and only if $ n \equal{} 3,a_1 \equal{} 2,a_2 \equal{} 3, a_3 \equal{} 6, x \equal{} 1.6248\cdots$,

which is unique real root of the following irreducible polynomial over $ \mathbb{Q}:$

$ 3x^5 \plus{} 159x^4 \plus{} 3570x^3 \plus{} 24464x^2 \plus{} 39456x \minus{} 145152.$

consider two cases $a_1(a_1-1)\le{x^2} $ and  ${x^2}\le{a_1(a_1-1)}$
the first case is easy , use AM-GM first then use cauchy ,the second part is more difficult,using  cauchy first\[\left(\sum_{i=1}^n\frac{1}{a_i^2+x^2}\right)^2\le\frac{1}{2}\cdot\frac{1}{a_1(a_1-1)+x^2} \]
$\[\left(\sum_{i=1}^n\frac{1}{a_i^2+x^2}\right)^2\le$$(\sum_{i=1}^n\frac{1}{a_i})$$(\sum_{i=1}^n\frac{a_i}{(a_i^2+x^2)^2})$
the rest part is easy


Find all function $ f : \mathbb{R^{*}} \to \mathbb{R}$ such that
\[g(xy) \ \equal{} \ \sqrt[2009]{y} g(x) \plus{}\sqrt[2009]{x} g(y), \quad \forall  x,y  \in    \mathbb{R^{*}},\]
and $ g(x)$ is continuous at $ x \equal{} \minus{} 2008$.

[quote="nguyenvuthanhha"][i]Find all function $ f : \mathbb{R^{*}} \to \mathbb{R}$ such that :

    $ g(xy) \ \equal{} \ \sqrt [2009]{y} g(x) \plus{} \sqrt [2009]{x} g(y) \ \forall \ x;y \ \in \ \mathbb{R^{*}}$

  and $ g(x)$ is continuous at $ x \equal{} \minus{} 2008$[/i][/quote]

Let $ h(x)\equal{}g(x)x^{\minus{}\frac 1{2009}}$. The equation becomes $ h(xy)\equal{}h(x)\plus{}h(y)$ with $ h(x)$ continuous at $ x \equal{} \minus{} 2008$, so continuous all over $ \mathbb R^*$

And so $ h(x)\equal{}a\cdot\ln(|x|)$

and so $ g(x)\equal{}a\sqrt [2009]{x}\ln(|x|)$


Two questions

1. If I say that my parents will go with me (on the registering thingy), but for some reason (me telling them not to go a billion times), they change their mind(s)...can it be undone?

2. What should the picture you upload be like? Like..how to take it and such?

The Mathcounts people will send you an email with a confirmation code that can be used to change your info when you want to.

As for the picture, I belive I just used the same one that I used for last year's yearbook since it was the only one I had on the computer I was on  :P .

the only thing that they do with the picture is putting it in the yearbook (as far as I know).  There's not req for what the photo has to be, most people have a close up on their face, but some people had full body, or them doing an action

[quote="math154"]I believe you can do something at the end that gives you a registration code which you can use to change the stuff: they'll send you the information to the email you put in. The picture doesn't really matter too much as far as I know; you can just put whatever you want.[/quote]

Oh, they do? Cool, thanks! So like..if I don't put in my picture now..I could edit it later?

xpmath: lazy :P

Quickster94: So..if I do a close up, I'd guess it'd be better for someone else to take it.

I'm very unphotogenic.


factorial
LaTeX
modular arithmetic
articles
number theory

[quote="MichaelFaraday"]Can someone please explain why $0! = 1$ is true?[/quote]

Because it is defined like that.

the reason it is defined to be that is so there doesn't have to be some weird exception when 0 is involved in $\binom{n}{k}$.   for example, $\binom{5}{5}=\frac{5!}{5!\cdot0!}$ which is undefined if $0!=0$ but $1$ if $0!=1$

Hi mf!

Factorial is 'how many ways to order objects' for example 5! is asking how many ways can you arrange 5 objects on a shelf. you have 5 choices for the first then 4 for the second 3 for the third then 2 then 1. this can be written in math terms as 5*4*3*2*1 or 5!.

To wrap your mind around it try it WITHOUT the factorials for 3, 2 and 1. for three you have 3 ways for the first object then 2 then 1 so the ways are 123,213,312,132,231,321. for two you can have 21 or 12. for one there is (oo how exiting!!) one way to do it! To address your problem of 0!=1 make a "combinatorics model" or a word problem to make sense of your question. Lets say your mother (or parental unit) asks you to put nothing on a shelf or she (or he)  will be really mad. how many ways can you do this without making her mad? Using the logic that 0! is 0 or undifined you would be saying that no matter what your mom (or dad or other) will be upset. there is 1 way to not make them mad. Don't put anything on the shelf!! This says 0!=1.

Edit: im sorry for double posting. I caught MF online and was trying to answer the question. my second post (a quarter of a hour later)was meant to bump this topic and bring his attention to it. I consider ramanujan's post post spam because it is obvious he (MF) knew the definition of factorial had 0!=1 included because he made this post. I didnt write it meaning like it sounded (this answer i made got refreshed 2 times! i was annoyed but still wanted to help a friend) simply saying pleasedont post if yo have nothing to contribute to the conversation. I was not accusing him of attemptng to raise his post count (I couldnt care less about post count) but cluttering up the question with givin information. In my eyes i saw it as somone saying "trivial" or "obvious" or the all too common "self explanitory". once agian I am sossy to have annoyed you and dont want to start a flam war.

[quote="nicholaskramer"]Hi mf!

Factorial is 'how many ways to order objects' for example 5! is asking how many ways can you arrange 5 objects on a shelf. you have 5 choices for the first then 4 for the second 3 for the third then 2 then 1. this can be written in math terms as 5*4*3*2*1 or 5!.

To wrap your mind around it try it WITHOUT the factorials for 3, 2 and 1. for three you have 3 ways for the first object then 2 then 1 so the ways are 123,213,312,132,231,321. for two you can have 21 or 12. for one there is (oo how exiting!!) one way to do it! To address your problem of 0!=1 make a "combinatorics model" or a word problem to make sense of your question. Lets say your mother (or parental unit) asks you to put nothing on a shelf or she (or he)  will be really mad. how many ways can you do this without making her mad? Using the logic that 0! is 0 or undifined you would be saying that no matter what your mom (or dad or other) will be upset. there is 1 way to not make them mad. Don't put anything on the shelf!! This says 0!=1.

By the way Ramanujan your post is spam.[/quote]
No, ramanujan's post isn't spam. He is right, that is the definition of 0!
There was no reason for you to double post, your second was spam since you could've just edited it back in.
Anyways, 0! is defined as 1, just as a circle is defined to be the set of all points equidistant from one given point.

When a definition makes everything else consistent and happy, it is usually used. For example, negative and fractional exponents may not make a whole lot of sense on a concrete level, but they still behave the same way as exponents should, and so their definition stands.

$0!=1$ wraps up a lot of things neatly, like $\frac{n!}{(n-1)!}=n$

[quote="nicholaskramer"]Im sorry. I edited my post so please reread it. Also i am deleating the double post. :lol:  hope there is no harm done![/quote]
Sorry if I sounded overly critical, I didn't mean too. What radomdragoon says is true. One won't see such things in High school basics.

You could also consider $0!$ to be the empty product, or the product of zero numbers, which is 1.

http://en.wikipedia.org/wiki/Empty_product

Thank you all, this was very helpful, I forgot about the counting side of factorials

0! is used in so many places that if wasnt equal to 1, many formulas and equations wuld not be true.  like in combinatorics and taylor polynomials...

what would you do with a negative factorial?

like $(-5)!$ because wouldn't $-5!$ be 5 factorial negative?

-jorian

Factorials are defined [b]only[/b] for whole numbers.

Edit: Ignore what is written above. :lol:

[quote="aahmadi"]Factorials are defined [b]only[/b] for whole numbers.[/quote]
For positive integers, $n!=\prod_{k=1}^{n}k.$

For nonnegative integers, $n!=\Gamma(n+1).$ Since $\Gamma(1)=1,$ we know $0!=1.$

For half-integers, $\left(n+\frac{1}{2}\right)!=\sqrt{\pi}\cdot \prod_{k=0}^{n}{\frac{2k+1}{2}}.$

[quote="aahmadi"]Factorials are defined [b]only[/b] for whole numbers.[/quote]

[url]http://mathworld.wolfram.com/Factorial.html[/url]

 :)

Wow, yeah, I knew I should have checked before blurting that, :blush:. I have, however, always been told that was the case. I would even have put a lot of money on it (good thing I don't bet often, lol). Here is another cool thing to know:  $(p-1)!\ \equiv-\ 1 \ ({\rm mod}\ p)$ iff p is a prime number.  (Wilson's Theorem). There's a lot of other cool stuff on wikipedia's factorial page. Btw, would memorizing double factorial identities help me for any competitions?

Edit: In my original post, I screwed up Wilson's Theorem since I'm a $\LaTeX$ noob.

[quote="aahmadi"]Wow, yeah, I knew I should have checked before blurting that, :blush:. I have, however, always been told that was the case. I would even have put a lot of money on it (good thing I don't bet often, lol). Here is another cool thing to know:  $(p-1)!\ \equiv\-1 \ ({\rm mod}\ p)$ (Wilson's Theorem). There's a lot of other cool stuff on wikipedia's factorial page. Btw, would memorizing double factorial identities help me for any competitions?[/quote]
Well I searched for double factorial on this entire forum, and didn't find anything actually useful for contests.
And I've never seen a double factorial used on a contest, so I guess you don't need them.

wait, so whats $x!$ if $x$ is a negative number?
-jorian

[quote="bpms"][quote="i_like_pie"][quote="t0rajir0u"]You mean $(p-1)! \equiv-1 \bmod p$.[/quote]
What if $p=4?$

$3!=6\not\equiv-1\pmod{4}$[/quote]
Read what he said again ;)
The letter p is used instead of a or x for a reason[/quote]
Oh... :blush:

[quote="t0rajir0u"]http://mathworld.wolfram.com/GammaFunction.html[/quote]

the link's very confusing

-jorian

Yes.  That's because factorials aren't normally defined for negative integers.

Try this:
5!=120
4!=24= 5!/1
3!=6 = 4!/4
2!=2 = 3!/3
1!=1 = 2!/2
So logically,
0!=1!/1=1

That's a very interesting article, confusing though  :huh:


If
$\frac{1}{a}-\frac{1}{b}=\frac{1}{a+b}$

Then what is the value of
$(\frac{a}{b}+\frac{b}{a})^{2}$?


Original problem???

Let a, b, c be positive numbers.
Prove that   $ \frac{3a^2 \minus{} 2c^2 \plus{} bc}{b \plus{} c} \plus{} \frac{3b^2 \minus{} 2a^2 \plus{} ca}{c \plus{} a} \plus{} \frac{3c^2 \minus{} 2b^2 \plus{} ab}{a \plus{} b}\ge \ (a \plus{} b \plus{} c)$ .

hello, your inequality is equivalent with $ \frac{3(a^4\plus{}b^4\plus{}c^4\minus{}a^2b^2\minus{}a^2c^2\minus{}b^2c^2)}{(a\plus{}b)(a\plus{}c)(b\plus{}c)}\geq0$, this is true because the inequality $ a^4\plus{}b^4\plus{}c^4 \geq a^2b^2\plus{}a^2c^2\plus{}b^2c^2$ holds for all real numbers $ a,b,c$ and in this case for all positive real numbers.
The equal sign holds in the case of  $ a\equal{}b\equal{}c$.
Sonnhard.


The time now is 2006-Nov-28~21:33:06
The following quote was posted on 2006-Nov-21~12:53:41
[quote="rrusczyk"]Do not start discussing the problems until after the Math Jam.  We expect to put up Round 3 problems next week.[/quote]

Based on this post, I'm guessing some time this week.

[quote="rrusczyk"][quote="tjames"]Yeah, I need something to do during AP US History class  :P

That's actually when I do a lot of my USAMTS'ing  :lol:[/quote]

Now that's a good use of history class.

I learned a ton in history class when I was in high school.  A ton of math, that is.[/quote]

Hmm I actually like history class...
I do my learning at home :agent:

Haha, I do a lot of stuff for USAMTS during my math class.

wait I do too... OK, self-contradiction. Oh well.

[quote="knexpert"]When will the third round problems be up?[/quote]

According to the transcript for [url=http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=172]USAMTS Round 2 Math Jam[/url]:

DPatrick (20:25:17)
Round 3 problems should be available on Friday.

It's Friday but there are no problem :\

I guess we should be patient and grateful  :yup:

um YES THEY ARE UP!!!!!

and they look... wordy

I mean, Alice and Bob :huh:

Yeah they're like all different this time :D

I don't like word problems :P

Bob should just go on the internet and check the standings.

[quote="rnwang2"][quote="rrusczyk"][quote="tjames"]Yeah, I need something to do during AP US History class  :P

That's actually when I do a lot of my USAMTS'ing  :lol:[/quote]

Now that's a good use of history class.

I learned a ton in history class when I was in high school.  A ton of math, that is.[/quote]

Interesting...

I'll keep that in mind  :whistling:[/quote]
AP Us history is the toughest AP class at our school, because the way the teacher grades - a decent essay obviously deserves a "c", for that is what a c represents, average. Only 1-2 people recieve an A, maybe 3-5 a B, etc. She is really into us history, and is very dramatic in her in class lectures (we are expected to learn everything from the textbook on our own, her lectures are about extraneous info.)
It is a very tough class, though I hear that it is one of the easier AP's in other schools... (and our school boasts some very nice AP test scores in history, as a consequence)

needless to say, I don't do math in that class :P

[quote="Erin J. Schram"][quote="eyefragment"]Pfft, I think we all know that the real question in #4 is that if Alice encrypts her value of $m$ for $n$ players is using a 64-bit DES cypher before sending it to Bob, Bob sends his value of $k$ back to Alice in the same encryption, and Eve intercepts the encrypted values while they are being transmitted, how fast of a computer would Eve need to crack the value of $m$ before January 8th, 2007, allowing her to submit quite possibly the shortest USAMTS proof ever?[/quote]
You caught me.

Richard Rusczyk asked several people to review the problems before he published them. I suggested that the wording for problem #4 would be clearer if it were a conversation between Alice and Bob. I chose those two names out of habit. I am an cryptologic mathematician, and we cryptologists tend to use Alice and Bob as default names in describing cryptographic protocols.

Erin Schram
USAMTS grader[/quote]
I thank you greatly for this then. I have had a hard time with number four on the previous two rounds because it always is seemingly intangible.


at mathcounts, i've seen many adult volunteers as proctors, speakers, graders, etc.

but is it possible for a high school student to volunteer in mathcounts?
maybe as a grader?

I think it is possible. Last year, a high school freshman who did MC in 2007 volunteered as a proctor.

graders for sure. at my chapter, almost all (maybe all) the graders were highschoolers - they included athunder's sister, mathgeniuse^ln(x), and other aops/non aops members.

[quote="vallon22"]kool
thanks!

also, should i talk to the chapter/state coordinator?
or should i talk to the mathcounts coach for our team first?[/quote]
Coordinators are kinda hard to reach, so maybe coach first? IDK
Just trying to help  :D

Ha, the one day I look at the MC forum...

This year, as a sophomore, I volunteered at states for MA--I graded, and was going to be the timekeeper for CD, but had to leave early...All I did was show up, and talk to the coordinator there. They won't turn away extra help.

Though, I did know someone else who was going as well and had already done it, so I knew it would be alright. To be safe, you might want to find out beforehand.

You can get your hours from MC for high school 2!

It was more likely a robot made to post that on other sites. Anyways,
if you have a sibling in it, can you proctor?

I volunteered at chapter, state, or both for all four years of high school.  I did both proctoring and grading, whichever they needed me for the most (although I recused myself from grading any tests from my former team).  Just contact the coordinator and they should give you all the info.  Some of them even keep an email list of volunteers, so they can contact you in future years.

Oh, and having friends in the contest never precluded me from proctoring, so siblings probably wouldn't either.


I have been thinking of the solution of this inequality for many days, but I couldn't find the solution

So I need help

$ 2 \sqrt{ab \plus{} bc \plus{} ca} < \equal{} \sqrt {3} \times [(a \plus{} b)(b \plus{} c)(c \plus{} a)]^{1/3}$

1. Try to expand and prove that for $ a,b,c > 0$, $ \frac{(a\plus{}b)(b\plus{}c)(c\plus{}a)}{8} \ge \frac{(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)}{9}$.
2. Try to show that $ \frac{(a\plus{}b\plus{}c)^2}{3} \ge ab\plus{}bc\plus{}ca$.


Is there a way to tell the cout object to not print a number under a certain condition (i.e., if the number is less than 0, do not print it)?

You want something like
[code]
int a=-1;
cout<<a<<endl;
[/code]
not to print anything, right?

Well, you could modify the corresponding entry from iostream.h   :)  If you use Borland C 3.1, then just F4 to the line with [b]cout[/b] and then F7. You'll probably reach an [i]inline[/i] function.

If you want to use the << operator with some class, if you could define a function like
[code]
ostream& classname::operator<<(ostream& os,classname x)
{
...
return os;
}
[/code]

Thank you all for your responses. I've solved my problem.


One-fourth of a bridge's length lies on the land to the east of a river, and one-sixth of the bridge's length lies on the land to the west of the river. The width of the span between the two sides of the river is 840 feet. How many feet are in the total length of the bridge?


algebra
polynomial
real analysis
complex analysis
complex analysis unsolved

I am looking for the lower bound for the minimum value of a given polynomial of even order $f(x)=a_{n}x^{n}+\ldots a_{1}x+a_{0}$ where $a_{n}>0$.  Any existing results for this?

I don't have an answer, but the question appears interesting. I suggest moving it to the (more active) real analysis forum, where it really belongs.


let ABC a triangle, $ \Gamma$ his circumcircle, N the minpoint of the arch BC that not contain A and M the midpoint of the arch BC that contain A. Let AM meet BC on K and MK meet $ \Gamma$ in A'. The tangent to $ \Gamma$ at A' meet the tangent to $ \Gamma$ at A in T and the perpendicolar line from A to AK meet the perpendicolar line from A' to A'K in R. Prove that K,T,R are collinear.

Dear Mathlinkers,
KTR is the pascal's line of the degenerated hexagon MTaNA'Ta'M where Ta and Ta' are resp. the tangent to the circumcircle of ABC resp. at A, A'.
Sincerely
Jean-Louis

[color=darkblue]$ AM$ and $ A'N$ are exterior bisectors of the angles $ BAC$ and $ BA'C$ respectively. Then we have

$ \dfrac{AB}{AC}\equal{}\dfrac{KB}{KC}\equal{}\dfrac{A'B}{A'C}$

Thus, the quadrilateral $ ABA'C$ is harmonical and $ BC$ is symmedian in triangle $ ACA'$, but $ CT$ is also, so $ K,B,C,T$ are collinear.

Easy get that $ RA$ and $ RA'$ are bisectors of the angles $ A$ and $ A'$, but, from the fact that quadrilateral $ ABA'C$ is harmonical, we get that they meet on $ BC$.[/color]

[b]Equivalent problem:[/b]
[color=darkblue][b]Let $ ABC$ be a triangle inscribed in circumcircle $ (O)$. $ M,N$ are the midpoints of arc $ BC$ ($ N$ is on the different side wrt $ A$ through $ BC$). $ AN\cap BC\equiv \{K\}$, $ MK\cap (O)\equiv \{A'\}$. Let $ T$ be the intersections of the tangents from $ A$ and $ A'$ wrt $ (O)$. $ S$ be the intersection of the line through $ A$ perpendicular to $ AK$ and the line through $ A'$ perpendicular to $ A'K$. Prove that: $ T,K,S$ are collinear.[/b][/color]

[i]Proof[/i]:

First we have a notice that $ AA'$ passes through the Pole of $ BC$ wrt $ (O)$, you can see a problem related to this result at: http://www.mathlinks.ro/viewtopic.php?t=276014. Let $ X$ be the pole of $ BC$  wrt $ (O)$. Then, $ TX$ is the polar of $ K$ wrt $ (O)$. Now, it is easy to notice that the line through $ A$ perpendicular to $ AK$ passes through $ M$ and the line through $ A'$ perpendicular to $ A'K$ passes through $ N$. Hence $ \{K\}\equiv MA'\cap AN\Longrightarrow$ $ K,S$ are [i]conjugated[/i] wrt $ (O)$. $ AA'$, $ MN$ intersect at $ X$ $ \Longrightarrow KX$ is the polar of $ S$ wrt $ (O)$.
We have the polar of $ \{T,K,S\}$ wrt $ (O)$, respectively are $ \{AA',TX,KX\}$, but $ AA',TX,KX$ have $ X$ is their common point, which implies that $ T,K,S$ are collinear.
Our proof is completed then.


1 ) sabet konoid agar tole yek ertefa'e mosalasi radikal 2 barabare sho'aye dayereye mohitiye an bashad , An vaght khate raste vasele  paye amodhaye vared az paye in ertefa bar zel'hayi ke An ra dar bar darand ,  az markaze dayereye mohitiye mosalas migozarad .

2 ) farz konim D yek noghte darone mosalase ba zaviyeye hade bashad  be torike
\[ AC.BD \equal{} AD.BC,\widehat{ADC \equal{} }\widehat{ACB} \plus{} 90
\]
 SABET KONID KHOTOTE MOMASE DAR NOGHTEYE C BAR DAYEREYE MOHITIYE MOSALASEHAYE ACD , BCD  BAR HAM AMODAND .

3 ) MOSALASE ABC RAKE DAREYE ZAVIYEY HADE AST RA DAR NAZAR MIGIRIM . NIMSAZE ZAVIYEYE DAKHELIYE A RA RASM MIKONIM TA ZEL'E BC RA DAR NOGHTEYE L VA DAYEREYE MOHITIYE MOSALASE RA DAR NOGHTEYE N GHAT'E KONAD .
AZ L AMODHAYI BAR AZLA'E AB VA AC RASM MIKONIM VA PAYE 2 AMOD RA K VA M MINAMIM . SABET KONID
\[ S_{\vartriangle ABC}  \equal{} S_{AKNM}
\]

[quote="SoNYDvDPro"]3 ) MOSALASE ABC RAKE DAREYE ZAVIYEY HADE AST RA DAR NAZAR MIGIRIM . NIMSAZE ZAVIYEYE DAKHELIYE A RA RASM MIKONIM TA ZEL'E BC RA DAR NOGHTEYE L VA DAYEREYE MOHITIYE MOSALASE RA DAR NOGHTEYE N GHAT'E KONAD .
AZ L AMODHAYI BAR AZLA'E AB VA AC RASM MIKONIM VA PAYE 2 AMOD RA K VA M MINAMIM . SABET KONID
\[ S_{\vartriangle ABC} \equal{} S_{AKNM}
\]
[/quote]

khob avval az hame deghat konid ke $ \triangle AKL \equal{} \triangle AML$ pas $ \angle ALM \equal{} \angle ALK \equal{} \alpha$...

pas dar 4zelie $ AKLM$ ghotre $ AL$ az in 4zeli nimsaze zavayaye rooberooye $ \angle A,\angle L$ ast,pas in 4zelie ye kayte pas ghotrash bar ham amoodand yeni $ KM\perp AL$ pas $ KM\perp AN$...

khob hala farz konid $ KM,AN$ yekdigar ra dar $ H$ ghat konand dar natije chon $ \triangle LMK$ motasaviosagheine khahim dasht $ HM \equal{} HK$ hamchenin darim:

$ S_{AKNM} \equal{} S_{AKN} \plus{} S_{AMN} \equal{} \frac12.KH.AN \plus{} \frac12.MH.AN \equal{} \frac12.KM.AN$

pas:

$ \boxed{S_{AKMN} \equal{} \frac12.KM.AN} (I)$

khob midoonim har noghte rooye nimsaz az do zele zavie be ye faselas pas $ LM \equal{} LK \equal{} h$ khob hala dar mosalase ghaemozavie $ \triangle LMH$ darim:

$ \sin \alpha \equal{} \frac {HM}{LM} \equal{} \frac {KM}{2h}$

$ \Rightarrow h \equal{} \frac {KM}{2\sin\alpha}$

hala mitoonim benevisim:

$ S_{ABC} \equal{} S{AKLM} \plus{} S_{BKL} \plus{} S_{LMC}$

$ \equal{} \frac12.KM.AL \plus{} \frac12.h.BK \plus{} \frac12.h.MC$

$ \equal{} \frac12.KM.AL \plus{} \frac12.h.(BK \plus{} MB)$

$ \equal{} \frac12.KM.AL \plus{} \frac12.\frac {KM(BK \plus{} MC)}{2\sin\alpha}$

$ \Rightarrow\boxed{S_{ABC} \equal{} \frac12.KM.AL \plus{} \frac12.KM.\frac {BK \plus{} MC}{2\sin\alpha}} (II)$

khob hala tebghe ravebete $ (I),(II)$ hokme masale tabdil mishe be inke sabet konim:

$ \frac12.AN.KM \equal{} \frac12.KM.AL \plus{} \frac12.KM.\frac {BK \plus{} MC}{2\sin\alpha}$

ke age sadash konim moadel mishe ba:

$ LN \equal{} \frac {BK \plus{} MC}{2\sin\alpha}$ (*)

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

khob hala dar mosalase ghaemozavie $ \triangle BKL$ darim:

$ \cos B \equal{} \frac {BK}{LB}\Rightarrow BK \equal{} LB.\cos B$

amma az tarafi chon $ AL$ nimsaze zavie $ \angle A$ hastesh darim $ BL \equal{} \frac {ac}{b \plus{} c}$

pas:

$ BK \equal{} \frac {ac.\cos B}{b \plus{} c}$

amma az tarafi tebghe ghazie $ \cos$ha dar mosalase $ \triangle ACB$ midoonim:

$ ac.\cos B \equal{} \frac {a^2 \plus{} c^2 \minus{} b^2}2$

pas:

$ BK \equal{} \frac {a^2 \plus{} c^2 \minus{} b^2}{2(b \plus{} c)}$

be tarighe moshabeh natije mishe $ MC \equal{} \frac {a^2 \plus{} b^2 \minus{} c^2}{2(b \plus{} c)}$

pas hokmemoon yeni hamoon rabete (*) moadel mishe ba:

$ LN \equal{} \frac {\frac {a^2 \plus{} c^2 \minus{} b^2}{2(b \plus{} c)} \plus{} \frac {a^2 \plus{} b^2 \minus{} c^2}{2(b \plus{} c)}}{2\sin\alpha}$

pas bad az sade sazi kafie sabet konim:

$ LN \equal{} \frac {a^2}{2(b \plus{} c)\sin\alpha}$

khob inam ziad sakh nis,avval az hame deghat konid ke tebghe $ \triangle LMA$ natije mishe $ \alpha \equal{} 90 \minus{} \frac A2$ pas $ \sin\alpha \equal{} \cos \frac A2$...

hamchenin dar mosalase $ \triangle BLN$ darim:

$ \frac {\sin LBN}{LN} \equal{} \frac {\sin LNB}{BL}$

$ \Rightarrow \frac {\sin \frac A2}{LN} \equal{} \frac {\sin C}{\frac {ac}{b \plus{} c}}$

$ \Rightarrow LN \equal{} \frac {\sin \frac A2.ac}{(b \plus{} c).\sin C}$

pas age ino too hokmemoon jagozari konim kafie sabet konim:

$ \frac {\sin \frac A2.ac}{(b \plus{} c).\sin C} \equal{} \frac {a^2}{2(b \plus{} c).\cos \frac A2}$

$ \Rightarrow \frac {2\sin \frac A2.\cos \frac A2}{a} \equal{} \frac {\sin C}{c}$

$ \Rightarrow \frac {\sin A}{a} \equal{} \frac {\sin C}{c}$

ke inam daghighan hamoon ghazie $ \sin$has dar $ \triangle ABC$...

[quote="SoNYDvDPro"]
2 ) farz konim D yek noghte darone mosalase ba zaviyeye hade bashad  be torike
\[ AC.BD = AD.BC,\widehat{ADC = }\widehat{ACB} + 90
\]
SABET KONID KHOTOTE MOMASE DAR NOGHTEYE C BAR DAYEREYE MOHITIYE MOSALASEHAYE ACD , BCD  BAR HAM AMODAND .

[/quote]

fek konam too neveshtane soorate soal eshteba kardi,ghaedatan bayad bashe $ \angle ADB = 90 + \angle ACB$... :maybe:

dar in soorat sharte avvalet ziadie va hich niazi besh nis... :huh:

chon bebin age esme dayere mohitie $ \triangle ADC,\triangle BDC$ ro be tartib $ C_1,C_2$ bezarimo farz konim momas dar $ C$ bar $ C_1$ dayere $ C_2$ ro dar $ K$ va momas bar $ C_2$ dar $ C$ dayere $ C_1$ ro dar $ L$ ghat kone oon moghe darim:

$ \angle DCK = \angle DAC,\angle LCD = \angle DBC$...

khob pas:

$ \textrm{zavie beine do momas } = \angle LCK = \angle LCD + \angle DCK = \angle DBC + \angle DAC$...

pas kafie sabet konim $ \angle DBC + \angle DAC = 90^\circ$...

ke inam ba tavvajoh be sharte $ \angle ADB = \angle ACB + 90$ vazehe cherake dar 4zelie $ ACBD$ darim:

$ \angle ADB = \angle ACB + \angle DBC + \angle DAC$

$ \Rightarrow \angle DBC + \angle DAC = \angle ADB - \angle ACB = 90^\circ$...

yeni hamoon chizi ke mikhastim...
 :maybe:

[quote="SoNYDvDPro"]1 ) sabet konoid agar tole yek ertefa'e mosalasi radikal 2 barabare sho'aye dayereye mohitiye an bashad , An vaght khate raste vasele  paye amodhaye vared az paye in ertefa bar zel'hayi ke An ra dar bar darand ,  az markaze dayereye mohitiye mosalas migozarad . [/quote]
in rahe shayad yekhorde sakh be nazar biad chon kolli mohasebe dare,vali be har hal ye rahe dge... :D

khob ma too esbate in masale az ghazie zir estefade mikonim ke besh migan C-A-C ba tavvajoh be shekli ke gozashtam soorate ghazie ine:

[b][u]ghazie:[/u]$ X,Y,Z\textrm{ hamkhatan }\iff \frac {m}{n} = \frac {k}{t}$[/b]

khob hala farz konid $ AH$ ertefamoon basheo $ AH = h = R\sqrt {2}$ va paye amoode rasm shode az $ H$ bar $ AB,AC$ ra be tartib $ M,N$ benamid,khob hala az $ M,N$ bar $ AC,AB$ amood konid ta anha ra dar $ M',N'$ ghat konand ($ M'$ rooye $ AC$ va $ N'$ rooye $ AB$ gharar darad)...

khob hamchenin vasate zele $ AB,AC$ ra be tartib $ P,Q$ benamid,khob hala tebghe C-A-C kafie sabet konim:

$ \frac {M'Q}{QN} = \frac {MP}{PN'}$

(havasetoon be tartibe gozashtane parekhatta bashe,chon in yekhorde ba soorate C-A-C fargh dareo yekhorde bayad deghat konid)
khob hala vase inkar kafie toolaro mohasebe konim:

$ M'Q = AQ - AM' = \frac b2 - AM'$

khob hala too mosalase ghaemozavie $ \triangle AM'M$ darim $ AM' = \cos A.AM$ (*)

hamchenin ba tavvajoh be do mosalase ghaemozavie $ \triangle AMH,\triangle ABH$ natije mishe $ \angle AHM = \angle B$ pas too mosalase $ \triangle AMH$ darim:

$ AM = \sin B.AH = \sin B.R\sqrt {2}$

khob amma tebghe ghazie $ \sin$ha dar $ \triangle ABC$ darim:

$ \frac {b}{\sin B} = 2R\Rightarrow R.sin B = \frac b2$

ke ba jagozari natije mishe:

$ AM = \sin B.R\sqrt {2} = \frac {b\sqrt {2}}{2}$

ke age ino too (*) gharar bedim natije mishe:

$ AM' = \cos A.AM = \cos A.\frac {b\sqrt {2}}{2}$

pas:

$ M'Q = \frac b2 - AM' = \frac b2 - \frac {b.\sqrt {2}.\cos A}{2}$

pas:

$ \boxed{M'Q = \frac b2\left(1 - \sqrt {2}.\cos A\right)}$

khob hala vase mohasebe $ QN$ darim:

$ QN = AN - AQ = AN - \frac b2$

khob hamchenin too mosalase ghaemozavie $ \triangle AHN$ darim $ \angle AHN = \angle C$ va darim:

$ AN = \sin C.AH = \sin C.R.\sqrt {2} = \frac {c\sqrt {2}}2$

pas:

$ QN = AN - \frac b2 = \frac {c\sqrt {2}}2 - \frac b2$

pas:

$ \boxed{QN = \frac {c\sqrt {2}}2 - \frac b2}$

khob be tarighe moshbeh darim:

$ \boxed{MP = \frac {b\sqrt {2}}2 - \frac c2}$

$ \boxed{PN' = \frac c2\left(1 - \sqrt {2}.\cos A\right)}$

khob pas hokmemoon tabdil mishe be:

\begin{eqnarray*}\frac {\frac b2\left(1 - \sqrt {2}.\cos A\right)}{\frac {c\sqrt {2}}2 - \frac b2} = \frac {\frac {b\sqrt {2}}2 - \frac c2}{\frac c2\left(1 - \sqrt {2}.\cos A\right)}\end{eqnarray*}

ke age yekhorde sadash konim mishe:

$ \frac {b(1 - \sqrt {2}.\cos A)}{c\sqrt {2} - b} = \frac {b\sqrt {2} - c}{c(1 - \sqrt {2}.\cos A)}$

khob hala ino tarafein vasatein konid mishe:

$ bc(1 - \sqrt {2}.\cos A)^2 = 3bc - \sqrt {2}.(b^2 + c^2)$

$ \Rightarrow 2bc.\cos^2 A - 2bc.\cos A.\sqrt {2} = 2bc - \sqrt {2}.(b^2 + c^2)$

amma tebghe ghazie $ \cos$ha dar mosalase $ \triangle ABC$ darim:

$ - 2bc.\cos A = a^2 - b^2 - c^2$

pas age jagozari konim kafie sabet konim:

$ 2bc.\cos^2 A + (a^2 - b^2 - c^2)\sqrt {2} = 2bc - \sqrt {2}(b^2 + c^2)$

$ \Rightarrow 2bc.\cos^2 A + a^2.\sqrt {2} = 2bc$

$ \Rightarrow a^2.\sqrt {2} = 2bc(1 - \cos^2 A)$

$ \Rightarrow a^2.\sqrt {2} = 2bc.\sin^2 A$

$ \Rightarrow \frac {a^2}{\sin^2 A} = \frac {2bc}{\sqrt {2}}$

khb amma dar mosalase ghaemozavie $ \triangle AHC$ darim $ \sin C = \frac {AH}{AC} = \frac {\sqrt {2}.R}{b}$ pas:

$ \frac {c}{2R} = \frac {\sqrt {2}.R}{b}$

$ \Rightarrow \frac {2bc}{\sqrt {2}} = 4R^2$

ke age ino too tasavie bala jagozari konim hokmemoon tabdil mishe be:

$ \frac {a^2}{\sin^2 A} = 4R^2$

ke inam tebghe ghazie $ \sin$ha dar mosalase $ \triangle ABC$ vazehe...


[quote="rachid"]Let $ a,b,c \geq 0$ such that $ ab \plus{} bc \plus{} ca \equal{} 3$.Prove that:

$ \frac {\sqrt {4 \minus{} ab} }{a \plus{} b} \plus{} \frac {\sqrt {4 \minus{} bc} }{b \plus{} c} \plus{} \frac {\sqrt {4 \minus{} ca} }{c \plus{} a} \geq \frac {3\sqrt {3}}{2}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=28175

[quote="rachid"]Let $ a,b,c \geq 0$ such that $ bc \plus{} ca \plus{} ab \equal{} 3.$ Prove that:

$ \frac {\sqrt {4 \minus{} bc} }{b \plus{} c} \plus{} \frac {\sqrt {4 \minus{} ca} }{c \plus{} a} \plus{} \frac {\sqrt {4 \minus{} ab} }{a \plus{} b}\geq \frac {3\sqrt {3}}{2}.$[/quote]$ 4 \minus{} bc \minus{} \left[2 \minus{} \frac {65bc}{12(11 \plus{} 6\sqrt {3})} \minus{} \frac {b^2c^2}{3(19 \plus{} 11\sqrt {3})} \minus{} \frac {b^3c^3}{12(7 \plus{} 4\sqrt {3})}\right]^2$

$ \equal{} \frac {bc(3 \minus{} bc)(bc \minus{} 1)^2[b^2c^2 \plus{} (1 \plus{} 4\sqrt {3})bc \plus{} 32(2 \plus{} \sqrt {3})]}{144(97 \plus{} 56\sqrt {3})}\geq0\Longrightarrow$

$ \sum{\frac {\sqrt {4 \minus{} bc} }{b \plus{} c}}\geq\sum{\frac {1}{b \plus{} c}\left[2 \minus{} \frac {65bc}{12(11 \plus{} 6\sqrt {3})} \minus{} \frac {b^2c^2}{3(19 \plus{} 11\sqrt {3})} \minus{} \frac {b^3c^3}{12(7 \plus{} 4\sqrt {3})}\right]}$

$ \geq\frac {3\sqrt {3}}{2}\Longleftarrow$

$ \left\{\sum{\frac {1}{b \plus{} c}\left[2\left(\frac {bc \plus{} ca \plus{} ab}{3}\right)^2 \minus{} \frac {65bc}{12(11 \plus{} 6\sqrt {3})}\left(\frac {bc \plus{} ca \plus{} ab}{3}\right) \minus{} \frac {b^2c^2}{3(19 \plus{} 11\sqrt {3})} \minus{} \frac {b^3c^3}{4(7 \plus{} 4\sqrt {3})(bc \plus{} ca \plus{} ab)}\right]}\right\}^2$

$ \minus{} \frac {(bc \plus{} ca \plus{} ab)^3}{4}$

$ \equal{} \sum\frac {a^2(b \minus{} c)^2}{648(c \plus{} a)^2(a \plus{} b)^2}\bigg(a(b \plus{} c \minus{} a)^2\Big\{31a^2(b \plus{} c)$

$ \plus{} 4a\left[8\left(b^2 \plus{} c^2\right) \plus{} \left(243\sqrt {3} \minus{} 368\right)bc\right] \plus{} \left(2484\sqrt {3} \minus{} 4183\right)bc(b \plus{} c)\Big\}$

$ \plus{} bc\Big\{a^2\left[9\left(492\sqrt {3} \minus{} 847\right)\left(b^2 \plus{} c^2\right) \plus{} 2\left(4914\sqrt {3} \minus{} 8479\right)bc\right]$

$ \plus{} 9a(b \plus{} c)\left[\left(479 \minus{} 276\sqrt {3}\right)\left(b^2 \plus{} c^2\right) \plus{} 8\left(89 \minus{} 51\sqrt {3}\right)bc\right] \plus{} 96bc\left(b^2 \plus{} c^2\right)\Big\}\bigg)$

$ \geq0.$

See also : http://www.mathlinks.ro/viewtopic.php?t=192323


Do people at your school think you're very smart?

Not when I forget things like what period it is.   :P

[quote="archimedes1"]I see there's no poll option for "I'm the greatest supergenius there ever was, and everyone thinks so, too."[/quote]

Hmmmm. I did enter in Yes, all the time as another option, but I must have forgotten to add it.

So, I guess if they think you are just push sometimes.

I go to a small school (pre-k to 12, 200 kid total), and nearly everyone from 5th-12th there knows I'm very smart. Not [i]thinks[/i], [i][b]KNOWS[/b][/i]!!

Where is that option?

People think I'm smart just because I do well on tests and don't talk much

Hold on... it says 9% not at all, 90% sometimes...

Where did the other 1% go  :o !

People at my school generally know me to be relatively good at math and science.  There are a lot of people at my school who are really smart though, I guess they just consider me to be the authority for math/physics...

I did not vote, because the option "I'm the greatest supergenius there ever was, and everyone thinks so, too." is not on there. Pity. Everyone worships the ground I walk on, so that proves how smart they think I am  :D

I teach my class instead of the teachers. :noo:

Heh, I still have mod powers here. Poll fixed.

i'm surprised only 2 people said yes...there's gotta be more than 2 smart people in AoPS...

That's because the yes option was only edited in (by me) today.

whatever...sorry i can't exactly remember. oh well you thought 3 was even or something last year so be quiet.

[quote="AIMEman"]It was a probability problem, and it had spinners.  You know I hate those. :blush:[/quote]

whatever. i hate water bottles.

and @cognos: i have no clue (im kidding andrew. i love you. but not literally, because that would be awkward.)

[quote="cognos599"]do you and michael really like hate eachother or is it just like friend-to-friend insults. anyway ppl don't think i am the smartest in my school even though i pwn everyone else at the math contests. everyone think our 2nd place mathcounter is smarter than me cuz he's in precalc. and im only in geometry.[/quote]

eh. there's a guy at my school who pwns me in grades.. b/c i don't really care about grades...
but everyone still accepts me as the smartest..
see posts WAYYY before...

As of right now, my principal doesn't think I'm very smart...

actually i'm in college and offcourse they think i'm smart :D  but more than that they think i'm funNy :P

Because of my slightly crappy grades people make up things like "oh he must be bad at history" to make themselves feel better, so I checked "sometimes".

I think that common sense outweighs knowledge.
You can be smart, but your behavior can get you a B (like my friend). Sad.

Yea guy don't get your teachers you hate you...currently most of my teachers hate me and uh...I got my math quiz dropped from a B to an F because the teacher took off points that I didn't actually lose, and she regularly tries to take random points off my tests...bad bad bad.

The thing I really hate is when they hear "math" and "olympiad problem", dumb people say "you think people in BC calc can't do them?".

It's not right to say people in BC calc can't solve them, some can. Be like ccy and argue for points (on like everything - at least in 8th grade)

[quote="alanchou"][quote="cognos599"]everyone think our 2nd place mathcounter is smarter than me cuz he's in precalc. and im only in geometry.[/quote]

whether he understands that stuff is the question :wink:

Two words:
I
fail

(ok, I lied, maybe not "fail", and certainly not 2 words)[/quote]
dude of course I understood the stuff ;)

meh yeah. I'm the 2nd-place-mathcounter-was-in-precalc-now-in-calc-BC-who-pwnt-cognos-on-AIME person :D

and er whoever said all Korean people are ugly on the second page of this topic should go die in a hole or something.. just kidding, but really - that's really stereotypical/insulting..

I first became known for (generally) clever humor, understandable explanations, practical solutions, and having a good reason for whatever I did. I don't think I ever came across as the study-nerd or smart~~~ type.

But something pretty funny arose as a consequence of that. A lot of people didn't know my name, and then suddenly they'd realize who I was, and they'd be like... wait, so did [i]you[/i] do that math whatever-thing? And I'd be like... uhuh... And they'd be like aahhhh... gotcha... didn't know you were into that stuff. And I'd be like, sure thing -- it's good stuff.

Although in School I'm very tired but I don't think I'm very smart, because .......... I love my School!  :D  :D  :D