I live in Romania - Bucharest to be maore specifically.

Host of the IMO for five times now, the initiating country of this great competition, Romania has proven to be of great value for the Mathematical World :)

It is a country of great landscapes, and, ussually, great hospitality. Low prices, many pretty girls and an English speaking population, plus a lot of ways to have fun makes Romania a paradise for those whishing for a great holiday.

more coming up soon! :)

well.. about trains we have a lot of them!  you can use http://www.bahn.de to find a train between any city in Romania! now this site is pretty useful for every1, even for Romanian's.. just take a look and see what I mean!
And also a lot of buses!

about there beer.. I am not saying anything because I don't like it! We have good wine's!!! :D

Romania is a beautiful country to go on a holiday, I have some British friends that come in Campulung and they like ROmania so much that they come here every year (not only in Campulung), they like the Black Sea and the Mountains too!

bye!

we have a lot of interesting features in our country, and for certain alchool is one of them :)

the beer is quite nice, it's better than in Japan anyway (and belive me I've drinked some there :D), and in fact in most of the countries I've been too ...

Most of the people come in Romania for the girls :) ... so  you won't have any problems at that chapter either - in plus now it's the sea-side season, and it's very nice at the sea-side now. The conditions are not like in Grecee or Turkey (although they are almost at the same prices) but the nightlife is very popular - you should definitely try it!

there are a lot of transportation methods; in Bucharest we have 4 subway lines, 50 tram lines, 30 trolley lines and about 200 bus lines; all these for a city with less than 2,000,000 people - so I think you won't have any problems with the transportation

About inter-city transport - we have a lot of trains going on and to; you can check the prices and timetables at [url=http://www.cfr.ro/infofer/default_engleza.asp?lng=2]http://www.cfr.ro/infofer/default_engleza.asp?lng=2[/url]

Any other questions ? :)

hum, for the girls, I guess it won't be a good idea for me.  My girlfriend wouldn't like me to chase other girls.   :(

Anyway, I'll think about Romania for next year's vacation

Hmm....in my opinion....for someone who comes to Romania generally it's a plesure.I've met many foreigners who loved it here...though sincerely...I'd like to forget about this place...I live in a pretty small town...so I can't have the same opinion as someone's who lives in Bucharest for example...though...as I said...many people apreciate our country for it's beauty and for the fact that people here treat them well...Valentin said our country has a lot of mathemathics lovers and it's true...I'd only wish all maths teachers would apreciate them...I can't say I'm very good at maths though I've been several times at the National Olimpiads....but my teacher doesn't care at all about me(or maybe she does when she receives her paiment!!!)...and most of all...I can't stand she doesn't even know many important things...that's why I'd really like to live somewhere else...so I could have another chance...because being here in thin place forgotten by every one just makes my life miserable....sorry if I made you sad...or if I bored you....I just wanted someone else to know about what is happening around here...thanks for listening(or reading....just the same)...Bye!

well Diadi, guess what - for those instances you described I have "invented" Mathlinks :) so be free to ask questions around, solve problems and mingle in the community here.

I had almost the same problem as you did: I run out of math materials while preparing for IMO 2001. Yup, I had no internet connection back then, at least not a stable one (Xnet Dial-up which sux, when ever you have the luck to make it connect :D:D romanians know all about it :P) and I had already read all the books I could get my hands on.

The teachers couldn't help me a lot, because themselves didn't had any materials.

But now I think we are passed that problem, so it only remains a small language barrier and an internet connection stuff :D.

Where are you from anyway ?
Tell us about your part of the country ... (you guys woudnl't want me to talk about the holes in the streets in Bucharest, or the bad traffic, would you? :D:lol:).

Well,I'm from some part Transilvaniaand I'm in the 10th grade.You said you had some problems when you prepaired yourself for the IMO(I wish I had problems prepareing my self for the IMO,unfortunately I can only dream about getting there,I had my county part of the NMO today,and I'm not very pleased of what I've done :(...but I just wanted to say in my post that not everyone has the luck of being prepared by good teachers...cause my school teacher had no idea about Cauchy's functional ecuation(and many many others-this was just an example),or when we have to demonstrate something I'm the one who has to go in front of the class and explain to the others...she even told me once when the break came she told me sha had to go(yeah really!)but that I stay at the blackboard to finish a problem and then to explain it to my class-mates.Do you want to know she told me last year after I qualified myself at the NMO(she didn't think I could do it...)?She told me she isn't paied for the extra hours with me and that I have to work a lot,but by myself... and you know...learning maths all alone in my opinion is not the easyest thing(by the way if you didn't understand at scool we learn maths by copying from our manual...and another thing no one wanted to prepare me because SHE was my teacher... :(  :( ...you can say that again!).The list can go on....but I think you got the picture...and I have to say I was really happy when I discovered your site...because it helped me a lot...really...well the see you next time...! :D

Hey now DiaDi, nice to meet you. Don't be too harsh on yourself. I had a similar situation as yours when I was in high school. I can't say I was smarter than the teacher but the olympiads stuff that I was doing she didn't even what tha heck was it. The only out-of-curriculum thing that she taught me was Pick's Theorem!!! It's funny but hey I learned to cope up with her. My teacher and I worked together on some of the math problems and we got the problem-solving going. Although your situation is certainly different than mine, search the path to the olympiads. I remember I had some problems with my status in the US and I was not qualified to take the USAMO because only legal residents of US can officially take it. The instant I learned about that, I nearly gave up mathematics. But now that I think about it, I should have done my best in AIME and pretend that I qualified for the USAMO. I could have made it to the USAMO but no, I was goofing off and did not even do well on the AIME at all (got 5/15 and 7/15, anyone got lower than this?). So don't get frustrated and we are grateful to have you here in MathLinks. And hopefully we see you next time because here is the place for you to practice your problem-solving skills. Bye.

PS:I would like to apologize for my ridiculously boring reply and that it isn't about Romania.  :blush:  Hey Valentin, I like this redface a lot  :blush:  :blush:  :blush:

Thats right, not anyone has the chance to have good teachers!That was also my situation with my teacher in high school.I don't remember her ever showing me how to solve a difficult problem.Now i am at the university of Galati, and here are some teachers who are getting me nervous!I think i'll move to Bucharest in the second year of college.

What is the salary of math teacher in high school in Romania ?
How many hours of teaching each week ?
How many students in a classroom ?
---------------------------------------------------
Same questions for university
thx

salary - well it depends on what age in work, and city, school he works in. usually it's around 120$ / month (maybe less?)

working some 18 hours / week, with classes of 20-35 students each (depending again on the school, city, etc.).

not much fun to be a high-school teacher in romania ...  :?

In Germany teachers earn about 1500$ / month. And Professors on University earn up tp 5000$ / month.

in romania the univ. teachers can earn up to 1000$ month (legally ;)).
so you see the difference :D

In Romania, McDonalds must be a real luxus-restaurant  :D

I mean, do you, for example, often eat there? I must admit, I am almost daily in McDonalds, because I like the food that much --- AND I am not thick (51kg, 177cm).  :)

I should mention, that I sit there always alone --- solving math problems (or thinking about solutions). :D

well as a matter of fact is really was at the beginning. Nowadays is more custom, but still very expensive. I usually go there to eat because it's a [i]fast food[/i] restaurant (ie grab my food, eat it, and be done with it in 10 mins). But some people (high-school students mostly) like to hang out in Macs that are close to their high-schools.

Also, when we go into camps in the province, we almost always use the Mac toilets because usually they are the only clean ones in the town (including those ones in hotels, motels, etc.) ;)

Daniel, you mean to say that you are not fat, because I'm 1,78 and 71 kilos and people say I'm thin :D:D

[quote]
Daniel, you mean to say that you are not fat, because I'm 1,78 and 71 kilos and people say I'm thin  :D  :D
[/quote]

No, you don't know what "thin" means  :D  :D

I know very well what "thin" means :D to put it simply, thin is the opposite of fat ;)

but still you are 90% right, because you wrote "thick" and I quickly misread it :D "thick" means excessive, so it might  be taken as a substitute for "fat" :D

[quote="Daniel Gutekunst"]AND I am not thick (51kg, 177cm).  :)
:D[/quote]
but he already declared he is not [b]thick[/b] so what's he worrying about? It's a good sign I believe but yeah it has been 'shown' that fast food increases body mass (fat as well, yuck) :P .

It follows, that I better keep eating fast food, or I will shrink like the man in Stephen King's [i]Thinner[/i]  :D

[quote=gery]Actually, I never thought of Romania as a possible destination for holiday.  Do you have good beer? I mean, not the kind of horrible beer they have in the USA.  And what about travelling?  Do you have plenty of trains? or buses?

I'll think about it...[/quote]

Wait your old enough?

He/she is definitely old enough now. You revived a 20 year old thread.

[quote name="piphi" url="/community/p18995735"]
[quote=Joe__2020]The OP here is now an AOPS class instructor.[/quote]

The OP here is more than that, he's the founder of MathLinks, the company that came before AoPS but later merged with them.  He also plays poker  :thumbup:
[center][img]https://upload.wikimedia.org/wikipedia/commons/2/2d/Valentin_Vornicu_2016wsop.jpg[/img][/center]

-[url=aops.com/wiki/index.php/User:Piphi][font=Segoe Print][size=200]πφ[/size][/font][/url]
[/quote]

Oh jeez thats weird
Oh thats so weird
:wacko:

[quote=Wukong]1) why did sheriff delete piphi's post
2) why that weird[/quote]

It had a big signature which isn't allowed.

[quote=Wukong]1) why did sheriff delete piphi's post[/quote]

Because it revealed personal info

[quote=SpaceHead1AU][url=https://en.wikipedia.org/wiki/Valentin_Vornicu]I think it is.[/quote]

That's where I got the picture from.  It's him, look at [url=https://artofproblemsolving.com/community/c7h225575p1253003]this[/url] post.

[quote=SBose]also that isnt Valentin Vornicu.

The poker player is a completely different person. They dont look alike at al[/quote]

or do they?
[img]https://wiki-images.artofproblemsolving.com//8/84/Vvornicu.png[/img]

-[url=aops.com/wiki/index.php/User:Piphi][font=Segoe Print][size=200]πφ[/size][/font][/url]

Valentin Vornicu is the first user on AoPS

[quote=pezzeepenguinstalker]I'm pretty sure rrusczyk would've come before that.[/quote]

No, MathLinks came before AoPS.  Also, Valentin is user #2  and rrusczyk is #1163

-[url=aops.com/wiki/index.php/User:Piphi][font=Segoe Print][size=200]πφ[/size][/font][/url]

[quote=jasperE3][quote=middletonkids][quote=pezzeepenguinstalker]I'm pretty sure rrusczyk would've come before that.[/quote]

rrusczyk is the $1163$th user (counting deleted and no-existing accounts)

And Valentin Vornicu is [i]technically[/i] the second official user.[/quote]

Who's the first.[/quote]

Probably a test account. It doesn't exist now.

[quote=SBose]also that isnt Valentin Vornicu.

The poker player is a completely different person. They dont look alike at al[/quote]

it is. I asked him on class :hehe:


Here are 3 geometry true/false questions:
1.The union of 2 non-collinear rays is an angle.
2.A set of 8 noncollinear rays with a common endpoint will always form 28 angles.
3.A line perpendicular to a plane is perpendicular to every line in the plane.

1. True
2. True; $ \binom{8}{2} \equal{} 28$ angles.
3. True

Well actually...

1 is true only if the rays share an endpoint.

2 is false if it doesn't matter which side you start measuring from. *(If an angle is formed by rays A and B, and if it doesn't matter if you measure from A to B or B to A, and you must measure the part less than 180 degrees, then this is false.)

3 is false because the two lines could be skew.


limit
calculus
integration
calculus computations

Show :D  :o  :wink:  :maybe:
$ \lim_{n \minus{} > \plus{} \infty}\sqrt[n]{\sum_{k: k > n(1 \plus{} a)}\frac {n^k}{k!}} \equal{} \left(\frac {e}{a \plus{} 1}\right)^{a \plus{} 1}$, $ n,k \in \mathbb{N}$ and $ a > 0$.

[quote="rocomath"]What course are you taking?  I've never seen a problem like this :-\[/quote]
it's easy. We have already solved something like this $ \sum^{n}\frac {n^{k}}{k!}$
just you need to note that is't some integral and.. later I will post solution.


modular arithmetic
number theory proposed
number theory

Let p be a prime number, m is a positive integer and GCD(m , p -1 ) = 1. Suppose that for any positive integer n we have

n^m - m is not divisible by p. Prove that p -1 is a power of 2

Is there something wrong with this:
Since $ \gcd(m,p\minus{}1)\equal{}1$, pick $ a$ such that $ am\equiv 1\pmod{p\minus{}1}$. Take a primitive root $ g\pmod p$. Write $ m\equiv g^b$ and take $ n\equal{}g^{ab}$.

Then, $ n^{m}\equiv g^{abm}\equiv (g^{am})^b\equiv g^b\equiv m \pmod p$. So there are no such $ (p,m)$


limit
calculus
derivative
algebra
polynomial
function
calculus computations

It's hard to explain what I'm asking so hopefully someone will understand what I'm saying.

$ f'(x)$ is to $ (a_{n\plus{}1} \minus{} a_{n})$ as $ \lim_{n \rightarrow \infty}f^n(x)$ is to what?

I'm going to guess that by $ f^n(x),$ you meant the $ n$th derivative of $ f.$ That's more commonly written as $ f^{[n]}(x)$ - either brackets or parenthesis to hint that that $ n$ isn't really an exponent. Or you can talk about differentiation as the operator $ D$ such that $ Df(x)\equal{}f'(x).$ In that case, the $ n$th derivative would be $ D^nf(x).$

(If you did mean $ n$th power rather than $ n$th derivative, then the entire rest of my post would be irrelevant and I wouldn't have understood your question.)

Note that $ \lim_{n\to\infty}f^{[n]}(x)$ just isn't normally a nicely behaved object. If $ f$ is polynomial, then it eventually vanishes, but generically, you don't have any good reason to expect it to turn out well. Which means that we shouldn't expect all that much of the sequence analogue either.

For the sequence $ (a_n),$ we can define the [i]finite difference operator[/i] $ \Delta:$
$ \Delta a_n\equal{}a_{n\plus{}1}\minus{}a_n.$

Now we can try iterating that operator:

$ \Delta^2a_n\equal{}\Delta(\Delta a)_n\equal{}\Delta a_{n\plus{}1}\equal{}\Delta a_n$

$ \equal{}(a_{n\plus{}2}\minus{}a_{n\plus{}1})\minus{}(a_{n\plus{}1}\minus{}a_n)\equal{}a_{n\plus{}2}\minus{}2a_{n\plus{}1}\plus{}a_n.$

We can continue this:

$ \Delta^3a_n\equal{}a_{n\plus{}3}\minus{}3a_{n\plus{}2}\plus{}3a_{n\plus{}1}\minus{}a_n$

$ \Delta^4a_n\equal{}a_{n\plus{}4}\minus{}4a_{n\plus{}3}\plus{}6a_{n\plus{}2}\minus{}4a_{n\plus{}1}\plus{}a_n$

$ \Delta^5a_n\equal{}a_{n\plus{}5}\minus{}5a_{n\plus{}4}\plus{}10a_{n\plus{}3}\minus{}10a_{n\plus{}2}\plus{}5a_{n\plus{}1}\minus{}a_n$

And so on. Note the binomial coefficients.

If $ a$ is a polynomial sequence, this vanishes for large enough $ n,$ but in general, we don't have any reason to think that much good will come of taking a limit in $ n.$

If I did want to take $ \lim_{x\rightarrow\infty} \Delta^x a_n$, it would converge into an infinite series. What would this series be?

[quote="epkid08"]it would converge into an infinite series[/quote]  Why do you think this?  Just as with any limit, it might converge, diverge to infinity, or oscillate in a variety of ways.  Convergence is the least likely of these options.  Just pick a few sequences for yourself to see!  (Examples: polynomials; $ a_n \equal{} a^n$ for $ a \equal{} \minus{} 1, 0, 1, 2, 3$; the Fibonacci numbers; and so on.)

The sequence $ \Delta^n a_0$ can be quite interesting, though.

Edit: fixed typo

[quote="JBL"][quote="epkid08"]it would converge into an infinite series[/quote]  Why do you think this?  Just as with any limit, it might converge, diverge to infinity, or oscillate in a variety of ways.  Convergence is the least likely of these options.  Just pick a few sequences for yourself to see!  (Examples: polynomials; $ a_n \equal{} a^n$ for $ a \equal{} \minus{} 1, 0, 1, 2, 3$; the Fibonacci numbers; and so on.)

The sequence $ \delta^n a_0$ can be quite interesting, though.[/quote]

I didn't mean to assume that it would converge, but I was just wondering how to write $ \lim_{x\rightarrow\infty} \Delta^x a_n$ in terms of a series, be it convergent or not.

If you look at the coefficients at $ a_{n\plus{}1}$ and subsequent terms, you will see that the limiting series you ask about just does not exist as a simple series whose terms are $ a_n$ with some coefficients (which is not surprising, because the limit of the derivatives, which you bring up as its continuous analogue, is undefined for most functions). Of course, every limit $ \lim_{n\to\infty} B_n$ can be written as the series $ B_1\plus{}(B_2\minus{}B_1)\plus{}(B_3\minus{}B_2)\plus{}\dots$ and in this particular case, it is not hard to find the explicit formulae for $ B_{n\plus{}1}\minus{}B_{n}$ in terms of $ a_k$. But this "series representation" is completely useless for any practical purpose.

Do you have any particular reason to ask such a strange question, or it is just something that came to your head out of nowhere? Also, do you have some particular answer in mind? If so, I doubt anybody here will be sufficiently interested in trying to guess it but it may be interesting to see it to understand whether it makes sense.

At the time I was thinking that if $ a_n \equal{} D^{n \minus{} 1}[f(x)]$, then maybe $ g_n(f) \equal{} \lim_{k\rightarrow \infty} \Delta^k a_n$ could possibly shed some light on finding the general form for finding the nth derivative of any function $ f(x)$. It was just a stupid idea I never really thought through. :D

Looking at the series I wanted again, I think $ \lim_{x \rightarrow \infty}\Delta^xa_n$ transforms into the series $ \lim_{N \rightarrow \infty}\sum_0^N (\minus{}1)^{n\minus{}1}a_{n\plus{}N}{N \choose n}$, but I can't make much more sense of it than this.


If $ x$, $ y$ and $ z$ are positive reals then prove inequality:

$ \sqrt{x(x\plus{}y)(x\plus{}z)}\plus{}\sqrt{y(y\plus{}x)(y\plus{}z)}\plus{}\sqrt{z(z\plus{}x)(z\plus{}y)}\geq 2\sqrt{(x\plus{}y)(y\plus{}z)(z\plus{}x)}$

Can constant $ 2$ be changed by better one, $ \frac{3\sqrt{2}}{2}$ :?:

[quote="delegat"]If $ x$, $ y$ and $ z$ are positive reals then prove inequality:

$ \sqrt {x(x \plus{} y)(x \plus{} z)} \plus{} \sqrt {y(y \plus{} x)(y \plus{} z)} \plus{} \sqrt {z(z \plus{} x)(z \plus{} y)}\geq 2\sqrt {(x \plus{} y)(y \plus{} z)(z \plus{} x)}$

Can constant $ 2$ be changed by better one, $ \frac {3\sqrt {2}}{2}$ :?:[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=48825  http://www.artofproblemsolving.com/Forum/viewtopic.php?t=50390  http://www.artofproblemsolving.com/Forum/viewtopic.php?t=48311  http://www.artofproblemsolving.com/Forum/viewtopic.php?t=46428  http://www.artofproblemsolving.com/Forum/viewtopic.php?t=27481


symmetry
combinatorics proposed
combinatorics

Hi, i'm searching for a formal solution of this problem:

[i]Let G be a graph with 8 vertices and without 4-cycle. Prove that the maximum number of its edges is 11.[/i]

I draw a graph with an 8-cycle e with 3 3-cycle... Maybe it is a graphical solution, but what about a formal proof?
 :blush:
Thanks in advance.

If you draw graph with 11 vertices having desired property, you don't need to prove that it does not contain a cycle.
What you need to prove is that any graph with 8 vertices and at least 12 edges contains 4-cycle.

[quote="mavropnevma"]See [url]http://www.mathlinks.ro/viewtopic.php?p=859306#859306[/url] for discussions on a more difficult, but identical otherwise question (my own problem in a Romanian TST 2007). There is a general result on the maximal number of edges, presented by Pierre Bornsztein, but it is not tight. The exact margin needs more detailed analysis, hard to post here. A full solution, with detailed method and also extra considerations is in the Romanian Mathematical Competitions RMC 2007 - available at the leader of your country's IMO team in 2007.[/quote]

I have read that topic... My problem is: how can i "convince" my auditor that in my construction of a graph with $ n$ vertices and $ \lambda$ edges there is not a $ C_4$?
I know how to prove [i]ab absurdo[/i] that a graph has a $ C_4$, but the inverse is not so clear to me...

For example, i'm able (thanks internet  :blush: ) to construct a graph with $ 11$ vertices, $ 18$ edges and no $ C_4$, but how can i persuade the others?

Most "small" examples don't need convincing, nor persuasion - there is probably some symmetry involved, and it's just a matter of exhaustive verification. It is the proof for the rest of the cases that needs be correct, and convincing. For example, can you prove that any graph on $ 11$ vertices and $ 19$ edges [b]must[/b] contain a $ C_4$ (which is true)?

Suppose we have a graph $ G$ with $ E$ edges, $ V$ vertices, and $ G$ has no 4-cycles.

let $ n$ be the maximum valency in the graph (number of edges meeting at a vertex).

Select a point $ P$ on the graph with valency $ n$.
Let $ B$ be the set of points which are nearest neighbours of $ P$ connected by an edge. The number of vertices in $ B$ is $ |B| \equal{} n$ otherwise there would be a 2-cycle.
Let $ C \equal{} G \minus{} \{P\} \minus{} B$ be the set of points not connected to $ P$ so $ |C| \equal{} V \minus{} n \minus{} 1$

Write $ v(S)$ for the total valency summed over a set $ S$ of vertices in the graph.

The the total valency $ v(G) \equal{} 2E$ because each edge joins two nodes.
Since $ n$ is the maximum valency per node we have $ v(C) \le (V \minus{} n \minus{}1)*n$.

So $ v(B) \equal{} v(G) \minus{} v(\{P\}) \minus{} v(C) \ge 2E \minus{} (V \minus{} n)*n$

Now let $ D$ be the subset of vertices in $ C$ that are directly connected by an edge to vertices in $ B$

No vertex in $ D$ can be connected to more than one vertex in $ C$ since otherwise there would be a 4-cycle.

So $ |D| \equal{} v(B) \minus{} n \ge 2E \minus{} n \minus{} (V \minus{} n)*n$

So the total number of vertices is $ V \equal{} 1 \plus{} |B| \plus{} |C| \ge 1 \plus{} |B| \plus{} |D| \ge 1 \plus{} 2E  \minus{} (V \minus{} n)*n$

So $ E \le \frac{1}{2}(V \plus{} (V \minus{} n)n \minus{} 1)$

but $ (V \minus{} n)n \le \frac{1}{4}V^2$

So $ E \le \frac{1}{2}(V \plus{}  \frac{1}{4}V^2 \minus{}1)$

For $ V \equal{} 8$ this gives $ E \le 11$ as required.

Of course this is just an upper bound for $ E$. I can't find a graph that meets this bound.
The bound could also be improved a little for larger $ V$ by using the fact that $ n \le \frac{2E}{V}$

Oh, the model for $ 8$ vertices and $ 11$ edges exists alright.


prove that \sum\frac\{1}{2jC2n}_{j=1\rightarrow n}=\frac{2n+1}{2^{2n+1}}.\sum\frac{2^j}{j+1}_j=0\rightarrow \{2n+1}


analytic geometry
graphing lines
slope
rotation
function
MATHCOUNTS
counting

1. A point (x, y) meets the condition that its distance from the origin is equal to the slope of the line through the point and the origin.  Find the set of all possible values of x.

2. If |a| + |b| = 1 and 2b - a = :rt3:, compute both possible ordered pairs, (a, b).

3. How many positive composite integers n are there such that (n - 1)! is not divisible by n?

4. How many paths are there on gridlines from the point (0,0) to (8,5) composed only of moves up and right that do not go above the line y = x?  (That is, they may touch the line, but not go above it.)

5. Three professors and ten students are to sit in a row of 13 chairs.  The professors take their seats first, before the students enter the room.  If the professors agree among themselves that each will sit between two students, how many ways can they choose their seats?  (Professors are distinct.)

6. Vadim wished to rotate a point in the Cartesian plane 120:^o: counter-clockwise around the point (3,4).  First, he reflected the point across the line y = 4, then he reflected it across the line   y = mx + b.  If the point ends up where he wanted it, compute (m, b).

7. If b*x^(log[size=75]b[/size]x) = x^(5/2), b > 0, find all possible values of x in terms of b.

1. [hide] the value of x is given by r/:sqrt:(r:^2:+1), where r is the distance from the origin. This is a continuous function, as the thing inside the square root can never be negative or zero. So, we want to maximize r/:sqrt:(r:^2:+1), and then we will have the minimum value by symmetry. x(r)=r/:sqrt:(r:^2:+1)=:sqrt:(r:^2:/(r:^2:+1)) (that's allowed, we're finding the MAX value and that'll be positive) =:sqrt:(1/(1+1/r:^2:)). This is an increasing function of r, and lim (r -> :inf:) of x(r) is :sqrt:(1/(1+0))=1. Thus the set is the open interval (-1,1). correct?[/hide]

2. [hide]the graph of the first equation (a horizontal axis) is a diamond, the graph of the second equation is a line with positive slope (b=(a+:rt3:)/2). This will intersect the diamond in the first and second quadrants. In the first quadrant, b=1-a => (a,b)=((2-:rt3:)/3,(1+:rt3:)/3). In the second quadrant, b=1+a => (a,b)=(:rt3:-2,:rt3:-1). Correct?[/hide]

3. [hide]n is composite => n=ab, 1<a:le:b<n. If a<b, then a and b are both in (n-1)!, as a,b:le:n-1. If a=b, then n=a:^2:. 2a<a:^2: for a>2, so we have 2 factors of a in (n-1)! for any a:ge:3. This means only 4 is a potential problem. Indeed, 4 does not divide 3!=6. Every other non-square positive number does, and every square greater than 4 does, so I get that there is 1 positive composite integer that satisfies that requirement. Correct?[/hide]

4. I only have a clumsy addition-recursion method similar to what Osiris did recently in a mathcounts thread.

5. [hide]we glue each professor to two students, one on each side. now, we want to place 3 (distinguishable) units in 7 places in P(7,3)=210 ways.[/hide]

6. [hide]shift the whole plane by (-3,-4), so that what was originally (3,4) is now the origin. Now, he wants to rotate 2:pi:/3 about the origin, but instead reflects across the x-axis and then across (some line). Taking the origin, we see that this line must go through the origin as rotation does not change the point, and neither does reflection through the x-axis. That means, in the new plane, the line has the form y=ax. Taking the point 1, it is rotated into e^2i:pi:/3. Meanwhile, it is reflected into 1. That means, we need a line through the origin, such that 1 is reflected through it into e^2i:pi:/3. This will be the perpendicular bisector of those two points. Since the angle this line makes with the horizontal is :pi:/3, it's slope will be tan(:pi:/3)=:rt3:. Going back to the normal plane (without the funky translation applied), we want a line with slope :rt3: that goes through (3,4). This will be y=x:rt3:+(4-3:rt3:). Thus (m,b)=(:rt3:,4-3:rt3:). correct?[/hide]

7. [hide]taking log base b of both sides, we have 1+logbx*logbx=5/2*logbx. Making the substitution logbx=y, we have 2y:^2:-5y+2=0. This factors into (2y-1)(y-2)=0, or logbx=2 or 1/2 => x=b:^2: or :sqrt:b. correct?[/hide]

[b]1)[/b]
Do you only want real number solutions Joel?

y/x = :sqrt:(x:^2: + y:^2:)
y:^2:/x:^2: = x:^2: + y:^2:
y:^2: = x:^4:  + y:^2:x:^2:
x:^4: + y:^2:x:^2: - y:^2: = 0
x:^4: + y:^2:(x:^2: - 1) = 0

Since x:^4: :ge: 0, then y:^2:(x:^2: - 1) :le: 0.
x cannot be 0 since the slope would be undefined.  x:^2: - 1 cannot equal zero either, since x > 0, making x:^4: + y:^2:(x:^2: - 1) not equal 0. Now -1 < x:^2: - 1 < 0, resulting 0 < x:^2: < 1 after simple subtraction. 0 < x < 1 and -1 > x > 0 is the domain of all possible values of x, or {x|-1>x>0, 0<x<1}.

[b]2)[/b]
2b - :sqrt:3=a
|a| + |b| = 1

|2b - :sqrt:3| + |b| = 1
|b| = 1 - |2b - :sqrt:3|

b = 1 - |2b - :sqrt:3|
|2b - :sqrt:3| = 1 - b
2b - :sqrt:3 = 1-b
3b = 1 + :sqrt:3
b = 1/3 + :sqrt:3/3
a = 2/3 - :sqrt:3/3

2b - :sqrt:3 = b - 1
b = :sqrt:3 - 1
a = :sqrt:3 - 2

b = |2b - :sqrt:3| - 1
b + 1 = |2b - :sqrt:3|
2b - :sqrt:3 = b + 1
b = 1 + :sqrt:3
1 + :sqrt:3 would not be a valid solution because there couldn't be a corresponding value of a since b > 0.

2b - :sqrt:3 = -b - 1
3b = :sqrt:3 - 1
b = :sqrt:3/3 - 1/3
a = -:sqrt:3/3 - 2/3
a doesn't equal 2b - 3 in this case so this solution wouldn't work.

The ordered pairs that are solutions are (2/3 - :sqrt:3/3, 1/3 + :sqrt:3/3), (:sqrt:3 - 2, :sqrt:3 - 1).

in reply to Yannik's #1: a slope of 0 (and a distance of 0) will give you the x-coordinate of 0. technically, the formulas don't work (i.e. y/x is not defined), but i think that is still a valid solution.

Well, if x was 0, wouldn't you think the slope would be well, infinity?  I mean, (0,0) just seems too trivial.

*Anything divided by 0 is infinity...*

There is not a uniquely defined line passing through (0, 0) and (0, 0), and thus the slope of a line passing through those "two" points would be undefined, so x = 0 is not okay.  Yannik got it.

You both look set on number 2.

Mystic's number 3 is correct, and I'd just like to say that I liked that problem very much.

Number 5, while technically correct, is (a) left in horrible form and (b) done in a much less nice way than necessary.  Keep looking for the nicer solution.

yes, matfan, that is what I was referring to. hrmph doesn't seem too elegant though. :(.

O.K., I finished [i]the blob[/i] (referring to the monster chart thingy).  It would actually only take about 90 seconds to do, it just took a looong time to show everything.  It's especially hard because the alignment doesn't turn out very well (the forum automatically takes out more than one space, I think.

Mystic, you changed your number 5 and now it's wrong.

for 5.

[hide]

i think if we imagine the 10 students in a row before the professors, with 1 space in between them so that there are 9 spaces, we'll have P(9,3) ways to seat the professors.  Is that right?[/hide]

Rocking.  Another way to consider that is to "glue" each professor to a student on his/her left, and then put one student in the rightmost seat.


On a square gameboard that is divided into n rows of n squares each, k of these squares lie along the boundary of the gameboard. Which of the following is a possible value for k?
A) 10, B) 25, C) 34, D) 42, E) 52
I don't know how to solve it.

Thanks for help.

There are four sides to the board, and the same number of boundary pieces on each side. What does this mean about the number of boundary pieces?


geometry
3D geometry
algebra
polynomial
function

Find the sum of the cubes of the roots of x^10 + x^9 + x^8 .....  X :^2: + x +1 = 0

hint:  [hide]try multiplying both sides by (x-1) you get an expression which is easier to work with, and you only have to remember to subtract 1 from the sum of the cubes of the roots.[/hide]

Thanks

I was trying to work with the coefficients as in 1 equals the sum taken 1, 2, 3, 4 ... at a time, didn't quite work out

Newton sums would be really, really unpleasant for a 10th degree equation.

why?  I thought it would be the same amt. of work as finding the sums of cubes of a 3rd degree polynomial.

You have a really nice polynomial there.  zscool's method is so pretty ... why would you want to go about manipulating sums with 55 terms in them?  I mean, it's doable, but you'd spend more time writing things down then you would spend on the other method altogether, ignoring the bits where you have to stop to think.

[quote="rrusczyk"]Why would Newton's sums be so rough here?  (Though admittedly, they lack the elegance of zscool's approach.)[/quote]

Well frankly, don't you think that 10 levels of recursion involving equations involving up to 10 terms is a pain?

Perhaps we're talking about different things?

Using Newton's sums, the answer pops out in roughly 3 seconds.

1s1 + 1 = 0, so s1 = -1
1s2 + 1s1 + 2 = 0, so s2 = -1
1s3 + 1s2 + 1s1+3 = 0, so s3 = -1.

Done.  Not super elegant, but no special knowledge about the roots of unity is required (as there is in the other solution, though you could couple zscool's approach with Newton's sums to be even faster than the above without using roots of unity).

Blah, sorry, I recalled the problem incorrectly. In any case, yeah, I see its not bad at all (even if we wanted to look for the sum of high degrees of the roots) with all the coefficients equal to 1.

Maybe I missed it, but it seems as though you are assuming (without saying so) that the sum of the cubes of the roots is equal to the sum of the roots.  This is true, but the way I see to prove it is to view the roots of x^11-1 as a multiplicative group. Since the group is cyclic, the function f(x)=x^3 is an automorphism, and since f(1)=1, f(x) is still an automorphism when restricted to the roots of x^10+x^9+x^8 +... + x + 1.  Then we can apply one of the methods above.  Also, what are Newton sums?  I've never heard that term before.

--Dan

petra - That is the advanced machinery I was referring to when I mentioned having to know something about roots of unity when using zscool's method.

Newton's sums are a way of relating the coefficients of a polynomial to the sums of powers of the roots of the polynomial.  Specifically, if our polynomial is

anxn + ... + a0,

and we let sk be the sum of the kth powers of the roots, then

ans1 + an-1 = 0

ans2 + an-1s1 + 2an-2 = 0

ans3 + an-1s2 + an-2s1 + 3an-3 = 0

and so on.  So you can see how we can find the sum of the cubes of any polynomial pretty quickly.


probability
counting
derangement
search
probability and stats

A box contains m tickets numbered 1, 2, ...m.The tickets are drawn
from the box one at a time at random and after each draw, the
ticket is put aside and not replaced in the box. Let Ai be the event
"the number on the ith ticket drawn is i”. Find the probability
P (∪Ai , from i=1 to m) when m = 3, 4and the limiting probability as m → ∞.

The complementary event is that the $ i$th draw is not equal to $ i$ for any $ i$.  The resulting permutation of tickets is called a [i]derangement[/i], their number is a [i]derangement number[/i], [i]rencontres number[/i] or [i]subfactorial number[/i], and the problem of counting them is called the "derangement problem" or "problème des rencontres."  Search for any of these terms on the forum or the broader internet.

consider n=m,
we know that P(A1)=P(A2)=P(A3)=.....=P(Am)=1/m....consider P(Ai's) are independently happening.
so sum of all P(Ai's) is 1=1/1!.
also,P(A1 intersection A2)=P(A2 intersection A3 )=..........=P(Ai intersection Aj)=1/{(m)*(m-1)}
as there are in total of mC2 combinations of two events taken at a time.therefore sum of all P(Ai's intersection Aj's) is mC2/{(m)*(m-1)}
that is 1/2=1/2!.
similarly if you continue in the same manner then you will find that for intersection of 3 events taken at a time probability is 1/6 or 1/3!.
so for m events taken at a time probability is 1/m!.
hence using equation 1 for m events,we get total probability as P=1/1! - 1/2! + 1/3! - 1/4! +-+- ......... -(-1)^m*1/m!
putting m=3, P(A1UA2UA3)=1-1/2+1/6 = 2/3.
put m=4, P(A1UA2UA3UA4)=1-1/2+1/6-1/24 = 15/24 =5/8.
As m tends to infinity, P(A1UA2UA3U.....UAm)=1-1/e=(e-1)/e or simply 0.6321

to be frank,i didn't get exact proof...i simply took m=3,4 and made few random observations and ended up with this technique(trial and error).  hope you are convinced by this solution.


let    d=gcd(a_n,a_(n+1))
       where, a_n=n^2+20
   show that d always divides 81 completely.[n is any natural number]
   please help quickly....

Bash with the Euclidean Algorithm...

EDIT: Beaten by stephen (that rhymes!)


Let n be Natural number  such that n satisfies one of the congruences listed below.
Then  n = x3+ y3  has no solution in Z:
1. n ≡ 3 or 4 mod 7
If I check case1 x>0 and y>0
               case2 x>0 and y=0 (or x=0 and y>0)
               case3 x>0 and y<0 (or x<0 and y>0) by |x|>|y|,Is it correct?
Please recommend me. :?

Umm... apparently x3 and y3 are special variables I am not aware of.  Could you please elucidate?  :blush:

lol that makes sense :P

In that case, I think the intended solution would be to consider x^3, y^3 (mod 7).  Possible values would be 0, 1, and 6.  The rest shouldn't be too difficult.


[quote="AIME15"]The length of rectangle R is $ 10$ percent more than the side of square S.  The width of the rectangle is $ 10$ percent less than the side of the square.  The ratio of the areas, R:S, is:

$ \textbf{(A)}\ 99: 100 \qquad \textbf{(B)}\ 101: 100 \qquad \textbf{(C)}\ 1: 1 \qquad \textbf{(D)}\ 199: 200 \qquad \textbf{(E)}\ 201: 200$[/quote]
[hide]Side length of square=x
Length of rectangle R=$ 1.1x$
Width of rectangle R=$ .9x$
Area of rectangle R=$ .99x^{2}$
Area of square=$ x^{2}$
Ratio of area R:S=$ .99x^{2}: x^{2}$=$ 99: 100$ or A[/hide]

Let $ s$ the side of the square.
The area of the rectangle is then $ \frac {11}{10}s \cdot \frac{9}{10}s \equal{} \frac{99}{100}s^2$.
The area of the square is $ s^2$, so the ratio of the areas is $ \frac{99}{100}\implies \boxed{A}$.


Given $ a_1,a_2,a_3,...,a_n \ge 0$ and $ 0 \le a_1 \plus{} a_2 \plus{} a_3 \plus{} ... \plus{} a_n \equal{} t \le 4$ and $ t \equal{} const$Find the maximum value
$ (a_1^2 \plus{} 1)(a_2^2 \plus{} 2)(a_3^2 \plus{} 3)...(a_n^2\plus{}n)$

Equality occur if and only $ a_1\equal{}t$.$ a_2\equal{}a_3\equal{}...\equal{}a_n\equal{}0$


we have $ \frac{s^2}{2(p\minus{}a)}\equal{}r_a$ from wich $ p\equal{} \frac{s^2}{2r_a} \plus{} a \equal{} \frac{s^2}{2r_a} \plus{} \frac{s^2}{h_a}$ that can be easly costruct with Euclide.
So we have $ h_a$, $ r_a$ and $ p$ and the problem it's equal to [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=215050]this[/url]


[hide="Solution"]
(a) Adding $ \frac{1}{9}$ to both sides, we obtain

$ x^2 \plus{} \frac{2x}{3} \plus{} \frac{1}{9} \minus{} \frac{1}{3} \equal{} \frac{1}{9} \implies \left(x \plus{} \frac{1}{3}\right)^2 \equal{} \frac{4}{9} \implies x \equal{} \minus{}\frac{1}{3} \pm \frac{2}{3} \equal{} \frac{\minus{}1 \pm 2}{3} \implies \boxed{x \in \left\{\minus{}1, \frac{1}{3}\right\}}.$

(b) Factoring out $ 2$, dividing by $ 2$, and adding $ \frac{9}{16}$ to both sides, we have

$ 2\left(x^2 \minus{} \frac{3x}{2} \minus{} \frac{1}{2}\right) \equal{} 0\implies \left(x \minus{} \frac{3}{4}\right)^2 \equal{} \frac{17}{16} \implies x \equal{} \frac{3}{4} \pm \frac{\sqrt{17}}{4} \implies \boxed{x \in \left\{\frac{3 \plus{} \sqrt{17}}{4}, \frac{3 \minus{} \sqrt{17}}{4}\right\}}.$
[/hide]

$ x^2 \plus{} 2x/3 \minus{} 1/3 \equal{} 0\longrightarrow(x \plus{} \frac {1}{3})^{2} \equal{} 4/9$.

$ x^{2} \minus{} 3x/2 \minus{} 1/2 \equal{} 0\longrightarrow(x \minus{} \frac {3}{4})^{2} \equal{} 1/2 \plus{} 9/16 \equal{} 17/16$.

EDIT: egh sunehra answered first...

To complete the square, simply move $ c$ to the other side of the equation, then add $ (\frac {b}{2})^2$ to the equation.

#1

$ x^2 \plus{} \frac {2x}{3} \minus{} \frac {1}{3} \equal{} 0$
$ x^2 \plus{} \frac {2x}{3} \equal{} \frac {1}{3}$

$ (\frac {\frac {2}{3}}{2})^2 \equal{} \frac {1}{9}$

$ (x \plus{} \frac {1}{3})^2 \equal{} \frac {4}{9}$

$ x \plus{} \frac {1}{3} \equal{} \pm\frac {2}{3}$
$ x \equal{} \pm\frac {2}{3} \minus{} \frac {1}{3}$

---------------------------------
#2

Here we must first divide by $ 2$ since completing the square is for equations with the leading coefficient being $ 1$.

$ 2x^2 \minus{} 3x \minus{} 1 \equal{} 0$

$ x^2 \minus{} \frac {3x}{2} \minus{} \frac {1}{2} \equal{} 0$

$ x^2 \minus{} \frac {3x}{2} \equal{} \frac {1}{2}$

$ (\frac {\frac { \minus{} 3}{2}}{2})^2 \equal{} \frac {9}{4}$

$ (x \minus{} \frac {3}{2})^2 \equal{} \frac {11}{4}$

$ x \minus{} \frac {3}{2} \equal{} \pm\frac {\sqrt {11}}{2}$

$ x \equal{} \frac {3}{2}\pm\frac {\sqrt {11}}{2}$

EDIT: arrggh  and forget the second part i messed up with the fractions somewhere..


induction
vector
projective geometry
analytic geometry

Use Pappus's formula and mathematical induction to show that the equation for more than two plates holds for any positive integer n>2.

c = (m1c1 + mc2) / (m1 + m2)
c = (m1c1 + m2c2 + ... + m_n c_n) / (m1 + m2 + ... + m_n c_n)

the c's are vectors

Can you explain the context of this problem a little bit more? As it's posted now, I think you're asking an ambiguous question.

Have you looked at the definition of Pappus' Thm.?

[b]Pappus’s Formula[/b]

Pappus knew that the centroid of the union of two nonoverlapping
plane regions lies on the line segment joining their individual centroids.
More specifically, suppose that $ m_{1}$ and $ m_{2}$ are the masses of
thin plates $ P_{1}$ and $ P_{2}$ that cover nonoverlapping regions in the xyplane.
Let $ \vec{c}_{1}$ and $ \vec{c}_{2}$ be the vectors from the origin to the respective
centers of mass of and Then the center of mass of the union
of the two plates is determined by the vector
(5) $ \vec{c}\equal{}\frac{m_{1}\vec{c}_{1} \plus{} m_{2}\vec{c}_{2}}{m_{1}\plus{}m_{2}}$
Equation (5) is known as Pappus’s formula. For more than two
nonoverlapping plates, as long as their number is finite, the formula generalizes to
(6) $ \vec{c}\equal{}\frac{m_{1}\vec{c}_{1} \plus{} m_{2}\vec{c}_{2}\plus{}...\plus{} m_{n}\vec{c}_{n}}{m_{1}\plus{}m_{2}\plus{}...\plus{}m_{n}}$
This formula is especially useful for finding the centroid of a plate of
irregular shape that is made up of pieces of constant density whose
centroids we know from geometry. We find the centroid of each piece
and apply Equation (6) to find the centroid of the plate.

51. Derive Pappus’s formula (Equation (5)). (Hint: Sketch the plates
as regions in the first quadrant and label their centers of mass as
$ (\bar{x}_{1}, \bar{y}_{1})$ and $ (\bar{x}_{2}, \bar{y}_{2})$ What are the moments of about the $ P_{1} U P_{2}$
coordinate axes?)

[b]
52. Use Equation (5) and mathematical induction to show that Equation
(6) holds for any positive integer[/b]

I'm now concerned with 52. Please help, thanks!

[quote="DiscreetFourierTransform"]In a more general sense, you're trying to prove the idea of [url=http://en.wikipedia.org/wiki/Superposition_principle]Superposition[/url].

The link there is very helpful, and should give a bit of an insight on how to attack this problem.[/quote]Agreeing with DFT: The momentum of the whole total mass is the sum of the momentums of the individual masses.


I would like to take the AMC 8 during the 2009-2010 school year as an 8th grader. The problem is that my school doesn't give out the AMC 8, and I'm pretty sure they aren't planning to, either. Does anyone have any suggessions for me if I can take it as a "homeschooled" student, even though I am not homeschooled (in another words, can I take it at home, or is the AMC 8 only offered through schools?)

[quote="RoFlLoLcOpT"] That's what I did (of course I didn't tell my principal that I was leaving at 11:30 just to take a math test :P)[/quote]

Should your principal feel happier if he/she knew you are taking a math challenge yourself?

Well he might not be happy that I'm missing half a day of classes just to take a math test

[hide]maybe move this to [url=http://www.artofproblemsolving.com/Forum/index.php?f=133]here[/url]?[/hide]

[quote="RoFlLoLcOpT"]I think it would be possible to leave school to go to another school to take it. That's what I did (of course I didn't tell my principal that I was leaving at 11:30 just to take a math test :P)[/quote]
Maybe. I don't know of any local schools that give it out, but I'll look.
Is the homeschool thing possible though?

Same problem here, but with AMC 10 and 12. Any suggestions?

Also, in the forum title, in parentheses, it says AMC 8. My question could be moved, but it is very similar to this question.


It had 2 real solutions $x=\pm \sqrt{2005}-\epsilon$. But solution can not be expession by nice formul.


Prove that the set $ \mathbf{B} \equal{} (B_{s}(a);s \in S$ and $ a$ is a non negative real number) where $ (S,\rho)$ is a metric space is a basis for topology in $ S$.

The author gives a proof where in he says $ B_{s}(0) \equal{} \emptyset$ so null set is generated by the set $ \mathbf{B}$ also $ \bigcup_{s \in S}B(s,a) \equal{} S$ which of course are true.
then what we have left to prove is:
$ B_{1}(s_{1},a_{1}) \cap B_{2}(s_{2},a_{2}) \equal{} \bigcup_{B \in \mathbf{B_{1}}}B$ where $ \mathbf{B_{1}} \subset \mathbf{B}$.
In proving this author uses union of balls of non zero radius.My doubt is why cant we just use the balls of zero radii and where $ \mathbf{B_{1}} \equal{} \mathbf{B}$ as used in proving that the null set is also generated by these sets

$ B(s,a)$ refers to the set ,whose elements are intersection of the  ball of radius $ a$ around the point $ s$ with the set $ S$.

The notation being used here implies that all the "balls of zero radius" are just empty sets.  The union of any collection of empty sets is the empty set, so we certainly couldn't describe $ B(s_1,a_1)\cap B(s_2,a_2)$ (or any other nonempty set) with a union of only balls of zero radius.  I may not be understanding your question, if so please clarify.

It puzzles me why the author would even introduce this odd notion of zero radius open balls.  I would much rather write the same basis as $ \mathbf{B} \equal{} \lbrace B_s(a): s\in S,a\in \mathbb{R}\cap (0,\infty)\rbrace\cup \lbrace \emptyset\rbrace$ to avoid this confusion, but in any case the notation doesn't change the argument.

ah! i get it by zero radius the author meant empty sets i had this in my mind(even while writing the previous question) i had in my mind that balls of zero radius referred to the point about which the ball was being made.Thanks!


number theory
greatest common divisor
calculus
integration
number theory proposed

Determine $ \gcd (n^2\plus{}2,n^3\plus{}1)$ for $ n\equal{}9^{753}$.

[hide="Solution"]
We will consider a general $ 1<n\in \mathbb{Z}$ and determine $ (n^2\plus{}2,n^3\plus{}1)$. Notice that $ (n,n^3\plus{}1)\equal{}1$, so we have $ (n^2\plus{}2,n^3\plus{}1)\equal{}(n^3\plus{}2n,n^3\plus{}1)\equal{}(n^3\plus{}2n\minus{}(n^3\plus{}1),n^3\plus{}1)\equal{}(2n\minus{}1,n^3\plus{}1)$. Because $ (2,2n\minus{}1)\equal{}1$, we may proceed as follows: $ (2n\minus{}1,n^3\plus{}1)\equal{}(2n\minus{}1,2n^3\plus{}2)\equal{}(2n\minus{}1,2n^3\plus{}2\minus{}n^2(2n\minus{}1))\equal{}(2n\minus{}1,n^2\plus{}2)$.

Again we can apply that $ (2,2n\minus{}1)\equal{}1$ to find: $ (2n\minus{}1,n^2\plus{}2)\equal{}(2n\minus{}1,2n^2\plus{}4)\equal{}(2n\minus{}1,2n^2\plus{}4\minus{}n(2n\minus{}1))\equal{}(2n\minus{}1,n\plus{}4)$. Applying this fact again, we find $ (2n\minus{}1,n\plus{}4)\equal{}(2n\minus{}1,2n\plus{}8)\equal{}(2n\minus{}1,9)$.

Thus, we have the formula $ (n^2\plus{}2,n^3\plus{}1)\equal{}(2n\minus{}1,9)$ for all integral $ n>1$. In the case $ n\equal{}9^{753}$, we find that $ (n^2\plus{}2,n^3\plus{}1)\equal{}(2n\minus{}1,9)\equal{}(2\cdot 9^{753}\minus{}1\minus{}9(2\cdot 9^{752}\minus{}1),9)\equal{}(8,9)\equal{}1$.
[/hide]


geometry
perimeter
inequalities
inequalities unsolved

Let a,b and c be positive real number with a+b+c=1,and Let p,q and r be the sides of triangle with perimeter 1 .

prove that 8abc =< ap+bq+cr

[quote="erdugan"]Let $ a,b$ and $ c$ be positive real number with $ a \plus{} b \plus{} c \equal{} 1$,and Let $ p,q$ and $ r$ be the sides of triangle with perimeter $ 1$ .

prove that $ 8abc \equal{} < ap \plus{} bq \plus{} cr$[/quote]
Let $ p \equal{} y \plus{} z,$ $ q \equal{} x \plus{} z,$ $ r \equal{} x \plus{} y$ and after homogenization use the following my inequality:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=258055
See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=216223


Find the total number of different integers the function

\[ f(x) = \left[x \right] + \left[2 \cdot x \right] + \left[\frac{5 \cdot x}{3} \right] + \left[3 \cdot x \right] + \left[4 \cdot x \right] \]

takes for $0 \leq x \leq 100.$

[hide]It is clear that the number of integers generated in the intervals $ [0,3),[3,6),[6,9),[9,12),\ldots,[96,99)$ are all equal. So now we consider the number of integers generated in the interval $ [0,3)$. Note that the value of $ f(x)$ change when at least one of $ [x],[2x],\left[\frac{5x}3\right],[3x],[4x]$ changes its value. It is easy to see that these values change when the value of $ x$ equals
\[ 0,\frac14,\frac13,\frac12,\frac35,\frac23,\frac34,1,\frac65,\frac54,\frac43,\frac32,\frac53,\frac74,\frac95,\frac2,\frac94,\frac73,\frac{12}5,\frac52,\frac83,\frac{11}4.
\]
So there are 22 different integers generated in the interval $ [0,3)$. Therefore there are $ 33\times22\equal{}726$ different integers generated in the interval $ [0,99)$. Now we calculate the number of integers generated in the interval $ [99,100]$, which equals to the number of integers generated in the interval $ [0,1]$. Analogously, we can easily see that there are 8 integers generated in this interval. Therefore there are $ 726\plus{}8\equal{}734$ different integers generated by this function for $ 0\le x\le 100$.[/hide]


abstract algebra
group theory
number theory proposed
number theory

For each $n$, define $L(n)$ to be the number of natural numbers $1\leq a\leq n$ such that $n\mid a^{n}-1$. If $p_{1},p_{2},\ldots,p_{k}$ are the prime divisors of $n$, define $T(n)$ as $(p_{1}-1)(p_{2}-1)\cdots(p_{k}-1)$.
a) Prove that for each $n\in\mathbb N$ we have $n\mid L(n)T(n)$.
b) Prove that if $\gcd(n,T(n))=1$ then $\varphi(n) | L(n)T(n)$.

(b) suggests that in (a) you actually meant $\varphi(n)|L(n)T(n)$.

(a)

In general, in a finite abelian group (it's true for non-abelian groups as well, but it should be significantly harder to prove), if $t\ |\ |G|$, then the number of elements $x$ such that $x^{t}=1$ is a multiple of $t$. We apply this to the group $G=(\mathbb Z/n\mathbb Z)^\times$ of invertible residues modulo $n$ with $t=(n,\varphi(n))$.

If $n=p_{1}^{a_{1}}\cdot\ldots\cdot p_{k}^{a_{k}}$, then by the above, $p_{1}^{a_{1}-1}\cdot\ldots\cdot p_{k}^{a_{k}-1}\ |\ (n,\varphi(n))\ |\ L(n)\ (*)$. After multiplying the leftmost and the rightmost terms in $(*)$ by $T(n)$ we get $\varphi(n)|L(n)T(n)$.

(b)

In this case, $a^{n}=1$ (in the group $G$) iff $a^{\frac{\varphi(n)}{T(n)}}=1\ (\#)$. $G$ is the direct product of a group of order $\frac{\varphi(n)}{T(n)}$ and one of order $T(n)$, and these orders are coprime. This means that the solutions to $(\#)$ are precisely the elements of the former group, so there are $\frac{\varphi(n)}{T(n)}$ of them. The desired result follows.

I think there is a strongly result:

$L(n)T(n)=\varphi(n)\prod_{i}\left(p_{i}-1,\frac{n}{p_{i}^{e_{i}}}\right)$ where $n=\prod_{i}p_{i}^{e_{i}}$

Someone confirm it?

How to prove this:

[quote="grobber"]
In general, in a finite abelian group (it's true for non-abelian groups as well, but it should be significantly harder to prove), if $t\ |\ |G|$, then the number of elements $x$ such that $x^{t}=1$ is a multiple of $t$.[/quote]

a)
Note that that if $p_i\mid a^n-1$ $\implies$ $v_{p_i}(n)\mid a^n-1$ simply because of LTE and the fact that order is coprime with $p_i-1$.Note that by previous once we found the solution set modulo each prime the rest is just shifting by $\text{rad}(n)$ $\implies$ $\frac{n}{\text{rad}(n)}\mid T(n)$  and so the conclusion.

b)
We show that $L(n)T(n)=L(n)\frac{n}{\text{rad}(n)}\prod_{i=1}^k \left(\frac{n}{v_{p_i}(n)},p_i-1 \right)$ which will effectively end it.$a^n=1$ in $\mathbb{F}_{p,{\times}}$ and so $\text{ord}(a)\mid \frac{n}{v_p(n)}$ but all such numbers are solutions to $X^{\frac{n}{v_p(n)}}-1$ which has $\left(\frac{n}{v_{p_i}(n)},p_i-1 \right)$ solutions and now using CRT and the note from a) we're done.

@[b]Boll[/b], I agree: By CRTing roots of $x^n-1$ modulo prime-power divisors of $n$, and the fact that $(\mathbb Z/p^k\mathbb Z)^\times$ is cyclic under multiplication:
$$L(n)=\prod \gcd\left(n,\varphi(p_i^{e_i})\right)=p_i^{e_i-1}\gcd\left(\frac{n}{p_i^{e_i-1}},p_i-1\right)$$
At which point we are basically done.


calculus
integration
algebra
polynomial
derivative
real analysis
real analysis unsolved

This is for sure a nice problem, point 2 especially :wink:  :)

Let $ I_1,...,I_n$ be disjoint closed nonempty subintervals of R.
1. Prove that if P is a real polynomial of degree less than n such that: $ \int_{I_j}^{}P(x)dx = 0, %Error. "foreach" is a bad command.
j=1,...,n$ (*), then P = 0
2. Prove that there is a nonzero real polynomial P of degree n that satisfies (*)

$ \int_{a_j}^{b_j} P(x) dx\equal{}0$ implies $ \exists x_j \in(a_j,b_j)$ such that $ P(x_j)\equal{}0$

n disjoint intervals leads to n zeros leads to any nonzero P having degree n or greater

hence P=0

let $ Q(x)$ be a n+1 degree polynomial

all we have to do is find such a Q with

$ Q(a_j)\equal{}Q(b_j)$ for all j, and then its derivative is the desired P

or notice that the intersection of $ n$ hyperplane of $ \mathbb{R}_{n}[X]$ (linear space of polynomial of degree at most $ n$) cannot be empty.


geometry
incenter
perimeter
number theory
Diophantine equation
national olympiad

1.Suppose $a>b>0$, ${f(x)=\frac{2(a+b)x+2ab}{4x+a+b}}$. Show that there exists unique $x>0$, such that $f(x)=(\frac{a^{\frac13}+b^{\frac13}}{2})^{3}$.

2.In $\triangle ABC$, $\angle ABC=90^\circ$, and $D$, $G$ are two points on $CA$. Let $E$, $F$ be the projection of $A$, $G$ on $BD$ respectively. Suppose $BE=EF$. Show that $\angle ABG=\angle DFC$.

3.A set of poker has 52 cards with 13 cards in each suit (diamonds, clubs, hearts, spades). Each suit of cards are marked $2, 3, \cdots , 10, J, Q, K, A$. Calculate the number of subsets with 13 cards of the poker, where $2, 3, \cdots , 10, J, Q, K, A$ all appear and no two consecutive cards ($A$ and $2$, $2$ and $3$, …, $K$ and $A$) of the same suite both appear.

4.For any positive integer $n$, let $a_{n}$ be the real number solution of $x^{3}+\frac{x}{n}=1$. Show that (1) $a_{n+1}>a_{n}$; (2) $\sum_{i=1}^{n}\frac{1}{(i+1)^{2}a_{i}}<a_{n}$.

5.In $\triangle ABC$, $\angle A=60^\circ$. The incentre is $I$. The inscribed circle of $\triangle ABC$ meets $AB$, $AC$ at $D$, $E$ respectively. Line $DE$ meets line $BI$ and line $CI$ at $F$, $G$ respectively. Show that $FG=\frac12 BC$.

6.Find out the smallest real number $m$, such that for all positive real number $a,b,c$ with $a+b+c=1$, $m(a^{3}+b^{3}+c^{3})\geq 6(a^{2}+b^{2}+c^{2})+1$.

7. (1) Find out the number of positive integer solutions in $m$, $n$ and $r$ of the Diophantine equation $mn+nr+rm=2(m+n+r)$.
 (2) Fix an integer $k>1$. Show that the Diophantine equation $mn+nr+rm=k(m+n+r)$ has at least $3k+1$ positive integer solutions in $m$, $n$ and $r$.

8.Given a positive integer $n$, let $P_{n}$ be the smallest positive integer such that there are $P_{n}$ points $A_{1}, A_{2}, \cdots , A_{P_{n}}$ on a circle of perimeter $n$ satisfying the following condition: for any integer $m$ in $1,2, \cdots , n-1$, there are two points $A_{i}, A_{j}$ ($1\leq i,j \leq P_{n}$), and one arc of the circle with endpoints $A_{i}$ and $A_{j}$ has length $m$. In this case, we call the sequence of the lengths $a_{1}, a_{2}, \cdots , a_{P_{n}}$ of the arcs of adjacent points a splitting sequence of $n$, written as $T_{n}=( a_{1}, a_{2}, \cdots , a_{P_{n}})$. For example, we have $P_{13}=4$ where the corresponding splitting sequences of 13 are $T_{13}=(1,3,2,7)$ or $(1,2,6,4)$. Find $P_{21}$ and $P_{31}$, and give a splitting sequence of $21$ and a splitting sequence of $31$.

Problem 1: http://www.mathlinks.ro/Forum/viewtopic.php?p=586686#586686

Problem 2: http://www.mathlinks.ro/Forum/viewtopic.php?p=586687#586687

Problem 3: http://www.mathlinks.ro/Forum/viewtopic.php?p=586688#586688

Problem 4: http://www.mathlinks.ro/Forum/viewtopic.php?p=586689#586689

Problem 5: http://www.mathlinks.ro/Forum/viewtopic.php?p=586690#586690

Problem 6: http://www.mathlinks.ro/Forum/viewtopic.php?p=586691#586691

Problem 7: http://www.mathlinks.ro/Forum/viewtopic.php?p=586693#586693

Problem 8: http://www.mathlinks.ro/Forum/viewtopic.php?p=586694#586694

Problem Authors:
Problem 1: Shenghong Li

Problem 2: Bin Xiong

Problem 3: Pingsheng Tao

Problem 4: Shenghong Li

Problem 5: Pengcheng Zhang

Problem 6: Bin Xiong

Problem 7: Hanhui Yuan

Problem 8: Pingsheng Tao

[quote="ychjae"]It seems that Mr.Xiong Bin proposes both the hardest question(Q.2) and the easiest one(Q.6).[/quote]
I think that problem 1 is also very easy......

Anyway, here is a PDF file for downloading and printing.


superior algebra
superior algebra solved

I am very much a beginner when it comes to proofs and would greatly appreciate any help. Thanks in advance.

Suppose :beta: is an integer. :beta: is prime and greater than 0. :beta: cannot be written as a sum of two squares. Show that :beta: is irreducible in Z[i].

Factor $\beta = xy$ and assume $x=a+bi$. By taking the norm on both sides of the equation you get $\beta^2=|\beta |=|xy|=|x||y|$. Now $|x|=1;\beta;\beta^2$, but we can exclude the case $a^2+b^2=|x|=\beta$ because $\beta$ isn't the sum of two squares, and $|x|=1$ gives that $x$ is an unity and $|x|=\beta^2$ gives $|y|=1$ so in that case $y$ is an unity.


trigonometry
geometry
circumcircle
inequalities
function
inequalities proposed

[color=purple] Let$A_{1}A_{2}....A_{n}$ is the n-gon in plane inscribe (O,R) out the (I,r)
  Prove that
        $\frac{R}{r}\geq \frac{1}{\cos\frac{\pi}{n}}$
 With n=3,n=4 we have the well -known $R\geq 2r$ and $R\geq\sqrt{2}r$
     Have fun !  :D  [/color]

[color=purple]No one like this problem?
        Thanks Gac Kiem! Nhung nhac' ban 1 ti' nha' anh em Viet Nam co' ung? ho nhau cung~ dung` post bai` kieu? nay` nhe' co' may' thang` noi' ra noi' vao` day' [/color]

I didn't want to reply anything in topics of Vietnamese, but here really is a nice problem!
ok the following is solution
Called $O$ is circumcenter of that polygon and let $\angle A_{i}OA_{i+1}=a_{i},n+1\equiv 1$ with $\sum_{i=1}^{n}a_{i}=2\pi$ by law of sine $A_{i}A_{i+1}=2R\sin\frac{a_{i}}{2}$ we let $S$ be area of polygon then
$S=\frac{1}{2}(\sum_{i=1}^{n}A_{i}A_{i+1})r=\frac{1}{2}R^{2}(\sum_{i=1}^{n}\sin a_{i})$
but $\sum_{i=1}^{n}A_{i}A_{i+1}=2R(\sum_{i=1}^{n}\sin\frac{a_{i}}{2})$ thus we have
$\frac{R}{r}=2\frac{\sum_{i=1}^{n}\sin\frac{a_{i}}{2}}{\sum_{i=1}^{n}\sin a_{i}}$
Now WLOG $a_{1}\ge a_{2}\ge .. a_{n}$ by Chebyshev's inequality
$n(\sum_{i=1}^{n}\sin a_{i})=2n(\sum_{i=1}^{n}\sin\frac{a_{i}}{2}\cos\frac{a_{i}}{2})\le 2(\sum_{i=1}^{n}\sin\frac{a_{i}}{2})(\sum_{i=1}^{n}\cos\frac{a_{i}}{2})$
Thus $\frac{R}{r}\ge \frac{n}{\sum_{i=1}^{n}\cos\frac{a_{i}}{2}}(1)$
But $\cos x$ is concave on $[0,\frac{\pi}{2}]$ function by Jensen's inequality we get
$\sum_{i=1}^{n}\cos\frac{a_{i}}{2}\le n\sum\cos(\sum_{i=1}^{n}\frac{a_{i}}{2})=n\cos\frac{\pi}{n}(2)$
Combine result $(1),(2)$ we are done!

[color=purple]  Ok it's my solution for this problem,too. Now do you have the stronger? [/color]

[color=purple]Let $A_{1}A_{2}.........A_{n}$ is the convex polygon with n side .M in plane .Let $B_{1},B_{2}...B_{n}$ is the ortho projection of M on the side $A_{1}A_{2}$.......... $A_{n}A_{1}$.Prove that
       $MA_{1}+MA_{2}.........+MA_{n}\ge \frac{1}{\cos\frac{\pi}{n}}(MB_{1}+MB_{2}+......+MB_{n})$
     [/color]


The negation of the statement "all men are honest," is:
$ \textbf{(A)}\ \text{no men are honest} \qquad\textbf{(B)}\ \text{all men are dishonest} \\
\textbf{(C)}\ \text{some men are dishonest} \qquad\textbf{(D)}\ \text{no men are dishonest} \\
\textbf{(E)}\ \text{some men are honest}$

[hide="Solution"]If $ P \equal{} \text{all men are honest}$, then $ \neg P \equal{} \boxed{\textbf{(C)}\ \text{some men are dishonest}}$.

Edit: You're right, that does make more sense.
[/hide]

[hide="Actually..."]Saying "not all men are honest" implies that there's at least one dishonest man, not that all men are dishonest. Thus, the answer is C, some men are dishonest.[/hide]
[/hide]
Oops, also found several typos and fixed.


number theory
prime factorization
combinatorics proposed
combinatorics

A [i]perfect partition[/i] of the positive integer $n$ is a finite sequence $a_1 \geq \cdots \geq a_k$ of positive integers such that each integer $m \in \{1,2,\ldots,n\}$ can be [i]uniquely[/i] written as the sum of $a_i$'s (two representations which differ only by a permutation are considered identical). Determine the number of perfect partitions of $n$ in terms of the prime factorization of $n+1$.

Enjoy this one!  ;)

--Vesselin

I think it needs some clarification. For example, is the sum $a_1+a_1+a_2$ valid? Another question: can two of the $a_i$'s be equal? I mean, you said that sums are different up to permutations of the terms, but did you mean permutations of the set $\{a_{k_1},\ldots,a_{k_m}\}$ or permutations of the indices of $a_{k_i}$?

Grobber is able to find questions on the empty place!  :lol:  :lol:  :lol:

:lol:  Of course you cannot take $a_1+a_1+a_2$ as a sum (how else would we have the uniqueness?). Regarding your second question, I mean permutation of the [i]multiset[/i] $\{a_{i_1},\ldots,a_{i_s}\}$ (and [i]not[/i] of the indices). In other words, equal $a_i$s are considered indistinguishable.

[b]Example. [/b] $(1,2,2,6)$ is a perfect partition of $11$. Indeed, we have the unique representations $1,2,1+2,2+2,1+2+2,6,1+6,2+6,1+2+6,2+2+6,1+2+2+6$.

[b]NB: [/b] I forgot to explicitly mention something which is understood by the word 'partition.' A perfect partition $a_1 \geq \cdots \geq a_k$ must also be a partition of $n$, i.e. $a_1 + \cdots + a_k = n$. :lol:

--Vesselin

I have solved it but I don't know how to write the closed form of my answer :?

Answer is the number of distinct decompositions of $n+1$ (two decompositions which differ only by a permutation are considered [i]distinct[/i]).

Indeed, suppose $\underbrace{a_1,a_1,...,a_1}_{k_1\textrm{ times}},...,\underbrace{a_l,a_l,...,a_l}_{k_l\textrm{ times}}$ is a perfect partition, $a_1<a_2<...<a_l$. Then it is clear $(k_1+1)...(k_l+1)=n+1$. So each perect partition determine a decomposition of $n+1$.

Suppose now $(k_1+1)...(k_l+1)=n+1$. Then it is clear $a_1=1$, $a_2=k_1+1$, $a_3=(k_1+1)(k_2+1)$, $a_l=(k_1+1)...(k_{l-1}+1)$. Thus each decomposition of $n+1$ determine a unique perfect partition.

well i had a recursive answer but i think Myth's is equivalent(or perhaps neater).

if $p(n)$ denotes the number of perfect partitions of $n$ then it is easy to get
$\displaystyle p(n)=\sum_{d|n+1,d<n} p(d-1)$.
this is quite easy since if the last distinct part is $a$ then the sum of all the parts $< a$ must be $a-1$,so that $a|n+1$. conversely if $a|n+1$ then for each perfect partition of $a-1$ one can append some $a$'s to get a perfect partition of $n$.


I've found one abstract algebra [url=http://www.amazon.com/dp/1441451579/]book[/url] at Amazon (published in January 13, 2009) by Nicolas Bourbaki  :idea:  How could this be possible :?:

Nicolas Bourbaki is not a real person.  Bourbaki is a pen name and belongs to a [url=http://en.wikipedia.org/wiki/Nicolas_Bourbaki]large group[/url] of mathematicians whose membership has changed over time.

Edit:  The book also doesn't seem to actually be by Bourbaki.  I can't really tell to what extend "Nicolas Bourbaki Junior" is a joke person or a joke group.

Nicolas Bourbaki was one of Napoleon's generals. His name was used for the French group which started in the 30s.

this seems like a joke to me. read the title for god's sake!


[quote="math92"]i'm having trouble on this one:

$\frac{6x^2+2x}{3x^2+4x+1}+\frac{4x}{3x+3}$[/quote]

[hide]$\frac{6x^2+2x}{3x^2+4x+1}+\frac{4x}{3x+3}$

Factor the first fraction's denominator and take out the 2x in that fraction's numberator...

$\frac{2x(3x+1)}{(3x+1)(x+1)}+\frac{4x}{3(x+1)}$

$\frac{2x}{x+1}+\frac{4x}{3(x+1)}$

$\frac{6x}{3(x+1)}+\frac{4x}{3(x+1)}$

$\frac{10x}{3(x+1)}$[/hide]

I'm not sure about an easy way, but here's my solution:

[hide]Multiply the first one by three (denominator and numerator to get $\frac{18x^2+6x}{9x^2+12x+3}$. The denominator is $(3x+1)(3x+3)$. Isn't one of the terms the denominator of the second one? I hope you how to add fractions with unlike terms, so multiply the second term, numerator and denominator by $3x+1$ to get

$\frac{12x^2+4x}{9x^2+12x+3}$.

$\frac{18x^2+6x}{9x^2+12x+3}+\frac{12x^2+4x}{9x^2+12x+3}=\frac{30x^2+10x}{9x^2+12x+3}=\frac{10}{3x+3}$
 [/hide]

[quote="236factorial"]I'm not sure about an easy way, but here's my solution:

$\frac{10}{3x+3}$
[/quote]
you made a tiny mistake.  3x+3 can be simplified to 3(x+1)

math, thats not really simplifying, as if there was more simplifying to do, you would have to make it 3x+3 again to simplify it.  However, there is no more simplifying.  I think 236 factorial was right :D

I don't know about everybody else, but I found Drunken Math's method way easier, I don't really get what 236factorial is saying  :oops:

YEah, I started to get ahead of this forum, just read Drunken_MAth's post.  :blush:


if $a$ and $b$ are two real positive numbers such that $ a \leq b $ and $ a+b > 0 $
prove that for all reals $ x_1,x_2,....x_n$ in $ [a,b]$
we have the inequality

 \[ \frac{\sum_{1} x_i}{n} \geq  \frac{ab}{a+b}+ \frac{ \sum_{1} x_i }{n(a+b)} \]

Since $a \le x_i \le b$, $(x_i - a)(b - x_i) \ge 0$, which expands as $(a + b) x_i \ge ab + x_i^2$.  Summing over $1 \le i \le n$ and dividing by $n(a + b)$ gives the result.


abstract algebra
vector
Galois Theory
Ring Theory
superior algebra
superior algebra unsolved

Let D be a division ring with characteristic p > 0 and let Z be the center of D.Suppose that $ a \in D \backslash Z$ such that $ a^{p^{n}} \equal{} a$ for some integer n > 0.Then there exists an  $ x \in D$ such that $ xax^{\minus{}1} \equal{} a^i \neq a$ for some integer i .

I actually had to rework [url=http://en.wikipedia.org/wiki/Skolem%E2%80%93Noether_theorem]Skolem-Noether[/url] for this one.
Please stay with me. It's a quite long proof.

$ K\equal{}Z(a)$ is a subfield of $ D$ with $ K/Z$ finite.
Let $ f(X)$ be the minimalpolynomial of $ a$ over $ Z$
and $ g(X)$ be the minimalpolynomial of $ a$ over the prime field $ \mathbb{F}_p$.
Then $ f(X)|g(X)|X^{p^n}\minus{}X$ and $ \deg f>1$.
But $ \mathbb{F}_{p}(a)$ is a finite field, so $ \mathbb{F}_{p}(a)/\mathbb{F}_p$ is galois with galois group
generated by the [url=http://en.wikipedia.org/wiki/Frobenius_automorphism]Frobenius automorphism[/url] $ \mathbb{F}_{p}(a)\to\mathbb{F}_p(a),x\mapsto x^p$.
So the algebraic conjugates of $ a$ over $ \mathbb{F}_p$ are $ a,a^p,a^{p^2},a^{p^3},\ldots \in K$
and $ g(X)$ is a product of coprime linear factors with those roots.
Then $ f(X)|g(X)$ and $ \deg f>1$ gives some root $ \not\equal{}a$ of $ f(X)$ of the form $ a^{p^i}\in K$ for some $ i\in\mathbb{N}$.

And now we get the $ Z$-isomorphism
$ \sigma: K\equal{}Z(a)\to Z[X]/(f)\to Z(a^{p^i})\subseteq D, a\mapsto X\plus{}(f)\mapsto a^{p^i}$
and I want to extend $ \sigma$ to an inner automorphism of $ D$ using Skolem-Noether.
But there is just the problem $ \dim_Z D$ not necessarily finite, only $ \dim_Z K$ finite.

The key is to consider the ring $ R\equal{}D\otimes_Z K$.
We have the extension $ R\supseteq D\otimes 1\cong D$.
So any $ R$-module is also a $ D$-module by left-multiplication and for any
decreasing sequnece $ R\supseteq I_1\supsetneq I_2\supsetneq \cdots$ of $ R$-submodules
we have $ \dim_D R\ge \dim_D I_1>\dim_D I_2>\cdots$
with $ \dim_DR\equal{}\dim _D D\otimes_Z K\equal{}\dim_Z K$ finite.
(modules over division rings are just like vector spaces over fields, i.e. free and [url=http://en.wikipedia.org/wiki/Semisimple_module]semisimple[/url])
So $ R$ must be an [url=http://en.wikipedia.org/wiki/Artinian_ring]artinian ring[/url].

Also using the fact, that $ A\otimes_Z B$ is a simple ring for any simple rings $ A,B$ with $ Z\equal{}Z(A)\subseteq Z(B)$
we see, that $ R$ is a simple artinian ring, and therefore by [url=http://en.wikipedia.org/wiki/Artin-Wedderburn_theorem]Artin-Wedderburn[/url] we have
$ R\cong \mathcal{M}_n(\tilde{D})$ with some division ring $ \tilde{D}$ and $ n\in \mathbb{N}$, which has up to isomorhism only one simple module $ \tilde{D}^n$.

But now $ D$ becomes an $ R$-module in two different ways by defining
$ (x\otimes y)d\equal{}xdy$ and $ (x\otimes y)d\equal{}xd\sigma(y)$ for $ x,d\in D,y\in K$.
Let's call these modules $ D_1,D_2$. Then $ \dim_D D_1\equal{}\dim_D D_2\equal{}1$,
so $ D_1,D_2$ are simple $ R$-modules and the above gives $ D_1\cong_R D_2$.
And now any $ R$-iso $ \phi: D_1\to D_2$ gives $ x: \equal{}\phi (1)\not\equal{}0$ with
$ ax\equal{}(a\otimes 1)\phi(1)\equal{}\phi((a\otimes 1)1)\equal{}\phi(a)\equal{}\phi((1\otimes a)1)\equal{}(1\otimes a)\phi(1)\equal{}x\sigma(a)$
and $ x^{\minus{}1}ax\equal{}\sigma(a)\equal{}a^{p^i}\not\equal{}a$.


probability
combinatorics proposed
combinatorics

An IPod automatically selects a song to play from a playlist consisting of m songs. The respective probabilities of selecting each of the m songs are p1, p2, p3 … pm such that $ \sum_{i \equal{} 1}^m$ pi = 1. Assume that the songs are played continually and each selection is independent of the other. Let X be the total number of songs required to be played before all of the m songs have been played at least once.  What is the probability P(X=n)? You must express the answer in p1,p2,..pm, m and n.  :)

Oops, the title needs to be revised. Can the moderators please revise the title as "Probability Problem" instead of "Probality Problem"

I am afraid your answer is wrong. Can you please explain your solution, then we will be able to see what went wrong (or not).

To make the question more clear:  Assume that each song played is an independent trial. And let X denote the number of trials needed until each song has been played at least once. The question is, what is the probability that after playing exactly n songs, all of the m songs have been played atleast once?

[u]Hint:[/u] A similar analogy would be to ask, what is the probability that after 5 flips of a fair coin we have observed both heads and tails at least once?

$ P(X\equal{}n) \equal{} C_n^m\prod_{j\equal{}1}^mp_j \minus{} P(X\leqslant n\minus{}1)$

$ P(X\leqslant n\minus{}1) \equal{} C_{n\minus{}1}^m\prod_{j\equal{}1}^mp_j$

So:

$ P(X\equal{}n) \equal{} C_n^m\prod_{j\equal{}1}^mp_j \minus{} C_{n\minus{}1}^m\prod_{j\equal{}1}^mp_j$

So far it looks quite reasonable to me :)

taxofonas wrote:
[quote]
P(X\leqslant n-1) = C_{n-1}^m\prod_{j=1}^mp_j
[/quote]

This is incorrect.

What I meant was that
$ P(X\leqslant n\minus{}1) \equal{} C_{n\minus{}1}^m\prod_{j\equal{}1}^mp_j$

is incorrect.

KRIS17, there's an edit button on each of your posts (on AoPS skin, in the lower right near the quote button) -- you can always go back and change what you've written (for up to 24 or 48 hours) instead of making double posts.  Also, when posting, there's a preview button so you can check in advance that what you're written is what you intended.

This is essentially a coupon collector problem.  It shouldn't be too hard to write down a big unpleasant sum for the answer using inclusion-exclusion, and I don't see any reason to suppose that a nicer answer exists.

$ P(X\leqslant n \minus{} 1) \equal{} C_{n \minus{} 1}^m\prod_{j \equal{} 1}^mp_j \minus{} P(X\leqslant n \minus{} 2)$ and so on until we rech m.

Then:

$ P(X \equal{} n) \equal{} \sum_{k\equal{}m}^{n}(\minus{}1)^{n\minus{}k}C_k^m\prod_{j \equal{} 1}^mp_j$

KRIS17
Your second formulation of the problem differs from the first one because of the
"required" condition you used (in the first).
In other words, in the first formulation of the problem you asked the probability
of hitting all the m song exactly at the n-th step; in the second one you asked
the probability of hitting all the songs at some step $ s\leq n$.

vl4d wrote:
[quote]
KRIS17
Your second formulation of the problem differs from the first one because of the
"required" condition you used (in the first).
In other words, in the first formulation of the problem you asked the probability
of hitting all the m song exactly at the n-th step; in the second one you asked
the probability of hitting all the songs at some step .
[/quote]

You are correct. So according to my second formulation of the problem you should be able to get the P(X=n) using the principle of inclusion-exclusion, but there is a trick involved. It is not possible to calculate P(X=n) directly, but it is possible to write an expression involving multiple sums for P(X>n) using the principle of inclusion-exclusion. Then write $ P(X\equal{}n) \equal{} P(X>n\minus{}1) \minus{} P(X>n)$ to get the answer.


geometry
geometric transformation
reflection

With the economy in turmoil and Congress rushing to reach an agreement, McCain has "suspended his campaign" and wants to postpone the debate tomorrow (to the date of the VP debate).  Some argue this shows that McCain is trying to show he is in touch with the people and wants to fix the crisis, others that since he is down in the polls, he wants to stall Obama's momentum.

What do you think?

I think "suspending his campaign" is supposed to mean he's surrendering the election to his opponent. However, people are still willing to vote for him, and people are still arguing over him...even his opponent, Obama, doesn't believe that he's pulling out.

Believe me, "suspending his campaign" does not mean that he's pulling out.  It means he's stalling.  If he pulled out, people would go crazy because of what he did to "Sarah".

"Suspending" his campaign is entirely a political stunt. He kept running ads, and spent the time on interviews, speeches, and the like.
Oh, and here's the next absurdity: he's got web ads out declaring that he won the debate which hasn't happened yet.

To, jmerry, I hope you read this before the debate tonight because I'd really like you to provide a link to such an ad.

I'm really looking forward to this debate.  Will Barack be left there alone or will Mac show some guts and show up?  As Obama pointed out, a president must multi-task, although it might be hard for a 72-year-old.

Here's an interesting little tidbit: In a recent poll, over 50% of those polled said Palin is ready to be VP, but only 49% thought she was prepared to be president.

All right. Here's a [url=http://voices.washingtonpost.com/thefix/2008/09/mccain_wins_debate.html]link[/url]. The screenshot there comes from about 10:15 AM today (Eastern?).

That's pretty absurd, assuming it's real.

Oh wow, either that's fake, or McCain is incredibly stupid...

That sort of disturbs me considering a fair number of people will probably see that and just assume it is legitimate. I also didn't think that debates are considered "won" or "lost." It is up to the individual to decide for themselves which was better.

Speaking of the debate, what were your thoughts?  McCain came across as pretty angry and combative, often times, refusing to even look at Obama.

For next time --

[u]Obama needs to:[/u]
0. Work on time management within his responses. Really.
1. Find a way to respond to long series of short statements than trying to cut in and say McCain's lying
2. Find a better way of cutting McCain off, and recognize when he can't
3. Figure out what to do about McCain's long anecdotes. Anything McCain says that Obama doesn't object to, including off topic anecdotes, is going to win him points
4. Think about if it really helped him to say "John" instead of "Senator McCain", and stuff like that

[u]McCain needs to:[/u]
0. Get a brain. While he did make several meritorious arguments, he often sounded like a complete idiot
1. Not act like a little kid that's mad because his mommy won't buy him ice cream
2. Relax
3. Not look/sound like a creeper when he laughs
4. Not use the word "maverick" without backing it up properly, and things like that

I thought both candidates did fine but neither did very well; I'd say the result is status quo.  If I had to point to one thing I thought was a mistake for either candidate, it was McCain once again invoking Reagan as if he were some kind of political and administrative godsend.  This might appeal to his base, but it surely alienated many left-leaning undecided votes.  For Obama, it was his continued use of the 'we should have never entered Iraq' strategy.  It came off as his trying to capitalize on an exhausted issue and, as McCain immediately pointed out, one that has little bearing on how to proceed now that we're already there.  Besides, any points he might have gained with those jab were neutralized by McCain's insistent distancing himself from Bush, a perception Obama tried to but was unable to counter.

Both candidates did badly when it came to being direct on answering questions.  Often it was too obvious they weren't at all interested in what Jim Lehrer had just asked, and focused instead on going one by one through their prepared points.  This became especially clear when the moderator tried to get the candidates to talk about how the financial crisis is going to change their previously outlined plans on tax and spending policy.  Both clearly had no good answer prepared, and just stuck to their own unrelated talking points (worse yet, most of these talking points were just attacks on the other guy).  After about half a dozen attempts to rephrase the question, the visibly frustrated Lehrer gave up and moved on.  As far as domestic issues, both of these guys lost in my eyes.

[quote="TZF"]
1. Find a way to respond to long series of short statements than trying to cut in and say McCain's lying
[/quote]
Well wasn't McCain deforming the truth A LOT? How is this even remotely acceptable

I know that this is kind of random, but who do you guys want to be president. I think McCain would be a better president, but he is old, and his vice president is very young, and Obama has a better vice president.

I agree with JoeBlow's analysis of how they handled the issues. I'm hoping the next debate is a bit juicier.

[quote="Kalle"]Well wasn't McCain deforming the truth A LOT? How is this even remotely acceptable[/quote]

Of course he was, and of course it wasn't acceptable. What I'm saying is that Obama needs to be better at handling it. At times, especially near the end, he seemed to break his usual level of composure when he tried to cut in going "Now, John, you know that's not accurate", or whatever. Sometimes, he even stuttered a lot, trying to wedge his comments in.

I liked it much better the way he handled the thing about the 42 million for troop supplies, when he waited, and then in his response said, I don't know why you keep saying this, you voted against one bill b/c it had a timetable and I voted against one bill b/c it didn't have a timetable, etc. He looked calm and poised, and his message was clear, and he really made John McCain look bad.

Obama needs to nail all of McCain's half-truth statements like he did with that one, because otherwise, he just makes himself look bad.

This is what Republicans do-they distort facts and then say things like "If you're not with us, you're anti-American" and the Democrats are spineless so they just say "Okay mommy, I agree" and just go with it.  Obama needs to jump on McCain'd distortion of facts and nail him on his lack of knowledge.  McCain said there is a different between a "strategy" and a "tactic" in a war.  Listening to him, one would think that the plan for Iraq was to f-up for the first 4 years, send in an additional 120,000 troops, and say "Look, we're geniuses".  Wait for the economics debate-McCain will sound like a bumbling idiot.

And where was McCain's lapel pin?

The quote about the difference between the words strategy and tactic was a jab McCain took at Obama after the latter accidentally used the incorrect word -- Obama said tactic where he meant to say strategy.  That's all.

After that tongue-in-cheek remark, if you bothered to listen, the candidate also proceeded to explain exactly what the strategy in question was, exactly what he believes the mistakes were that were made earlier in the war, exactly why he believes the new strategy that was undertaken did work, and why he believes that reflects positively on him.  This issue was one of the few on which McCain spoke clearly and decisively.  And, if you were listening, Obama [u]agreed[/u] with McCain on the success of the strategy.  The only relevant retort Obama had was that we shouldn't have been there in the first place.

In this instance it is only you who is taking words, the meaning of which you obviously haven't bothered to think about, and distorting them to fit your own prejudices about a candidate's position.  Considering the opening line of your tirade, that's incredibly hypocritical.

[quote="JRav"]This is what Republicans do-they distort facts and then say things like "If you're not with us, you're anti-American"[/quote]Maybe Rush Limbaugh does it but no one sane. You're making a humongous generalization. Most republicans are sane, just like most democrats.

Most Republicans are normal, just with some views I disagree with.  Some, like Sarah Palin, are not.

At least they're funny  :) Even Palin thinks they're funny

I think they should stop with them.  They lose their edge after however many sketches.  I haven't heard that Palin thinks they're funny.

[quote="JRav"]I just watched the latest SNL sketch of Palin.  It was funny, but even I think they portrayed her to be more stupid than she is (and I don't think highly of her).[/quote]

did you just complain about a satirical television show not providing an accurate portrayal of an individual?

To be perfectly honest, the bailout section of the latest SNL skit sounded quite accurate, both in style and content.

I can't wait for the vice-presidential debates!

That skit wasn't just accurate in style and content- it included a lot of [i]exact quotes[/i]. The original interview was already horrifically funny.

[quote]did you just complain about a satirical television show not providing an accurate portrayal of an individual?[/quote]

I can't stand Palin and I thought the first two were funny.  I just think that they lose their edge after however many sketches.  If, on the other hand, Palin continues to deliver material, then by all means they should continue with the sketches (of course, as long as Tina Fey is willing to do them).

To worthawholebean:

"Either you are with us, or you are with the terrorists."

This was said by George W. Bush, our President, in a speech addressing the Congress, not by Rush Limbaugh.  It's not *exactly* the quote you were talking about, but it really is on the same level of thoughtfulness.


geometry
IMO Shortlist
circumcircle
quadrilateral
reflection
similar triangles

The point $M$ is inside the convex quadrilateral $ABCD$, such that

\[ MA = MC, \hspace{0,2cm} \widehat{AMB} = \widehat{MAD} + \widehat{MCD} \quad \textnormal{and} \quad \widehat{CMD} = \widehat{MCB} + \widehat{MAB}. \]

Prove that $AB \cdot CM = BC \cdot MD$ and $BM \cdot AD = MA \cdot CD.$

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

Let A' = DM^BC, from <CMD = <MCB+<MAB it follows that ABA'M is ciclyc.
Let C' = BM^AD, from <AMB = <MAD+<MCD it follows that DMCC' is ciclyc.

As supplementary of the same angle, it holds that <DMA = <CBA. From [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=145011#p145011]collinearity problem[/url] we have that <ACB = <DAM. Then DM/AM = AB/BC and finally DM/CM=AB/BC.

As both supplementary of the equal angles <C'MC = <C'DC, it holds <BMC = <ADC.  Furthermore, see again [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=145011#p145011]collinearity problem[/url], <MCB = <MCA+<ACB = <MAC+<DAM = <DAC. Then BM/MC=CD/AD or BM/MA=CD/AD.

[b]OFFICIAL SOLUTION:[/b]

Construct the convex quadrilateral $PQRS$ and the interior point $T$ such that $\triangle PTQ \equiv \triangle AMB, \triangle QTR \sim \triangle AMD$ and $\triangle PTS \sim \triangle CMD$. It follows that

\[
TS = \frac{MD \cdot PT}{MC} = MD, \hspace{0,2cm} \frac{TR}{TS} =
\frac{MD \cdot MB}{MA \cdot MD} = \frac{MB}{MC}
\]

and $\widehat{STR} \equiv \widehat{BMC}$, therefore $\triangle RTS
\sim \triangle BMC$. The assumption on angles leads to

\[
\widehat{QPS} + \widehat{RSP} = \widehat{QPT} + \widehat{TPS} +
\widehat{TSP} +\widehat{TSR} = \widehat{PTS} + \widehat{TPS} +
\widehat{TSP} = 180^{\circ}
\]

and

\[
\widehat{RQP} + \widehat{SPQ} = \widehat{RQT} + \widehat{TQP} +
\widehat{TPQ} +\widehat{TRS} = \widehat{QTP} + \widehat{TQP} +
\widehat{TPQ} = 180^{\circ},
\]

so $PQRS$ is a parallelogram. Hence $PQ = RS$ and $QR = PS$, that
is

\[
AB = \frac{BC \cdot TS}{MC} = \frac{BC \cdot MD}{MC}
\hspace{0,2cm} and \hspace{0,2cm} \frac{AD \cdot QT}{AM} =
\frac{CD \cdot TS}{MD} = CD.
\]

The conclusion is now obvious.

[quote="orl"]The point $ M$ is inside the convex quadrilateral $ ABCD$, such that
\[ MA \equal{} MC, \hspace{0,2cm} \widehat{AMB} \equal{} \widehat{MAD} \plus{} \widehat{MCD} \quad \textnormal{and} \quad \widehat{CMD} \equal{} \widehat{MCB} \plus{} \widehat{MAB}.
\]
Prove that $ AB \cdot CM \equal{} BC \cdot MD$ and $ BM \cdot AD \equal{} MA \cdot CD.$[/quote]
[hide="Solution"]
Invert around $ M$ with radius $ MA$ and let point $ X$ map to $ X'$.  Notice that \[ \angle AD'C\equal{}\angle MD'A\plus{}\angle MD'C\equal{}\angle MAD\plus{}\angle MCD\equal{}\angle AMB\] so we  can consider point $ T$ on $ MB$ so that $ \triangle AD'C\sim \triangle AMT$.  This means that a spiral similarity maps $ \triangle AMD'\sim \triangle ATC$.  Without loss of generality, let $ \angle TAM\ge \angle BAM$.  These similar triangles give that $ \frac{AT}{AM}\equal{}\frac{AC}{MD'}$.  This, along with the ratio properties of inversion, yields that \[ \frac{AT'}{T'C}\equal{}\frac{AT}{TC}\equal{}\frac{AM}{MD'}\equal{}\frac{MD}{CM}\] Furthermore, we have that the spiral similarity gives that $ \frac{AM}{AT}\equal{}\frac{AD'}{AC}$, so $ AC\cdot AM\equal{}AT\cdot AD'$.  Furthermore, notice that the spiral similarities give that $ \frac{CD'}{MT}\equal{}\frac{AD'}{AM}$ and $ \frac{TC}{MD'}\equal{}\frac{AT}{AM}$.  Thus, \[ AD'\cdot AT\equal{}(\frac{CD'\cdot AM}{MT})(\frac{AM\cdot TC}{MD'})\equal{}(\frac{CD'\cdot MC}{MD'})(\frac{TC\cdot CM}{MT})\] The inversion gives that $ \frac{CD}{CD'}\equal{}\frac{MC}{MD'}$ and $ \frac{T'C}{TC}\equal{}\frac{CM}{MT}$.  Thus, \[ CD\cdot T'C\equal{}AT\cdot AD'\equal{}AC\cdot CM\] Then, $ \frac{CD}{AC}\equal{}\frac{CM}{CT'}\equal{}\frac{MD}{AT'}$ so $ \triangle CMD\sim \triangle CT'A$.  This means that \[ \angle BAM\plus{}\angle BCM\equal{}\angle CMD\equal{}\angle AT'C\equal{}\angle TAM\plus{}\angle TAC\] the last one because of the inversion.  We conclude that \[ \angle TAM\minus{}\angle TAB\plus{}\angle TCM\minus{}\angle TCB\equal{}\angle BAM\plus{}\angle BCM\equal{}\angle TAC\plus{}\angle TAM\] This means that $ \angle TAB\plus{}\angle TCB\equal{}0$.  Yet, since $ \angle TAM\minus{}\angle BAM\ge 0$,  we see that $ \angle TCM\minus{}\angle BCM\ge 0$ since $ T$ lies on $ BM$.  Then, $ \angle TAB\plus{}\angle TCB\ge 0$, so the only way for equality to occur is if $ T\equal{}B$.  Above, we had that $ \frac{AT}{TC}\equal{}\frac{MD}{MC}$, so $ AB\cdot MC\equal{}BC\cdot MD$.  Similarly (it is not shown above, but there are mutiple ways to get this: use the symmetry of the diagram, use the ratios from the inversion and spiral similarities, or we can repeat the phantom point process for $ D$ instead of $ B$), $ BM\cdot AD\equal{}MA\cdot CD$.  [/hide]

[hide]
Let $M_1$, $M_2$ be points such that $\triangle BM_1 C \sim \triangle CDA$ (indirectly) and $\triangle AM_2 D \sim \triangle CBA$ (indirectly). Let $M_2 '$ be the point such that $\triangle BM_2 'C \sim \triangle AM_2 D$ (directly), so that $CM_2' B M_1$ is similar to $ABCD$, and therefore

$AM_2 = BM_2 ' \cdot \frac{AD}{BC} = CM_1$

Also, $\angle M_2 AC = \angle DAC - \angle M_2 AD = \angle M_1 CB - \angle ACB = \angle M_1 CA$.

If $M_1$ and $M_2$ coincide we are clearly done, since this point satisfies all the properties required (it's easy to prove $\angle BMC = \angle D$ and $\angle AMD = \angle B$ are equivalent to the given angle conditions, and that at most one point $M$ can exist for a given quad, since $\angle BMC = \angle D$ and $\angle AMD = \angle B$ already show only 2 can exist and we get contradictions if we want both to be equidistant from $A$ and $C$ and within the quad).

If $M_1$ and $M_2$ don't coincide then no $M$ can exist because otherwise such an $M$ would have the following properties: $AMM_2 \cong CMM_1$, $AMM_2 D$, $BMM_1 C$ cyclic with equal radii, and a bunch of other stuff.  At this point there's so much information many obvious contradictions follow, for example that $AC$ must contain the centers of the two circumcircles of $AMM_2 D$ and $BMM_1 C$, but then $M$ can not lie inside the quad.
[/hide]

[hide]
Let $\ell$ be the interior angle bisector of $\angle AMC$.  Reflect $D$ over $\ell$ to $D'$.

Construct point $C'$ such that $\triangle C'D'M\sim \triangle BCM$.  Thus, \[\frac{MC'}{MD'}=\frac{MB}{MC} \Rightarrow \frac{MC'}{MB}=\frac{MD'}{MC}\] Hence, $\triangle C'BM\sim \triangle D'CM$ by SAS.

Note that \begin{align*}\angle AMD'&=\angle MAB+\angle MCB\\&=\angle MAB+\angle MD'C'\end{align*}
so $D'C'||AB$.  Also, $AD'||BC'$.  Thus, $D'C'=AB$ and $AD'=BC'$.

Since $\triangle C'BM\sim \triangle D'CM$,
\[\frac{C'B}{MB}=\frac{D'C}{MC}\]
so \[C'B\cdot CM=D'C\cdot MB\] which means that
\[CD\cdot AM= BM\cdot AD\]
 Similarly, $AB\cdot CM=BC\cdot MD$ using the fact that $\triangle C'D'M\sim \triangle BCM$, as desired.

edit: Fixed some typos and fixed a few equations due to the fact that I misread the problem.
[/hide]

Let $l$ be the perpindicular bisector of $AC$.  We note that the exterior angle of quadrilateral $AMCD$ at $M$ is equal to the sum of the interior angles at $A,C,D$, which leads to $\angle BMC=\angle ADC$.  Similarly $\angle AMD=\angle ABC$.

Now construct $B'$ so that $\triangle AB'C \sim \triangle AMD$.  Then $\angle AB'C=\angle AMD=\angle ABC$.  But the construction will also make $\triangle ADC \sim \triangle AMB'$, so $\angle AMB' =\angle ADC=\angle BMC$.  Since $l$ bisects $\angle AMC$, it bisects $BMB'$ as well.  So $B'$ must be the reflection of $B$ in $l$, which makes $\triangle CBA \simeq \triangle AB'C \sim \triangle AMD$, and the desired result quickly follows.

Hint: $ACB \equiv AMD$.
[hide]Notice $\angle B = \angle AMD, \angle D = \angle BMC$. Take $B' $ such that $\triangle ACB' \equiv \triangle AMD$. Then $ABB'C$ is cyclic and note that by rotohomothety, $\angle CMB = \angle ADC = \angle AMB'$. Since $MA=MC$ this implies $B'$ is the reflection of $B$ across the perp. bisector of $AC$. Now some quick computations solve the problem.[/hide]

I have solved it in a different way, I have used an inversion with center $M$.
Here is my solution :
Consider an inversion with center $M$ and radius $r=MA=MC$.
Let $X'$ be the image of $X$ under that inversion.
We know that $A'=A$ and $C'=C$.
Since $\angle{AMB} = \angle{MAD} + \angle{MCD}$, So $\angle{A'MB'} = \angle{MD'A'} + \angle{MD'C'} \Longrightarrow \angle{A'MB'} = \angle{A'D'C'}$
And since $\angle{CMD} = \angle{MCB} + \angle{MAB}$, So $\angle{C'MD'} = \angle{MB'C'} + \angle{MB'A'} \Longrightarrow \angle{C'MD'} = \angle{A'B'C'}$
We need to prove that:
$$AB \cdot CM=BC \cdot MD \Longleftrightarrow \frac{r^2 A'B'} {MA' \cdot MB'}\cdot \frac{r^2}{MC'}=\frac{r^2 B'C'} {MC' \cdot MB'}\cdot \frac{r^2}{MD'}$$
$$\Longleftrightarrow \frac{A'B'}{B'C'} =\frac{MA'}{MD'} \Longleftrightarrow \frac{A'B'}{B'C'} =\frac{MC'}{MD'}$$
We also need to prove that :
$$BM \cdot AD=MA \cdot CD \Longleftrightarrow \frac{r^2 A'D'} {MA' \cdot MD'}\cdot \frac{r^2}{MB'}=\frac{r^2 D'C'} {MC' \cdot MD'}\cdot \frac{r^2}{MA'}$$
$$\Longleftrightarrow \frac{A'D'}{D'C'} =\frac{MB'}{MC'} \Longleftrightarrow \frac{A'D'}{D'C'} =\frac{MB'}{MA'}$$
So the problem is equivalent to this problem:
The point $M$ is inside the quadrilateral $ABCD$ such that $MA=MC, \angle{CMD} =\angle{ABC} \quad \textnormal{and} \quad \angle{AMB} =\angle{ADC} $.
Prove that $\frac{AD}{DC} =\frac{MB}{MA}$ and $\frac{AB}{BC} =\frac{MC}{MD}$
------------------------
Let $T$ be a point such that $\triangle ABM \sim \triangle ACT$ ($ADTC$ is a cyclic quadrilateral because $\angle{ATC} =\angle{AMB} =\angle{ADC} $)
So $A$ is the center of spiral similarity that takes $BM$ to $CT$, so it will takes $BC$ to $MT$  $\Longrightarrow \triangle ABC \sim \triangle AMT$
$$ \Longrightarrow \angle{AMT}=\angle{ABC} =\angle{DMC} ...(1)$$
Since the perpendicular bisector of $AC$ bisects $\angle{AMC}$ so from $(1)$ it bisects $\angle{DMT}$ and since the reflection of $D$ in the perpendicular bisector of $AC$ lies on $(DACT)$ so it is obvious that $T$ is the reflection of $D$ in the perpendicular bisector of $AC$.
$$\Longrightarrow \triangle CAD \simeq \triangle ACT$$
But since $\triangle ACT \sim \triangle ABM$ so we have $\triangle CAD  \sim \triangle ABM$
$$\Longrightarrow \frac{AD}{DC} =\frac{MB}{MA}$$
And then we also have $\triangle ABC \sim \triangle CMD \Longrightarrow \frac{AB}{BC} =\frac{MC}{MD}$
As desired...
$\blacksquare$

Let $DM$ intersect $BC$ at $P$ and $BM$ intersect $AD$ at $Q$. Let $B_0$ be on the same side as $B$ of $AC$ such that $\triangle AB_0C\sim \triangle AMD$. Note that $\angle MPC=\angle DMC-\angle MCP=\angle MAP$ so $MABP$ is cyclic. Additionally, $\angle DQM=\angle AMB-\angle MAD=\angle MCD$ so $MDQC$ is cyclic.
\\$~$\\
Now, $\angle AB_0C=\angle AMD=180^\circ-\angle AMP=\angle ABP=\angle ABC$, so $ABB_0C$ is cyclic. Thus, $A$ is the Miquel Point of quadrilateral $MDCB_0$. We have
\[\angle AMB_0=\angle ADC=180^\circ-\angle QDC=180^\circ-\angle QMC=\angle BMC\]
By symmetry across the perpendicular bisector of $AC$, $ABB_0C$ is an isosceles trapezoid, so $\triangle ABC\sim \triangle DMA$. We also have $\angle ADC=\angle BMC$ and $\angle DCA=\angle MB_0A=\angle MBC$ so $\triangle MBC\sim \triangle DCA$. In particular,
\begin{eqnarray*}
\frac{DM}{AM}=\frac{AB}{BC}&\implies& AB \cdot CM = BC \cdot MD \\
\frac{BM}{MC}=\frac{CD}{AD}&\implies& BM \cdot AD = MA \cdot CD
\end{eqnarray*}
as desired.


geometry
circumcircle
incenter
geometric transformation
homothety
geometry proposed

Let $U$ be the incenter of a triangle $ABC$ and $O_{1}, O_{2}, O_{3}$ be the circumcenters of the triangles $BCU, CAU, ABU$ , respectively. Prove that the circumcircles of the triangles $ABC$ and $O_{1}O_{2}O_{3}$ have the same center.

Funny :)
$\angle BUC = 90^{o}+\frac{\angle A}2$, so $\angle BO_{1}C=180^{o}-\angle A$, thus points $O_{1},O_{2}$ and $O_{3}$ ARE on the circumcircle of triangle $ABC$.
Bye

I think this problem is very nice.  :)
I will call the incenter $I$:

We know that $I$ is the orthocenter of the excentral triangle of $\Delta ABC$.
Because $O_{1}O_{2},O_{2}O_{3},O_{1}O_{3}$ are the perpendicular bissectors of $CI,AI,BI$ respectively, we conclude that the homothecy $h$ with center $I$ and factor $2$ maps $\Delta O_{1}O_{2}O_{3}$ onto the excentral triangle of $\Delta ABC$.

Let us call $O'$ the circumcentre of $\Delta O_{1}O_{2}O_{3}$, and it now follows that $h(O')=V$, the Bevan point of $\Delta ABC$.
It is well known that $O$ is the midpoint of $[IV]$, so we conclude that $O' \equiv O$

[img]http://img504.imageshack.us/img504/4933/nttuanbevanpointzn2.jpg[/img]

This problem is rather easy than nice, but you made it look hard :)
Just one question, what is the Bevan point of triangle?
edit: I found it on the net, never mind :)

Let $I_A, I_B, I_C$ be the excenters of the triangle $ABC$, we know that $U$ is the orthocenter of $I_AI_BI_C$ and $A, B, C$ are the feet of the altitudes of the triangle, thus the nine points circle of $I_AI_BI_C$ is the circumcircle of $ABC$ and it pass through the midpoints if segments $UI_A$, $UI_B$, and $UI_C$. As $UBCI_A$ is cycle quadrilateral of diameter $UI_A$, its center is the midpoint of $UI_A$, so it is equal to $O_1$, then the circumcircle of $ABC$ pass through $O_1$,  simmilary it pass through $O_2$ and $O_3$. $\square$

Just note that: $O_1$, $O_2$, $O_3$ are midpoint of $\stackrel\frown{BC}$ not containing $A$, $\stackrel\frown{CA}$ not containing $B$, $\stackrel\frown{AB}$ not containing $C$ then: $O_1$, $O_2$, $O_3$ $\in$ ($ABC$)

Does anyone ever do the daily problem on the HSO+College feed, and is it the same for everyone?

By Fact 5, $O_1,O_2,O_3$ lie on $(ABC)$, done.


Let's say I have:
[asy]import graph;
size(300);
Label f;
f.p=fontsize(6);
xaxis(-2,2,Ticks(f, 4, 0.5));
yaxis(-2,4,Ticks(f, 6, 0.5));
real f(real x)
{
return x^2;
}
draw(graph(f,-2,2),red+linewidth(1));
label("$x^2$",(1.5,2.25),SE);[/asy]

How can I put arrows at the end of the parabola?

[asy]import graph;
size(300);
Label f;
f.p=fontsize(6);
xaxis(-2,2,Ticks(f, 4, 0.5));
yaxis(-2,4,Ticks(f, 6, 0.5));
real f(real x)
{
return x^2;
}
D(graph(f,-2,2),red+linewidth(1), Arrows);
label("$x^2$",(1.5,2.25),SE);[/asy]

[asy]import graph;
size(300);
Label f;
f.p=fontsize(6);
xaxis(-4,4,Ticks(f, 4, 0.5));
yaxis(-8,8,Ticks(f, 8, 0.5));
real f(real x)
{
return x^3;
}
D(graph(f,-2,2),blue+linewidth(1), Arrows);
label("$x^3$",(1.5,2.25),SE);[/asy]


?诸?推??一本大陆数学竞赛期刊： 数学竞赛之窗（2005年首次创刊?）  http://www.sxjszc.com/

大陸中學數學刊物滿多的，我校圖書館就有7種
說實話，新刊物?能沒什麽優勢

有點奇怪，我一直?能在那?註冊，它的asp程?似乎有點?題

[quote="siuhochung"]有點奇怪，我一直?能在那?註冊，它的asp程?似乎有點?題[/quote]

说实?，那里的确问题挺多的，我有时?了帖?也总是传?上去。还有，我记得我?传上去两张帖?，资料上?说我以?传了10张了。?能是它刚?了系统的缘故?。

大陸中學數學刊物滿多的，我校圖書館就有7種
說實話，新刊物?能沒什麽優勢

中国的期刊虽多，但专门关于竞赛的?有两?，一个是我说的，?一个是《中等数学》。

?能这两个都是“?“?的? :D

[quote="paganinio"][quote="bugzpodder"]晕，这个网?上的Pythagorean Theorem Applet 跟artofproblemsolving（就在我现在左边的）那个一摸一样[/quote]

這種抄襲或者説是借用，?有我們這些國際化的網??能發?，??它應該標明出處（是?是AoPS?）阿。[/quote]

早就在那里?现了，一看就是盗用的?。

若是在大陆地区读高中，可以看看《中等数学》
请问有人看吗这里？

[quote="Connie"]若是在大陆地区读高中，可以看看《中等数学》
请问有人看吗这里？[/quote]
我没看过...
也许你可以介绍一下?

中等数学
可以在邮局定购到。

我定了２００５年的《中等数学》

[quote="chuyue"]我定了２００５年的《中等数学》[/quote]
应该是订了 :)

我们学校图书馆有十几种杂志吧，可惜学校不鼓励学生搞竞赛，好好的书都白白的浪费掉了，实在可惜呀

今年年初很激动地订了《数学竞赛之窗》，结果让人失望。还是《中等数学》好。

中等数学是我们天津市主编的，蛮好的！