Princeton
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topology
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complex analysis

There have already been a couple of discussions here about the Princeton Math department. You should try a search to read them.

The question however is very vague. What does "measure up" mean? Or rather I should ask, what are the "measures" that are you are thinking about?

[quote="neoscient"]how does the undergrad math program at princeton measure up to the others around the country??[/quote]The interesting question is How do you measure an undergrad or grad math program by itself or related to others in the country?

[quote="manuel"]just look at some of the big names in the faculty:
Conway
Wiles
Seymour
 And every professor at princeton is a top researcher in his or her area.[/quote]For better or worse, this has very little to do with the quality of the [i]undergraduate[/i] math program.

It has long been my opinion that, at the undergraduate level, there is very little difference between most of the top 20 or 30 programs (and Princeton is certainly in this list).

But does that necessarily mean that they're good teachers? And as an undergraduate, how important should research credentials be to you?

btw, I had Prof. Wiles for both a class and a seminar at Princeton. I thought he was a very good teacher. He includes a lot of useful historical information and motivation for the theorems in his lectures.

I never had Conway or Seymour for classes, but from what I heard some people really liked Conway as a teacher and some didn't. As you know if you've ever met him, he has his own unusual style.

I've listened to at least five or so lectures (or talks, rather) by Conway, on two different occasions, and I didn't fall in love with his style.

There are definitely a few mediocre math professors at princeton as far as teaching goes.  The first year math courses are notorious for being badly taught, especially 201 and 203.  Although I hear Elias Stein is good and so is Gunning.

First: Being a good researcher and being a good teacher are two very different things. A professor can be both, one or none. I think that it is a lot more likely to be a Princeton tenure if you are an excelent researcher than if you are an excelent teacher (if you are both, great!), and I agree that the researchers on Princeton are first line, top top.

Nevertheless, I honestly agree with DPatrick, for the undergrad program you need more good teachers than good researchers, while in the grad program, probably it is the other way around!

The four semester analysis sequence, Analysis I-IV, are fairly new at Princeton and are the child of Prof Elias Stein. He is one of the modern giants of analysis.

When I was as Princeton Prof Stein used to bemoan the fact that undergraduates weren't getting enough of a unified vision of analysis. So he created this sequence of courses and wrote the textbooks for them. From what I hear they are now one of the highlights of the Princeton education. So if you're looking to get a solid basis in analysis, Princeton might be good choice.

When I was a sophomore I jumped at the chance to take an analysis class from Prof Stein, but I did not have sufficient preparation for the course and had to drop it. (By the way, I recommend taking Topology before Real Analysis. It makes a lot of the ideas of Analysis more clear).

From what I saw of him though, he was a good teacher. (I hope to god he wrote a better book than the one we were using for the class. Royden's Analysis book should be burned forever!). Unfortunately, he must be getting pretty old by now so I imagine he could retire at any moment.

For those who are really interested, here's a link to the descriptions of those Analysis courses from the Undergraduate Announcement: http://www.princeton.edu/pr/catalog/ua/03/256.htm They're courses MAT 330-333.

[quote="gauss202"]Royden's Analysis book should be burned forever![/quote]

I have a copy of it on my shelf.  I guess I won't be leafing through it any time soon  :)

[quote="blahblahblah"]The set of texts that they wrote for the Analysis 1-4 sequence there look pretty neat (you can find them on amazon, they're by Stein and Shakarchi)[/quote]I used the one of Complex Analysis (the number 2 of the collection) for a course last year. The book is not bad, has a high level, but sometimes is too boring (it spend some times more time than necessary in simple calculations). But it is pretty good. I have listened similar comments about the others, but I haven't read them myself.

[quote="gauss202"]Royden's Analysis book should be burned forever![/quote]Can I ask what is your complain about Royden's book. I consider it a good book, kinda basic but good one (if I have to teach a grad course in Real Analysis, I would probably take Folland's book) but most of the time is very easy tu understand, and in general, pretty complete.

If you want a book on Analysis hard to read and hard to learn from, pick the Dieudonnee, it is one of the most complete books in general analysis that I have seen, but you want to puke when you are reading it (my personal opinion!)

[quote="djimenez"]
Can I ask what is your complain about Royden's book? I consider it a good book, kinda basic but good one (if I have to teach a grad course in Real Analysis, I would probably take Folland's book) but most of the time is very easy tu understand, and in general, pretty complete.

If you want a book on Analysis hard to read and hard to learn from, pick the Dieudonnee, it is one of the most complete books in general analysis that I have seen, but you want to puke when you are reading it (my personal opinion!)[/quote]

First of all, I didn't think it was an appropriate book as an introduction to undergraduate Real Analysis (not that that's a complaint against the book itself, since he states in the Preface that its intended for graduate study). But I found many of the propositions to be not sufficiently motivated before their proof. A lack of examples/counter-examples (which are so crucial to learning real analysis) also hurts the book. And his writing style (the little that there was) was immensely dry to me.

I found that I learned Analysis much better from a combination of Rudin's [u]Real and Complex Analysis[/u] and Wheeden and Zygmund's [u]Measure and Integral[/u]. And I think I finally understood the [i]motivation[/i] for Lebegue Measure and Integration from reading the first chapter of Dym and McKean's [u]Fourier Series and Integrals[/u] (a very good book by the way).

I wish that they'd had Pugh's book when I was an undergraduate, as I've heard that it's quite good for an introduction to Real Analysis.

I haven't looked at the book on complex analysis by Stein and Shakarchi much, but I've gone through a significant portion of the book on Fourier analysis, and I think it is amazing. For one thing, it's sufficiently well-written that I can just sit down and read it! (The problems, of course, are another story.) There aren't very many books at that level that can be understood that easily. (Another one is Pugh's book on real analysis, by the way.) Ideas in Stein and Shakarchi are also well-motivated. I highly recommend it.

[quote="gauss202"]
I never had Conway or Seymour for classes, but from what I heard some people really liked Conway as a teacher and some didn't. As you know if you've ever met him, he has his own unusual style.[/quote]

Hee Hee  :D , Conway definetely has his own style. I was only in his class for a few days, but he was very entertaining. I will always remember: "Linear dependence is where you have $\lambda_1, \lambda_2, \lambda_3, \cdots$ [size=200][b][u]NOT ALL ZERO[/b][/u][/size] (at the top of his voice)

Yeah, but I agree with DPatrick in that although Princeton may have (arguably) the best researchers in Math, that does not necessarily mean that they are the best teachers. However, you cannot say that they are terrible teachers. When it comes to top ranked math universities, Princeton is among them, but there is not much of a difference (for undergrad) between Princeton and other top math schools. For grad schools, then the individual professors become more important, so I would say that Princeton is one of the best in the world (in terms of people)

I tend to disagree.  I think that the very best math schools have a much better planned math curriculum than, say, the 'rest' of the top 25.  mcalderbank, your frame of reference pretty much consists of only Princeton - not every university offers, say, a four course analysis sequence with texts designed specifically for the courses.  Not every university begins one's math education with a sequence of analysis - linear algebra - analysis on manifolds.  You'll find that once you go down the list a bit, these courses are replaced with 'honors calculus' and 'honors multivariable calculus' and whatnot.  And it's just.not.the.same.


Prove that $x^{n}+y^{n}=z^{n}$ doesn't have nontrivial integer solutions when $n>2$.
 :D  :D  :D  :D  :D  :D  :D  :D  :D  :D

Well, why didn't you post it into the 'proposed and own problems' section? It would be more funny...

Pierre.

I think we should move it to Getting Started.
 :lol:  :lol:  :lol:  :lol:  :lol:  :lol:  :lol:

$n$ odd

$x=t$
$y=-t$
$z=0$

does that count as non-trivial?  :lol:

[quote="Myth"]If THIS is non-trivial for you, blahblahblah, then what in such case is "trivial" for you?  :lol:  :lol:  :lol:  :lol:[/quote]

Even negative integers.

[quote="blahblahblah"][quote="Myth"]If THIS is non-trivial for you, blahblahblah, then what in such case is "trivial" for you?  :lol:  :lol:  :lol:  :lol:[/quote]

$x,y,z\in \{0,1\}$  :lol:[/quote]
I thought $x=y=z=0$  :lol:  :lol:  :lol:

??? I thought that one cannot post as a guest with the new forum...

Pierre.

It was my post. I sent it from aops forum.
So we see one more great bug!

Pierre! You should tell about it to Valentin.

[quote="Armo"]Okay guys I am glad that you like my joke.  :)  :)  :)  :)  :)  :)
By the way do you believe that[b] Fermat[/b] found an elementar and easy solution for the problem?[/quote]
well...there may be ;)
[url]http://jamesjoe.51.net/news/news13.html[/url]
Anyway, here a website, just read it if you know chinese. I'm really interested in finding out how simple his proof is. :D

[quote="nntrkien"]The solution Fermat ...  :?: Maybe he did one and it could be false. An advice for you: Go to France and ask him.[/quote]
Nntrkien send Mecrazywong to the other world :?

Yeah, that the math world of Fermat, Mecrazywong can see  the steps of our mathematician, and then he can get the Fermat's solution.
Finally, it is a joke  :D

Sorry i forget to login, I am NNtrkien. :P

Hi, it's a greeting from the "other world" :D
Umm....I think it's possible for the existence of a simple proof, maybe this is very tricky and the solution style is different from any number theory problems in the world :P
Anyway, just a [i]maybe[/i].

This problem is called Fermat's last theorem, and is proved by Andrew Wiles in 1993. More information you can find on [url]http://mathworld.wolfram.com/FermatsLastTheorem.html[/url].

Are you sure his proof is correct? As far as I know many mathematicians don't trust to new proofs of famous hard problems, especially if these proofs take hundreds of pages and use another results, which proofs also take hundreds of pages...

Yeah, this one is much easier than Fermat's Last Theorem. :D:D.

Oh, yes.
I have invented nice solution for Anh Cuong's problem, but solution doesn't fit to this small text-filed...

[quote="Armo"]Actually [b]Myth[/b] as I know when [b]Andrew Wiles [/b] solved it at the first time his solution was incorrect but there isn't still found any mistake in his second solution.By the way I saw the book with solution of Fermat's "problem" few years ago and I can say that it is rather "faaaaat".    :)  :)[/quote]
QED

[quote="Anh Cuong"]That means we still have chance to solve it with another nice solution. Did you have it ,Myth  :) ?
As far as I remember,my teacher gave us a problem that until now, he hasn't told us the solution yet.
Prove that the equation
  [tex] x^p+y^p=z^p [/tex] doesn't have positive integer roots where $ p $ is a prime. :)[/quote]
Anh Cuong, either I remember wrongly or your teacher gave you another "joke": As far as I remember, if this is proved, then Fermat's Last Theorem can be deduced immediately in an easy and elementary way.

EDIT:
Indeed, it is.
If this is proved, then for any positive integer n=pq, we have $(x^q)^p+(y^q)^p=(z^q)^p$.

Hihi, that is the reason why Myth's solution doesn't fit the text-filed.... However, just wait for me.   :D


geometry
cyclic quadrilateral
geometry proposed

assume that ABCD is cyclic quadrilateral,and AC ,BD don't pass through the center of cercumcircle(P is the intersection of AC and BD) .we draw four circles with diameters AB ,BC,CD ,AD,assume that the area of the four crescent are F1,F2,F3,F4,and also assume the midpoints of AB,BC,CD,DA are M1,M2,M3,M4. suppose H1,H2,h3,h4 are the pedal from P to AB,BC,CD,DA prove Mi(1<=i<=4) and Hi are on a circle if F1+F2+F3+F4=[ABCD]

any idea? it's nice problem with nice points in its heart ;)

Is P an arbitrary point? And what do you mean by "crescent"?

when you draw four circles with diameter AB,BC,...,four crescent are formed like a part of moon  :)

Actually, I kew that, but I was wondering if you were considering the crescents on the outside or the ones on the inside of the circle. This was what I meant to ask, I just didn't make it clear :)

I also asked about P, but now I see what it was: $P=AC\cap BD$

(it's great to be able to use Tex for every single little and meaningles thing :D:D)

I knew that I am disabled in implying my concept :D
about your question I mean the outside one


The average age of a group of teachers and students is 20. The average age of the teachers is 35, and the average age of the students is 15. What is the ratio of teachers to students? Express your answer as a common fraction.

you,simply take the distance away of the average age of each sub-group to the total average and make a fraction out of it. The closer it is to the overall average,the more influential it is.

So, in that case teachers' age is 15 away from 20 while students' age is 5 away. Switch the larger number for the dominant group (more influential) and you get a fraction of $ {5/15}$ which reduces to $ \boxed{1/3}$


y varies directly as x. If Y=6 when x=3/4, find x when y=-24

[hide]A direct relationship means that

$\frac{\frac{3}{4}}{6}=\frac{x}{-24}$

$-18 = 6x$

$x = -3$[/hide]


integration
real analysis
real analysis unsolved

Prove that for $0<|a|<1$ and $\Sigma=\{x\in\mathbb R^3: |x|=1\}$. Prove that : \[ \int_\Sigma \frac{1-|a|^2}{|x-a|^3}dS=4\pi  \]
[hide=" Hint"] Use Divergence Theorem[/hide]
Could you generalize the equation to $\mathbb R^n$?


function
algebra
functional equation
algebra proposed

Hallo dear MLinkers !

  Well , I'm trying to creat a [u]simple functional equation [/u]with two variables x , y , which
    a) after some special values for x or y leads to a relation , for example of the form $f^2 (x) = x^2$ and then
     b) we can find the function(s) f  , using again the initial equation.

[b]Who can make such an example - or a similar one - to send me ?[/b]

[i]I'm trying to teach the fact that $f(x) g(x) = 0$ does not y[color=red]iel[/color]d  $f(x) = 0$  for every x , or $g(x) = 0$  for every x and I need a good problem for this - not olympiad level !.[/i]
   Thank you

[u]  Babis[/u]

try $x^{\sqrt{2}}$. What means yeald (sotty for my unknowledge)?; if it means "implicates" then try modular equation mod $ab$, with $a>1$,$b>1$

The words are "yield" and "implies", not "yeald" and "implicates", and yes - in this case the original poster means "implies".

Finally I able to make such an example ! See

              http://www.mathlinks.ro/Forum/viewtopic.php?p=409004#409004

   [u]Babis [/u]

 [i] I'm still waiting with pleasure more and better examples.[/i]


[hide]
The hands start 90 degrees apart. Each minute, the minute hand moves 6 degrees away from the hour hand, and the hour hand moves 1/2 degree closer to the minute hand. Since the hands move clockwise, the equation is:
$ 332\equal{}90\plus{}6x\minus{}1/2x$
The reason I didn't use 28 on the left side of the equation is because that would require the hands to move backwards.
$ 332\equal{}90\plus{}6x\minus{}1/2x$
$ 332\equal{}90\plus{}5.5x$
$ 242\equal{}5.5x$
$ 44\equal{}x$

44 minutes
[/hide]
Probably wrong, but eh...

the hour hand moves 1/3 degrees every minute, which is 4 degrees for the minute hand. they go 11/3 degrees apart every minute. at 9:00, they are 90 degrees apart and getting closer. 90-28=62 62*3/11 rounds to $ \boxed{17}$

[hide]

So they start 90º apart. The minute hand moves 6 degrees a minute and the hour hand moves 1/2 degree a minute.

Since the minute hand is ahead of the hour hand, make it 270º

since the minute hand moves 5.5º faster a minute, do 270/5.5 and you get

49.09 with the 09 repeating

to make it exact, since it doesn't tell you to round, change it to 49 and 1/11 minutes[/hide]

I get 44 minutes exaclty which is an exact answer (maybe this is the reason why it doesnt tell you to round?-since its going to be an exact answer)?

[quote="pascal_1623"]The time is 9:00. On a 12-hour clock, how many minutes elapse before the hour hand and minute hand are 28 degrees apart for the first time?[/quote]
At 9:00, the two hands form a 90 degree angle.
Each minute, the minute hand moves 6 degrees, while the hour hand moves $ \frac {1}{2}$ degrees.
This means that the minute hand will move $ 5\frac {1}{2}$ degrees closer to the minute hand each minute.
And since they start out at 270 degrees apart, and you want to get to 28 degrees apart,
$ \cfrac{270 \minus{} 28}{5\frac {1}{2}} \equal{} \cfrac{242}{5\frac {1}{2}} \equal{} \frac {484}{11} \equal{} 44$

That doesn't work out.  The minute hand doesn't move backwards. :wink:

forgot about the 28 minutes apart thing. I did it until they were right on top of each other ;)

[quote="CheeseIsGood"]4 degrees per minute?

That means it takes 90 minutes for the hand to go around the clock, but aren't there only 60 minutes on it? Also if the hour hand moves 1/3 degrees a minute, that means it takes 90 minutes to go one hour on the clock.  :ninja:[/quote]
6 degrees per minute and 1/2 degrees per minute. my bad. then it becomes 11/2. and its the other way around so its (270-28)*2/11 which equals 44

Err... To find the angle between the two hands, calculate $ |30H\minus{}5.5M|$, where $ H$ denotes the number of hours and $ M$ denotes minutes.  Assuming that $ H\equal{}9$, then $ |270\minus{}5.5M|\equal{}28$, OR $ |270\minus{}5.5M|\equal{}332$
Solving for $ M$ in the first equation yields: $ M\equal{}\frac{242*2}{11}\equal{}44$
Second equation yields $ M\equal{}\frac{62*2}{11}$
$ M\equal{}\frac{124}{11}$

I THINK.

[quote="mathcrazed"]Err... To find the angle between the two hands, calculate $ |30H \minus{} 5.5M|$, where $ H$ denotes the number of hours and $ M$ denotes minutes.  Assuming that $ H \equal{} 9$, then $ |270 \minus{} 5.5M| \equal{} 28$, OR $ |270 \minus{} 5.5M| \equal{} 332$
Solving for $ M$ in the first equation yields: $ M \equal{} \frac {242*2}{11} \equal{} 44$
Second equation yields $ M \equal{} \frac {62*2}{11}$
$ M \equal{} \frac {124}{11}$

I THINK.[/quote]

I do not understand your solution. Maybe I'm stupid. But I'm still fairly sure that the answer is 44, if it wasn't supposed to be a whole number, it would tell you the form to put your answer in.

[quote="mathcrazed"]Err... To find the angle between the two hands, calculate $ |30H \minus{} 5.5M|$, where $ H$ denotes the number of hours and $ M$ denotes minutes.  Assuming that $ H \equal{} 9$, then $ |270 \minus{} 5.5M| \equal{} 28$, OR $ |270 \minus{} 5.5M| \equal{} 332$
Solving for $ M$ in the first equation yields: $ M \equal{} \frac {242*2}{11} \equal{} 44$
Second equation yields $ M \equal{} \frac {62*2}{11}$
$ M \equal{} \frac {124}{11}$

I THINK.[/quote]
My question is why would you do 332? All we want is 28 I think..

[quote="darkmatter47"]28 + 332 = 360, the number of degrees in a circle, like a clock. Taking the absolute value lets you go either way (backwards or forwards) on the circle/clock.[/quote]
I know that, but that's for going backward, I think.

[quote="Z-coli"]You guys, this problem is pretty trivial.  You just have to derive and solve the equation:
270+30(m/60)=28+6m==>m=44.[/quote]

First off, it doesn't help to say the problem is trivial.  I doubt that would be the case on a national sprint #24.  Also, what do you mean by "derive and solve the equation"?  It doesn't make that much sense, and deriving a formula isn't pretty trivial.

OK, I'm sorry I said that it was trivial :/ .  This is how you derive, or make, the equation:
We know that the minute hand will be before the hour hand when the hands make a 28-degree angle.  Hence, let the number of minutes at which this can occur be m.  So, we know that in m minutes, the hour hand has moved m/60 of the way along towards the 10.  Since we are talking about degrees, we know that the hour hand will be past 360(9/12)=270 degrees, and the m/60 will contribute to m/60(360/12)=m/60(30)=m/2 degrees.  Hence, after m minutes, the angle degree that the hour hand makes is 270+m/2.  Since after every minute the minute hand moves 360/60=6 degrees, we have that after m minutes, it has made an angle of 6m degrees between the minute and the 12-hour symbol.  Thus, our equation is:
270+m/2=6m+28==>m=44

Sorry people.  I accidentally messed something up =).  In the second equation, it should be $ 5.5M\equal{}602$, which yields a completely different solution.  The first time is at 44 minutes.  Second time would be when $ 270\minus{}5.5M\equal{}\minus{}28$

=D

[quote="mathcrazed"]Sorry people.  I accidentally messed something up =).  In the second equation, it should be $ 5.5M \equal{} 602$, which yields a completely different solution.  The first time is at 44 minutes.  Second time would be when $ 270 \minus{} 5.5M \equal{} \minus{} 28$

=D[/quote]
Oh, now I see. Thank you.

[quote="mathcrazed"]Err... To find the angle between the two hands, calculate $ |30H \minus{} 5.5M|$, where $ H$ denotes the number of hours and $ M$ denotes minutes.  Assuming that $ H \equal{} 9$, then $ |270 \minus{} 5.5M| \equal{} 28$, OR $ |270 \minus{} 5.5M| \equal{} 332$
Solving for $ M$ in the first equation yields: $ M \equal{} \frac {242*2}{11} \equal{} 44$
Second equation yields $ M \equal{} \frac {62*2}{11}$
$ M \equal{} \frac {124}{11}$

I THINK.[/quote]

With the absolute value bars, shouldn't there be two solutions for each equation?

The minute hand moves 6 degrees every minute and since there are 12 hours on a 12-hour clock, the hour hand moves 1/12 of that, or 1/2 a degree. Every minute, the minute hand moves 5 1/2 degrees more than the hour hand. In the beginning, the hour hand and minute hand are 90 degrees apart. Every minute, that number increases by 5 1/2 degrees. When the number reaches 332 (360-28), the hands are 28 degrees apart from each other, and the problem is solved. Now, the minute hand has to move 242 (332-90) degrees more than the hour hand. Dividing 242 into 11/2 gives the answer, [b]44[/b].

P.S. Please tell me if my answer is wrong.


given that x(t) = 2+ 3 sint +4 cost find the time it takes to travel 100metres.

I know how to to use transformation and find its amplitude and angle but what i still dont understand is how to find time take for it travel 100 metres, i need a dummies explanation for that part thanks

could you please define how you understand "travelling 100 metres" for an oscillator?

lol... ill try

moving back and fourth 100m between the end points? im not sure

Maybe he means travel a total distance of 100m?


How do you prove the following identity?
\[ \sum_{j\equal{}0}^{k\minus{}1}\cos \left(2^j x\right)\equal{}\frac{\sin \left(2^k x\right)}{2^k \sin\left(x\right)}\]

$ \sin x\cos 2^J x\Longrightarrow product\ to\ sum\ \Longrightarrow Telescope$


geometry
geometric transformation
IMO
IMO 2006

Hello!

Last year I found the Spanish guide quite useful. Is it possible to post some Slovenian words here? It would be part of our Slovenian experience  :D .

Sebastian

Good idea, I'll make this an announcement :)

I will try to make a guide like the Spanish one from last year.

[b]Basic Supervivence [/b]

Good morning = Dobro jutro

Good afternoon = Dober dan

Good evening = Dober večer

Good night = Lahko noč

Bye = Adijo

Goodbye! (more formal) = Nasvidenje!

Do you speak english? = Govoriš angleško?   more formal: Govorite angleško?

I can't speak Slovene very well. = Ne znam dobro slovensko.

Thank You = Hvala

Please = Prosim

Hi! My name is (your name here)= Zdravo! Ime mi je (your name here).

I am from (your country here)= Sem iz (your country here)

What time is it?= Koliko je ura?

What's your name?= Kako ti je ime?

How did you do in the test?= Kako ti je šlo na testu?

Where is the bathroom?= Kje je kopalnica?

I want some cviček please!= Cviček prosim! (cviček is a special slovene reddish wine)

I want some beer please!= Pivo prosim! or Laško pivo prosim!

Could you help me please, I am lost= Mi lahko prosim pomagate, izgubil sem se.

Where is the dinner?= Kje je večerja?

I am in room (your room number)= Sem v sobi (room number)

I love Slovenia= Slovenija mi je zelo všeč.

I am a Mathlinker, my nickname there is (your nickname)= Sem Mathlinker, moj vzdevek je (nickname).

Where is boarding school Ivana Cankarja? = Kje je Dijaški dom Ivana Cankarja?  (you will stay there)

We need a wake up service at 7 am= Naročili bi bujenje za ob sedmih (=7) zjutraj.

I can't find my sit for the exam= Ne najdem svojega sedeža za tekmovanje.

How much is it?= Koliko to stane?

Please give me a (name of product you wanna buy)= Prosim, dajte mi (name of product)

[b]More advanced stuff[/b]

Hi girl, you are very good looking= Hej, rad se potapljam

Can you give me your phone number?= Naj pokličem mamico?

So we can party tonight?= Lahko danes častim?

Did you score 21 today?= Si videl 21 (=enaindvajset) žiraf?

I Love You = Ustrelil sem konja

1=ena
2=dva
3=tri
4=štiri
5=pet
6=šest
7=sedem
8=osem
9=devet
10=deset
11=enajst
12=dvanajst
13=trinajst
14=štirinajst
15=petnajst
16=šestnajst
17=sedemnajst
18=osemnajst
19=devetnajst
20=dvajset
21=enaindvajset
30=trideset
40=štirideset
50=petdeset
60=šestdeset
70=sedemdeset
80=osemdeset
90=devetdeset
100=sto
101=sto ena
200=dvesto
600=šeststo
1000=tisoč

If you want to know something else, please ask  :)

Thank you!

I believe there is no "tricky pronouncing" involved, so there was only one word I didn't know how to pronounce: the name "Cankarja". How do you read the "c" at the beginning?

C is pronounced "ts."  C-with-a-hat is pronounced like the "ch" in English.  [url=http://www.ff.uni-lj.si/publikacije/sft/alphabet.htm]Slovene alphabet and pronunciation guide.[/url]  I think pronunciation is pretty straight-forward -- just a couple of tricky things to get used to (like words with no vowels -- "vrh" for peak of a mountain, "Trg Francoske Revolucije" for French Revolution Square in Ljubljana (There's a nice breakfast place there, and an obelisk.)).

[quote="majaa"]

Did you score 21 today?= Si videl 21 (=enaindvajset) žiraf?

I Love You = Ustrelil sem konja

[/quote]

Now, do you really expect that we will believe that "žiraf" means anything else than "giraffe"? And "konj" actually means horse, I don't think you can find the word "horse" in the sentence "I love you".

[quote="majaa"]

[b]More advanced stuff[/b]

Hi girl, you are very good looking= Hej, rad se potapljam

Can you give me your phone number?= Naj pokličem mamico?

So we can party tonight?= Lahko danes častim?

Did you score 21 today?= Si videl 21 (=enaindvajset) žiraf?

I Love You = Ustrelil sem konja

If you want to know something else, please ask  :)[/quote]

Haha, mnogu smesno! Ma nie ne jademe treva!

This place is a place where many people are going to form themselves an opinion about Slovenia and Slovenians. It looks as if you want to make it a negative opinion. What is really the point of giving wrong translations?

Somebody better clarify this quickly  :?

I guess [b]portoseb[/b] is right since, AFAIK, in Slovenian, "I love you" is "Ljubim te"..  :?

Yeah it's clear he is right, but does that mean the whole guide is wrong? If so I am curious why majaa did this ...

I think that only the "more advanced stuff" is wrong.

Anyway, see here ;) : http://wikitravel.org/en/Slovenian_phrasebook .

Sorry, it could be a little joke. The first version was right but it was changed later. (Basic Supervivence is absolutely right.)
Now, I owe you (right) translations.

[b]More advanced stuff  v.2 [/b]

Hi girl, you are very good looking= Hej, dobro izgledaš

Can you give me your phone number?= Mi daš telefonsko?

So we can party tonight?= Torej, lahko imamo danes zabavo?

Did you score 21 today?= Si dobil danes 21 točk?

I Love You = Ljubim te

Sorry = Oprosti (or simply "sori")

So, sori...

Okay  ;) I can understand the joke now ...

[quote="Tiks"]Guys I guess that slovenian is very similar to russian and I think it will be  easy ther for them hwo know russian![/quote]

That is what I thought too. Like Russian written in Latin characters.  ;)

For me it was funny reading those "translations", although I don't speak Slovenian, but understanding, that is the other thing. If Slovenians could speak just a little slower...  :)

On my opinion, the winner is "Si videl enaindvajset žiraf?"

I have just remembered: the toughest slovenian word I know is:

ČMRLJ

(a bumblebee)

Try to pronounce it!

[quote="js131090"]could say to me which is the climate in Slovenia?  or in  Ljubljana ???
it rains or it does not rain? ???
the climate is dry or humid? ????
I must wear dress coat or no?

->thanks...... :)  :blush:[/quote]
Hi! Last week was about 30°C, now about 25°C because it was rain. I think it will be hot, but we will see :wink: .

Slovenia´s official money is TOLAR (SIT). 239 SIT=1 euro. Soon our oficcial money will be euro and all prices are in euros too. I think that you can pay in euros, but I am not sure.

[quote="RAKAR"]Hi !!!I don't think that language will be problem.....There are also signes,sentiments.....I think that more dificult will be the first 2 days untill we learn some easy thinks,like names,and some frases....After that I think there will be no problem....Does anybody know which is slovenia's official money?Did they accept euro? or dollar?

RALLIS :D[/quote]

Actually, it is tolar.

[quote="RAKAR"]Hi !!!I don't think that language will be problem.....There are also signes,sentiments.....I think that more dificult will be the first 2 days untill we learn some easy thinks,like names,and some frases....After that I think there will be no problem....Does anybody know which is slovenia's official money?Did they accept euro? or dollar?

RALLIS :D[/quote]

The official Slovenian currency is Slovenian tolar (SIT). Exchange rate: Slovenia Tolar - Euro :  0,0042. Euro is widely accepted as a way of paying.


find the smallest number with 24 divisors.

if $ n\equal{}\prod_{i\equal{}1}^{k}p_i^{\alpha_i}$ has exactly $ 24$ divisors, then $ \prod_{i\equal{}1}^{k}(\alpha_i\plus{}1)\equal{}24$.
Now the problem reduced to considering several possible values of $ k$,afterwards,to pick up the minimum.

[quote="Erken"]if $ n \equal{} \prod_{i \equal{} 1}^{k}p_i^{\alpha_i}$ has exactly $ 24$ divisors, then $ \prod_{i \equal{} 1}^{k}(\alpha_i \plus{} 1) \equal{} 24$.
Now the problem reduced to considering several possible values of $ k$,afterwards,to pick up the minimum.[/quote]
Yes Erken,Having Used your hint I found $ n\equal{}360$


calculus
integration
function
calculus computations

f is defined and continuous in [a,b] (a<b) satisfying $ \int_a^bf(x)dx\equal{}o$.
prove that there exists $ c \in (a,b)$ such that $ f(c)\equal{} 2009 \int_a^cf(x)dx$

Let $ F(x) \equal{} \int_{a}^{x} f(t) \, dt$ and consider $ g(x) \equal{} e^{\minus{}2009x}F(x)$. Since $ g(a) \equal{} g(b) \equal{} 0$, there must exist a point $ c$ in $ (a, b)$ such that $ g'(c) \equal{} 0$. This completes the proof.

how can u just   give  a single  function  and gneralise? :o


$ c = a^2 + b^2$
Express:
a) $ 4c$ b) $ 2c$ c) $ 20c$ d) $ 3c$
as a [b]SUM[/b] of [b]TWO[/b] squares in the terms $ a,b$

[hide="Look after you tried"][hide="are you sure you tried"][hide="suresuresure?"]Im not sure the fourth one has a solutiion or not[/hide][/hide][/hide]

Sorry

[quote="RobRoobiks"]$ c = a^2 + b^2$
Express:
a) $ 4c$ b) $ 2c$ c) $ 20c$ d) $ 3c$
as a >>  product << of squares in the terms $ a,b$

[hide="Look after you tried"][hide="are you sure you tried"][hide="suresuresure?"]Im not sure the fourth one has a solutiion or not[/hide][/hide][/hide][/quote]

Even if it is the sum of squares in terms of $ a, b$, there is no mention of the limit of terms
to be used to be squared and added.

a) $ 4c = 4(a^2 + b^2) = 4a^2 + 4b^2 = (2a)^2 + (2b)^2$

b) $ 2c = 2(a^2 + b^2) = 2a^2 + 2b^2 = a^2 + a^2 + b^2 + b^2$

c) $ 20c = 20(a^2 + b^2) = 20a^2 + 20b^2 = 16a^2 + 4a^2 + 16b^2 + 4b^2 =$
$ (4a)^2 + (2a)^2 + (4b)^2 + (2b)^2$

d) $ 3c = 3(a^2 + b^2) = 3a^2 + 3b^2 =$
$ a^2 + a^2 + a^2 + b^2 + b^2 + b^2$

The original question needs an edit/clarification.

Sorry
For example
b) $ (a\plus{}b)^2\plus{}(a\minus{}b)^2$
$ \equal{}2a^2\plus{}2b^2$
$ \equal{}2c$

d. definitely has NO solution. It is well known that if a number can be written as a sum of squares, all its factors which are 3 mod 4 are to an even power. however, one of c, 3c has 3 to an odd power, so one of those two cannot be written as a SOS.

unless a=b=0


calculus
derivative
geometry
geometric transformation
rotation
reflection
real analysis

Suppose $f : \mathbb{D}\to \mathbb{D}$, where $\mathbb{D}$ is the unit disk. If $f$ is continuously differentiable (i.e. the first order partial derivatives of $f$ exist and are continuous), and $|f(x,y)| = |(x,y)| = \sqrt{x^{2}+y^{2}}$ for all $(x, y) \in \mathbb{D}$, prove that $f$ is a rotation or that $f$ is a reflection about the x-axis followed by a rotation.

A counterexample in complex notation: $f(z)=e^{i|z|}z$


function
logarithms
integration
real analysis
real analysis unsolved

Let $u: E^{n}\rightarrow\mathbb{R}$ be defined by $u(x)=\log\left(\log\left(1+\frac{1}{|x|}\right)\right)$.
Show that $u\in W^{1,p}$ for $p>1$, meaning that $\left(\int_{E^{n}}|Du|^{p}dx\right)^{1/p}<\infty$.
$\left(E^{n}=\{x\in\mathbb{R}^{n}: |x|<1\}\right)$

if $f(r) = \ln (\ln(1+\frac{1}{r}))$ and you work in $\mathbb{R}^{n}$, you have to check that for all $p>1$,
$\int_{0}^{1}|f'(r)|^{p}r^{n-1}dr = \int_{1}^{\infty}\frac{u^{2p+1-n-2}}{ln(1+u)^{2p}}du < \infty$, which seems false if $2p+1>n+2$ ? but it's late, so ..  :lol:

$p$ should be restricted to $1\le p\le n$, where $n\ge 2$ .

This is a classical example to show that in dimensions $n\ge 2$ the Sobolev space $W^{1,n}$ is not contained in $L^{\infty}$.  It is interesting because for $1\le p<n$ we have $W^{1,p}\subset L^{p^{*}}$ with $p^{*}=\frac{np}{n-p}$. This embedding breaks down at the limiting case $p=n$, and is replaced with the Moser-Trudinger inequality.


Matt recieved change totalling $\$$1.85 in nickels, dimes, and quarters, He received at least one coin of each type. What is the least number of coins he could have received?

[hide="What's wrong with this?"]So he had at least 5 pennies. That makes it to 1.80, then having 1 nickel he now has 1.75. Then thats 6 quarters to have a total of 12 coins. But I'm still wrong. The anser was 10. Can someone help?[/hide]

[quote="i_like_pie"][quote="ckck"]So he had at least 5 pennies. That makes it to 1.80, then having 1 nickel he now has 1.75. Then thats 6 quarters to have a total of 12 coins. But I'm still wrong. The anser was 10. Can someone help?[/quote]
Who said anything about pennies?

[quote="ckck"]Matt recieved change totalling $\$$1.85 in nickels, dimes, and quarters, He received at least one coin of each type. What is the least number of coins he could have received?[/quote]
[hide]6 quarters, 3 dimes, 1 nickel.

$\boxed{10}$[/hide][/quote]
Oops. I mis-read the question. I usually go to fast and thought the dimes said pennies.


i hate to be impatient, but where are the prizes? i'd like some closure on something i spent so much time on...

The NSA is handling the end of Year 15, so we at AoPS don't know the answer to that question.  I will look into it, though.

It looks like many people received prizes several weeks ago.  However, I still haven't; is there anyone else who still hasn't gotten anything?

I recommend sending a letter to the address where you sent the solutions last year.

I also did not recieve my bronze prize. (I know it's not actually bronze) Would emailing the old USAMTS address work, or is sending a letter via the postal service the only way?

You can try emailing, but I would try old-fashioned hard copy as well.

By the way... what are the prizes? A plaque?


[quote="b555"]An author writes a good book with a probability of 1/2 .

If it is good it is published with a probability of 2/3.If it is not, it is published with a probability of 1/4.

Find the probability that he will get atleast one book published if he writes two.[/quote]
A random book will be published with probability $ \frac{1}{2} \frac{2}{3} \plus{} \frac{1}{2} \frac{1}{4} \equal{} \frac{11}{24}$. The probability of him writing two books none of which gets published must therefore be $ (1\minus{}\frac{11}{24})^2$. The probability that he we will get at least one published is therefore: $ 1\minus{}(1\minus{}\frac{11}{24})^2 \equal{} \frac{407}{576}$


Prove that all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ satisfying, for all real $x$,
\[ f(x)=f(2x)=f(1-x)\]
are periodic.

when i conclude correctly $f$ always has at least a period of $\frac 12$:

$g(x):=2x;\;g^{-1}(x)=\frac 12 x;\;h:=1-x\\
f(x)=f(g(x)) \Rightarrow f(g^{-1}(x))=f(g(g^{-1}(x)))=f(x)\\
f(x)=f(h(x))=f(g(h(x)))=f(h(g(h(x))))=f(g^{-1}(h(g(h(x)))))=f\left(\frac 12\left(1-(2(1-x))\right)\right)=f(x-\frac12)$

peeta

See here also http://www.artofproblemsolving.com/Forum/viewtopic.php?f=38&t=289832


Let a point $Q$ lie on the circumcircle of a triangle $ABC$. Prove that the feet of the perpendiculars from the point $Q$ to the sidelines $BC$, $CA$, $AB$ of the triangle $ABC$ are collinear.

[The line through these feet is called the [i]pedal line[/i] or the [i]Simson line[/i] of the point $Q$ with respect to triangle $ABC$.]

Please give a complete reference of that problem....Not only 'book'...

Pierre.

Just use some basic angle-chase to prove it...

Hi all.
 There is a great relation between the Simson line with Ptolemy theroem.
  Let ABC is a triangle and M is on the circumcircle of ABC. Let A',B',C' are points lying on BC,CA,AB satisfyng MA',MB',MC' are perpendicular to BC,CA,AB. We have A',B',C' are collinear.
  By computing A'B',B'C',C'A' we have Ptolemy's thereom.
  We also have the Ptolemy's inequality.
  By applying the cos theroem to A'B'C' we have a very beautiful formular
     m^2*n^2= a^2*c*2+b^2*d^2 - 2abcd cos(A+C).
  here m=AC, n = BD, a= AB, b= BC, c= CD, d= DA.
   This is Breidsnaider 's theroem.
  :D

Here it is the Simson line extended to an angle $ \omega$
[color=darkblue][b]Let $ M$ be a point on the circumcircle of $ ABC$ and $ A' \in BC$,$ B' \in CA$,$ C' \in AB$.If $ \angle MC'A \equal{} \angle MB'C \equal{} \angle MA'C$ then $ A',B',C'$ are colinear.[/b][/color].
[b]Proof[/b]
The quadrilaterals $ ABCM$,$ AB'MC'$,$ A'B'MC$ are inscriptible.Therefore $ \angle C'B'M \equal{} \angle C'AM \equal{} \angle MCB$ $ \Rightarrow \angle A'B'M \plus{} \angle MB'C' \equal{} \angle A'B'M \plus{} \angle MCB \equal{} \pi$ from where $ A',B',C'$ are colinear.

$ \odot$[b]Here is a nice property for this line:[/b]
[color=darkblue][b]Prove that the Simson line $ xy$ for a point $ M$ from the circumcircle of a triangle related to a triangle is the Simson Line for point $ M$ related to an infinity of triangles inscribed in the same circle.[/b][/color]

[b]Proof 1[/b]

Take $ E$ on $ xy$.
Trace the line $ AEC \bot ME$ ($ A$ and $ C$ are on the circle).On $ MC$ as a diameter build a circle who is intersecting $ xy$ in $ E$ and $ F$.
$ CF$ intersects the circle in $ B$
$ xy$ is the Simson line for point $ M$ related to $ \Delta ABC$
[b]
Proof 2[/b]
Let $ B$, $ C$ be points on the circle.The perpendicular from $ M$ to $ BC$ is intersecting the circle in the point $ N$ and the parallel line thru $ N$ at $ xy$ is intersecting again the circle in point A.An infinity of possibilityes !

[b]Referece :[/b] G.Titeica "Geometry problems"


probability
combinatorics proposed
combinatorics

Let A,B two tennis players. We know that P(A)=40% to win a set. Which is the probability for A to win the game?
 We say that A wins the game if A wins 3 sets.

You can draw a scheme of all the possible states of the game (there are only 3*3 non-ending states) and calculate the probabilities manually. See the picture. We start in left/top point, score is 0:0 (set score). If A wins a set, we move down, if B wins, we move right. Small-font numbers are probabilities of set winning/losing in some state (that means, moving in the scheme), the large numbers are probabilities that you end up in some state. States when A wins the whole game are on the bottom - when we have moved three times downwards and not more than 2 times rightwards. If you sum up the probabilities, you get 992/3125, which is 31 * (2/5)^5 = 31 * (40%)^5 = 31.744%.


[b]Problem: [/b]Let $ x,y$ and $ z$ be positive real numbers. Such that $ x \plus{} y \plus{} z \equal{} xyz$
Prove that
$ \frac{x}{y} \plus{} \frac{y}{z} \plus{} \frac{z}{x} \ge \frac{{x(1 \plus{} y^2 )}}{{y(1 \plus{} x^2 )}} \plus{} \frac{{y(1 \plus{} z^2 )}}{{z(1 \plus{} y^2 )}} \plus{} \frac{{z(1 \plus{} x^2 )}}{{x(1 \plus{} z^2 )}}$

[quote="daigiaga1994"][b]Problem: [/b]Let $ x,y$ and $ z$ be positive real numbers. Such that $ x \plus{} y \plus{} z \equal{} xyz$
Prove that
$ \frac {x}{y} \plus{} \frac {y}{z} \plus{} \frac {z}{x} \ge \frac {{x(1 \plus{} y^2 )}}{{y(1 \plus{} x^2 )}} \plus{} \frac {{y(1 \plus{} z^2 )}}{{z(1 \plus{} y^2 )}} \plus{} \frac {{z(1 \plus{} x^2 )}}{{x(1 \plus{} z^2 )}}$[/quote]
It's equivalent to $ \sum_{cyc}\left(\frac {x}{y} \minus{} \frac {x \plus{} y}{y \plus{} z}\right)\geq0,$ which after expanding gives:
$ \sum_{cyc}(x^4z^2\plus{}x^3y^3\minus{}x^3y^2z\minus{}x^2y^2z^2)\geq0,$ which is obviously true.

[quote="arqady"][quote="daigiaga1994"][b]Problem: [/b]Let $ x,y$ and $ z$ be positive real numbers. Such that $ x \plus{} y \plus{} z \equal{} xyz$
Prove that
$ \frac {x}{y} \plus{} \frac {y}{z} \plus{} \frac {z}{x} \ge \frac {{x(1 \plus{} y^2 )}}{{y(1 \plus{} x^2 )}} \plus{} \frac {{y(1 \plus{} z^2 )}}{{z(1 \plus{} y^2 )}} \plus{} \frac {{z(1 \plus{} x^2 )}}{{x(1 \plus{} z^2 )}}$[/quote]
It's equivalent to $ \sum_{cyc}\left(\frac {x}{y} \minus{} \frac {x \plus{} y}{y \plus{} z}\right)\geq0,$ which after expanding gives:
$ \sum_{cyc}(x^4z^2 \plus{} x^3y^3 \minus{} x^3y^2z \minus{} x^2y^2z^2)\geq0,$ which is obviously true.[/quote]
Who can prove it by the different way ?

See problem [b](2)[/b] here :
http://www.mathlinks.ro/Forum/viewtopic.php?t=5263


counting
derangement
search
combinatorics unsolved
combinatorics

Suppose you have an alphabet of 26 letters.
A letter in the alphabet is said to be fixed if the encryption of the letter is the letter itself. How many simple substitution ciphers are there that leave:

(i) no letters fixed
(ii) at least one letter fixed
(iii) exactly one letter fixed
(iv) at least two letters fixed

(i) is the derangement problem.  It has appeared here many times.  Search for derangement on the forum.  (ii) and (iii) are easy to solve once you know how to solve (i).  (iv) is easy once you've solved (i) and (iii).


hi,

i have a simple question :

-what does "homogeneous inequality" mean ?  :blush:

thanks in advance.

A homogeneous inequality is an inequality where all the terms are of the same degree

[quote="toetoe"]-what does "homogeneous inequality" mean ?[/quote]
See also [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=54263[/url].


I need some help with these two problems, so I decided to post here after trying to solve them in vain...

David and Richard each roll a pair of fair dice. Compute the probability that the sum of the numbers shown on David's dice is equal to the sum of the numbers shown on Richard's dice.

Express in simplest form:

${ { \frac{\sqrt{x+\sqrt{x^2-1}} + \sqrt{x-\sqrt{x^2-1}}}{\sqrt{x+\sqrt{x^2-1}} - \sqrt{x-\sqrt{x^2-1}}}}}^2$ for all $x$ greater than or equal to 1

[hide]I know the answer is $\frac{x+1}{x-1}$ (which was on the site's answer page), but I am not sure how they got that, because only the answers were shown)[/hide]

[hide]

${({ \frac{\sqrt{x+\sqrt{x^2-1}} + \sqrt{x-\sqrt{x^2-1}}}{\sqrt{x+\sqrt{x^2-1}} - \sqrt{x-\sqrt{x^2-1}}}}})^2 $

ignore the square and compute the inside first, then square our result.

multiply the numerator and denomenator by ${\sqrt{x+\sqrt{x^2-1}}+ \sqrt{x-\sqrt{x^2-1}}}$

you get $\frac{x+1}{\sqrt{x^2 - 1}}$
with a lot of difference of squares.
then you squre that
you get $\frac{(x+1)*(x+1)}{x^2 - 1}$
$ = \frac{(x+1)*(x+1)}{(x-1)*(x+1)}$
which gets to your answer
$ \frac{x+1}{x-1} $
[/hide]

[hide]
sample space: 6^4
there are 1 way to have sum = 2:   1,1 for both players. so 1^2
2 ways to have sum = 3: 1,2 and 2,1   for both players. so 2^2
so on till sum = 7 1,6and6,1and 2,5and 5,2 and 3,4 and 4,3 that gives6^2
when sum = 8 we can have 2,6 and 6,2 .... 4,4  there are 5^2 ways
sum = 9  ... 4^2
that goes back to sum = 12 : 6,6    1^2
so
answer should be
$\frac{1^2+ 2^2 + 3^2+4^2+5^2+6^2+5^2+4^2+3^2+2^2+1^2}{6^4}$
$ = \frac{146}{1296} = \frac{73}{648}$
[/hide]


Euler
geometry
geometric transformation
homothety
geometry proposed

Let triangle $ABC$ is given. Point $A_a$ is center of a excircle that touches segment $BC$. $B_b$ and $C_c$ are defined similarly. Proof that Euler lines of triangles $ABC_c, BCA_a,CAB_b$ intersects in one point.

$AI_a, BI_b, CI_c$ are heights in $I_aI_bI_c$.
Hence we need to prove that Euler lines for triangles formed
by two projections from two vertexes and third vertex are concurent.
This is well known fact proven here:

http://forumgeom.fau.edu/FG2004volume4/FG200401index.html

It is also worth to mention that the longest distance from the concurrency

  point of the E-lines to the verticies of ABC is equal to the sum of the other two.

   T.Y.

   M.T.

Besides this concurrency point of the E-lines is the image by homothety with center the centroid and factor -2 of the touch point between incircle and the nine point circle of ABC.


I'll assume you mean 2 standard die.

[hide="major hinto"]
Let's do complementary counting
[/hide]

[quote="abacadaea"]what is the probability that the product of the top faces of 6 standard die is even when rolled?[/quote]
[hide="With six dice..."]

Let's see if I can do this type of problem or if it's too early in the morning

First, consider the rule that any number times an even number will be even.

e.g., 7 * 4 = 28, 9 * 16 = 144

Therefore, for the product to be even, at least one die must be even.

Let's consider the case where no die is even, thus the product is odd.

There is a 3 in 6, or 1/2 probability that a die will be odd.

$ \frac {1}{2} * \frac {1}{2} * \frac {1}{2} * \frac {1}{2} * \frac {1}{2} * \frac {1}{2} \equal{} \frac {1}{2^6} \equal{} \frac {1}{64}$

This is the chance that the product would be odd.

So, to be even, the probability is $ 1 \minus{} \frac {1}{64} \equal{} \boxed{\frac {63}{64}}$[/hide]

Just fyi, in problem threads it's good to hide your answers.

EDIT: And I believe your answer is incorrect.
With my hint and some common sense [hide="Im pretty sure the answer is "]$ \frac {3}{4}$[/hide]

EDIT: 6 die? 6??? Nobody ever asks about six, so I assumed two.  :(   In that case, it's [hide]$ \frac{63}{64}$[/hide]

Yeah, it's
[hide]$ 1\minus{}\frac{1}{4}\equal{}\frac{3}{4}$ since the only way you don't get an even is if you roll 2 odds.[/hide]

Um, guys, Ritty is right. $ \frac {3}{4}$ is the correct answer when there are [i][b]2[/b][/i] dice. However, the problem says [i][b]6[/b][/i] dice. Which is what Ritty answered.

By the way, you can generalize this problem: What is the probability of getting an even number as the product of the top of $ n$ dice?

[hide="Answer"]$ \frac {2^n \minus{} 1}{2^n}$[/hide]

Anyone want to prove it? (it's not too hard :) .)

Sorry, I read circle's post about 2 die
[hide]It's $ 1 \minus{} \frac {1}{2^{n}} \equal{} \frac {2^{n} \minus{} 1}{2^n}$, by complimentary counting.[/hide]
Thanks for the tip :)


Let an acute triangle ABC incribed in (O,2p) (p is a prime number) . The lines OA , OB , OC intersect BC , CA , AB at $ A_1 , B_1 , C_1 $ . $ OA_1 , OB_1 , OC_1 $ is integers . Find AB , BC , CA

I think this was used in a Romanian TST some time ago. It might also have been posted before, but I'm not sure.

Can you post your solution here , Grobber ?

I think it was pretty long.. If I can't find it, I don't think I'll be thinking about it any time soon. Sorry.. I don't have too much time on my hands right now :(


How to proof, that
$tg \frac{3 \pi}{36} \cdot tg \frac{ 5 \pi}{36} \cdot tg \frac{ 7 \pi}{36}= tg \frac{ \pi}{36}$

I haven't taken the time to do this all out, but one way may be to write everything in terms of the sine and cosine and then use the product to sum formulas and simplify.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=87514

no need to post in two forums


To the nearest second, what is the first time between 4:00 and 5:00 that the minute and hour hands of a clock are perpendicular to each other?  Express your answer in the form hr:min:sec.

These are from past nats right??? And I don't really have a calc... so yah

[quote="Syntax Error"]ah, okay

but yeah, these are past nats.

im trying to appease tare for not doing much[/quote]
well
a.) I'm not (well, maybe a bit :twisted: )
b.) This is the parody of "Wierd clocks..." by Chinaboy
c.) I think the answer can only be expressed as fractions unless you round it to the nearest second.


$ a,b,c,A,B,C$ are nonnegative numbers satisfying $ a\plus{}A\equal{}b\plus{}B\equal{}c\plus{}C\equal{}k$
So prove $ aB\plus{}bC\plus{}cA\le k^2$

$ k^3\equal{}abc\plus{}ABC\plus{}k(aB\plus{}bc\plus{}CA)\geq k(aB\plus{}bC\plus{}cA)$ Done.


[quote="garci2k4"]he states of winsconsin, michigan, illinoise, indiana and ohio made up the north west territory, the land or ordenants of 1785 divided this territory into townships.each township was a square six miles on a side and was divided into 36 sections. one of these sections, the sixteenth, was reserved to support public schools, if a square piece of land contained 2304 square miles, how many sections were reserved for a public schools?[/quote]

Wait, a 6 mile by 6 mile square is 2304 square miles? Wha...?

[quote="Teki-Teki"]no - the piece of land is 2304 sq. mi. it contains multiple townships which are 6. mi by 6. mi each.

1/16 of the total land is for schools (i think) * 2304 sq mi =

144 sq mi?[/quote]

Ahh. Could you please be clearer next time, garci2k4?

errrr... don't you have to divide the 144 by 36? So that way, $\boxed{4}$ sections are used for the school.

[quote="garci2k4"]he states of winsconsin, michigan, illinoise, indiana and ohio made up the north west territory, the land or ordenants of 1785 divided this territory into townships.each township was a square six miles on a side and was divided into 36 sections. one of these sections, the sixteenth, was reserved to support public schools, if a square piece of land contained 2304 square miles, how many sections were reserved for a public schools?[/quote]

I can't possibly read that...so I rewrote it.

"The states of Wisconsin, Michigan, Illinois, Indiana, and Ohio made up the Northwest Territory. The Land Ordinance of 1785 divided this territory into townships; each township was a square, six miles on a side, and was divided into 36 sections. One of these sections, the 16th, was reserved to support public schools. If a square piece of land contained 2304 square miles, how many sections were reserved for public schools?"

And the answer is simply $2304/36=\boxed{64}$, I believe.

For everyone confusing the word "the 16th":
[quote="wise Wikipedia writers"]The ordinance was also significant for establishing a mechanism for funding public education. Section 16 in each township was reserved for the maintenance of public schools. [/quote]

Now I'm confused... if there are 36 6x6's, there are 36^2 or 1296 square miles... where do those 1008 extra square miles come from?

There are actually $64$ of these $6 \times 6$ land... so this land is $8 \times 8$ of $6 \times 6$. (?)

I don't think anyone said there were 36 6$\times$6's...

From experience, I know that townships are numbered from 1 to 36, so "the 16th" is most likely referring to one of the 36 sections.


A problem of my own:

Consider a frictionless sphere: $ x^2\plus{}y^2\plus{}z^2\equal{}R^2$. We place a covering (that is allowed to slide) of uniform surface density $ \sigma$ on the 4th of the sphere where $ x\le 0$, $ z\ge 0$. Connect this object to a string at $ (0,0,R)$. Then place a block of mass $ m$ at $ (R,0,0)$ and connect it to the string. Given that the mass-covering system balances, what is $ m$ in terms of $ \sigma$ and $ R$?

I'm not sure I've got the problem. One end of the string is attached to the 'sheet' at $ (0,0,R)$ and the other end to a block of mass $ m$ at $ (R,0,0)$
Is that correct?

Yes, sorry if that was a bit unclear. The idea is to balance the sheet with the block, and they are connected by a string.

[asy]size(80);
real R=2;defaultpen(fontsize(8));
draw(circle((0,0),R));
draw((0,R)--(0,0)--(-R,0));
draw(arc((0,0),R,0,90),linewidth(2));
fill((0,0)--(0,R)..arc((0,0),R,90,180)--cycle,gray);
dot((R,0));label("$+x$",(R,0),(1,0));label("$+z$",(0,R),(0,1));[/asy]im guessing we can treat the sheet as a single block of mass $ \frac{4\pi R^2\sigma}{4}\equal{}\pi R^2\sigma$, located at the point $ \left(\minus{}\frac{R}{\sqrt{2}},0,\frac{R}{\sqrt{2}}\right)$.. then we have that $ mg\equal{}\frac{1}{\sqrt{2}}\pi R^2\sigma g\implies m\equal{}\frac{1}{\sqrt{2}}\pi R^2\sigma$

Is that right? I don't see how you deduce that. There isn't that type of symmetry because the gravity is down and you have a quarter of a circle.


大家都知?在?等教育中有10年制和11年制和12年制三?学制。我们中国在上世纪七八??年代中是10年制，5年?学和5年中学。到了???年代时?改为12年制，大部分地区是6年?学和6年中学，上海等地是5年?学和7年中学。
?我所知，在世界上其他国家，比如说俄罗斯为10年制，1年级到11年级，?包?4年级；
越?为11年制，5年?学和6年中学；
哥伦比亚是11年制，?样为5年?学和6年中学；
此外?加利亚等一些东欧国家也大多采?10年制或11年制。

大家觉得三?学制哪?更??？

我说一下我的看法，我觉得无论如何12年制肯定是???的。我想，政府和媒体??声声说我们的民?是勤劳智慧的优秀民?，那既然外国人能在10年或11年内学完我们12年?能学完的知识，那我们为什么?好缩短学制呢？难?中国人天生就比外国人笨？有人说中国的教??难，但最近几年的改?几乎删掉了教??中的所有难点。根?我所知?，罗马尼亚的教??中是有群论，数域和微积分（似乎?是?步）的，因为这是他们数学竞赛中的考点，许多国家的教??中包括?等数论和??几何，而这些内容我们全没有。所以我认为“学制?缩短，教育??命??（毛主席语）

完全?以缩短两年。?三浪费了一年，高三?浪费了一年。而且现在的数学书简?过头了，还是1980年代的课本好

在下认为: 缩短教育时间是没有太大必?的,甚至在有些时候,还是有害的.
    因为在中国的竞争?学的大环境下,如果缩短为11年或10年的教育制,?能给学生带?更大的压力.他们等于?在少一年的情况下完?与他们的学长们相?的任务! 我在北京的一所课业负担在全国都应该算是数一数二的[b]少[/b]的高中,也许有点站?说??腰疼,但?我的别的高中的?学将,他们的压力和负担?是大的惊人.所以,如果缩短为11年制的?,在这个僧多粥少的现实社会中.他们的压力会更大!所以学制缩短,切??心!

    而且我的物??师说,课本好象?修改了,?把高数放到高中,高中的数学放到?中,课本难度加大,?是时间问题而已了......

但在八??年代?样是??年制，?学压力也??，那一代的人?也这样过?了么。
而且缩短了学制，学校也就没有?由在?三高三给学生进行中高考?习了，??而让学生?轻负担。

中国的教育现状是僧多粥少,因此想上高一级的学校,接?高一级的教育,就必须?过激烈的竞争!
??,?消?三高三,?能把?习课挤到高二,而?能真正?轻学生负担!

[quote="shobber"]而且我的物??师说,课本好象?修改了,?把高数放到高中,高中的数学放到?中,课本难度加大,?是时间问题而已了......[/quote]
that is what i expected

哪??志能介?一下八??年代的数学书?
?惜我没赶上呀......


factorial
floor function
number theory
prime factorization

How many zeros are at the end of the binary expansion of $ 100!$?

[hide]This is a common problem.

20 of the first 100 consecutive integers are divisible by 5
4 of these are also divisible by 25
None are also divisible by 125, or any higher power of 5.

Thus the exponent on 5 in the prime-factorization of 100! is 20+4=24

Since 50 of the first 100 consecutive integers are divisible by 2, the exponent on 2 is well over 24.

Pairing 5's with 2's, there are 24 factors of 10, and thus [b][color=blue]24 trailing zeros[/color][/b].

In general, 5's are the "limiting reagent", so it suffices to count those. In general, if $ T(n)$ is the number of trailing zeros at the end of $ n!$, then:

$ \boxed{ T(n) \equal{} \sum_{k \equal{} 1}^{\infty} \left \lfloor \frac {n}{5^k} \right \rfloor}$[/hide]

Edit: I read decimal. Shoot me.

We are working in base 2, not base 10. Thus, it is the number of factors of 2, not 5, in 100!, which is ${ \sum_{k = 1}^{\infty} \left \lfloor \frac {100}{2^k} \right \rfloor}$


[quote="Mamat"]Rezultaty nevernoe budut ochen' plohimi, nikto ne uveren v svoih resheniyah. Ya lichno zabolel i ne smog skoncentrirovatsya, tak chto u mena' vsego lish budet mansion ( navernoe ). A da Said vozmojno reshil 6-yu zadachu, razve eto ne kruto?[/quote]

English please

[quote="Mamat"]Rezultaty nevernoe budut ochen' plohimi, nikto ne uveren v svoih resheniyah. Ya lichno zabolel i ne smog skoncentrirovatsya, tak chto u mena' vsego lish budet mansion ( navernoe ). A da Said vozmojno reshil 6-yu zadachu, razve eto ne kruto?[/quote]
translation:
The results probably will be very bad, nobody is sure of their solutions. I got sick and could not concentrate, so I will probably get only honourable mention(probably). By the way, Said maybe solved problem 6, isn't that cool?

[quote="idioteque"][quote="Mamat"]Rezultaty nevernoe budut ochen' plohimi, nikto ne uveren v svoih resheniyah. Ya lichno zabolel i ne smog skoncentrirovatsya, tak chto u mena' vsego lish budet mansion ( navernoe ). A da Said vozmojno reshil 6-yu zadachu, razve eto ne kruto?[/quote]

Which language is this? I'm asking beacuse I speak Croatian and understand some of these (nobody's sure about his/her solutions, you couldn't concentrate..).[/quote]

It's Russian

Hello, Mamat. How are you?
When will primary results? Today?

[quote="rustam"]HI Mamat!
How are you?
Have you seen our guys from UZB?  What they  did?[/quote]
 Hello Rustam The guys from UZB has solved two problems yet as I talked with them today. Maybe at least three of them, I do not or I am confusing thez can take bronze because there were contestant who solved three or in half.

Vsyo u mena' net sil, ya rastroen. U nas net medali v etom godu. U mena' 11 ballov, u Evgeniya 10 i td. V pervy' den' ya silno bolel i sdal 2 stranicy i poluchil 8 ballov. Vo vtoroy mne bylo luchshe i ya napisal na vse tri zadachi i nashol horoshie rezultaty, no pluchil 3 balla. Kak eto tak. Pochemu kogda my reshem zadachu na polovinu, my poluchem 1 ili 2 balla. .... nam nujny horoshie uchitelya i horoshiy leader..........................
           Mamat is off line

[quote="Mamat"]Vsyo u mena' net sil, ya rastroen. U nas net medali v etom godu. U mena' 11 ballov, u Evgeniya 10 i td. V pervy' den' ya silno bolel i sdal 2 stranicy i poluchil 8 ballov. Vo vtoroy mne bylo luchshe i ya napisal na vse tri zadachi i nashol horoshie rezultaty, no pluchil 3 balla. Kak eto tak. Pochemu kogda my reshem zadachu na polovinu, my poluchem 1 ili 2 balla. Otvet prost lider prosto ...... Konechno ego nelza obvenat, no vsyo taki nam nujny horoshie uchitelya i horoshiy leader..........................
           Mamat is off line[/quote]
Translation(again :lol: )
That's it, I can't take this anymore, I am disappointed. We did not get a medal this year. I got 11 points, Evgenyi got 10, etc. On the first day I was badly sick and handed in 2 pages and got 8 points. On the second day I was feeling better I wrote up all three pages and got good resluts, and got three points. How come. The answer is simple: the leader just......Of course I can't blame him, but nevertheless we need good teachers and a good leader.................
Mamat vishel is interneta :rotfl:

[quote="Mamat"]Vsyo u mena' net sil, ya rastroen. U nas net medali v etom godu. U mena' 11 ballov, u Evgeniya 10 i td. V pervy' den' ya silno bolel i sdal 2 stranicy i poluchil 8 ballov. Vo vtoroy mne bylo luchshe i ya napisal na vse tri zadachi i nashol horoshie rezultaty, no pluchil 3 balla. Kak eto tak. Pochemu kogda my reshem zadachu na polovinu, my poluchem 1 ili 2 balla. Otvet prost lider prosto ...... Konechno ego nelza obvenat, no vsyo taki nam nujny horoshie uchitelya i horoshiy leader..........................
           Mamat is off line[/quote]

A kak KGZ4 V 2005 godu nabral 1 ball posle 8ballov v 2004....
Jal' 4to BALAY malo nabral....
Lana zabeite na eto vse eto huinya....

For others:sorry, but i very like russian language in english forums :)

Please!!!! It's an English Forum!!!!!!!!!!

[quote="rem"][quote="w13"]

A kak KGZ4 V 2005 godu nabral 1 ball posle 8ballov v 2004....
Jal' 4to BALAY malo nabral....
Lana zabeite na eto vse eto huinya....

For others:sorry, but i very like russian language in english forums :)[/quote]
Translation:
And how come KGZ4 in 2005 got 1 point after 8 points in 2004...
Too bad Balay got a low score....
Oh well, ignore this, this is all *bleep*.... :rotfl:

Для других: извините, мне очень нравится русский язык в английских форумах :)
 :rotfl:[/quote]

Спасибо rem, ты их по ходу просвещаешь :D

Maybe they don't speak English enough to understand your message? :P

Маубэ тгэе дон'т спек Эн... nah, doesn't look quite russian :(

[quote="Arne"]I repeat: this is an English forum. Do NOT post in Russian here.[/quote]
OK, I will try :)
Bat my dont spaek engliw gud... :)


inequalities
real analysis
inequalities proposed

Because $a^{3}+b^{3}+c^{3}-3abc\geq 0$ and $(a-b)+(b-c)+(c-a)=0$, WLOG we can suppose that $a-b>0,b-c<0,c-a<0$, also $b<c<a$. Denote $x=\frac{c}{b},y=\frac{a}{b}$. We need prove $y^{3}+1+x^{3}-3xy\geq 4(y-1)(1-x)(x-y)$ for $1<x<y$.

Fix $y>1$, we need prove $f(x)=y^{3}+1+x^{3}-3xy-4(y-1)(1-x)(x-y)\geq 0$ for all $x\in (1,y)$. Now, we need use real analysis, i think so :)

Only interesting case to check is $a\geq c\geq b$ (for other is $RHS\leq 0\leq LHS$). Let $a=b+x+y$ and $c=b+x$ where $x,y\geq 0$.

Setting this in inequality yealds $x^{3}+y^{3}+3by^{2}+3x^{2}y+3xy^{2}+3bx^{2}\geq 4x^{2}y+4xy^{2}+3bxy$ or
$x^{3}+y^{3}+3bx^{2}+3by^{2}\geq x^{2}y+xy^{2}+3bxy$. If $x,y>0$ we have strict inequality (just AM-GM or some triviality with zeros). If excaltly  one of $x,y$ is $0$ there is strict inequality, and if $x=y=0$ we have equaity.

[quote="chien than"]Prove that:$a^{3}+b^{3}+c^{3}-3abc \geq 4(a-b)(b-c)(c-a)$
where $a;b;c$ are non-negative[/quote]
See also:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=103040
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=101916


Let a,b positive real numbers such that

 a^2001+b^2001 =a^1999 +b^1999

Prove that a^2+b^2 \leq 2.

It is also true if we suppose

a2001 + b2000 = a2000 + b1999.

This makes the problem more interesting :)

I know a simple neat solution. If you are interested I can post it.

I hope my solution is enough neat.

It is obvious that a^1999(a^2-1)=b^1999(1-b^2) or A=B.

If a=1 then b=1 so a^2+b^2=2
If a>1 we have A>0 and then 0<b<1.
Therefore

          a^2-1<a^1999(a^2-1)=b^1999(1-b^2)<1-b^2

The case for a<1 is similar.
Arne, if your solution is neater , can you please post it ?

[quote="Arne"]It is also true if we suppose
a2001 + b2000 = a2000 + b1999.
[/quote]
[quote="Manlio"] It is obvious that a^1999(a^2 - 1) = b^1999(1 - b^2) [/quote]

That is not true, Manlio, but the idea is nice

Sorry, Arne.
My solution is only for the initial inequality. I have not a neat solution for your inequality. Can you please post your solution for your inequality?

Your solution is very neat and beautiful.
Really a very nice solution.


Suppose we have a 4x4 matrix.  In this matrix, the only entires are +1 and - 1, and they are positioned in such a way that the sum of each column is zero and the sum of each row is zero.
How many different matricies like this are their?

I was curious if anybody would know how to go abouts solving a problem like this?

This is problem #8 from the 1997 AIME.  It is solved here:

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=393645#p393645


How many integers between 1 and 200 are multiples of both 3 and 5 but not of either 4 or 7?

Equivalently, we can find the multiples of $ 15$ less than $ 200$ that are not also divisible by $ 4$ and $ 7$. There are $ 13$ multiples of $ 15$ less than $ 200$. There are $ 3$ multiples of $ 4$ and $ 15$, and $ 1$ multiple of $ 15$ and $ 7$. So, we have $ 13\minus{}3\minus{}1\equal{}\boxed{9}$.


number theory
relatively prime
prime numbers
number theory proposed

Prove that each even number $2k$ is for each number $m$ a difference of two natural numbers coprime with $m$.

we are given $m, k$.
we need to find an integer $x$ such that $x+k$ and $x-k$ are both relatively prime to m, i.e., that their product $x^2-k^2$ is relatively prime to $m$.
Let $\{p_1, p_2, \cdots,p_n\}$ be the set of all _odd_ prime numbers dividing $m$.
For each i, (i=1,2,...,n), pick an integer $s_i$ with $k^2 \not\equiv s_i^2 \text{ mod } p_i$.
By chinese remainder theorem, there is an integer $x$ satifying all of:
(1) $x > k$
(2) $s_i \equiv x \text{ mod } p_i$ for each i.
(3) $x \not\equiv k \text{ mod } 2$
For this x, the number $x^2-k^2$ is relatively prime to m.

(the condition (3) is there for the case  m is even)
(the condition (1) is there to make $x-k$ a _natural_ number)

_________________
One minute prayer for peace in korea


Solve the following equations
2^x=1-y
2^y=1-x

[quote="phuchung"]Because I don't remember, the problem may be
"Solve the folllowing equations
2^x=1+y
2^y=1+x"[/quote]

I guess the question should be the former one .The equations is symmetry so WLOG let x>y so 2^x > 2^y

But 2^x - 2^y = y-x < 0 contradict . So the question should be 2^x = 1-y and 2^y = 1-x . If the question is the later one then i would say that the only answer is x=y=0 .

[quote="shobber"]$x=y=1$[/quote]

Oh ya I forget the case where x=y ......how silly .... :oops:  :oops:

I've been wondering about equations like this; equations where the variables lie in both the base and the exponent.  Besides this problem, are there ways of solving equations like $a^x=bx$ for $x$, or can they only be done numerically?


Given $ a, b, c \geq\ 0$. Prove that:
$ \sqrt {\frac {a}{b \plus{} c}} \plus{} \sqrt {\frac {b}{c \plus{} a}} \plus{} \sqrt {\frac {c}{a \plus{} b}} \plus{} \frac {ab}{(a \plus{} b)^2} \plus{}$$ \frac {bc}{(b \plus{} c)^2} \plus{} \frac {ca}{(c \plus{} a)^2} \geq\ \frac {9}{4}$
Equality occurs when $ a \equal{} b; c \equal{} 0$ or many cyclics  :)
@can_hang: [hide]Anh ko dc lam` bai` nay` dau nhe' :P[/hide]

[quote="nguoivn"]Given $ a, b, c \geq\ 0$. Prove that:
$ \sqrt {\frac {a}{b \plus{} c}} \plus{} \sqrt {\frac {b}{c \plus{} a}} \plus{} \sqrt {\frac {c}{a \plus{} b}} \plus{} \frac {ab}{(a \plus{} b)^2} \plus{}$$ \frac {bc}{(b \plus{} c)^2} \plus{} \frac {ca}{(c \plus{} a)^2} \geq\ \frac {9}{4}$
Equality occurs when $ a \equal{} b; c \equal{} 0$ or many cyclic.
[/quote]
We can prove it by "only Am-Gm"  :)
And besides, it's similar to Iran96's ineq because Iran96 can be rewrite:
$ \frac{a}{b\plus{}c} \plus{} \frac{b}{c\plus{}a} \plus{} \frac{c}{a\plus{}b} \plus{} \frac {ab}{(a \plus{} b)^2} \plus{}\frac {bc}{(b \plus{} c)^2} \plus{} \frac {ca}{(c \plus{} a)^2} \geq\ \frac {9}{4}$
 :)

Here is my proof.
Assume $ a \ge b \ge c$.We have
$ \sqrt{\frac{c}{a\plus{}b}}\plus{} \sqrt{\frac{b}{a\plus{}c}} \ge \sqrt{\frac{b\plus{}c}{a}}$.
 and we will to show that
$ \sqrt{\frac{a}{b\plus{}c}}\plus{} \sqrt{\frac{b\plus{}c}{a}} \plus{} \frac{ab}{(a\plus{}b)^2}\plus{} \frac{bc}{(b\plus{}c)^2}\plus{} \frac{ca}{(c\plus{}a)^2} \ge \frac{9}{4}$
We have $ \frac{ab}{(a\plus{}b)^2}\plus{} \frac{bc}{(b\plus{}c)^2}\plus{} \frac{ca}{(c\plus{}a)^2} \ge \frac{ab}{(a\plus{}b)^2}\plus{} \frac{ca}{(a\plus{}b)^2} \ge \frac{a(b\plus{}c)}{(a\plus{}b\plus{}c)^2}$
Finally,we will to show that
$ \sqrt{\frac{a}{b\plus{}c}}\plus{} \sqrt{\frac{b\plus{}c}{a}}\plus{}\frac{a(b\plus{}c)}{(a\plus{}b\plus{}c)^2}$
or $ \frac{a\plus{}b\plus{}c}{\sqrt{a(b\plus{}c}}\plus{} \frac{a(b\plus{}c}{(a\plus{}b\plus{}c)^2} \ge \frac{9}{4}$
By AM-GM we have
$ \frac{a\plus{}b\plus{}c}{8\sqrt{a(b\plus{}c}}\plus{} \frac{a\plus{}b\plus{}c}{8\sqrt{a(b\plus{}c)}}\plus{} \frac{a(b\plus{}c)}{(a\plus{}b\plus{}c)^2} \ge \frac{3}{4}$
$ \frac{3(a\plus{}b\plus{}c)}{4\sqrt{a(b\plus{}c)}} \ge \frac{3}{2}$
this proof is completed.Equality occur if and only a=b,c=0 and cyclic permulation.
[hide] không hiểu sao em bị ban nick tận đến 25-5 năm 2011 anh ạ[/hide]

Nice proof, secrets. If you want to see a shorter and simpler solution, please wait to see in our book  :)
@secrets: [hide]Anh se~ trao doi? voi' thay` Khanh' ve` chuyen. 1 so thanh` vien bi ban nick. Trong luc ay', em co the? creat 1 cai' nick moi' de dung`tam. nhe  :) '[/hide]

[hide]Em cứ chép bài và giải bài vậy,chắc khoảng làm đuợc 300 bài thì em sẽ thử viết sách mặc dù kém,hy vọng anh sẽ giúp đỡ[/hide]

[quote="nguoivn"]Given $ a, b, c \geq\ 0$. Prove that:
$ \sqrt {\frac {a}{b \plus{} c}} \plus{} \sqrt {\frac {b}{c \plus{} a}} \plus{} \sqrt {\frac {c}{a \plus{} b}} \plus{} \frac {ab}{(a \plus{} b)^2} \plus{}$$ \frac {bc}{(b \plus{} c)^2} \plus{} \frac {ca}{(c \plus{} a)^2} \geq\ \frac {9}{4}$
Equality occurs when $ a \equal{} b; c \equal{} 0$ or many cyclics  :)
[/quote]
And an intersting question:
Find all k for this ineq is trues ($ a, b, c \geq\ 0$):
$ \sqrt[k]{\frac {a}{b \plus{} c}} \plus{} \sqrt[k]{\frac {b}{c \plus{} a}} \plus{} \sqrt[k]{\frac {c}{a \plus{} b}} \plus{} \frac {ab}{(a \plus{} b)^2} \plus{}$ $ \frac {bc}{(b \plus{} c)^2} \plus{} \frac {ca}{(c \plus{} a)^2} \geq\ \frac {9}{4}$
 :)


[quote="paladin8"]I tried faxing my entry form and permission form to 1-619-445-2379 (1 for long distance) and it gave some office of some person... is the fax number correct?[/quote]

That's the right number - perhaps you mistyped it when you dialed?

I had a similar problem. I faxed my entry form, but it says they didn't get it. :?:

[quote="paladin8"]I tried it again today, with the same number (I'm almost 100% sure I dialed the right number the first time, I did it 3 times and got the same office each time) but it worked fine. Strange, but ok, hopefully it went through.[/quote]We got it.


Does there exist a infintely many sequence of perfect squares fulfill following conditions:
i/Arithmetic mean of two consecutive tearms is also  a perfect square.
ii/Two consecutive tearms are co-prime.
iii/This squence is strictly increasing.

I want to ask one more time the above question


geometry
rectangle
inequalities
graph theory
combinatorics proposed
combinatorics

How many coins can be placed on a $10 \times 10$ board (each at the center of its square, at most one per square) so that no four coins form a rectangle with sides parallel to the sides of the board?

I can say it's less than or equal to 35.But I still can not find an example for 35.

But I find the larest example is 33.So I need to find a contradiction to 34 and 35.

I think you are wrong. Taking for example ($i,j\in\{1,..,10\}$) all the points $i+j\in\{6,10,11,13,19\}$, then you have 34 points already...?

Sorry all...

But can anyone spare 5 minutes to tell me why its 34? Thanx to whoever does.

mmm, I didnt knew this problem and i proposed this to our semifinal:

Prove for a $5\times 5$ board the maximun is 12.

If no one gives a proof I can give a proof of the general case.

Nice problem, it needs inequalities for the general case. But even even if we know what's the possible maximum it is neccesary to give an example.

Someone has an example for 35 :?:

[quote="Peter VDD"][quote="M4RI0"]Someone has an example for 35 :?:[/quote]No, the maximum is 34, and I gave an example of 34. We only miss the proof om maximality. (but 35 is not possible)[/quote]

For the general case, let x be the maximum number of coins and we have a N x N board, then:

$x \leq N(\frac{1+ \sqrt{4N-3} }{2})$

From this we have 35 should be the maximum, if not, could someone help me proving it?

[quote="Arne"]The answer is 34 indeed, (I think) I've got a proof. :)[/quote]Well then, could you please post your proof Arne?

mmm, check this:

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=83994[/url]

Construct a bipartite graph, with vertex sets columns and rows, and an edge between a row and a column if the intersection square is selected...

There you get the relation.

For the case $n=10$, we can say that row $i$ contains $d_i$ selected squeres, we must then have $\sum_{i=1}^{10}\binom{d_i}{2}\leq \binom{10}{2}$ So we must maximize $\sum d_i$ with the given restrictions.

It is easy to observe that the best we can do is to have all numbers as close as we can. So the best we can get is having sequences like: $4,4,4,4,3,3,3,3,3,3$. Which shows in fact that the maximun is 34


Let $m$ and $n$ be positive integers. If $x_1$, $x_2$, $\cdots$, $x_m$ are positive integers whose arithmetic mean is less than $n+1$ and if $y_1$, $y_2$, $\cdots$, $y_n$ are positive integers whose arithmetic mean is less than $m+1$, prove that some sum of one or more $x$'s equals some sum of one or more $y$'s.

I think it is a pretty nice problem though I didn't have a solution for it.
My first idea towards it is induction,if we use induction hypothesis we can deduce it to $ \sum x_i \ge mn$ similar for $ \sum y _ i$but I can't continue with it.
My second idea is pigeon principle.
That is if we can find two of
$ x_1,x_1 \plus{} x_2,...x_1 \plus{} x_2 \plus{} ... \plus{} x_m,y_1 \plus{} x_1...y_1 \plus{} x_1,y_1 \plus{} x_1 \plus{} ...x_m,y_1 \plus{} y_2 \plus{} x_1,...y_1 \plus{} y_2 \plus{} x_1 \plus{} ... \plus{} x_m,...y_1 \plus{} ... \plus{} y_n \plus{} x_1 \plus{} ... \plus{} x_m$ are equal then we are done.So I want to use some ineqs prove it.But I failed :blush:
Can you continue with it,or present your ideas or solutions :)

I finally found a solution(I wish it is correct)!
But I want to see some other approaches or ideas before posting my solution :D
I will post my solution this weekend :)

[quote="hxy09"]That is if we can find two of
$ x_1,x_1 \plus{} x_2,...x_1 \plus{} x_2 \plus{} ... \plus{} x_m,y_1 \plus{} x_1...y_1 \plus{} x_1,y_1 \plus{} x_1 \plus{} ...x_m,y_1 \plus{} y_2 \plus{} x_1,...y_1 \plus{} y_2 \plus{} x_1 \plus{} ... \plus{} x_m,...y_1 \plus{} ... \plus{} y_n \plus{} x_1 \plus{} ... \plus{} x_m$ are equal then we are done.[/quote]
The induction proof is nice, but we can also find a solution along these lines (I think EmptySet showed me this). Let $s_i=x_1+\cdots+x_i$ and $t_j=y_1+\cdots+y_j$ ($s_0=t_0=0$). WLOG $s_m>t_n$ (if they're equal we're done). Now consider $s_i+t_j\pmod{s_m}$ for $i=1,2,\ldots,m$ and $j=0,1,\ldots,n$; since $m(n+1)>s_m$, there exist distinct pairs $(i_1,j_1)$ and $(i_2,j_2)$ in this range of indices (WLOG $j_1\ge j_2$) such that $s_{i_2}-s_{i_1} \equiv t_{j_1}-t_{j_2}\pmod{s_m}$. Since $i_1,i_2>0$ and $t_n<s_m$, neither side vanishes (or else $(i_1,i_2)=(j_1,j_2)$). Thus $j_1>j_2$. If $i_2>i_1$, we get $t_{j_1} - t_{j_2} = s_{i_2} - s_{i_1}$; if $i_2<i_1$, we instead have $t_{j_1} - t_{j_2} = (s_m-s_{i_1}) - (s_m-s_{i_2})$. Either way, we're done.

I'll post a solution ,but i'm not 100% sure if it's right.If there's any mistake,please point it out,thank you~ :)
$x_1<x_2<x_3<……<x_m<x_1+x_m<……<x_(m-1)+x_m<x_1+x_(m-1)+x_m<……<\sum x_i$
There are $\sum_{k=1}^n k=\frac{m(m+1)}{2}$ numbers in total and all of which are less than$mn+m$
Similarly,we replace the $x$ with $y$, and we get $\frac{n(n+1)}{2}$ sums in total and all of which are less than $mn+n$
WLOG we assume that $n\geq m$
thus we get $\frac{m^2+n^2+m+n}{2}$sums of $x_i$ or $y_i$ less than $mn+n$
but $\frac{m^2+n^2+m+n}{2}\geq mn+n$（Because it is equilavent to$(n-m)(n-m-1)\geq 0$ which is obvious
so there must be two which have the same value.
Q.E.D


calculus
integration
Putnam
trigonometry
geometry
calculus computations

Putnam 1980
Find the following:
$\int^{\frac{\pi}{2}}_{0}\frac{1}{1+(\tan x)^{\sqrt{2}}}dx$

[hide="Hint"] The $\sqrt{2}$ can be replaced by any constant.  Let $f(x) = \frac{1}{1+\tan^{\alpha}x}$, then show that $f\left(\frac{\pi}{2}-x\right) = 1-f(x)$.[/hide]

Use symmetry.  Prove the equavalence of points on either side of the graph.  Then you get to calculate the area under the graph using basic geometry. :)

Masoud Zargar

$t=\frac{\pi}{2}-x$.

It works out very quickly.


function
real analysis
real analysis unsolved

Does there exsist function f which is inverse of Rieman's function? For  the additional information , Rieman function is continuous in irrational set(R/Q)  and discontinuous in Q.

Rieman's function isn't $ 1-1$ mapping, so there doesn't exist function which is inverse of Rieman's function.


geometry
Euler
circumcircle
geometric transformation
reflection
geometry proposed

$ \triangle ABC$ is an equilateral triangle, $ M$ is the midpoint of $ BC$ and let $ P$ be a  point on the circumference with center $M$ and radius $MA.$ Show the following propositions:

$ \bullet$ $ B,C,$  the centroid and orthocenter of $ \triangle PBC$ are concyclic on a circle $ \mathcal{K}.$

$ \bullet$ Envelope of the Euler line of $ \triangle PBC$ while $ P$ varies on $ (M,MA)$ is a circle.

Let $ O, H, G$ be circumcenter, orthocenter and centroid of $ \triangle PBC.$ Let $ PG$ cut $ (O)$ again at $ Q.$
$ \frac{_3}{^4}BC^2 \equal{} MA^2 \equal{} MP^2 \equal{} \frac{_1}{^4}(2CP^2 \plus{} 2PB^2 \minus{} BC^2)$ $ \Longrightarrow$ $ CP^2 \plus{} PB^2 \equal{} 2BC^2.$
By Steiner theorem, moment of inertia of $ P, B, C$ WRT $ O$ is
$ 3R^2 \equal{} OP^2 \plus{} OB^2 \plus{} OC^2 \equal{} GP^2 \plus{} GB^2 \plus{} GC^2 \plus{} 3 GO^2 \equal{}$ $ \frac{_1}{^3}(BC^2 \plus{} CP^2 \plus{} PB^2) \plus{} 3GO^2 \equal{} BC^2 \plus{} 3GO^2.$
Power of $ G$ to $ (O)$ is then $ \overline{GP} \cdot \overline{GQ} \equal{} R^2 \minus{} GO^2 \equal{} \minus{}\frac{_1}{^3}BC^2 \equal{} \minus{}GP^2$ $ \Longrightarrow$
$ \overline{GQ} \equal{} \minus{}\overline{GP},$ $ G$ is midpoint of $ PQ$ $ \Longrightarrow$ $ OG \perp MG,$ where $ MG \equal{} \frac{_1}{^3}MP \equal{} \frac{_1}{^3}MA \equal{} \text{const}$ $ \Longrightarrow$ $ OG$ is tangent to circle with center $ M$ and radius $ MG.$
$ G$ is midpoint of $ PQ$ $ \Longrightarrow$ $ M$ is midpoint of $ GQ$ $ \Longrightarrow$ $ B, C, H, G$ are all on reflection of $ (O)$ in $ BC.$

Thanks for your solution Vladimir, your proof is quite shorter than mine. I proved [url=http://www.artofproblemsolving.com/viewtopic.php?t=292346]here[/url] an equivalent statement, which is: If $ B,C,H,G$ are concyclic, then locus of $ P$ is the circumference with center $M$ and radius $MA.$ The perpendicularity of the Euler line of $ \triangle PBC$ and $ PG$ also follows easily from the latter result.


Na eksetasete ean iparxoun apeiroi fisikoi N, oste na mporoume na svisoume ta prota k psifia tou N^2 (stin dekadiki anaparastasi) kai na paroume ton arithmo N, Opou k einai kapoios thetikos akeraios.

P.X 6^2=36 ara gia k=1 exoume mia lisi.

Stergios

Γεια σου Στέργιο .
 Η λύση μου στο attachment , κάνε login .

Distixos den mporo na "do" to attachment. Prepei na exei kapoio problima o browser mou.
Mporeis se parakalo na to grapseis sto forum ?

Stergios

Τι γίνεται ρε στέργιο (έλεγξε για ορθότητα!)
Λύση:

Το πήγα ψιλοεπαγωγικά!
625*625=390625 είναι λύση αν διώξουμε τα 3 πρώτα ψηφία.
Έστω Ν μία λύση ( θεωρώ ότι 10|(Ν-5) ). Θα δείξω ότι είναι λύση ο
αριθμός Ν' = Ν+ Ακ * 10^κ  όπου Ακ είναι ο πρώτος
μη μηδενικός όρος του Ν^2 - Ν (ξεκινώντας από τα δεξιά και κ η
θέση στην οποία βρίσκεται ο Ακ θεωρώντας 0 την αρχική δεξιά θέση)

Συνεπώς αρκεί να δείξω ότι 10^(κ+1) | Ν'^2 - Ν'.
Όμως  Ν'^2 - Ν' = Ν^2 + Ακ^2 * 10^(2κ)  + 2*Ν * Ακ * 10^κ - Ν - Ακ*10^κ
και άρα αρκεί  10 | (Ν^2 - Ν)/ (10^κ)   + (2*Ν-1) * Ακ

όμως  (Ν^2 - Ν) / (10^κ)  είναι ισουπόλοιπο με Ακ mod 10 και άρα
αρκεί   10 |  2*Ν* Ακ που ισχύει!

Άρα το Ν' είναι λύση, οπότε υπάρχουν άπειρες λύσεις.

Στέργιο μία ερώτηση! Για ποια κ δεν υπάρχουν λύσεις, όπου κ τα ψηφία που πρέπει
να σβήσουμε?

Στέργιο

 το διόρθωσα το πρόβλημα με το attachment , κάνε login και θα φανεί η λύση μου .

 Οκ?

Poli oraia, bravo kai stous dio sas!

jj-the-king i lisi sou einai poli oraia  :)

Niko exo merikes paratiriseis omos gia sena, "...Άρα αν Ν1 είναι μία λύση με τ ψηφία τα τελευταία (τ + 1) ψηφία του Ν2 θα είναι άλλη λύση..." den simfono akrivos me auto, p.x. 625^2=390625, ara me ta t+1 psifia kratas ton arithmo 0625 ?? fisika o 90625 einai lisi alla auto den problepete apo tin lisi sou (sosta ?). Nomizo oti i kataskeui sou thelei merikes tropopoiiseis... alla i geniki prossegisi einai sosti :)
(Episeis eno dineis ena logo giati afto pou grafeis douleuei, den fenete na dineis kamoia apodeiksi...giauto mallon proekipse kai to parapano problimataki. )

Stergios

Έχεις δίκιο Στέργιο

 παρέλειψα να γράψω ότι , αν το (τ + 1) ψηφίο είναι το 0 , τότε πάω στο επόμενο ψηφίο κτλ
 (όλα δεν γίνεται να είναι 0 ).
 Αυτό που ήθελα να πώ είναι ότι , αν έχουμε μία λύση Ν1 με τ ψηφία και αναζητούμε μία με (τ+ 1) ψηφία τότε το ατ +1 θα
 είναι το τελευταίο του (10ατ+1) + m , άρα το (ατ + 1) δεν επηρεάζει το καινούργιο ψηφίο , άρα αρκεί να βρούμε το
 επόμενο του Ν1 στο τετράγωνο (που είναι το m) .
 Στηρίζομαι πολύ στο ότι το ν-οστό ψηφίο του κ2 καθορίζεται μόνο απο τα τελευταία ν ψηφία του κ .

Έστω $m$ το πλήθος των ψηφίων του $n$. Τότε προφανώς $n<10^m$
Το ζητούμενο είναι να βρεθούν οι αριθμοί $n$ για τους οποίους ισχύει $\boxed{10^m \mid n^2-n} \ \ (1)$
(ή τουλάχιστον να δειχθεί ότι υπάρχουν άπειροι τέτοιοι)

Θα πάρω επαγωγή στο $m$.
Θα προκύψουν δύο ομάδες λύσεων και στην καθεμιά απ'αυτές υπάρχουν τιμές του $m$ για τις οποίες δεν έχουμε λύση.
Δεν είμαι όμως σίγουρος αν συγκεντρωτικά (και στις δύο ομάδες) υπάρχει λύση $\forall m$.
Ομοίως και για το $\kappa$ (το πλήθος των ψηφίων που αφαιρούνται).

Το τελευταίο ψηφίο του $n$ μπορεί να είναι $1,5$ ή $6$.
Η πρώτη περίπτωση απορρίπτεται. [hide="Πράγματι,"]$n=10b+1 \Rightarrow n^2-n = n(n-1) = (10b+1)10b$
Αυτός ο αριθμός πρέπει να διαιρείται από τον $10^m$.
Αν $n>1\ ($ δηλαδή $b>0)$ τότε έχουμε $10^m \mid 10b \Rightarrow 10^m \mid n-1$, άτοπο αφού $10^m>n-1>0$
Η μοναδική λύση λοιπόν είναι για $n=1,$ αφού τότε έχουμε $b=0$ και $10^1 \mid 0,$ η οποία δεν παρουσιάζει ενδιαφέρον (θα έπρεπε να πούμε ότι από το $n^2$ αφαιρέσαμε $0$ ψηφία και βρήκαμε το $n$)[/hide]
Επομένως έχουμε να ψάξουμε το $n$ ανάμεσα στα στοιχεία των κλάσεων $5 \mod 10$ και $6 \mod 10$

Θα σχηματίσουμε μία ακολουθία $\{y_m\}$ αριθμών που να ικανοποιούν την $(1)$, δηλαδή $10^m \mid {y_m}^2-y_m$
Ο αριθμός $y_m$ θα έχει το πολύ $m$ ψηφία, άρα $y_m<10^m$.
Λύσεις θα αποτελούν εκείνα τα $y_m$ που θα έχουν [u]ακριβώς[/u] $m$ ψηφία.

[hide="Ορισμός της ακολουθίας y_n"]
Έστω λοιπόν ένας φυσικός $m>0$ και ένας $n<10^m$ που έχει την ιδιότητα $(1)$, δηλαδή $10^m \mid n^2-n$
Τότε θα είναι $n^2-n = a \cdot 10^m$

Ορίζουμε τον αριθμό $N$ από τη σχέση $N = x \cdot 10^m+n$, όπου $x\in \{0,1,\cdots, 9 \}$
Προφανώς είναι $n\leq N<10^{m+1}$

Αν $x >0$, τότε ο $N$ έχει ακριβώς $m+1$ ψηφία
Αν $x=0$, τότε έχει λιγότερα από $m+1$ (είναι ίσος με το $n$)

Ας εξετάσουμε ποια πρέπει να είναι η τιμή του $x$ ώστε ο $10^{m+1}$ να διαιρεί τον $N^2-N$

$N^2-N = N(N-1) = (x \cdot 10^m+n)[x \cdot 10^m+n-1] =$
$=x^2 \cdot 10^{2m} + x \cdot 10^m(2n-1) + n(n-1)$

Προφανώς $10^{m+1} \mid 10^{2m}$ αφού $m\geq 1$. Έχουμε λοιπόν:

$10^{m+1} \mid N^2-N \Leftrightarrow$
$10^{m+1} \mid x \cdot 10^m(2n-1) + n(n-1) \Leftrightarrow$
$10^{m+1} \mid x \cdot 10^m(2n-1) + a\cdot 10^m \Leftrightarrow$
$10 \mid x(2n-1) + a$

Αρκεί λοιπόν να επιλέξουμε το ψηφίο $x$ έτσι ώστε να ισχύει $x(2n-1) + a \equiv 0 \mod {10}$
Θέτοντας $y_m=n$ ο επόμενος όρος είναι  $y_{m+1}=N$. Άρα έχουμε:
[/hide]

Η ακολουθία $\{y_m \}$ θα έχει τον αναδρομικό τύπο $\boxed{y_{m+1} = x \cdot 10^m+y_m}$,
όπου το $x \in \{0, \cdots, 9\}$ προκύπτει από την ισοδυναμία $x(2y_m-1) + a_m \equiv 0 \mod {10} \ \ (2)$
και το $a_m$ δίνεται από $a_m = \frac{y^2_m-y_m}{10^m} \in \mathbb{N}$

Παρατηρούμε ότι η ακολουθία $\{y_m \}$ διατηρεί σταθερό το τελευταίο ψηφίο αφού για να πάμε από το $y_m$ στο $y_{m+1}$ προσθέτουμε ψηφία στα αριστερά. Επομένως όλα τα $y_m$ είναι ισοδύναμα $modulo\ 10$.

Έχουμε δύο διαφορετικές ακολουθίες. Η μία είναι για $y_1=5$ και η άλλη για $y_1=6$
Στην πρώτη, η ισοδυναμία $(2)$ γράφεται $-x + a_m \equiv 0 \mod {10}$
ενώ στη δεύτερη,  $(2) \Rightarrow x + a_m \equiv 0 \mod {10}$

[hide="Το πρόβλημα που προκύπτει και η απάντηση"]
Αν ήταν πάντοτε $a_m \not \equiv 0 \mod {10}$ δε θα υπήρχε πρόβλημα, μιας και θα ίσχυε $y_{m+1}>y_{m}$ και θα είχαμε τότε άπειρους αριθμούς με τη συγκεκριμένη ιδιότητα.

Αν όμως ισχύει $a_m \equiv 0 \mod {10} \ \ (3)$ για κάποιο $m$,
τότε το $x$ που προκύπτει είναι αναγκαστικά μηδέν, άρα $y_{m+1}=y_m$
[color=brown]Θα πρέπει να εξετάσουμε αν η ακολουθία γίνεται μετά από κάποιο $m$ σταθερή.[/color]

Αν ισχύει η $(3)$ τότε έχουμε $x=0 \Rightarrow y_{m+1} = y_m \Rightarrow a_{m+1} = \frac{y^2_{m+1}-y_{m+1}}{10^{m+1}} = \frac{y^2_m-y_m}{10^{m+1}} = \frac{a_m}{10}$

Επομένως, όποτε το $a_m$ διαιρείται με το $10$, στο επόμενο βήμα θα παίρνουμε $a_{m+1} = \frac{a_m}{10}$

Επαγωγικά, αν $10^s$ είναι η μεγαλύτερη δύναμη του 10 που διαιρεί το $a_m$,
τότε σε $s$ βήματα θα έχουμε απαλλαγεί από αυτά τα 10άρια (το $s$ είναι προφανώς πεπερασμένος αριθμός).

Αν δηλαδή $a_m = 10^s \cdot \rho$ για κάποιο $m$, τότε όλα τα $y_{m+1}, \cdots , y_{m+s}$ θα είναι ίσα με το $y_m$ και έτσι δε θα παίρνουμε καινούριες τιμές.
Αλλά το επόμενο θα είναι $a_{m+s} = \rho \not \equiv 0 \mod {10}$
Επομένως το $x$ θα προκύψει μη-μηδενικό (modulo 10) και θα έχουμε $y_{m+s+1}>y_m$[/hide]
Άρα για κάθε $m$ θα υπάρχει κάποιος δείκτης $w>m$ τέτοιος ώστε $y_w>y_m \Rightarrow$ άπειρες τιμές

Για το $5 \mod 10$ παίρνουμε την ακολουθία των αριθμών $5,25,625,(625),90625,890625 \cdots$
Για το $6 \mod 10$ παίρνουμε την ακολουθία των αριθμών $6,76,376,9376,(9376),109376, \cdots$

(στις παρενθέσεις οι τιμές της ακολουθίας που επαναλαμβάνονται)

απ' αυτές τις τιμές βλέπουμε πχ ότι το $25^2=625$ για να γίνει $25$ πρέπει να αφαιρέσουμε ένα ψηφίο, ενώ από το $625^2=390625$ πρέπει να αφαιρέσουμε 3 ψηφία

Παρόμοια ισχύουν και στην άλλη ομάδα λύσεων $(modulo \ 6)$

Δεν υπάρχει δηλαδή κάποια "προφανής λογική" στις τιμές του κ, και φαντάζομαι ότι εξαρτάται από το δεκαδικό λογάριθμο του $n$.

Λουκάς

Λουκά , πολύ ωραία.

Άργησα να διαβάσω την αλληλογραφία. Νομίζω στο βιβλίο του  << ENGEL : Problem solving strategies >> έχω συναντήσει το πρόβλημα.Τώρα είμαι Σχολείο . Αν θυμηθώ , θα το κοιτάξω στο σπίτι ,μήπως υπάρχει και καμιά άλλη προσέγγιση , αν και δεν το νομίζω. Η λύση σου είναι πλήρης.

 Μπάμπης


Determine if $ 100,895,598,169$ is prime, along with your solution.

Using a prime tester from online:

 112303*898423

I'm too dumb too do it manually though...

How did the online prime tester figure it out then?

uhh testing either 1) all odd #s
or 2) all primes in a database

yeah thats probably how

ok my suggestion:
the number is a palindrome+168

[quote="stevenmeow"]uhh testing either 1) all odd #s
or 2) all primes in a database

yeah thats probably how
[/quote]

I would bet that's not how it was done. There are several (advanced) techniques for determining [url=http://en.wikipedia.org/wiki/Primality_testing]if a number is prime[/url]. They can tell you if a number is prime without explicitly determing any factors of the number. Then there are also advanced techniques for [url=http://en.wikipedia.org/wiki/Integer_factorization]actually factoring numbers[/url] (aside from trial division), but they are comparably slower, by computer standards, than the techniques for just testing for primality.

Mathematica can quickly verify that $ 2$ is a [url=http://en.wikipedia.org/wiki/Fermat_primality_test]Fermat witness[/url], so this number is not prime.  The computation using binary exponentiation is not as terrible as it might seem, but is still fairly prohibitive to do by hand.  What's the source of this problem?

[quote="mathwizarddude"]I got this problem from [url=http://www.mcs.csueastbay.edu/~malek/Mathlinks/Minds/Minds1.html]here[/url].[/quote]

Next time, I recommend when typing a url link, place the link as [code][url=LINK]other word[/url][/code]not as [code][url]LINK[/url][/code]
It's much better (also see the bug in action [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=230809]here[/url]). :)

[quote="mathwizarddude"]I got this problem from [url]http://www.mcs.csueastbay.edu/~malek/Mathlinks/Minds/Minds1.html[/url][/quote]

Fixed your [url=http://www.mcs.csueastbay.edu/~malek/Mathlinks/Minds/Minds1.html]link[/url]


if $\frac{sin^2x-2cos^2x+1}{sin^2x+2cos^2x-1}=4$ then find the value of $tan^2x$.

Dividing the numerator and denominator by $\cos^2 x$ yields $\displaystyle\frac{\tan^2 x - 2 + \sec^2 x}{\tan^2 x + 2 - \sec^2 x} = \displaystyle\frac{2 \tan^2 x - 1}{2 - 1} = 4$, which yields $\tan^2 x = \frac{5}{2}$.

$\LaTeX$ words of wisdom:

Use \sin, \cos, \tan, \csc, \sec, \cot for the trigonometric functions; it makes them look much nicer.

$\sin^2{x}+\cos^2{x} = \sec^2{x}-\tan^2{x} = \csc^2{x}-\cot^2{x} = 1$.


$f(f(n))=n$ proves $f$ is bijective.

Since $f$ is bijective we have $f(n+2) + 2 = f(n)$

$f(f(1))=f(0)$ so $f(1) = 0$.

$f(n+2) + 2 = f(n)$, $f(1)=0$, $f(0) = 1 \Rightarrow f(x) = -x + 1$  is the only solution ?


1 half-cylinder of radius R moves with constant velocity V, all the while referring to the movement of the rod 2. Determine the angular velocity of the rod depending on the angle $ \alpha$ , with $ OB \equal{} 3R$ .
http://i630.photobucket.com/albums/uu22/daogiauvang/DSC00753.jpg

I think that in the tangent point T, and $ V_T \equal{} V$
Projection of the velocity along the rod  $ V_ (p) \equal{} V \sin \alpha$ So,$ \omega \equal{} \frac {V \sin \alpha}{ OT}$ But how to determine the $ OT$

Another problem is the same ... I then write the full texts
http://i630.photobucket.com/albums/uu22/daogiauvang/DSC00754.jpg

Is $ OB$ the vertical?

By the way $ v_T \equal{} V$ is not correct

Is friction present anywhere?


Let $ ABC$ be a given triangle with incircle $ \gamma(I)$ and circumcircle $ \Gamma(O)$. Let $ Y$, $ Z$ be the tangency points of its $ A$-excircle with the sidelines $ AB$ and $ AC$, respectively, and denote by $ T$ the midpoint of $ YZ$. Let $ M$ be the midpoint of the arc $ BC$ of $ \Gamma$ , not containing the vertex $ A$, and furthermore, let the circle centered at $ M$ with radius $ MB$ intersect (again) the line $ AB$ at $ P$. If $ O_{a}$ is the center of the $ A$-mixtilinear incircle, prove that the points $ T$, $ O_{a}$, $ P$, $ B$ are concyclic.

[i]Note[/i]. The $ A$-mixtilinear incircle is the circle simultanously tangent to the sidelines $ AB$, $ AC$ and internally to the circumcircle $ \Gamma$ of $ ABC$.

[quote="limes123"]It's well known that the midpoint of YZ is the center of the incircle, so I=T, and C,T,P,B lie on the circle with center M and radius MB... How to prove that Oa also lies on this circle?[/quote]
A-excircle  is not incircle

See that AP = AC, then prove AT*AO_a = AB*AC (use inversion!). The fact that the tangency point of (ABC) and A-mistilinear circle, I and the midpoint of arc BC containing A are collinear might be helpfull...

Best regards,
sunken rock

Sorry for that mistake... Here (I hope) is a good solution.
[hide="Solution"]It's known that I lies on the circle with center M and radius MB. Let E be the center of A-excircle. We can prove, that E also lies on this circle. Note that I and T lie on the polars of A (incircle and A-excircle). Hence $ AO_a\cdot AT\equal{}AI\cdot AE\equal{}AB\cdot AP$ Q.E.D.[/hide]


algebra
polynomial
Vieta
number theory proposed
number theory

Let n be an integer greater than 1, and let

      [color=red]p(x) = x^n + a[n-1] x^(n-1) + ... a[1]x + 1[/color]          when a[i] are de coeficients

be a polynomial with nonnegative integer coeficients such that a[j] = a[n-j] for 1≤j≤n-1. Prove
that there exist infinitely many pairs of positive integers (a; b) such that a divide p(b) and b divide p(a)

This is very interesting problem, but I can't solve it.. who can post a solution?

I think something similar to classical Vieta jumping method works here as well:
Clearly $ (1,1)$ is a solution.
Main step: if $ (x, y)$ works with $ x<y$, then $ (y, \frac {P(y)}{x})$ also works.
To prove this, clearly $ \frac {P(y)}{x} | P(y)$. So it suffices to prove:
$ y | P ( \frac {P(y)}{x} )$
We have $ P( \frac {P(y)}{x} ) \equal{} \frac { P(y)^{n} \plus{} a_{n\minus{}1} x {P(y)}^{n\minus{}1} \plus{} ... \plus{} a_{j} x^{n\minus{}j} P(y)^{j} \plus{} ... \plus{} a_{1} x^{n\minus{}1} P(y) \plus{} a_{0} x^{n} }{x^n}$.
Since $ (y,x) \equal{} 1$, and since $ P ( \frac {P(y)}{x} )$ is clearly an integer, we can ignore the $ x^{n}$ on the denominator and, so it suffices to prove that the numerator is divisible by $ y$. Looking at it $ (mod$ $ y)$, since $ P(y) \equiv 1 (mod$ $ y)$, it follows that the numerator is equivalent to $ P(x)$ in $ Z_{y}$ (here we use the fact that $ a_{j} \equal{} a_{n\minus{}j}$). But since $ y | P(x)$, it follows that $ y$ is a factor the numerator, as required.


Hi guys I wonder if you could help me with the following inequality
If $ a,b > 0$ show that $ a^b \plus{} b^a > 1$.
I don't know whether you will find it easy or not but I cannot do it.I have also heard there is the following generalisation
For $ a_1,a_2,...,a_n > 0$ show that $ \displaystyle\sum_{i \equal{} 1}^n{{a_i}^{a_{i \plus{} 1}}} > 1$ where we set $ a_{n \plus{} 1} \equal{} a_1$.So if you could also help me with this one I would be grateful.
I am not sure I have posted it in the right subforoum... :(

If any of a,b are greater than 1, the inequality is trivial.
Otherwise, assume that $ 0<a,b<1$.

From [b]Bernoulli[/b] Inequality, we have

$ a^{1\minus{}b} \leq 1\plus{}(a\minus{}1)(1\minus{}b) \equal{} a\plus{}b\minus{}ab < a\plus{}b$.

This implies $ a^b > \frac{a}{a\plus{}b}$.
Similarly, we have $ b^a > \frac{b}{a\plus{}b}$.

Therefore, we have $ a^b \plus{} b^a > 1$.

The proof is similar for n variables:

We only need to prove it for $ 0< a_i < 1$. In this case,

$ a_i^{1 \minus{} a_i\plus{}1} \geq 1\plus{}(a_i \minus{} 1)(1\minus{}a_i\plus{}1 ) \equal{} a_i \plus{} a_i\plus{}1 \minus{} a_i a_i\plus{}1 < a_1 \plus{} .. \plus{} a_n$ which implies
$ a_i^{a_i\plus{}1} \geq \frac{a_i}{a_1 \plus{} a_2 \plus{} .. \plus{} a_n}$.
Summing up, we get the conclusion immediately.[/b]

Nice but I think you have a typo here: $ a_i^{1 \minus{} a_i\plus{}1} \geq 1\plus{}(a_i \minus{} 1)(1\minus{}a_i\plus{}1 ) \equal{} a_i \plus{} a_i\plus{}1 \minus{} a_i a_i\plus{}1 < a_1 \plus{} .. \plus{} a_n$ Shouldn't it be $ a_i^{1 \minus{} a_{i\plus{}1}} \leq 1\plus{}(a_i \minus{} 1)(1\minus{}a_{i\plus{}1} ) \equal{} a_i \plus{} a_{i\plus{}1} \minus{} a_i a_{i\plus{}1} < a_1 \plus{} .. \plus{} a_n$?


calculus
integration
limit
Euler
logarithms
function
real analysis

Suppose $ f$ is positive and decreasing in $ [1,\infty)$ and $ \displaystyle \int_1^{\infty} f(x)dx$ diverges.  Given the set $ 1\equal{}x_1<x_2<x_3\cdots\infty$, under what conditions will the difference:\[ \lim_{N\to\infty}\left\{\sum_{n\equal{}1}^{N} f(x_n)\minus{}\int_1^{x_N}f(x)dx\right\}\]
converge?  The Euler constant is one such example which converges:\[ \lim_{n\to\infty}\left\{1\plus{}\frac{1}{2}\plus{}\frac{1}{3}\plus{}\cdots\plus{}\frac{1}{n}\minus{}\log(n)\right\}\equal{}\gamma\approx 0.57721\]Would a different partitioning of $ 1/x$ also converge:
\[ \lim_{n\to\infty}\left\{1\plus{}\frac{1}{1.9}\plus{}\frac{1}{2.8}\plus{}\frac{1}{3.7}\cdots\plus{}\frac{1}{x_n}\minus{}\log(x_n)\right\}\]
Do all such partitions of $ 1/x$ converge?  Do all such limits converge?  Can you give a counter example such that:\[ \lim_{N\to\infty}\left\{\sum_{n\equal{}1}^{N} f(x_n)\minus{}\int_1^{x_N}f(x)dx\right\}\equal{}\infty\]
I don't know.

The error in a Riemann sum for $ \int_a^b f(x)\,dx$ with equal subintervals is at most the total variation of $ f$ times the length of those subintervals- "stack the boxes". In this case ($ 1 \plus{} \int_1^{N}\frac1x\,dx$), that variation is less than $ 1$ and the subintervals each have length $ 1$. The difference between the left endpoint sum $ 1 \plus{} 1 \plus{} \frac12 \plus{} \frac13 \plus{} \cdots \plus{} \frac1{N \minus{} 1}$ and the right endpoint sum $ 1 \plus{} \frac12 \plus{} \frac13 \plus{} \cdots \plus{} \frac1N$ is $ 1 \minus{} \frac1N$, and those are the maximum and minimum.

Convergence is guaranteed- you're looking at the tail for that, and the total variation in the tail goes to zero.

I think the question was does [i]any[/i] sequence $ 1\equal{}x_1<x_2<x_3<...\infty$ satisfy
\[ \lim_{N\to\infty}\left\{\sum_{n\equal{}1}^{N} \frac{1}{x_n}\minus{}\int_1^{x_N}\frac{1}{x}dx\right\}<\infty\]
If I understood correctly, then the answer is certainly no.

For example, let $ x_k\equal{}k^{\alpha}$ with $ \alpha>1$ and $ f(x)\equal{}\frac{1}{x}$ then  \[ \sum_{n\equal{}1}^N{f(x_n)}\minus{}\int_1^{x_N}{f(x)dx}<\zeta(\alpha)\minus{}\alpha\log n\]
which goes to $ \minus{}\infty$ as $ n\to\infty$.  Though jmerry's assertion is true for any $ f$ of bounded variation if you require in addition that  $ \sup_{k\in\mathbb{N}} \lbrace x_{k\plus{}1}\minus{}x_{k}\rbrace<\infty$.

What matters? It has to still be a Riemann sum for the integral: $ n\le x_n\le n \plus{} 1$. Some relaxation is possible, but doesn't make much sense. If you spaced them 0.9 apart, you would effectively be multiplying the integral by $ \frac {10}{9}$, so you'd have to subtract $ \frac {10}{9}\ln n$ for convergence.

For arbitrary $ x_n$ and general $ f$, you absolutely need $ \lim\frac {x_n}{n}$ to exist; that will scale the integral.

Let $ f$ be defined on $ [0,\infty)$ with bounded variation. If $ |x_n \minus{} cn|$ is bounded, $ \lim_{n\to\infty}\sum_{k \equal{} 1}^n f(x_n) \minus{} \frac1c\int_0^{x_n}f(x)\,dx$ is bounded. If additionally either $ f(x)\to 0$ as $ x\to\infty$ or $ x_n \minus{} cn\to a$ as $ n\to\infty$, the sequence converges.

Thanks guys.  That's very helpful.  This looks like an application of the fact that a bounded monotonic sequence is convergent.  Right?
The sequence \[ \displaystyle S_N\equal{}\left\{\sum_{n\equal{}1}^{N} \frac{1}{x_n}\Delta x\minus{}\int_1^{x_N}\frac{1}{x}dx\right\}<\textbf{V}(f)\Delta x\] is monotonic and bounded by $ \textbf{V}(f)\Delta x$ where $ \textbf{V}(f)$ is the variation of $ f$, and thus the limit:\[ \displaystyle\lim_{N\to\infty}\left\{\sum_{n\equal{}1}^{N} \frac{1}{x_n}\minus{}\frac{1}{\Delta x}\int_1^{x_N}\frac{1}{x}dx\right\}<\textbf{V}(f)\]which if the function is of bounded variation, is less than infinity.  I don't know, I may need a little more work on it though.  My original problem is how do I show directly, the limit for the Euler constant exists?  I think this would do it right?