summer program
Mathcamp
Euler
function
algorithm
number theory
Euclidean algorithm

Rules: No calculators, computers, or turing machines (those were the rules set down by mathcamp).

The last 3 digits of the 133rd power of a number are 007. What are the last 3 digits of the square of the number?

[hide="hint"]
use the Euler phi function (described in other topic).
[hide="second hint"]
You can find the inverse of a number by hand using the Euclidean algorithm for gcds.
[/hide]
[/hide]

[quote="Gyan"][hide]Easy(:)  to  see $207^{33}= 7 \bmod 1000$ so the answer is $849$ [/hide][/quote]

Ok so how do you "easily" see that? I can offer a much more systematic way if need be.

[quote]Ok so how do you "easily" see that? [/quote]
Actually very "easily".. much faster than  to write all the steps! :)
Here was my train of thought ...

(Most of the thoughts ran in parallel, hence you get the answer in a few secs)

[hide]1. Obviously the number is odd and not divisible by 5 (else 133 power would be even or will end in 0 or 5)  (so when I write $a$ it is understood it is co prime to 10)
2. $1000=8*5^{3}$  , so consider the problem in  Mod 8 and Mod 125.
3. $a^{2}\equiv 1 \bmod 8 \implies a^{133}\equiv a \equiv 7 \equiv-1 \bmod 8$
3. Knowing, $7^{2}\equiv 49 \equiv-1 \bmod 25$ and $\phi(25)=20$ we see  right away $7 = (-1)(-1)(7) \equiv 7^{10}7^{2}7^{1}\equiv 7^{13}\equiv 7^{20 \cdot 6+13}$ this implies our number is 7 mod 25(or last 2 digits are 07 - actually 7 mod 200 (25*8)
4. Now even trying  007, 207,407 ..807 etc is fast enough .. but even faster binomial expansion of   $(200k+7)^{33}$  (133 and 33 are same as phi(125) is 100)  gives right away k=1 .

(You know 7^32, last digit is 1 and you need just 2 terms etc.)

... As I said writing it  takes much longer ...:)
Hope this helps ...
[/hide]

Wow, nice. Could you do this for any residue mod 1000 or was 007 a special case?

For example, could you do similar reasoning if the last 3 digits of $a^{133}$ were 137?

[quote="K81o7"]I'd say just solve for the number by cubing what you have there, then finding the inverse of $343$ mod $1000$ and then squaring...[/quote]
That was what my team did at camp, but we squared before inverting, but it really doesnt' matter.

Some comments:

[hide]Didn't realize the exponent was $133$ (and $3\cdot 133 = 400-1$) so the answer was simply $( 1/7)^{3}$ and the answer s square of this. This of course, if one takes    mod 8  is $(-1)^{3}\equiv-1 \bmod 8$
and noticing $7*18 = 126 \equiv 1 \bmod 125$ so we find $18^{3}$ etc
We want to find the square,  all we have to find is $(1/7)^{6}$   it is 1 mod 8 and $18^{6}\equiv 99 \bmod 125$ so answer is $849$

If you don't have pen and pencil too, it is easy to see that $7^{2}\equiv-1 \bmod 25$ so answer is $(-1)^{3}\equiv-1 \bmod 25$ combine with 1 mod 8 you know it is something like x49 where x is even digit - not too hard to find.

[hide]Also if you happen to remember that $7 \cdot 11 \cdot 13 = 1001$ so all you have to do is to find $(11 \cdot 13)^{6}$ and again if you are in a fun mood, there are many short cut you can take (without pen or pencil)  to get the answer for example:

If you keep in mind $2^{7}\equiv 3 \bmod 125$ and
If you notice that $13^{2}= 169 \equiv 44 \equiv 11\cdot 4 \bmod 125$
and $11^{2}= 121 \equiv-4 \bmod 125$

we get $13^{6}\cdot 11^{6}\equiv 4^{3}11^{3}11^{6}\equiv 4^{3}\cdot 11 \cdot 11^{8}\equiv 4^{3}\cdot 11 \cdot 4^{4}\equiv 11 \cdot (2^{7})^{2}\equiv 11 \cdot 3^{2}\equiv 99 \bmod 125$

and you know it is 1 mod 8 so the answer is 849..
(Takes much long to write.. but all the calculations can be done in the head[/hide])[/hide]

You are given [b]x^133 [/b]is congruent to 7 mod 1000 . The phi of 1000 is 400, so x^400 is congruent to 1 mod 1000 by the Euler-Fermat theorem. It is easy from here on.


MIT college Harvard Princeton AMC USA(J)MO USAMO

It is almost time for me to apply to colleges. In preparation for this, I started looking for information about the application processes of various schools. I would like to investigate the validity of what one source told me.

According to this source, MIT starts out the admissions process looking for reasons to reject students -- low SAT scores, poor GPA, etc., unless you are lucky enough to have done some special activity, like make one of the national olympiads. In this case, they move you to a special list, in which they are looking for reasons to admit you.

Is this true? And if so, does anyone know of specifically which activities can get you on this special list?

Thanks,
sran

sort of, and mostly just rsi and the national olympiads, i think.

What is considered a poor GPA for those applying to say, MIT or Harvard?

I really hope they don't do that.  I am sure with the thousands of applications they have to have some methodical way of doing it, but classifying students by how well they do on a test..plainly..sucks.  Some people just aren't very good at taking tests.

SnowStorm, what you say makes good sense, which makes me doubt the validity of my information -- poor test-takers are unfairly judged in the cited system. On the other hand, a previous poster supported the description, so I am unsure what to think.

In terms of GPA for Harvard/MIT, I'm betting the only GPA that is good enough is a perfect or near perfect...right? In which case, I'm going to have a bit of trouble.

Its not quite that simple. Here's an explaination of the MIT admission process http://www.pclaunch.com/~kayton/MIT/InsideAdmissions.htm

I got into both MIT and Caltech (not princeton), but my average was 91.8, ranked 92nd out of 891 seniors.

was I something special? I dont know, but I only made USAMO this year.
It is quite stupid to give up your dreams because of 1,2 points on GPA or couple of questions on SAT.

Be yourself.
Show your passion.
Good luck for all of you who will be class of '10.

Actually, what I hear about the RECENT admissions process at MIT is a little bit different. I hear that these days the big concern is finding students who "fit" into MIT in terms of personality and interests. One way you can get a reality check on what you hear from nonofficial sources is to [url=http://www.google.com/search?q=Marilee+Jones+MIT]Google Marilee Jones[/url], the admissions director at MIT, and see what she says for the record about what kind of entering class MIT is looking for.

That said, I think it's important to compare both official and unofficial sources of information on college admission. Sometimes the admissions officers do, unawares, things that they don't realize they are doing.

By all means have courage to be yourself, and apply to the places you desire, even if you think your chances are low. Your chances are lowest of all if you don't apply.

[quote="tokenadult"] Your chances are lowest of all if you don't apply.[/quote]
Precisely. :yup:

Is a 98 average coupled with things such as RSI and USAMO?  Good for MIT and Harvard application?

if you have a 98 average and went to RSI and the USAMO, theres virtually zero chance you won't be accepted to harvard or MIT.  of course sat scores matter and you have to have a good legal record, but unless there's something really wrong in that application, you're in.

Yes, those are an impressive (if brief) set of credentials (although it wasn't clear to me if they were real or hypothetical).  However, ffd should stop making assurances about things that he personally have no control over.

don't be so fast to yell at me, JBL.  of course i don't mean that you will certainly no matter what get into MIT or harvard with those credentials as it all depends on the admissions officers reading your application.  but, if you have read "What It Really Takes To Get Into the Ivey League and Other Highly Selective Colleges" by Chuck Hughes (Harvard Admissions Officer), which is probably the most recommended book written by an admissions officer, it says that being a part of such programs as RSI, a finalist in the Intel Science Talent Search, a finalist in the Siemen's Westinghouse Competition, or any of the international olympiads in math and science makes the admissions office give extremely careful consideration to the applicant.  a combination of them would "almost assure" acceptance to the college.

i'm not sure where i heard this or how true it is, but there was something about 1600 SATs with math team experience not getting into MIT because the college was looking for more diversity in the student body... it would be pretty boring if everyone likes and does the same stuff wouldn't it??

colleges want one of two kinds of people.  you have to be nicely well rounded, the most common type of applicant, or you can be extremely lopsided e.g. IMO and Intel Finalist.

First, I was clearly not yelling at you, just chiding you gently.  Second, although you say [quote]of course i don't mean that you will certainly no matter what get into MIT or harvard with those credentials[/quote] you had said [quote]theres virtually zero chance you won't be accepted to harvard or MIT. ... unless there's something really wrong in that application, you're in.[/quote] and then went on to say [quote]a combination of them would "almost assure" acceptance to the college.[/quote]

Third, I'm glad you've read a book by a Harvard admissions officer.  That's wonderful.  I am a student at Harvard and have a pretty good feel for the student body Harvard admissions officers have chosen for their school.  I have no pretensions to knowledge of what happened during the admissions process, but I do know what the outcome of that process was by direct observation of my peers.

That said, neither of us is in a position to make any statements about who will or will not be accepted, especially based on [i]one single sentence fragment[/i] of information.  For example, you have no knowledge at all whether this person, real or hypothetical, is an American student -- that will have a lot of impact on an application process.  Indeed, if you give me any invented resume, I can design a person around it who gets rejected at a given school.

The college admissions process is probabalistic.  There are more students who have the credentials to get into Harvard than there are spots at Harvard.  What can be legitimately said is the following: having made USAMO and having an A average in school are strong credentials.  The latter is a weaker credential in the sense that there are a lot of high schools in the country and a 98 is worth more at some than at others.  The USAMO is a weak credential in the sense that it doesn't tell me about anything other than a very specialized ability (math contests), although it does make the case for unusual strength in that particular specialized area.  Together, they are stronger since the weakness in each is off-set by the strengths of the other.  RSI is a particularly strong credential (and, perhaps, a particularly good indicator of acceptance into a prestigious school) because it has its own rigorous applications process that is relatively similar to that of prestigious universities, compared with things like USAMO.  (MIT, after all, would never accept you based on your results on 40 math problems.)  All told, those three credentials are very impressive.  I think it is safe to say that they place their posessor at the top of the applicant pool.  I would say that I would not agonize so much about my chances for acceptance if I had those credentials.  I would [i]not[/i] say that those credentials guarantee, or almost guarantee, a spot at any particular school, although I would feel safe saying that those credentials are very likely to amount to a spot at [i]some[/i] prestigious university.

I would have to second what JBL is saying. We have to start discouraging these threads of "here are my credentials, will I get into xyz school?," because none of us here are the ones who get to make the decision. There's also more to an application than just test scores and lists of achievements (for example, your essay and recommendations are very important).

We all understand that applying to college is a stressful thing for students, and everyone wants as much information as they can get before applying - but ultimately, all you can do as a student is take advantage of the opportunities around you and put your best foot forward when the time comes on the application. Whether or not anyone here thinks you will get into xyz school isn't of any use to you. Just apply, keep your options open, and see what happens!

By the way I am only in ninth grade and have not achieved these things, they are my goals for the end of high school.

[quote="probability1.01"]Perhaps it would behoove us all if a general FAQ for college admissions or something were made as a sticky.[/quote]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=16376]It was done[/url] quite a while ago. Well, all right, I know you said "FAQ," which implies a slightly different format from a bibliography. But, really, the only way to advance your knowledge about the college admissions process, which JBL aptly characterized as "probabilistic," is to read EXTENSIVELY in the more recommended books on the subject, and then THINK about the gaps, implausible statements, and outright contradictions you will find in those books.

But of course it will always be germane to discuss here the college admissions process as it relates to math-liking young people. Mr. Rusczyk opened this forum, back before I was an AoPS member, because he wanted young people not to miss out on chances to succeed in study of mathematics due to mere ignorance of opportunities. One good piece of advice one might pick up on an AoPS thread is how to gain conspicuous achievements in mathematics while young. That dresses up an application more than merely getting straight 'A's in one of the 38,000 high schools in the United States. Another good piece of advice one might pick up is to apply to more than one different college, and to build your college application list from a good "safety" (= sure bet for admission) college on up. Getting into MIT offers wonderful opportunities to a math student, but not getting into MIT isn't the end of the world.

Good luck to all of you in high school classes of 2006, 2007, 2008, and beyond.

MIT has waitlisted someone who got an Axline scholarship at Caltech. So sometimes, even strong applicants get rejects. I like this following quote from a member of the student admissions committee at Caltech:

[quote]In any case, I wouldn't say MIT is becoming a rancid TTT, but it /is/ becoming a second-rate Harvard. There's no doubt that it is denying Axline-quality scientists to admit, in their place, girls and minorities of middling intellectual talent, plus singers, dancers, water polo players, etc. who would get mauled by many Caltech kids in a head-to-head academic match-up. Let me say, first, that there's nothing wrong with that in itself. It is certainly MIT's right as a university, and it makes for a diverse and vibrant community.
But let's not kid ourselves: Harvard is, and probably will be for a long time yet, the top (certainly most prestigious) dog in the "diverse and high-quality" niche. Its immense prestige is a huge draw in every field, especially outside the hard sciences. It will be a long time before MIT can beat them in competing head-to-head for the plum extracurricular/humanities kids. Maybe never.
Since it's destined to be #2 or worse at Harvard's game for a while, does it really make sense for MIT to become widely known for abandoning honestly meritocratic science and engineering-centered admissions (and ceding that crown to Caltech)? Isn't that how MIT made the name it's so famous for in the first place? It seems like a confused strategy to me -- losing the bragging rights of a community based on pure smarts (which it had for a long time) *AND* having to be crushed by Harvard at the diversity/prestige game.
This isn't bitterness, by the way. I've am at peace with the wonderful place that is MIT, and if they hired me as a consultant and tied my pay to their long-term success, I would tell them exactly what I said above.
[/quote][/quote]

TTT means something like "third-tier trashheap." In other words, a really lousy college when compared to the best colleges.

What school is an example of a TTT depends partly on what school you prefer. What one applicant calls a great chance to get a college education might be the same school that another applicant wouldn't dream of applying to.

Ben Golub, who I know personally and whose company I enjoy, is wonderful at spouting opinions of this sort that don't necessarily have any value to them.  MIT is several times larger than Caltech, so it's no surprise that Caltech's best should be better than MIT's average since they're drawing from the same student pool (more or less).  In any event, the idea that Caltech students would academically "maul" MIT students, while it does present an interesting image, is completely absurd, no matter how much Ben dislikes Affirmative Action.

The fact that MIT waitlisted an axline scholar, by the way, is just further evidence in support of my thesis that the college admissions process is highly non-deterministic.

[quote="JBL"]In any event, the idea that Caltech students would academically "maul" MIT students, while it does present an interesting image, is completely absurd, no matter how much Ben dislikes Affirmative Action.
[/quote]

Certainly, this year's Putnam scores would indicate otherwise.

certainly, a given school's putnam scores don't indicate anything about that school or it's student makeup. It indicates that a certain few of the super top at a given school can achieve a given score.

Maybe you didn't look at this year's Putnam scores.  MIT put 7 people in the top 26 and 19! at HM or better.  Granted, that represents a tiny fraction of the students at MIT, but it   certainly implies that MIT is admitting a significant proportion of the *best* students, and not just "girls and minorities of middling intellectual talent".  Moreover, you can infer that MIT has significantly more individuals of exceptional mathematical talent than Caltech does, whether it be minority, female, or whatever.

I disagree. I think beyond those however many students, the ability at MIT drops off significantly, while the case is not nearly as drastic at Cal-Tech. Look, I love MIT and I hate Cal Tech, so don't assume I'm biased. I just don't think one test where only the very elite compete says anything about the average student body.

Hiya. I'm Ben, originator of original quote. My friend, Joel's girlfriend, says I'm famous now since I've been quoted on AOPS. Just to let you know the import of this board :).

I don't want to make this a big deal, but I do want to correct a misreading of my post. I never said Caltech students [i]in general[/i] would maul MIT students. Certainly the top n%, for most n, at both schools are, to avoid reawakening a very old debate, comparable.

What I said was that MIT is

[quote]denying Axline-quality scientists to admit, in their place, girls and minorities of middling intellectual talent, plus singers, dancers, water polo players, etc. who would get mauled by many Caltech kids in a head-to-head academic match-up.[/quote]

That is, MIT takes some "diversity" kids "in place of" the top-level science students that MIT rejects. [i]Those[/i] kids, e.g., people who got in thanks, in part, to a letter from the water polo coach, would get eaten academically by the average Caltech student, and I don't think anyone will challenge me on that. They would [i]certainly[/i] get eaten academically by that Axline scholar that MIT denied.

All I was saying was that Caltech tries much harder not take scientifically weaker students above scientifically stronger ones.

And I agree with Joel that this process is hugely nondeterministic.

and I believe I read the quote correctly and agree with your statements. I also happen to think it's most unfortunate. Oh well.

I also happen to disagree with some of MITs admission decisions; but in their defense, the most important thing to look at isn't the quality of the students [i]entering[/i] the University, but the quality of the students [i]exiting[/i] the University. They aren't necessarily the same thing.

In particular, the question should be whether MIT is maximizing the entire "marginal educational value" of its student body through its admission policies. (i.e. maximizing: overall quality and potential to service of the students exiting - overall quality and potential to service of the students entering).

sorry, but i'm just in a very feministic mood this week...
[quote]denying Axline-quality scientists to admit, in their place, girls and minorities of middling intellectual talent, plus singers, dancers, water polo players, etc.[/quote]
what is wrong with girls getting into MIT???

Well, if you're going to treat it as an (academic) meritocracy, the girls at MIT don't have the same achievements as the guys do, just because overwhelming proportions of the really bright math/science students are male.

Don't forget about the attached prepositional clause: "girls ... of middling academic talent." Sorry if that was unclear. Admitting girls is not a problem (and MIT admits plenty of brilliant ones), but giving some girls who wouldn't otherwise make it a leg up due to their gender is, for some people, a problem.

Haha, hi, Ben.  You joined the board just to argue with me?  ;-)

Are university admissions harder for International students? How does it go (particularly for MIT)?

way, way, way harder.

I know one USAMO qualifier with 1600 SAT and great GPA still did not get into MIT, just because he was an international student.
The international students are extremely qualified.
my friends who got into MIT from korea are mostly IPhO, IChO or IOI medalists.
I'm sure if I applied as an international student, I wouldnt have gotten in.

Now i'm freaked out. I hope this doesn't apply to American schools...  :blush:

[quote="Geehoon"]I know one USAMO qualifier with 1600 SAT and great GPA still did not get into MIT, just because he was an international student.[/quote]
he was attending american school.

but again, they have to pick some anyway, your chance is not zero.
good luck.

[quote="Geehoon"]I know one USAMO qualifier with 1600 SAT and great GPA still did not get into MIT, just because he was an international student.[/quote]

 :huh: Oh no. Did your friend go to Seoul Foreign School? :P I'm in Shanghai American School.

I was talking about two differents groups of friends.

1. IPhO, IMO, IChO, IOI winner from Korea - they got into MIT
-> no, they did not go to Seoul Foreign High School.

2. USAMO qualifiers, 1600 SAT and got american education - he did not get into MIT
-> of course, he has never been to Korea.

you can draw a single conclusion from both 1 and 2 -> international students are really really well qualified.


...y es múltiplo de 3, 5, 7 y 11.     :lol:

$Tipe$

[color=blue]¿Ya se enteraron del mayor primo de Mersenne conocido hasta ahora en [url]http://www.mersenne.org[/url]?
El archivo que lo contiene es un bloc de notas de más de 9MB.[/color]


inequalities
symmetry
function
algebra
polynomial
absolute value
inequalities proposed

Given an integer $ n > 3.$ Let $ a_{1},a_{2},\cdots,a_{n}$  be real numbers satisfying $ min |a_{i} \minus{} a_{j}| \equal{} 1, 1\le i\le j\le n.$ Find the minimum value of $ \sum_{k \equal{} 1}^n|a_{k}|^3.$

And could you explain what happens if a1 is negative. Then the inequality between ai and a1+i+1 wont necessary be true if we take the module.

I don't see any problem when $ a_1$ is negative; could you please explain?

Notice that any two numbers among $ a_1, \ldots, a_n$ are at distance at least $ 1$, and also notice that moving any two consecutive numbers closer together can only decrease the result. So we may assume that the numbers $ a_1, \ldots, a_n$ are $ t \minus{} \tfrac{n\minus{}1}{2}, \ldots, t \plus{} \tfrac{n\minus{}1}{2}$ for some real $ t$. Also notice that if $ |2t| > 1$ we may transfer the endpoint of the sequence that is largest in absolute value to the other end and reduce the result; so we may assume $ |t| \leq \tfrac12$. In addition, because of symmetry, we may assume that $ t$ is nonnegative.

Thus for odd $ n \equal{} 2k \plus{} 1$ we need to find the obtainable minimum of
\[ f(t) \equal{} t^3 \plus{} \sum_{i \equal{} 1}^{k}\left(i \minus{} t\right)^3 \plus{} \left(i \plus{} t\right)^3
\]
and because the function is increasing (a polynomial with only positive coefficients) in $ t$, the minimum is at $ t \equal{} 0$. Similarly, when $ n \equal{} 2k$ is even, the sum is
\[ g(t) \equal{} \sum_{i \equal{} 1}^{k}\left(i \minus{} \frac12 \minus{} t\right)^3 \plus{} \left(i \minus{} \frac12 \plus{} t\right)^3
\]
and also the minimum is at $ t \equal{} 0$.

So the minimum is $ f(0) \equal{} \tfrac{(n^2 \minus{} 1)^2}{32}$ for odd $ n$ and $ g(0) \equal{} \tfrac{n^2 (n^2 \minus{} 2)}{32}$ for even $ n$.

I'm sorry but I don't understand anything you are trying to say. Note that I did NOT put modulus in $ a_i\ge a_1 \plus{} i \minus{} 1$. And $ \min\{|a_i \minus{} a_j|\} \equal{} 1$ holds for all $ i\neq j$ as well so I don't see any problem.

Quick outline-ish proof of a slicker way than smoothing:

If you arrange $a_1<a_2<...a_n$ increasing and do $2b_k=a_k-a_{n+1-k}$, you get the new sequence has sum of cubes of absolute values at most as large while maintaining the condition (to check that it maintains the condition, it's just triangle inequality and Holder's).  The new sequence has $b_k+b_{n+1-k}$, so it's easy to finish and get the answer.  If n is odd the middle element is $0$, and something $r$ elements later or earlier has magnitude at least $r$.  If $n$ is even the middle two elements have magnitude at least $\frac {1} {2}$, the next outermost $2$ have magnitudes at least $\frac {3} {2}$, and so on.


Show that for any positive integer $n$, \[\frac{\sigma(n!)}{n!}\ge \sum_{k=1}^{n}\frac{1}{k}.\]

[quote="Robert Gerbicz"]For all $ 1\leq k\leq n$ integers $ \frac{n!}{k}$ are distinct positive divisors of $ n!$, so $ \sigma(n!)\geq \sum_{k=1}^{n}\frac{n!}{k}$ Dividing this inequality by $ n!$ we can get the solution.[/quote]


integration
limit
real analysis
function
calculus
derivative

Σήμερα (επί 4 αρκετά κουραστικές  ώρες) ήταν ο διαγωνισμός επιλογής για την ομάδα του πολυτεχνείου για τον IMC.
Παραθέτω τα θέματα [b][i]ΟΠΩΣ ΤΑ ΘΥΜΑΜΑΙ[/i][/b] (μην φωνάξετε αν έχει λάθη και δν βγαίνει  :oops: )

ΘΕΜΑ 1:

Αν C το σύνολο των συναρτήσεων $ f: [0,1]\rightarrow R$ που ανήκουν στο $ C^1$ με επιπλέον $ f(0) \equal{} 0$ και $ f(1) \equal{} 1$, βρείτε την ελάχιστη τιμή της παράστασης
$ \int_0^1 \sqrt {1 \plus{} x^2}\cdot (f'(x))^2\,dx$
και για ποια συνάρτηση επιτυγχάνεται.

ΘΕΜΑ 2:

Αν $ \psi,\phi : R\rightarrow R$ και $ (b_n), n\in$ ακολουθία πραγματικών με $ b_n > 0$ και επιπλέον ισχύουν:
-H ψ συνεχής στο 0 με ψ(0)>0
-$ \forall t \in R, \phi(t)\leq 1$
-$ \forall t \in R, (1 \minus{} \phi(t))\leq C\cdot(1 \minus{} \phi(\frac {t}{2})), C > 0$
-$ \forall t \in R\lim_{n \to \infty} \phi^n(\frac {t}{b_n}) \equal{} \psi(t)$

Δείξτε ότι $ \lim_{n \to \infty} \phi(\frac {t}{b_n}) \equal{} 1$

ΘΕΜΑ 3:

Αν G σύνολο με n στοιχεία και $ F_1,...F_{n \plus{} 1}$ μη κενά υποσύνολά του, δείξτε ότι υπάρχουν ξένα ανα δύο $ L_1, L_2\subset{1,2,...n \plus{} 1}$ ξένα μεταξύ τους τέτοια ώστε:

$ \bigcup_{i\in L_1}F_i \equal{} \bigcup_{i \in L_2}F_i$

ΘΕΜΑ 4

Aν $ M_n(R)$ ο διανυσματικός χώρος των nxn πινάκων με πραγματικά στοιχεία δείξτε ότι:
i)tr(AB-BA)=0
ii)Ο υποχώρος Χ που παράγεται από τους πίνακες της μορφής ΑΒ-ΒΑ εχει διάσταση $ n^2 \minus{} 1$

Το πέμπτο θέμα ήταν κάτι με περίεργες κυκλικές υποομάδες και ίσα που πείραξα το πρώτο ερώτημα για να θυμάμαι ολόκληρη την εκφώνηση...

Για το (3) κοιτάξτε τα χαρακτηριστικά διανύσματα των $ F_1,\ldots,F_{n\plus{}1}$.

Για το (4), το (ι) είναι απλό. Για το (ιι) αν γράψουμε $ E_{ij}$ για τον πίνακα που έχει 0 σε όλες τις θέσεις εκτός από την $ (i,j)$ θέση που έχει 1, αρκεί να δείξουμε ότι όλα τα $ E_{ij}$ με $ i \neq j$ και όλα τα $ E_{11} \minus{} E_{ii}$ ανήκουν στον διανυσματικό χώρο. Αλλά $ E_{ij} \equal{} E_{ii}E_{ij} \minus{} E_{ij}E_{ii}$ για $ i \neq j$ και $ E_{11} \minus{} E_{ii} \equal{} E_{1i}E_{i1} \minus{} E_{i1}E_{1i}$

Το 1 νομίζω ότι λύνεται  εύκολα με Lagrange ή όχι;

Μια άλλη ιδέα από BCS

$ \int_0^1{\frac{1}{\sqrt[4]{1+x^2}^2}dx.\int_0^1{(f'(x)\sqrt[4]{1+x^2})^2dx\ge(\int_0^1{f'(x)dx})^2=(f(1)-f(0))^2=1}}$...
Η ισότητα ισχύει όταν...
$ f'(x)\sqrt{1+x^2}=c_1$
(εύκολη διαφορική, οι σταθερές από τα f(0),f(1))

[quote="r_boris"]Μια άλλη ιδέα από BCS

$ \int_0^1{\frac {1}{\sqrt [4]{1 + x^2}^2}dx.\int_0^1{(f'(x)\sqrt [4]{1 + x^2})^2dx\ge(\int_0^1{f'(x)dx})^2 = (f(1) - f(0))^2 = 1}}$...
Η ισότητα ισχύει όταν...
$ f'(x)\sqrt {1 + x^2} = c_1$
(εύκολη διαφορική, οι σταθερές από τα f(0),f(1))[/quote]

Ετσι ακριβως το ελυσα και εγω. Αν δεν εκανα λαθος καπου το ελαχιστο βγαζει:
$ \frac{1}{ln(\sqrt{2} + 1)}$

Και η συναρτηση
$ \frac{1}{ln(\sqrt{2} + 1)}ln(x + \sqrt{x^2 +1})$

Στη γραμμικη εκανα ενα πολυ κακο λαθος. Θεωρισα οτι οι χωροι X και Y = {A/ trA = 0} ταυτιζονται και βρηκα τη διασταση του Υ...

Στη συνδιαστικη δεν ειδα οτι λεει "τα Λ1, Λ2 να ειναι ξενα μεταξυ τους" και ελυσα ενα αλλο προβλημα...

Και εκανα και αρκετα πραγματα και στο 2 (βρηκα με limsup και supinf οτι το οριο 1 στα t στα οποια ψ(t) <> 0

[quote="kunny"]Could someone translate the context #1 into English?[/quote]

Let C be the set of the continious functions f in [0, 1] with continious first derivatives and $ f(0)\equal{}0, f(1)\equal{}1$
Find the smallest possible value of the integral
$ \int_{0}^{1}\sqrt{1\plus{}x^{2}}\cdot (f'(x))^{2}\,dx$
and the function f in C for which the smallest possible value of the integral is attained.

[quote="Durandal"][quote="Demetres"]Για το (ιι) αν γράψουμε $ E_{ij}$ για τον πίνακα που έχει 0 σε όλες τις θέσεις εκτός από την $ (i,j)$ θέση που έχει 1, αρκεί να δείξουμε ότι όλα τα $ E_{ij}$ με $ i \neq j$ και όλα τα $ E_{11} \minus{} E_{ii}$ ανήκουν στον διανυσματικό χώρο. Αλλά $ E_{ij} \equal{} E_{ii}E_{ij} \minus{} E_{ij}E_{ii}$ για $ i \neq j$ και $ E_{11} \minus{} E_{ii} \equal{} E_{1i}E_{i1} \minus{} E_{i1}E_{1i}$[/quote]

Γιατί δε χρησιμοποιείς απλώς τη συνθήκη του (i)?

Cheerio,

Durandal 1707[/quote]

Η (ι) μου λέει ότι ο διανυσματικός χώρος έχει διάσταση το πολύ $ n^2\minus{}1$. Πρέπει όμως να βρω και $ n^2\minus{}1$ γραμμικώς ανεξάρτητα διανύσματα.

[quote="Demetres"]Η (ι) μου λέει ότι ο διανυσματικός χώρος έχει διάσταση το πολύ $ n^2 \minus{} 1$. Πρέπει όμως να βρω και $ n^2 \minus{} 1$ γραμμικώς ανεξάρτητα διανύσματα.[/quote]

Είχα διαβάσει τη λύση σου πολύ βιαστικά και μου είχε φανεί ότι ξαναέδειχνες [i]και[/i] το $ \dim(X)\leq n^2\minus{}1$ ανεξάρτητα από το (i)... αλλιώς, προφανώς το (i) σου δίνει μόνο την ανισότητα $ \dim\langle AB\minus{}BA\rangle\leq n^2\minus{}1$ :)

Cheerio,

Durandal 1707

Η Ομαδα που θα εκπροσωπισει το ΕΜΠ στον IMC 2009 ειναι η εξης (με τυχαια σειρα):

1) Μαστακας Κων/νος (mast) - ΣΕΜΦΕ
2) Γεωργακοπουλος Αλεξανδρος (AlexandrosG) - ΣΕΜΦΕ
3) Κολλιοπουλος Νικολαος (NickNafplio) - ΣΕΜΦΕ
4) Ψυχας Κων/νος - ΣΗΜΜΥ
5) Λαμπροπουλος Λεωνιδας (lemonidas) - ΣΗΜΜΥ
6) Καναρη Λυδα - ΣΕΜΦΕ
7) Νακος Βασιλειος - ΣΗΜΜΥ
8 ) Γουλεακης Θεμιστοκλης - ΣΗΜΜΥ
9) Τουφεξης Φιλιππος - Πολιτικος Μηχανικος


hi,

Let $ A_{1}, A_{2}, A_{3}$ be compact sets in $ \mathbb{R}^{n}$. Show that there is one plane $ P \subset \mathbb{R}^{n}$ that divides each $ A_{i}$ into two pieces of equal measure.
(i think this problem is well-know)


If $ A,$ $ G,$ and $ H$ are the arithmetic, geometric, and harmonic means respectively of three positive real numbers, prove that
\[ 3A\plus{}H\geqslant4G
\]

[hide]Let $ abc \equal{} 1$. We have to show $ a \plus{} b \plus{} c \plus{} \frac {3}{ab \plus{} bc \plus{} ac} \ge 4$
Let $ a \plus{} b \plus{} c \equal{} k$. Then $ ab \plus{} bc \plus{} ac \le \frac {k^2}{3}$. Also by AM-GM, $ k \ge 3$. So we need to show
$ k \plus{} \frac {9}{k^2} \ge 4$
By AM-GM,
$ \frac {k}{3} \plus{} \frac {k}{3} \plus{} \frac {k}{3} \plus{} \frac {9}{k^2} \ge 4 \sqrt [4] {\frac {k}{3}} \ge 4$.[/hide]


Putnam
college contests
contest solutions

I have a question concerning the style of presentation that putnam markers want.  During the putnam, I think I always write too much to be on the safe side, but that clutters my solution and wastes time.

I'm wondering if the amount of details that people put into solutions on this site would be sufficient for full marks on the putnam; alternatively, is the amount of detail in the solutions in the lastest putnam book by Poonen, Kedlaya and Vakil sufficient for full marks on the putnam?  Does anyone posting here actually mark the putnam or know this info?

Thanks.

Well, the solutions that people are posting on this site are generally not enough to earn a full mark on Putnam or any olympiad (well, except for Darij's :D). Then again, they are not intended to be.

The solutions in books, on the other hand, are alwayed polished. They don't expect to you to write up perfect solutions on a time-restricted contest (unless you are Reid Barton).

In my opinion, the general rule of thumb is - the more you write, the safer it is. Don't skip any steps, or you might risk the markers not understanding your logic and throwing your papers into the garbage bin  :lol: .

[url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_HowWrite.php]This article[/url] illustrates some of the ideas quite well.

From my experience, the grading scale on a Putnam is significantly harsher than most olympiads. A solution that would recieve full credit on a high school olympiad would usually recieve full credit on the Putnam- possibly one or two points off. A solution that would recieve half credit on a high school olympiad would recieve one or two points out of 10 on the Putnam. The Putnam graders give scores of 0,1,2,8,9,10 on a problem, and it is very easy to overestimate your score. The first time I took the Putnam, I came out thinking I had eight problems, and my final score was 40.

Clarity is far more important than length, and a solution that is too long can be hard to understand. You need to justify your arguments well, particularly on the easy problems, but don't mistake rambling and clutter for a necessary part of the solution.

Hi

I think Yufei might have misunderstood the intent of my question.  I was asking specifically about writing solutions for the putnam (where the marking is fairly distinctive and not transparent since we don't get our breakdowns).

With regard to the latest putnam book, the solutions are about half the length that I would write up (and usually much shorter than the solutions here on the website).  They are pretty clear but they skip many steps as long as they are routine.  The authors are all former putnam fellows so they would definitely know how to write up solutions for the putnam well, but they might have shortened the solutions considerably for the book.

How much is getting the right answer from the wrong reason worth?  What I mean is how much is the proof worth compared to the answer.  Am I right in assuming that if you put down the right answer and no proof you get 0 points?

I quote my friend Simon (ComplexZeta):  "The answer is correct, but this is mathematics. In mathematics, an answer without a proof is fairly meaningless."

It seems hard to tell though, for instance, to what level you can say that something is essentially trivial. For instance, it's pretty obvious that the int(2a) = 2*int(a) or 2*int(a) +1, but I'm not sure if I had to write a proof for it or not on the exam.


group theory
abstract algebra
superior algebra
superior algebra unsolved

The question, for those who find it really annoying to open these attachments:

Prove that there does not exist a group whose commutator subgroup is isomorphic to $ S_4$.

[hide="Solution"]
The relevant facts are that $ S_4$ is a complete group (no outer automorphisms, trivial center) which is not perfect (that is, the commutator subgroup of $ S_4$ is not $ S_4$ itself). If I'm not mistaken, any group which has these properties is never a commutator subgroup of anything. Here's why.

[b]Lemma[/b]: If $ K$ is a complete group and $ K \lhd G$, then $ G$ is the direct product $ K \times H$ of $ K$ by its centralizer $ H\equal{}C_G(K)$.

In other words, a complete group can be a normal subgroup only in the most trivial fashion: the large group is just a direct product of the normal group by something.

Proof: Let $ H \equal{} C_G(K)$ be the centralizer of $ K$ in $ G$, namely the set of all elements which commute with all elements of $ K$. $ H$ is a normal subgroup of $ G$, and $ H \cap K \equal{} Z(K) \equal{} 1$ since $ K$ has trivial center. Any element $ g \in G$ induces an automorphism $ \phi_g$ of $ K$ by conjugation: $ \phi_g(k) \equal{} g^{\minus{}1}kg$. But $ K$ has no outer automorphisms, so $ \phi_g$ must equal some inner automorphism of $ K$, that is, for some $ k \in K$, $ \phi_g \equal{} \phi_k$. Now conjugation by $ gk^{\minus{}1}$ does nothing to $ K$, so $ gk^{\minus{}1} \equal{} h \in H$. In other words $ g \equal{} kh$: every element of $ G$ is expressible as a product of an element of $ k$ and an element of $ H$. Since $ H$ and $ K$ commute, $ G$ is the direct product of $ H$ and $ K$. QED

Now suppose that $ G$ is some group such that $ K\equal{}G'\equal{}[G,G]$, the commutator subgroup, is such that $ K$ is complete and non-perfect. By the lemma, $ G \equal{} K \times A$ where $ A \cong G/K$ is an abelian group. So any element of $ G$ can be written as $ ka$ with $ k \in K$ and $ a \in A$, and moreover, $ ka\equal{}ak$ for any $ k\in K, a\in A$.

Consider a commutator $ c\equal{}xyx^{\minus{}1}y^{\minus{}1}$ in $ G$. Write $ x\equal{}ka$ and $ y\equal{}lb$. Since $ K$ and $ A$ commute, $ c\equal{}aba^{\minus{}1}b^{\minus{}1}xyx^{\minus{}1}y^{\minus{}1}$. Since $ A$ is abelian, the first part vanishes and $ c\equal{}xyx^{\minus{}1}y^{\minus{}1}$. So any commutator of $ G$ lies in the commutator subgroup of $ K$, and it follows that $ G' \equal{} K'$. Since $ K' \neq K$, $ G' \neq K$ as well.

It remains to show that $ S_4$ is complete and non-perfect.

1) $ S_4$ has trivial center: this is obvious. No permutation commutes with all other permutations.

2) $ S_4$ has no outer automorphisms. This is true for all $ S_n$ except $ n\equal{}2,6$. It's a standard result but if needed I can write down a proof.

3) $ S_4$ is not perfect. This is also obvious: for any two permutations $ \sigma, \tau \in S_n$, the commutator $ \sigma^{\minus{}1}\tau^{\minus{}1}\sigma\tau$ is an even permutation, so the commutator subgroup is contained in the alternating group $ A_n$. In fact the commutator group equals $ A_n$, but we don't need that here.

[/hide]

nice proof.

a simple modification: $ K' \subseteq G'$ is clear, and the other direction follows since $ G / K' \equal{} K/K' \times A$ is abelian.

Well done, randomgraph.
Here is a proof not using those well-known facts about $ S_4$. (though it's easy to derive them with it)

$ S_4$ has exactly $ 4$ Sylow $ 3$-subgroups $ P_1\equal{}\langle (234)\rangle,P_2\equal{}\langle (134)\rangle,P_3\equal{}\langle (124)\rangle,P_4\equal{}\langle (123)\rangle$,
where $ P_i$ is the only Sylow $ 3$-subgroup of the stabilizer of $ i$ in $ S_4$ for $ i\equal{}1,2,3,4$.
So $ \sigma P_i\sigma^{\minus{}1}\equal{}P_{\sigma(i)}$ for all $ \sigma\in S_4$ and $ i\equal{}1,2,3,4$.

Assume $ G'\equal{}S_4$ and take $ g\in G$.
The conjugation with $ g$ permutes $ P_1,P_2,P_3,P_4$,
so we find $ \rho\in S_4$ with $ gP_ig^{\minus{}1}\equal{}P_{\rho (i)}\equal{}\rho P_i\rho^{\minus{}1}$ for $ i\equal{}1,2,3,4$.
So $ h: \equal{}\rho^{\minus{}1}g\in G$ satisfies $ hP_ih^{\minus{}1}\equal{}P_i$ for $ i\equal{}1,2,3,4$.

Then for all $ \sigma\in S_4$ we have also $ h^{\minus{}1}\sigma h\in S_4$ and
$ P_{h^{\minus{}1}\sigma h(i)}\equal{}(h^{\minus{}1}\sigma h)P_i(h^{\minus{}1}\sigma h)^{\minus{}1}\\
\hspace*{0.55in}\equal{}h^{\minus{}1}\sigma hP_ih^{\minus{}1}\sigma^{\minus{}1} h\\
\hspace*{0.55in}\equal{}h^{\minus{}1}\sigma P_i\sigma^{\minus{}1} h\\
\hspace*{0.55in}\equal{}h^{\minus{}1}P_{\sigma(i)} h\equal{}P_{\sigma(i)}$
and therefore $ h^{\minus{}1}\sigma h(i)\equal{}\sigma(i)$ for $ i\equal{}1,2,3,4$.

This gives $ h\in C_G(S_4)$ and $ g\equal{}\rho h\in S_4C_G(S_4)$ for all $ g\in G$.
So $ G\equal{}S_4C_G(S_4)$ and $ |G: A_4C_G(S_4)|\le 2$.
But then $ A_4C_G(S_4)\unlhd G$ with abelian factor, so $ S_4\equal{}G'\le A_4C_G(S_4)$
and by Dedekind we get
$ S_4\equal{} A_4C_G(S_4)\cap S_4\equal{}A_4(C_G(S_4)\cap S_4)\equal{}A_4Z(S_4)\equal{}A_4$
since $ Z(S_4)\equal{}1$, as $ \sigma\in Z(S_4)$ would give $ P_i\equal{}\sigma P_i\sigma^{\minus{}1}\equal{}P_{\sigma(i)}$ and $ i\equal{}\sigma(i)$ for $ i\equal{}1,2,3,4$.
Contradiction!


$ON$ is the radius which is perpendicular to chord $AB$ in circle, center $O$. This radius $ON$ meets $AB$ at $M$. $P$ is any point on the major arc $AB$ which is not diametrically opposite $N$. The lines $PM$ and $PN$ determine respectively, $Q$ and $R$ on the circle and $AB$. Prove that $RN$ is always longer than $MQ$.

A splendid solution was published in Math.Mag. 49(1976), pag.217-218 by Mark Kleiman but I have not that issue so if anyone has it can please post that solution.

I have solved it by trigonometry but my solution is ugly so I would like to see a concise geometric solution.

Thank you very much.

I think it goes like this :

Let $H$ be the foot of the height from $M$ in $MQN$. Since $QMN < \frac {\pi} 2$, the point $H$ is interior to the segment $NQ$.

Note that $<MNP + <PQN$ intercepts an arc which is a semi circle, thus $<MNP + <PQN = \frac {\pi} 2$
We then have :
$<MQH = <PQN = \frac {\pi} 2 - <MNP$. It follows that the right-angled triangles $MQH$ and $NRM$ are similar. Therefore : $\frac {MH}{MQ} = \frac {MN} {NR}$.
But clearly, we have $MH < MN$ thus $MQ < RN$, as desired.

Pierre.

That is, without any doubt, beautiful, but I went through the trouble of thinking about the problem, so I'd like to post my solution.

Let $S$ be the antipode of $N$. $MQ\cdot MP=AM^2,\ NR\cdot NP=NM\cdot NS=AN^2$, so we must prove that $\frac{AN^2}{AM^2}>\frac{PN}{PM}$. Let the external bisector of $\angle MPN$ cut line $NS$ at $S'$. It's obvious (left as a trivial exercise to the reader :)) that $S'$ is outside $(SN)$, so $\frac {PN}{PM}=\frac {S'N}{S'M}<\frac {SN}{SM}=\frac {AN^2}{AM^2}$.


[hide] Wow this is easy.
$ab+bc=-ca$

$\frac{ab+bc}{-ca}= 1$

$\frac{ab+bc}{ca}=\boxed{-1}$ [/hide]

[quote="perfect628"]If a, b, and c are nonzero, and $ab+bc+ca=0$, evaluate $\frac{ab+bc}{ca}$[/quote]

[hide]
$ab+bc$ is the opposite of $ca$, so that $-ca+ca=0$
$\frac{-ca}{ca}=-1$
[/hide]

see, simple. now, a slightly more difficult, but still pretty simple, extension:If $\frac{ab+bc}{ca}=\frac{bc+ca}{ab}=\frac{ca+ab}{bc}$, and $ab+bc+ca\neq0$, evaluate $\frac{ab}{bc+ca}$

This is assuming a, b, and c are positive, right?

[hide="hint for the extension"]
Assume WLOTMG that $\frac{ab+bc}{ca}=1$[/hide]

[quote="The QuattoMaster 6000"]Ubemaya, I don't think that that's necessary:
[hide] Say that all of the ratios given were equal to a number n.  Since $\frac{ab+bc}{ac}=n, acn=ab+bc$.  Now, do the same for the rest of them to get that abn=ac+bc and bcn=ab+ac.  Now add everything up to get that abn+acn+bcn=n(ab+bc+ac)=2ab+2bc+2ac=2(ab+bc+ac).  Since ab+bc+ac is not 0, divide it from both sides to get that n=2, so $\frac{bc+ca}{ab}=2\Rightarrow \frac{ab}{bc+ac}=1/2.$ [/hide][/quote]

Yeah, I just realized that I screwd up in one of my steps... and I got a square root and stuff.

The weird thing is, I still got the same answer of $\frac{1}{2}$.

Yeah, that's the solution I got from others who I gave this to. I prefer my slightly more complicated solution, though. [hide="here it is"]Call the 3 equal expressions $k$. Multiplying them together yields ${k}^{3}=\frac{a(b+c)b(c+a)c(a+b)}{{(abc)}^{2}}$, or ${k}^{3}=\frac{(a+b)(b+c)(c+a)}{abc}$. So, ${k}^{3}=3k+2$, or ${k}^{3}-3k-2=0$. Factoring, we get ${(k+1)}^{2}(k-2)=0$, so $k=2$ or $k=-1$. If $k=-1$, then $ab+bc+ca=0$, so $k=2$, and our desired # is $\frac{1}{2}$[/hide]

expand the numerator, and factor it as $2abc+{a}^{2}(b+c)+{b}^{2}(c+a)+{c}^{2}(a+b)$. Then divide each term individually by $abc$.


probability
expected value
real analysis
real analysis unsolved

Suppose that the number of accidents in a certainfactory is Poisson-distributed with an average of 7 accidents per week.  What is the probability that it will be at least one day until the next accident occurs?

The raw data for a Poisson distribution is usually given as a rate - in this case, 7 accidents per 7 days - and a time or space period within which we are counting. In this case, that is one day. We're asking what the probability of at least one day elapsing before the next accident; that is the same as asking what the probability of at zero accidents on the first day. The first thing you have to do is match the rate to the space to determine what the expected value is. In one day, that expected value is one accident. The number of accidents in one day is a Poisson random variable with $\lambda=1.$ The probability of zero accidents on that day is then simply $e^{-\lambda}=e^{-1}\approxeq 0.37.$

If the rate is the same at all times of day, the elapsed time (measured continuously) until the first accident will be an exponentially distributed random variable with mean equal to $1/\lambda.$ But you don't need to know that to answer your question.


For  positive reals  x,y,z  show  that
Σ[x^3/(x^2+xy+y^2)]>=(Σx)/3
19  th  Tournament of the  towns , spring 1998
Thanks!


[quote="236factorial"]A light pole casts a shadow of 6m and is 21m high. A mailbox casts a shadow of 1.5m. How high is the mailbox?[/quote]

[hide]Say the height of the mailbox is $x$

$\frac{21}{6}=\frac{x}{1.5}$

$x=\frac{31.5}{6}=\frac{\frac{63}{2}}{6}=\frac{63}{12}=5\frac{1}{4}$[/hide]

i got this right or i wouldn't be posting.

i used proportions to figure out that for every 2 meters something is tall,
it casts a shadow of 7m. I then used a formula that i think i invented :lol:   (which i probably didn't) which is shown below.

if it is 1.5m tall, you divide by 2 to find what one part of this equation. you then multiply by 7 to get your answer.

in other words,

(x/2)*m=y where y is the hight of the object, and m is the shadow length.


modular arithmetic
number theory
relatively prime
geometric series
number theory unsolved

Let $ n$ be an integer such that $ (n,10) = 1$. Show that the decimal expansion of $ \frac {1}{n}$ is repeating.

[hide=" My proof"]
Since $ (n,10) = 1$, Euler's Theorem tells us that $ 10^{\phi(n)}\equiv 1 \pmod n$, where $ \phi(n)$ is the number of positive integers less than or equal to $ n$ that are relatively prime to $ n$.
Thus, by definition of congruence, $ n | 10^{\phi(n)} - 1$. This implies there exists some integer $ q$ such that $ nq = 10^{\phi(n)} - 1 \Rightarrow \frac {1}{n} = \frac {q}{10^{\phi(n)} - 1}$.

Note that the digits of $ 10^{\phi(n)} - 1$ consists of all nines, specifically it consists of $ \phi(n)$ number of nines. (e.g. $ 10^n - 1 = 9(10^0) + 9(10^1) + ... + 9(10^{n - 1}) \Rightarrow n$ number of nines)

Since $ \frac {q}{10^{\phi(n)} - 1} = q (\frac {1}{10^{\phi(n)} - 1}),$ we need to show that the general equation $ \frac {1}{10^k - 1}$ is always a repeating decimal.

This can be done by examining a repeating decimal of the form $ 0.\overline{000...0001}$, where the number of zero's after the decimal is equal to some integer $ k - 1$. Then
$ 0.\overline{000...0001} = \frac {1}{10^k} + \frac {1}{10^{2k}} + \frac {1}{10^{3k}} + ...$

(e.g. $ k - 1 = 2$, $ 0.\overline{001} = \frac {1}{10^3} + \frac {1}{10^6} + \frac {1}{10^9} + ...$)

We note that $ \frac {1}{10^k} + \frac {1}{10^{2k}} + \frac {1}{10^{3k}} + ...$ is simply the sum of an infinite geometric series equivalent to $ \frac {a}{1 - r}$ where $ a = \frac {1}{10^k}$ and $ r = \frac {1}{10^k}$. Thus $ \frac {1}{10^k} + \frac {1}{10^{2k}} + \frac {1}{10^{3k}} + ... = \frac {\frac {1}{10^k}}{1 - \frac {1}{10^k}} = \frac {1}{10^k - 1}$.

Thus we have shown that a fraction of the form $ \frac {1}{10^k - 1}$ is always a repeating decimal of the form $ 0.\overline{000...0001}$ with $ k - 1$ number of zero's.

This brings us back to the conclusion that for an integer $ n$ such that $ (n,10) = 1$,

\begin{eqnarray*}\frac {1}{n} & = & q(\frac {1}{10^{\phi(n)} - 1}) \\
& = & q (0.\overline{000...0001}) \\
& = & 0.\overline{000...000q}\end{eqnarray*}

where the number of zeros after the decimal is equal to $ \phi(n) - 1$ and $ q$ is the quotient of $ \frac{{10^{\phi(n)} - 1}}{n}$.
[/hide]

Is my proof correct? If there are any errors or ambigous parts please let me know how I can fix them!

Makes it easier, http://www.mathlinks.ro/viewtopic.php?t=220577


I think you mean
 :D
$\frac{1}{\sqrt{a^{2}+bc}}+\frac{1}{\sqrt{b^{2}+ca}}+\frac{1}{\sqrt{c^{2}+ab}}\ge 2\sqrt{2}$


I am having my first physic's olympiad in December . This is the first time my country is  organizing a Pysics's olympiad for     9-grade's students. I am confused of what do I learn for it. Can you suggest me something ( knowledges,problems  )

Thankyou

Assuming you are comfortable with most high school math (sans calculus), any college textbook without physics will probably be sufficient prep to do at least somewhat well.  The problems at the back of the chapter provide nice practice, just make sure you can solve the hard ones usually at the very end.

Note: Some of your spelling is pretty wrong so let me correct some of them: having, olympiad, organizing

Stupid mistakes , Is there any other (mistake) ?

:P [quote="Elio (n)"] physic's , knowledges[/quote]
shud be "Physics"
and "knowledge" :P


I've heard some people talking about printing the best answer found so far by a program before the time-limit is exceeded. How is it possible to do this?

Does this work: update the output file everytime I find a better answer? I mean, will I earn points if the program keeps running for more then the time-limit but the correct answer has been written to ouptut?

I guess it depends on the judging software for your suggestion. Most of the contests I know of use stdout, so that wouldn't work. The main way I know of doing what you want is just to use the system time - just have a test every so often to see if the time is too big, and if so, stop and output an answer (usually 0, or something).

And how do I use the system time? I program in Pascal.

One thing that should be pointed out is that calling the function that calls the system time has a very large overhead, so you might actually, ironically, just be killing your own program's speed (one that might otherwise be fast enough) with such a "safeguard."

In C and C++, you can call the alarm function to set an alarm after a certain amount of time has elapsed.


calculus
integration
trigonometry
calculus computations

[quote="coffeym"]Calculate

$\int^{1}_{0}{\frac{1}{\sqrt{1+x}+\sqrt{1-x}+2}}\,dx$

by using the substitution $x=\cos{t}$ or otherwise.

[hide="HINT"]
$\cos{t}\equiv2\cos^{2}{\frac{t}{2}}-1$[/hide][/quote]

My answer is
$\sqrt 2-\frac{\pi }{2}+Tan\frac{\pi }{8}=2\sqrt 2-\frac{\pi }{2}-1$
If this is correct, I will post the solution.

Good (I had done this question with different limits, and so thought your answer was wrong, sorry  :blush: )

[color=darkblue][b]Very nice this exercise :[/b] ${I\equiv\int^{1}_{0}{\frac{1}{\sqrt{1+x}+\sqrt{1-x}+2}}\ dx\ }\ .$ I"ll use the substitution $\boxed{\ x=\phi (t)=\sin 2t\ }\ :$

$I=\int_{0}^{\frac{\pi}{4}}\frac{2(2\cos^{2}-1)}{\sin t+\cos t+\cos t-\sin t+2}\ dt=$ $\int_{0}^{\frac{\pi}{4}}\frac{\left(\[2\left(\cos^{2}t-1\right)\right]+1}{\cos t+1}\ dt=$ $2\int_{0}^{\frac{\pi}{4}}\left[(\cos t-1)+\frac{1}{4\cos^{2}\frac{t}{2}}\right]\ dt=$
$2\left[-t+\sin t\right]_{0}^{\frac{\pi}{4}}+\left[\tan\frac{x}{2}\right]_{0}^{\frac{\pi}{4}}=$ $2\left(\frac{\sqrt 2}{2}-\frac{\pi}{4}\right)+\tan\frac{\pi}{8}=$ $\sqrt 2-\frac{\pi}{2}+\left(\sqrt 2-1\right)$ $\Longrightarrow$ $\boxed{\ \int_{0}^{1}\frac{1}{\sqrt{1+x}+\sqrt{1-x}+2}\ dx=2\sqrt 2-1-\frac{\pi}{2}\ }\ .$

[b]Remark.[/b] [/color]$0\le t\le\frac{\pi}{4}\Longrightarrow \cos t\ge \sin t\Longrightarrow \sqrt{1-\sin 2t}=|\cos x-\sin x|=\cos x-\sin x\ !$


many ways to solve this problem. i would first put the x's together

so we get $x+15/5x$ - $x-2/2x$ = $1/2$

instead of cross-multipyling and factoring,
there is an easier way in my opinion to do it

u could see that
$x+15/5x$, and $x=10$ would make it $1/2$

so easily we could say that the answer is less than 10.

one way in basic algebra is to try numbers.
(some times save times but TOO risky)

this is just one method

ANSWER Algebraically
[hide]
multiply 10x to both sides
we get
$2x+30-5x+10=5x$
from here we can get
$-3x+40=5x$
$8x=40$
$x=5$
[/hide]

we could find the LCM of the 5x,2 , and 2x =10xand multiply the equation by this cancelling out the fractions :)


Prove that for w, x, y, z :ge: 0,

w6x3 + x6y3 + y6z3 + z6w3 :ge: w2x5y2 + x2y5z2 + y2z5w2 + z2w5x2.

from AM-GM,

[tex]\displaystyle
w^6x^3+x^6y^3+x^6y^3 \ge 3\sqrt[3]{w^6x^{15}y^6}=3w^2x^5y^2.[/tex]

Sum this inequality in a cyclic manner to get the desired inequality (after cancelling the 3 from both sides).

Nicely (and quickly) done.

Now, can you prove it using Holder?

Sure!

[tex]
(w^6x^3+x^6y^3+y^6z^3+z^6w^3)^3
[/tex]

[tex]
=(w^6x^3+x^6y^3+y^6z^3+z^6w^3)\cdot(x^6y^3+y^6z^3+z^6w^3+w^6x^3)
[/tex]
[tex]
\cdot(x^6y^3+y^6z^3+z^6w^3+w^6x^3)
[/tex]

[tex]
\underset{\text{\tiny{from Holder}}}{\ge}
(\sqrt[3]{w^6x^{15}y^6}+\sqrt[3]{x^6y^{15}z^6}+\sqrt[3]{y^6z^{15}w^6}+\sqrt[3]{z^6w^{15}x^6})^3
[/tex]

[tex]
=(w^2x^5y^2+x^2y^5z^2+y^2z^5w^2+z^2w^5x^2)^3.
[/tex]

Cube-root both sides to get the desired inequality. (its practically the same proof, just using Holder instead of the cyclic sum)

[quote="zabelman"](its practically the same proof, just using Holder instead of the cyclic sum)[/quote]

Nice insight.

[quote="Fierytycoon"][quote="blahblahblah"]The triplet <6,3,0,0> majorizes <5,2,2,0>, so we can apply Muirhead and we are done.

At least I think we are; can someone make sure my application of muirhead is correct?[/quote]

No, we can't use Muirhead because the sums are not symmetric, only cyclc.[/quote]

I was afraid of that.


geometry
trapezoid
integration
function
ratio
real analysis
real analysis theorems

True or False:

If $ f$ is continuous on [a,b] and [a,b] is partitioned into n subdivisions, then the error of the midpoint riemann approximation is always less than or equal to the error of the trapezoid approximation on those partitions.

Try to think of a counterexample where $ f(a) \equal{} f(b) \equal{} 0$ and $ \int_a^b f \equal{} 0$ but where $ f \left( \frac {a \plus{} b}{2} \right)$ is large.  Now ask yourself whether it is enough that $ f$ is monotonic, or perhaps either concave or convex.

An interesting question is whether if is possible to construct a continuous function such that the error of the trapezoid approximation is less than that of the midpoint one for all $ n\ge 1$. ;)

I know how to make them almost exactly equal.
[hide="Hint"]Smooth periodic function[/hide]
Now, what can I do with the coefficients...

The "normal" state is that the midpoint error is about $ \frac12$ of the trapezoid error, in opposite directions. Simpson's rule is a weighted average that takes advantage of this.

I don't think anyone else will go after it, so here's an example for fedja's question:

Choose some positive number $ t$. Let
$ f(x)=\sum_{m=0}^{\infty} (-1)^m\frac{\cos(2^m x)\sinh(2^m t)}{\cosh(2^{m+1}t)-\cos(2^m x)}$
on an interval of length $ 2\pi$. The terms of this series decay rapidly, with the $ m$th term no greater than ${ \frac1{\sinh(2^m t)}}$ in absolute value. Consequently, the function is smooth.
Its Fourier series is $ f(x)=\frac12\sum_{n\in \mathbb{Z}\setminus 0} s(n)e^{-|n|t}e^{inx}$, where $ s(n)=\begin{cases}1&n\text{ is odd}\\-s(n/2)&n\text{ is even}\end{cases}$
... Yes, I started with the Fourier series before finding the other form.

Now, why does it work? First, note that $ \int_0^{2\pi} \sum_n a_ne^{inx}=2\pi a_0$, so $ \int_0^{2\pi} f(x)=0$. Consider an arbitrary $ 2\pi$-periodic function $ g$, with Fourier coefficients $ a_n$. By periodicity, the trapezoid rule is the same as the left endpoint rule, and $ \frac{2\pi}{N}\sum_{k=0}^{N-1} g\left(\frac{2\pi k}{N}\right)=\frac{2\pi}{N}\sum_{k=0}^{N-1}\sum_n a_n e^{2\pi ink/N}=\frac{2\pi}{N}\sum_n a_n\sum_{k=0}^{N-1}e^{2\pi ink/N}$. The inner geometric sum is $ N$ when $ N|n$ and zero otherwise, so the $ N$-interval trapezoid estimate for $ \int_0^{2\pi} g(x)\,dx$ is $ 2\pi\sum_{N|n}a_n=2\pi\sum_k a_{Nk}$. The midpoint rule is similar: $ \frac{2\pi}{N}\sum_{k=0}^{N-1} g\left(\frac{2\pi (k+1/2)}{N}\right)=\frac{2\pi}{N}\sum_n a_n\sum_{k=0}^{N-1}e^{2\pi in(k+1/2)/N}=2\pi\sum_n e(n)a_n$, where $ e(n)=\begin{cases}1&2N|n\\-1&N|n,2N\nmid n\\0&\text{otherwise}\end{cases}$; the geometric sum is zero unless its ratio is 1, and is either positive or negative in that case depending on the parity of $ \frac{n}{N}$. We can rewrite this last expression as $ 2\pi\sum_k (-1)^k a_{Nk}$.

Now, we apply this to $ f$. We get
$ T_N=\pi\sum_k s(Nk)e^{-|Nk|t}=2\pi\sum_{i=0}^{\infty}\sum_{j\text{ odd},j>0}s(2^i Nj)\exp(-2^i Njt)$
$ =2\pi s(N)\sum_{i=0}^{\infty}(-1)^i\sum_{j\text{ odd},j>0}\exp(-2^i Njt)$
and $ M_N=\pi\sum_k (-1)^ks(Nk)e^{-|Nk|t}=2\pi\sum_{i=0}^{\infty}\sum_{j\text{ odd},j>0}(-1)^{2^i j}s(2^i Nj)\exp(-2^i Njt)$
$ =2\pi s(N)\left(-\sum_{j\text{ odd},j>0}\exp(-Njt)+\sum_{i=1}^{\infty}(-1)^i\sum_{j\text{ odd},j>0}\exp(-2^i Njt)\right)$

$ T_n$ is a strictly decreasing alternating series in $ i$. $ M_n$ is that same series with the first term reversed in sign. By the alternating series error estimate, $ M_n$ is farther from zero.


Hello everyone!

This is a very famous problem. I am searching for it on the Net but I didn't have so much theory (and I didn't have any luck ... :( ) about it.

If ABCD is a quadrilateral, construct its diagonal triangle using compass only.
(Well, with 'straightedge' it is very easy :D).

Bye, Mathlinkers!

I am sorry but I don't get it what should be construct, crossing points of diagonal or???  :oops:  :roll:

Yeah... you can't physically draw a line using a compass lol.  You can construct the intersection of two lines though, and one way (maybe only?) is by inversion.

Try here http://www.cut-the-knot.org/do_you_know/compass.shtml you'll maybe find somethin interesting

Well, I want to say "Using compass", in the sense with the rule is prohibited only in this case (well, a triangle is determined by this three vertex, and it is my question: to determine these three points).
Well, in any case, many thanks!


number theory
prime numbers
number theory unsolved

[quote="greatestmaths"]let $ p$ and $ q$ prime numbers such that : $ 3 < p < q$
let $ a \equal{} p \plus{} q$.
[hide]you can see that $ a$ isn't prime because it's even . [/hide]

1.find the biggest  divisor  for  $ a$ and  # $ a$
2.prove that  all divisor for $ a$ are  small than $ q$.
3.prove that  all prime divisor  for $ a$ are  small than $ p$.[/quote]
1.if $ d|a,d<a\Rightarrow{a\equal{}kd,k\geq2\Rightarrow{a\equal{}kd\geq 2d}\Rightarrow {d\leq\frac{a}{2}}}$
therefore  $ \frac{p\plus{}q}{2}$  is the biggest divisor for $ a$
I dont know what mean #$ a$...can you define it please?
2.note that $ \frac{p\plus{}q}{2}<q$ because $ p<q$
3.I think the statement is false!let$ p\equal{}5,q\equal{}17$ then $ a\equal{}22$ but 11 is prime divisor for $ a$ and not small than $ p\equal{}5$


In the Village League, the team to win two of three softball games becomes the champion. If the probability of Team Alpha beating Team Beta is $ 60 percent$ for every game, what is the probability that Beta wins the championship? Express your answer as a common fraction.

Answer: [hide]$ 44/125$[/hide]

[hide="what, a solution for a ctdn problem?"]
Ways Beta can win (letter stands for team that wins)
BBB ($ \frac{8}{125}$)
ABB ($ \frac{12}{125}$)
BAB ($ \frac{12}{125}$)
BBA ($ \frac{12}{125}$)

$ \frac{44}{125}$

[/hide]

[hide]$ \binom{2}{2}\equal{}1$ way Beta can win in a sweep (BTW, it doesn't matter if Beta wins all 3 or if they win 2, as can be seen from Circle's solutions.
$ \binom{3}{2}\equal{}6$ ways they can win in 2 games. Calculating the probabilities too, the answer is
$ \frac{44}{125}$.[/hide]

Which competition level and year was this question from?


Check out

http://isabio.com

for interactive solutions to *both* the AMC 10A and AMC 12A (sorry I forgot to include the 12A part in my last post)

Also: Use internet explorer instead of firefox.

Does Safari work? I've only got Safari and Firefox. Grr.  :mad:

For safari, the videos work, but the website is screwed up.

The solution is pretty good. The guy is pretty funny though.

omg i love this guy!
he explains things so well.
he even explained that 16:30 was military time and was equal to 4:30 pm.
i even love his accent
 :)
thx. for posting the link, math ninja

Looks like we all like his funny accent. Explains well though, because I think almost everyone can understand what he means. I like the cursor fiddling  :lol: .

He used a pretty non-optimal solution on 24 IMO but whatever.

Funny guy.


geometry
geometric transformation
reflection

A small object is placed $30\text{cm}$ in front of a concave mirror of focal length $15\text{cm}$ . A glass of thickness $6\text{cm}$ is placed in between the mirror and object . Where will the image of the object be if the refraction index of the glass is $n=1.5$ ?

** I face problem when the image is displaced the second time by the glass went the light returns from the mirror . It seems confusing to me . Anyone have any idea ?

[quote="shadysaysurspammed"]Is the answer something like  $\frac{420}{13}$ from the mirror?[/quote]

that is what I get also if we only consider one time displacement by the glass , then what about the second time displacement by the glass ? It would be difference ? :o

according to me,we have to take the glass just once.You haven't even given the distance of the slab from the mirror,how can we calulate the second displacement and all then.Only once I think(Refer to someone because our class hasn't reached higher optics yet,I've just read it once)

Here's an example from my book...

A small object is placed $21cm$ in front of a concave mirror of focal length $10cm$ . A glass of thickness $3cm$ is placed in between the mirror and object . Where will the image of the object be if the refraction index of the glass is $1.5$?

Solution-

apparent displacement by glass=$1cm$(calculate it)

Thus image formed by glass is $21-1=20$cm from trhe mirror.

But 20cm is the radius of curvature
,hence the image formed byglass is at centre of curvature
therefore,the rays will fall perpendicularly on the mirror and hence,retrace their path,Hemnce final image is at the 20cm from the mirror.

The above was an example from my book in which theyu didnot consider the galss for the second time.I'm not sure either,refer to your teacher or someone.

well , maybe the book you use is same as my teacher :P . Anyway , my teacher also gives me the same solution as above .But when i ask him what about the second refraction thing , he said he havent thought of it before ...... :(

If your teacher said he didn't think of it before,you don't have a good teacher :|

Does he use H.C.Verma(its indian)

O.K,just a minute.You are saying you have doubts about what will happen when the light comes to the slab after reflection right,right?Well I think I've got the answer but my explaination will be a bit vague.....

See,first the onject has its image formed due to the glass.
Then  now we have to take this as the $object$ [i]for the mirror[/i] but if we consider that the light from this second object/image of object undergoes refraction through the glass slab when it gets reflected,why don't we think that before it reaches the mirror,won't it undergo another refraction due to glass slab? ;)

So that means,we would get another image,then again another image and so on so this will go on infinitely which is ofcourse,not possible,right?

So hence,we should not comnsider the glass slab afetr we have taken refraction to it once.

What I said above should be right(just thought it up).Ponder over it for a while,ask me the part you didn't understand


[quote="Treething"]http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=45&year=1986&p=394399

I believe that this answer is incorrect - I'd say it's 135, as $10^6$ is excluded due to the phrase "proper divisor".

Adding my name there would be nice as well :P .[/quote]

I think he took that into account already in the first line of the solution... the $26$'s at the end should be $23$'s though.

I think that usually for the purposes of AMC/AIME, "divisor" or "proper divisor" usually means all positive integer divisors, including one and the number itself.  If it doesn't, it's often clear which definition is intended.  There was a topic about this a while back [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=70542]here[/url].  (Hmm, I seem to have changed my opinion slightly since then.)

[quote="Phelpedo"]Just to be clear, does proper divisor include 1?[/quote]

That is totally irelavent to the problem because :

$\log_{10}1=0$

But I believe it does. I might be wrong.

I was under the impression that 1 DOES count as a proper divisor, but the number itself does not. But then other contests I've seen define proper divisor as not being one, so it's a wash.

[hide]
Yeah, the answer is 47*3, but if you include 10^6, it is equal to 49*3.[/hide]

[quote="calc rulz"]Positive divisor includes one and the number.

Proper divisor does not include the number.

Btw, I did the problem, it does come out to 141 (147 if you include $10^6$).

[size=200][color=red]EDIT[/color][/size] It does include one, I looked on MathWorld.[/quote]

At the end of the definition, Mathworld says that 1 is [i]sometimes[/i] included...so this makes it really confusing. Same thing happens if you look up natural numbers, counting numbers, and whole numbers.


pigeonhole principle
probability
expected value
combinatorics proposed
combinatorics

I have n pigeonholes and n letters, and i put each letter randomly in a pigeonhole. What is the chance that exactly r holes each have at least one letter in them?

this problem seems quite tricky, i havent worked it out yet. could someone shed some light on this and point me to something?

[quote="wootles"]$n^n$ total ways to distribute the letters
$\binom{n}{r}$ ways to choose the r holes that will have letters in, for each of which there are $(n-r)^n$ ways to put the letters in the chosen holes.

So $P(r) = \frac{\binom{n}{r}(n-r)^n}{n^n}$[/quote]

you're right about the sample space = $n^n$ and $\binom{n}{r}$ ways to choose the r holes to have letters in them, but at this point, if u just randomly put letters in, in some cases, you will put say all the letters in one hole and you won't have exactly r holes occupied i.e. some of the r holes willl be empty

Wow, this problem really is harder than it looks.

I get ${P(n,r) = \frac{1}{n^n}\binom{n}{r}\sum^{r}_{i=1}{(-1)^{i+r}\binom{r}{i}(i)^n}}$ ... seems to work OK for the cases I have tried. Can that be simplified at all?

[hide="method"]
Let f(n,r) denote the number of ways to split n distinct objects into r distinct non-empty subsets.
$f(n,1) = 1$
$f(n,2) = 2^n - \binom{2}{1}f(n,1)$ Since there are $2^n$ ways to split the objects into two sets, but these include the f(n,1) ways to split the objects into just one set (leaving an empty set to go in one of the two places).
$f(n,3) = 3^n - \binom{3}{1}f(n,2) - \binom{3}{2}f(n,1)$ There are f(n,2) ways to leave just one empty set, which goes in one of three places, and f(n,1) to leave two empty sets, which go in two of the three places....

In general: $f(n,r) = r^n - \sum^{r-1}_{i=1}{\binom{r}{i}f(n,i)}$

We get $f(n,r) = \sum^{r}_{i=1}{(-1)^{i+r}\binom{r}{i}(i)^n}$. (I just wrote out f(n,1), f(n,2), f(n,3), f(n,4) and the pattern became obvious)

Now for each combination of r holes, there are f(n,r) ways to put the letters in so that each hole gets at least one letter. In total, there are $n^n$ ways to put the letters in the holes.

So the probablity ${P(n,r) = \frac{\binom{n}{r}f(n,r)}{n^n} = \frac{1}{n^n}\binom{n}{r}\sum^{r}_{i=1}{(-1)^{i+r}\binom{r}{i}(i)^n}}$
[/hide]

yes, that was pretty much what i got, but i wonder if that can be simplified. there seems to be another approach using multichoose [url=http://magnet.amt.canberra.edu.au/forums/viewtopic.php?p=1335&sid=9ff1dbc9d4f459ce7e1e9a9564dc8858#1335]here[/url] but i don't quite get it...

its interesting that the expected value of r as n tends to infinity is $(e-1)/e$ so about 63% of the boxes on average will be occupied

I tried a multichoose approach but it doesn't work here. We are forced to work with the sample space $n^n$ because of how the letters are distributed, and since this space gives permutations, we have to consider the letters distinct when evaluating the number of ways they can be distributed.

For example, if n = 5, r = 2, we could have 4 letters in the first hole, and 1 in the second. There are 5 ways to do this that we have considered in our sample space (because there is one of 5 letters in the second hole). But we could also have 3 letters in the first hole, and 2 in the second, and there are are 5C2 = 10 ways to do this. But the 'multichoose' method only considers the [i]size[/i] of each set, so for the cases above it would count 2 ways instead of 15 - it's never going to work because it just said the two were equally likely when clearly they aren't.


geometry
3D geometry
tetrahedron
ratio
incenter
trigonometry
analytic geometry

[b]T is a regular tetrahedron. T' is a tetrahedron whose vertices are the centroids of the faces of T. The ratio of the volume of T' to the volume of T is m/n, where m and n are relatively prime positive integers. Find m+n.[/b]

[hide="what i tried"]assume a side of T is x. if we connect two centroids of two adjacent faces, we get an isosceles triangle with two congruent sides that are the length of the incenter of T. the angle between the two sides is 60. we can use the law of cosines to find the length of the third side which is the length of a side of T'. we can then find the ratio of the sides, cube that, and we have the ratio of the volumes. unfortunately, for some reason, when i do all this, m and n are not both integers. i get a radical answer. where did i go wrong?[/hide]

[hide="hint"]
Embed the larger tetrahedron in a four-dimensional coordinate space
[/hide]
[hide="Interesting solution"]
Embed the larger tetrahedron in four space. Its coordinates are as follows.

$ (0,0,0,1), (0,0,1,0), (0,1,0,0), (1,0,0,0)$
The length of each side is $ \sqrt {2}$

The centroid of each side is the average of the points, meaning that the centroids are

$ (0,\frac {1}{3},\frac {1}{3},\frac {1}{3}), (\frac {1}{3},0,\frac {1}{3},\frac {1}{3}), (\frac {1}{3},\frac {1}{3},0,\frac {1}{3}), (\frac {1}{3},\frac {1}{3},\frac {1}{3},0)$

The length of each side is

$ \sqrt {\left(\frac {1}{3}\right)^2 \plus{} \left(\frac {1}{3}\right)^2 \plus{} 0^2 \plus{} 0^2} \equal{} \frac {\sqrt {2}}{3}$

The ratio of side lengths is $ \frac {\sqrt {2}/3}{\sqrt {2}} \equal{} \frac {1}{3}$

The ratio of volumes is $ \left(\frac {1}{3}\right)^3 \equal{} \frac {1}{27}$

$ m \plus{} n \equal{} 1 \plus{} 27 \equal{} 28$
[/hide]

[hide="Wait.."]
If we take a cross section as shown, doesn't it follow that the height of the inner tetrahedron is $ 1/3$ that of the outer one, so the volume ratio is $ 1/27$?
[/hide]

thanks for the help guys!
i just have a question, can anyone explain where my method went wrong?

[quote="chickendude"][quote="ckck"][quote="chickendude"][hide="hint"]
Embed the larger tetrahedron in a four-dimensional coordinate space
[/hide]
[hide="Interesting solution"]
Embed the larger tetrahedron in four space. Its coordinates are as follows.

$ (0,0,0,1), (0,0,1,0), (0,1,0,0), (1,0,0,0)$
The length of each side is $ \sqrt {2}$

The centroid of each side is the average of the points, meaning that the centroids are

$ (0,\frac {1}{3},\frac {1}{3},\frac {1}{3}), (\frac {1}{3},0,\frac {1}{3},\frac {1}{3}), (\frac {1}{3},\frac {1}{3},0,\frac {1}{3}), (\frac {1}{3},\frac {1}{3},\frac {1}{3},0)$

The length of each side is

$ \sqrt {\left(\frac {1}{3}\right)^2 \plus{} \left(\frac {1}{3}\right)^2 \plus{} 0^2 \plus{} 0^2} \equal{} \frac {\sqrt {2}}{3}$

The ratio of side lengths is $ \frac {\sqrt {2}/3}{\sqrt {2}} \equal{} \frac {1}{3}$

The ratio of volumes is $ \left(\frac {1}{3}\right)^3 \equal{} \frac {1}{27}$

$ m \plus{} n \equal{} 1 \plus{} 27 \plus{} 28$
[/hide][/quote]
You mean they equal 28 right?  :P[/quote]

Yes... thank you =)
I fixed it

Also, I do not believe the angle you specified is equal to 60 degrees[/quote]
you mean to tell me that two adjacent faces of a tetrahedron do not meet at a 60 degree angle? that's weird. what angle do they meet at?

At the base, the faces of the tetrahedron do indeed form a $ 60^{\circ}$ angle (of course). However, at the midpoint of the edge, the angle formed between faces is different. If we connect the midpoint of one side of the tetrahedron to the two other vertices of the tetrahedron, we form an isosceles triangles. By Pythagorean, we find that the leg length is $ \sqrt{x^2 \minus{} (x/2)^2} \equal{} (\sqrt{3}/2)x$; and the base of the triangle is clearly $ x$. We now use the law of cosines on a triangle with sides $ \frac{\sqrt{3}}{2}x, \frac{\sqrt{3}}{2}x, x$, and we get $ \theta \equal{} \cos ^{\minus{}1} (1/3) \approx 70.5 ^{\circ}$.


Let $f: \mathbb{Z} \rightarrow \{-1,1\}$ be a function such that \[ f(mn) =f(m)f(n),\ \forall m,n \in \mathbb{Z}. \] Show that there exists a positive integer $a$ such that $1 \leq a \leq 12$ and $f(a) = f(a + 1) = 1$.

[quote="john111111"][quote="codyj"]$n=0\implies f(m)=1$...?[/quote]
Not necessarily.Either this or $f(0)=0$.[/quote]

By the condition, it cannot be $0$

My solution:
Assume the contrary. Since $f(1) = f(1)^2$ we have $f(1) = 1 \Longleftrightarrow f(2)=-1$. Similarly we get the values of $f(3)$,..., and observe that $f(4)=-1$, and $f(8) = -1$. Hence $f(8) \neq f(4)f(2)$, a contradiction.

Note that the fuctional value of any perfect square is 1 since f(m^2) = (f(m))^2 so f(1) =1  It implies f(2) = -1 otherwise we are done.  since f(4) = 1 therefore f(3)=f(5)=-1 now f(9) = 1 and f(10) =1 and we are done

Setting $m=0$ we get $f(n)\equiv 1.$ Q.E.D.


function
topology
superior algebra
superior algebra solved

$X$ a compact set, $C^0(X,\mathbb{C})$ the ring of continuous function from $X$ to $\mathbb{C}$. Is there a simple description of the maximal ideals of this ring. (remember it is simple in the case of $\mathbb{R}$)

Of course- the maximal ideals are simply the sets of functions which are zero at a single point. These are clearly maximal ideals. If we have an ideal $I$ and a finite set of functions $f_1,f_2,\dots,f_n\in I$, then $f=\sum_i f_i\cdot \overline{f_i}\in I$ and the zero set of $f$ is the intersection of the zero sets of the $f_i$. If $f$ is nowhere zero, $\frac1f$ is continuous and $I$ is the whole ring.
Now, for each element $g$ of $I$, let $S(g)$ be the closed set on which $g$ is zero. We have that the intersection of any finite number of these $S(g)$ is nonempty; by compactness, the intersection of all of the $S(g)$ is nonempty. Therefore, $I$ is a subset of one of the maximal ideals we have already named, and we are done.

Why would an ideal be closed under conjugation?

It doesn't need to be. $f_i$ belong to the ideal, so $gf_i$ also do, for every $g$. Now take $g=\overline{f_i}$.


Numbers $ 1,2,...,1997^2$ are written in the cells of a $ 1997\times 1997$ table. It is allowed to apply the following transformations: exchange places of any two rows or two columns, or reverse a row or a column. (When a row or column is reversed, the first and last entry exchange their positions, so do the second and next-to-last ones, and so on. ) Is it possible that after some transformations, some two numbers exchange their places, and all the other numbers stay in the original positions?

It was posted but not solved at http://www.artofproblemsolving.com/Forum/viewtopic.php?t=184757


limit
trigonometry
calculus
calculus computations

suppose, we have $ \lim_{n\to\infty}\ ((1 \plus{} sinn^2)/(3n \plus{} 4sqrt(n) \plus{} 5)$ and I need to proove that the limit is 0 by using definition. Can I just cross out 4sqrt(n) + 5 or 3n and then write (2/3n) < $ \epsilon$ ? or there are some limitations that I can't cross anything (that the fraction would be bigger) I want?

$ \lim_{n \to \infty}\frac {1 \plus{} \sin {n^2}}{3n \plus{} 4.\sqrt {n} \plus{} 5}\equal{}0\Leftrightarrow$
$ \Leftrightarrow [(\forall)\varepsilon > 0; \ (\exists)n(\varepsilon)\in \mathbb{N}: \ (\forall)n > n(\varepsilon)\Rightarrow \frac {1 \plus{} \sin {n^2}}{3n \plus{} 4.\sqrt {n} \plus{} 5} < \varepsilon].$
$ \mbox{Or: } \frac {1 \plus{} \sin {n^2}}{3n \plus{} 4.\sqrt {n} \plus{} 5} < \frac {2}{3n \plus{} 5} < \varepsilon\Leftrightarrow 3n \plus{} 5 > \frac {2}{\varepsilon} \Leftrightarrow n > \frac {2 \minus{} 5.\varepsilon}{3.\varepsilon}\Leftrightarrow \\
\Leftrightarrow (\forall) \ n > n(\varepsilon) \equal{} \left[ \frac {2 \minus{} 5.\varepsilon}{3.\varepsilon}\right]\in\mathbb{N} \mbox{ if } \varepsilon < \frac {2}{5} \mbox{ and }n(\varepsilon) \equal{} 0 \mbox{ for }\varepsilon\ge\frac {2}{5}.$

[b]REMARK:[/b] $ \frac {1 \plus{} \sin {n^2}}{3n \plus{} 4.\sqrt {n} \plus{} 5} < \frac {2}{3n \plus{} 5} < \frac {2}{3n} < \varepsilon\Leftrightarrow n > \frac {2}{3.\varepsilon} > \left[ \frac {2}{3.\varepsilon}\right] \equal{} n(\varepsilon).$

[quote="mihai miculita"]$ \lim_{n \to \infty}\frac {1 \plus{} \sin {n^2}}{3n \plus{} 4.\sqrt {n} \plus{} 5}\Leftrightarrow$
$ \Leftrightarrow [(\forall)\varepsilon > 0; \ (\exists)n(\varepsilon)\in \mathbb{N}: \ (\forall)n > n(\varepsilon)\Rightarrow \frac {1 \plus{} \sin {n^2}}{3n \plus{} 4.\sqrt {n} \plus{} 5} < \varepsilon].$
[/quote]
I think that mihai miculita intended to state this: $ \lim_{n \to \infty}\frac {1 \plus{} \sin {n^2}}{3n \plus{} 4.\sqrt {n} \plus{} 5}\equal{}0\Leftrightarrow$

$ \Leftrightarrow [(\forall)\varepsilon > 0; \ (\exists)n(\varepsilon)\in \mathbb{N}: \ (\forall)n > n(\varepsilon)\Rightarrow \frac {1 \plus{} \sin {n^2}}{3n \plus{} 4.\sqrt {n} \plus{} 5} < \varepsilon].$


Hello there,

For the following problem, I am arriving at an answer different from the given one so I would appreciate any help or hints.

Thank you. My work is shown below.

---

1. A slightly smaller plastic pipe is inserted inside a second plastic pipe to produce an air column, open at both ends, whose lengths can be varied from 35 cm to 65 cm. A loud speaker, connected to an audio frequency generator, is held over one of the open ends. As the length of the air column is increased, resonance is heard first when the air column is 38 cm long and again when it is 57 cm long.

a) Calculate the wavelength of the sound produced by the audio frequency generator.

---

Since the first resonant length is 38 cm long:

$ L \equal{} \frac {\lambda}{2}$

$ 0.38 m \equal{} \frac {\lambda}{2}$

$ \lambda \equal{} 0.76 m$

However, my textbook says that it is [hide]0.38 m[/hide].

You forgot to consider the second resonance, the one at $ 57\;cm$.

In an open air column, the antinodes (points of maximum vibration of the medium) are at the open ends, so the “expanding pipe” has antinodes at $ 38\;cm$ and $ 57\;cm$. These antinodes are $ 19\;cm$ apart, so we could simply consider an air column of the same length. The wavelength would then be

$ 0.19\;m \equal{} \frac {\lambda}{2} \Rightarrow \lambda \equal{} 2(0.19)\;m \equal{} \boxed {0.38\;m}$.

This works, because the distance between antinodes is equal to one-half of the wavelength of the wave passing through it. Remember that the wave passing through the pipe has the same frequency and wavelength throughout the “experiment.”

If you're still confused, try reading these pages:
[url=http://www.glenbrook.k12.il.us/GBSSCI/PHYS/CLASS/waves/u10l4c.html]Nodes and Anti-nodes[/url]
[url=http://www.glenbrook.k12.il.us/GBSSCI/PHYS/Class/sound/u11l5c.html]Open-end Air Columns[/url]

(Notice that the frequency of the wave is the second harmonic of the $ 38\;cm$ pipe, and the third harmonic of the $ 57\;cm$ pipe.)


geometry calculus integration derivative invariant trigonometry rotation

well this is a supposed fallacy which i and some of my friends had been debating on for months(with some proffesors from our nearby institute of Physics) and finally reached a conclusion last November

well here it goes
[b]Question[/b]:
This example is a standard one in all books .U have a metallic rail of resisitance $ R$ and on it a metallic conductic wire whic is moved with a constant speed $ v$ w.r.t earth and u have a magnetic field say perpendcular to the plane of the rail over which the wire is sliding.What is the current produced,emf induced and so on??

Proof:
well there are supposedly two proofs to it ..one from the faraday's law(supposedly) which is of our major concern
at any time the flux $ \Phi \equal{} Blx$ l is the  length of the wire
so we have $ e \equal{} \frac {d \Phi}{dt} \equal{} Bvl$
and there ends our proof and begins the problem...if u go through the correct statement of

[b]Faraday's Law[/b] it states that "if the net flux of the magnetic field through a [b]fixed[/b] loop i schanging then an emf is induced in the coil whose magnitude satisfies the relation ship.
$ E \equal{} \frac {d\Phi}{dt}$
actually these have origins in from the more fundamental Gauss's Law which states
$ \boxed{\nabla \times E \equal{} \frac {\delta B}{\delta t}}$
integrating both sides and an application of stokes theorem gives us the faraday's law statement iff the loop is constant otherwise the $ \frac {\delta}{\delta t}$ can't come out.
So the Faraday's law statement should hold true only for constant loops how come it also holds tru for changing loops.Here clearly the flux is only changing because u r changing the loop.
[b]Hypothetical Situation[/b]
this is for those who argue that farday's law also holds for changing loops and not only for changing fields
suppose u have a room with magnetic field inside it and say u make a [b] hypotheical[/b] coil in ur mind inside the room and start expandng it [b]hypothetically[/b] so will EMF be induced and an electric field created...this means by sitting here and by applying manetic field inside a room u can make any [b] stationary[/b]  charge move by justing thinking of a loop :o

how come there is such a fundamental connection between the two cases and i know veveryone must have solved hundreds of problems using this concept

If you use wrong formulas the result will be wrong.

$ \nabla \times \mathbf{E} \equal{} \minus{} \frac {\partial \mathbf{B}} {\partial t}$

You are misunderstanding Faraday...

First of all, state your problem with Clear Language. “say u make” and “ur mind” are not good enough.

Faraday Law is clear. There is a variable magnetic field? Yes? There is so a Electric Field and it’s rotational is equation to the negative of the variation of the magnetic field due the time.

Begin Clearing the problem and what you want. Make things with rigor. How is the magnetic field in the room?! After that is easy. The Electric field induced don’t need a medium to exist. Those teachers teach as a Loop of wire. Etc…

The coil Expanding? Sure you are forgetting some law of conservation. If the area of the coil gets big, the number of turns get small.

You are of that kind of those old-fashioned physicist that believe that you can create energy from nothing? Because I thing that you want to prove that with this “Experiment” that is not clear enough.

If you know that something is impossible, and you get the solution proving that is possible, your reasoning is wrong. By this principle, make your problem better. What Changing Loop?

"this is for those who argue that farday's law also holds for changing loops and not only for changing fields "

Is not clear what you meant

ok i missed out the negative sign :huh: but that doesn't atke any hell out of my discussion.
i am not concerned with how the coil gets made...but making things clearer for u
suppose i have a square coil(imaginary) in a room filled with magentic field(time independent) created by 2 poles of a very strong magnet and suppose now i expand the hypothetical loop in my mind so will a electric field be induced??
hope u get it

@EulerIntegral

you are getting over-enthusiastic...

faraday's law was never wrong...
What pardesi tried to point out was the completely absurd use of faraday's law in school textbooks...

But I still don't get the second part of pardesi's fallacy..."Hypothetical situation"

Would you mind elaborating a bit..??

well it is a consequence of the blind application of faraday's laws.
like we used in the problem above where the loop was expanding but the magnetic field was constant and yet we applied faraday's laws

Uh Oh!!!  :D
Shre , Chaps , Shrids , Chemrock escape.........  :D
We seem to have stumbled upon a group of professionals.  :rotfl:

Am still reading this but perhaps this can be moved to general physics section for more views and a better debate.

Madness, surely if physics isn'y your forte why did you post that??

Well that can be deleted later  :D    :rotfl:
But do you seem to understand it honestly  :P

Hmm....interesting. I believe the other method we get BLV is as follows.

$ q(vxB)\equal{}qE
E\equal{}Bv
dV/dl\equal{}Bv
V\equal{}BvL$

which is much more from the fundamentals.

Hmm...The question is that why does faraday's law hold good. The ans is maybe because of the electric field.

there is an induced electric field in the conductor

Integral (e.dl) =do/dt. (sorry I dunno how to bring about integral)

Hmm...This balances the force on e- due to mag field. and since both formula work that basic idea, it works. Hmm... am not really sure. will post when I am able explain with more clarity. :D

yes the other proof is by lorentz force which has nothing to do with Maxwell's equations
this was naother point we were discussing
this is an uncanny similarity that holds everywhere...Lorentz force law and Maxwell's equation are both independent and fundamental...so having such a similarty everywhere actually surprised us  :)

Let me see if I can add to the confusion here just a bit.  :lol:

In its original form Faraday's law stated that an [i]electromotive force[/i] (voltage) is induced in a circuit that is proportional to the time rate of change of magnetic flux through a fixed loop in said circuit.

If the magnetic flux through one such loop is $ \phi$ then it should be self-evident that the total flux through $ N$ loops is $ \Phi \equal{} N\phi$.

The time rate of change for the total flux is then given by:

$ \frac {d\Phi}{dt} \equal{} \frac {dN}{dt}\phi \plus{} N\frac {d\phi}{dt}$

and

$ \phi \equal{} \iint_S \mathbf{B}\cdot\hat{\mathbf{n}}\, da$

Now from Leibniz's integral rule (differentiation under the integral):

$ \frac {d\phi}{dt} \equal{} \iint_S \frac {\partial\left(\mathbf{B}\cdot\hat{\mathbf{n}}\right)}{\partial t}\, da \plus{} \mathbf{B}\left(a_2\right)\cdot\hat{\mathbf{n}}\frac {\partial a_2}{\partial t} \minus{} \mathbf{B}\left(a_1\right)\cdot\hat{\mathbf{n}}\frac {\partial a_1}{\partial t}$

The units of the magnetic field, $ \mathbf{B}$, are such that the constant of proportionality is $ \minus{} 1$, i.e.

$ \mathcal{E} \equal{} \minus{} \left\{\frac {dN}{dt} \iint_S \mathbf{B}\cdot\hat{\mathbf{n}}\, da \plus{} N\left\{\iint_S \frac {\partial\left(\mathbf{B}\cdot\hat{\mathbf{n}}\right)}{\partial t}\, da \plus{} \mathbf{B}\left(a_2\right)\cdot\hat{\mathbf{n}}\frac {\partial a_2}{\partial t} \minus{} \mathbf{B}\left(a_1\right)\cdot\hat{\mathbf{n}}\frac {\partial a_1}{\partial t}\right\}\right\}$

Faraday stated his law in way that was consistent with his experiments, but as can be seen by examining the above expansion, the implications of explaining this phenomenon via the calculus are much more broad.

The magnetic flux through a coil is given by the famous integral.

$ \Phi_m \equal{} \int \!\!\! \int \mathbf{B} \cdot d\mathbf S\,$

We know that $ \mathbf{B}$ is independent of the time, but $ \mathbf S$ is dependent. So...

$ \mathcal{E} \equal{} \minus{} N{{d\Phi_B} \over dt} \equal{} \minus{} N \ { d \over dt }   \int   \mathbf{B} \cdot d\mathbf{S}$

Supposing the surface line on a plane perpendicular to the magnetic field. We Simplify to.

$ \Phi_m \equal{} \minus{}B N S'(t)$

And this formula is true, since the coil is expanding etc.

But if we take out the coil, what is expanding to generate emf?
It would be the space.

Just like Relativity.

A variable magnetic flux trought a flat space is the same of a constant magnetic flux trought a variable space.

But the question is, the space is variable? Clear the answer is not, unless you are near a variable gravitational field.

Since the space is constant, there is no eletric field.

Can you see that expand the coil is the same as expand the space?

we all know integrating...
 so no problem with that just answer this
[quote="pardesi"]suppose u have a room with magnetic field inside it and say u make a  hypotheical coil in ur mind inside the room and start expandng it hypothetically so will EMF be induced and an electric field created...this means by sitting here and by applying manetic field inside a room u can make any  stationary charge move by justing thinking of a loop[/quote]
so will a electric field be induced or not

No, we cannot move a charge stationary in S (where $ \vec E$ = 0) by expanding an imaginary loop. The movable charge has to be physically forced into the moving frame S', to see the net positive and/or the net negative charges in the solenoid or in a permanent magnet or in whatever is producing the time constant magnetic field. This forcing can be done by the potential well, which keeps movable charges (electrons) confined in a moving conductor.

yes the obvious anwer is no..because that would in real sense contradict conservation of energy...but my real doubt is wats the difference in applying the 'Faraday's Law'here and in the case of the standard example here
[quote='pardesi']
Question:
This example is a standard one in all books .U have a metallic rail of resisitance R and on it a metallic conductic wire whic is moved with a constant speed v w.r.t earth and u have a magnetic field say perpendcular to the plane of the rail over which the wire is sliding.What is the current produced,emf induced and so on??

Proof:
well there are supposedly two proofs to it ..one from the faraday's law(supposedly) which is of our major concern
at any time the flux $ \Phi \equal{} Blx$ $ l$ is the length of the wire
so we have $ e \equal{} \frac {d \Phi}{dt} \equal{} Bvl$
[/quote]

So you mean to say that we should state Faraday's law as The rate of change of Magnetic flux through a surface bound by a loop in the frame of the magnetic field is equal to the induced emf in the loop [i]in the loop's frame[/i].

Have I got it right?

no not that......
Let's not think about the frame changing and relevant transformations now...lets fix ourselves to earth frame....
what happens here is our major ccorncern....why are we getting contradictory results while using the same method in case of a 'real' and a hypothetical loop :)

If the electric and magnetic fields are $ \vec E \equal{} (E_1, E_2, E_3), \vec B \equal{} (B_1, B_2, B_3)$ in the frame S and if a frame S' is moving with $ \vec v \equal{} (v, 0, 0),$ the electric and magnetic fields do not transform separately into S', but as components of antisymmetric tensor satisfying Maxwell equations.

$ \vec E' \equal{} [E_1, \gamma(E_2 \minus{} \beta c B_3), \gamma(E_3 \plus{} \beta c B_2)],$
$ \vec B' \equal{} [B_1, \gamma(B_2 \plus{} (\beta/c)E_3), \gamma(B_3 \minus{} (\beta/c)E_2)].$

If $ \vec E \equal{} (0, 0, 0)$ and $ \vec B \equal{} (0, 0, B_3)$ in the room frame S, then $ \vec E' \equal{} (0, \minus{} \gamma \beta c B_3, 0)$ and $ \vec B' \equal{} (0, 0, \gamma B_3)$ in the rail frame S'. Using $ \vec E'$ in the rail frame S', the induced voltage is

$ V' \equal{} \oint_L \vec E' \cdot \text d\vec l' \equal{} E_3'l \equal{} \minus{} \gamma \beta c B_3 l \equal{} \minus{} \gamma B_3 v l \equal{} \minus{} \gamma B_3 \frac {\text dS}{\text dt} \equal{} \minus{} \gamma \frac {\text d\Phi}{\text dt}$

The components of force perpendicular to $ \vec v$ transform form the rail frame S' back to the room frame S as $ F_\perp \equal{} \frac {F'_\perp}{\gamma},$ therefore $ V \equal{} \frac {V'}{\gamma} \equal{} \minus{} \frac {\text d\Phi}{\text dt}$ in the room frame S, as expected.

now $ V\equal{}\frac{\minus{}V^{'}}{\gamma}$ implies there is an emf induced in the earth frame along the loop..but  isn't this some kinda 'miracle' ....
u can then move a stationary charge in a time invariant non-uniform magnetic field just by thinking of an expanding loop :o

No, you cannot. All this reasoning and calculations depend on the movable charge being in the rail frame S', moving with respect to whatever generates the magnetic field, to see the non-zero electric field. By thinking about the loop, without the moving rail, the stationary charge will not go there, it will not see the electric field, and the constant magnetic field alone will not move a stationary charge.

ok ...what u basically have done is used lorentz force law which requires the presence of 'real charges'...
but what about this kinda approach to the case of real 'loop' and an 'imaginary' loop

Just the flux $ \Phi$ at any time is $ \Phi\equal{}Blx$ where $ x$ is it's length so $ \frac{d\Phi}{dt}\equal{}Bvl$ ...nowhere in the above proof one uses the fact that the flux is due to a real loop or an imaginary one

No, I did not. I showed you, where do the real charge densities come from and why are they seen only by an observer or charge moving in a direction perpendicular to the constant magnetic field, but not seen by a stationary observer or charge.

ok i think i just about  get you....but the problem is not there...the problem is why does
[b]"Faraday's Law" apparently fail in case of 'imaginary loops'[/b]..probably this is the best possible way i can put it

I have never said that Faraday law fails. An imaginary loop simply does not drag any charge with it, which could see the electric fileld, but I am just repeating myself. Let's hear your explanation, the one you arrived to last November.

do u want the question or the 'answer' :)

also i am not telling that u said anything wrong what i am stressing on is why do we get such absurd results on applying faraday's law blindly on an imaginary loop

Can you give us some kind of explanation for this??

ok here is it....

see it's wrong to Apply Farady's  law blindly in case of  a 'changing loop'...though the formulaes are the same ...FARADAY'S LAW As it is holds TRUE FOR ALL CASES whether LOOP CHANGES or NOT...but the Reasons are different.

Here if u apply blindly to a changing loop then you would  get a contradiction in the case of a 'IMAGINARY LOOP'

So where's the mistake . :?:
the mistake lies on the origin of the equation $ \boxed{\frac { \minus{} d \Phi}{dt} \equal{} \int \vec{f_{e}}.\vec{dl}}$
this has two different origins in two different cases

[b]CASE 1-LOOP UNDER CONSIDERATION FIXED[/b]
we have by Maxwells's Equation
$ \frac {\delta \vec{B}}{\delta t} \equal{} \vec{\nabla} \times \vec{E}$
integrating over any loop and using stoke's law and the fact that the area under consideration is time invariant we have
$ \boxed{\frac { \minus{} d \Phi}{dt} \equal{} \int \vec{f_{e}}.\vec{dl}}$
Hence this equation has it's ORIGIN FROM MAXWELL's EQUATIONS.

[b]CASE 2-LOOP UNDER CONSIDERATION CHANGING WITH TIME[/b]
consider any loop which is moving by changing it's shape let $ \Phi(t)$ represent the flux thru it at any time $ t$ the we have after time $ t \plus{} dt$ the flux thru it $ \Phi(t \plus{} dt)$
now the cahnge in flux $ d \Phi \equal{} \Phi(t \plus{} dt) \minus{} \Phi(t)$ that is the flux thru the 'extra' region the loop has acquired.
Now as the loop move sthru the magnetic field the charge carriers that is the electrons will move along the wire due to a net unbalanced magnetic force say with speed $ \vec{u}$ w.r.t wire at time $ t$ then they posses a net velocity $ \vec{r} \equal{} \vec{v} \plus{} \vec{u}$
where $ \vec{v}$ is the velocity of the loop at time $ t$
so area of that small region covered in time $ dt$ can be written as $ \vec{da} \equal{} (\vec{v}dt \times \vec{dl})$ by the well known formulae.
so we have $ \vec{d \Phi}{dt} \equal{} \oint \vec{B}.(\vec{v} \times \vec{dl})$ since
where the integration is thru the 'extra area' covered in time $ dt$
also as the electrons move parallel to $ \vec{dl}$ we have $ \vec{u}\parallel{} \vec{dl}$
so we have $ \frac {d \Phi}{dt} \equal{} \oint \vec{B}.(\vec{r} \times \vec{dl})$ now using te standard STP formulae
we have $ \frac {d \Phi}{dt} \equal{} \minus{} \oint(\vec{r} \times \vec{B}).\vec{dl}$
but we have $ (\vec{r} \times \vec{B}) \equal{} \vec{f_{e}}$
so we have
$ \boxed{\frac { \minus{} d \Phi}{dt} \equal{} \int \vec{f_{e}}.\vec{dl}}$
HENCE Origin of the SAME Result In this case is FROM LORENTZ FORCE LAW which has Nothing to do with Maxwells's law
but that's pretty amazing isn't it :)