AMC USA(J)MO USAMO USAMTS HCSSiM MIT college

Browsing the USAMO grading results from 2007, I noticed that the question-wise points

aren't on the web.

Are the points by question revealed?

Or are they announced only to the test taker personally?

Or are they kept secret?

I've never really understood the point of that.  It seems like it wouldn't be that hard to enter scores by question, but it would definitely make a difference to the people taking the test.  Just for an example, over the two years I've taken it, I've solved 4 problems each year but gotten scores of 5 and 11, respectively.  I mean, with such a low average score per problem solved, it would really help my proof-writing if I could see what I was having problems with and what I was doing well.

Does anybody know what the reasoning is for not giving out the point breakdowns?

I agree with samath. It would be very helpful to actually know what went wrong with your score, what you got credit for etc. and that would promote math far more than keeping them secret. It seems like a strange policy to me...

[quote="moogra"]Wait then how do people know they got 7/7?[/quote]

if they got 14 points and only answered two problems.

[quote="samath"]Does anybody know what the reasoning is for not giving out the point breakdowns?[/quote]
If what I've heard is accurate, they used not to post the scores at all, until one year they posted them on the web by accident.  Before that, I believe they just told people their score interval, or maybe only their ranking interval.

If the AMC released individual scores for each question, I think people would actually be less happy, because then they would see what they got on a particular question, disagree with it, and want to start arguing.  There are usually people unhappy enough to have barely missed the MOP cutoff, etc.

In any case, if you want to improve your proofwriting skills, the most important thing is to practice a lot.  It is helpful to read well-written proofs to which you can compare yours.  It is helpful to have another person to check that you're really writing proofs.  You can find such persons in many ways :  you can do USAMTS or the Canadian math correspondence program, both of which are free, I think.  If you want to invest money, you can do WOOT, or Awesomemath's correspondence program.  Or you can go to a math program over the summer (e.g. HCSSiM, Math Camp).  Or you could find a mathematician who is willing to read your proofs, if you like.  The USAMO alone will not teach you how to write nice proofs.

Anyhow, remember that on the USAMO it is easy to fool yourself into thinking you've solved problems that you haven't.  Two years ago, I thought I had solved two problems, whereas I'd actually solved one.  Last year, I thought I had solved three problems, whereas I'd actually solved two.  In both cases, I got an appropriate score.

[quote="Boy Soprano II"]The USAMO alone will not teach you how to write nice proofs.[/quote]
Unless you do well enough to take MOP, where you'll take graded tests. Those give you a pretty good idea of the scoring.

Yeah, being a good proof-writer is a huge part of doing well on the USAMO.  Every year, you'll see people claiming they got 30 points on the forums before the scores are up, and only ending up with 10 because of bad proofwriting.  I'm usually the other way around - last year I thought I had at most 15 points, and I ended up with 17.

I guess there's no way to practice your proofwriting for this year's USAMO, but one good tip is to write up full proofs to any proof problem you solve for practice.  A lot of times the real challenge of a problem is not solving it but figuring out how to best write the proof.  Sometimes you'll even catch holes in your argument while writing it up that you wouldn't notice otherwise.

I'm not saying that we should release those results simply to teach problem solving; I know you need more than that. But I just don't think it makes [i]sense[/i] to give total score but not individual problem score... Just saying

[quote="MellowMelon"]Unless you do well enough to take MOP, where you'll take graded tests. Those give you a pretty good idea of the scoring.[/quote]
Well, this is true, but I stick to my statement, because first, MOP is not "the USAMO alone", and second, if you make it to MOP, you probably at least have some idea of how to write a proof.

My point is, in the past I've felt that, even after talking to other people who got higher schores, my solutions were correct, and I couldn't tell which questions I got points for and which I didn't.

It's not like I would start arguing with answers, I trust that the graders are doing a good job, it's just that apparently I'm not and I want to see which specific parts I'm doing wrong, so that I can fix them.

It's not that I expect to learn how to write good proofs through the USAMO, but I would like to use the opportunity to find weaknesses in my proof writing.

Just for comparison, for those of you who do USAMTS, I've always found it useful to see grader comments and scores so I know what I messed up.  Sometimes it's as simple as, say, not proving the case $ x\equal{}0$ or something, in which case just seeing the score breakdown might help me on.

I whole heartedly agree with samath. And the graders don't even have to write comments (we know they are in a hurry), but it definitely helps to at least see which problems we get credit for. Isn't that the point of such a competition anyway?

I think it may be to do more with a database issue, since graders probably just write a total score at the top and give it to a secretary or something. But it isn't that hard to write something liike 7/1/0/2/6/0 and put it in a database after you've read the whole paper anyway...

Don't they grade problems individually rather than one person's test, so that no one grader grades all six problems by one competitor?

If that's the case, it would REALLY be simple to put down the point breakdown.

The reason they don't, as mentioned, is that it would probably cause a mass wave of points protests none of which they would respond to anyway.

Is there not a thread on this site where people try to guess what the rubric is for scoring the problems? That's what it comes down to, on each problem the graders decide what level of progress constitutes a 1 pt solution, 2 pts, 5 pts, 6 pts, and then of course full credit. I left out 3 and 4 because there is usually a leap from 1 or 2 to 5 or 6 (or even 7) ... when you're grading these I hate to say it but the real question you are asking is "does the student understand why the statement is true?", not "did he accidentally flip a plus or minus sign somewhere". Most students who think they got a 5 or a 6 and actually get a 2, do so because they truly don't understand why their proof is wrong, not because they made 3 or 4 silly errors out of carelessness.

If you want to see what I'm talking about, ask your little brother to write 10 proofs and then guess what his scores are, and in the meantime grade his proofs.

If you really want to know where you lost points, you could share the proofs with someone locally who knows math (like maybe a professor) or just post it on AoPS and ask for feedback.

[quote="Boy Soprano II"]
If what I've heard is accurate, they used not to post the scores at all, until one year they posted them on the web by accident.  Before that, I believe they just told people their score interval, or maybe only their ranking interval.
[/quote]

False! Where do people come up with this stuff??

In 2002, at the one and only centralized USAMO (held at MIT), we had a quasi-vote (meaning it wasn't an official vote but it was clear that a lot of people raised their hand and supported this) and they decided to start releasing your score instead of your range because people declared that they didn't care about "getting their feelings hurt" (in other words, being told that you got a 0 instead of a 0-6).

How do you accidentally post the results on the web anyway? Terrible miscommunication? Some random troublemaking grader going "hmm let me add a couple of html tags and build a table and host this thing"?

[quote="darktreb"][quote="Boy Soprano II"]
If what I've heard is accurate, they used not to post the scores at all, until one year they posted them on the web by accident.  Before that, I believe they just told people their score interval, or maybe only their ranking interval.
[/quote]

False! Where do people come up with this stuff??[/quote]
I read something on these forums, [url=http://www.mathlinks.ro/viewtopic.php?p=496221#496221]here[/url], to be specific.  I suppose that post did not say that the scores were posted on the web by mistake, but rather that they were released by mistake.  In any case, humans are fallible, and possibly I don't know as much as you do about the internet; it sounded plausible to me.  Thank you for correcting me.

[quote="AMCDirector"]The scores were not released by accident, the scores were deliberately announced and posted.  If I recall correctly, I made the suggestion, with the  consensus and approval from the USAMO committee, we changed our practices.

In 2001, I had just barely started at the AMC, and was not familiar enough with the administration of the USAMO to make any changes or suggestions in its practices.  I think at that time and previously, scores were posted in "score groups", that is, USAMO participants  only found out that they had scored 0-6-7-13, 14-20, etc.  I think the reason was to avoid embarrassing participants with a score of 0 and also to cut down on the difficulties caused when  students got e.g. a 20 and the Honorable Mention level was e.g. 21.  This practice has its advantages and disadvantages.

At the 2002 USAMO when we had several hundred students present at MIT, dividing into score groups was one more step in a very tight schedule.  Furthermore, as an educator, it was my opinion that announcing scores in score groups had more disadvantages than advantages.  I believe in providing as much information as reasonable.  Finally, as darktreb recalls, a straw poll indicated that the students preferred to have the scores released and that this change would not cause problems.  So, in a considered way, we changed.  No accidents.  I won't say that we never have miscommunication or that we never slip up, but I certainly work hard to avoid it, and on something as important as the USAMO, I make doubly sure.

Steve Dunbar
MAA Director, American Mathematics Competitions[/quote]

of course back before this it seemed like the graders at MOP always had a copy of the scores and it was something of an unofficial tradition to find it and steal it. as i recall, you could see exactly how your score broke down for the current year, and if you took it the previous year you could see what your overall score was minus the breakdown for that year. good times


Is there positive integer $n$ such, that:
1) $gcd(n,6)=1$;
2) there isn't $a,b\in\mathbb{N}$ such, that $|2^a-3^b|=n$ :?:

[quote="ZetaX"]I'm sure I misunderstood something, but what speakes against $a=b=n=1$¿[/quote]
Excuse me :blush: I certainly had in mind:
Is there positive integer $n$ such, that:
1) $gcd(n,6)=1$;
2) [b]there is not[/b] $a,b\in\mathbb{N}$ such, that $|2^a-3^b|=n$.

Can anybody give answer on my question :?:

Iam sure that it was posted before (yes such number exists), but I can't find it at the moment.


This is not difficult.
Let a = x, b = y, c = z. Then abc = (xyz) = 1.
So we have to prove that S = \sum (a + b)/(a + ab + b) \geq 2.
Now note that S = \sum (a + b)(a - ab + b)/(a + ab + b).
I will prove that for all p, q > 0, we have (p - pq + q)/(p + pq + q) \geq 1/3.
Now (p - pq + q)/(p + pq + q) \geq 1/3 if and only if 3p - 3pq + 3q \geq p + pq + p.
This last inequality is equivalent to p + q \geq 2pq, which is true of course.
So S = \sum (a + b)(a - ab + b)/(a + ab + b) \geq 1/3 \sum (a + b).
Hence S \geq 2/3(a + b + c) \geq 2(abc)^(1/3) = 2 by AM - GM.

Good job! You are right it's not dificult but I posted it in hope that there will be more than one solution. That is my solution also but maybe there's another one.

Yeah, mine's the same: I proved that x^3+y^3/(x^2+xy+y^2) >= (x+y)/3 because x^2*y+y^2*x<=x^3+y^3. This was given at the national olympiad in 97 for the eighth graders.

Well I'm still working on another one.
I tried minimizing the left member like so:

a=x^3 b=y^3 c=z^3

Left member \geq (a+b+c)^3*2/3(2a^2+2b^2+2c^2+ab+bc+ca)
but I got no result. I'm still trying.

Well here's a solution a bit diferent from the other one at the begining but I guess the end is the same:

(a^3+b^3)/(a 2 +ab+ b 2 )=a+b- 2ab(a+b)/(a 2 + ab+ b 2 )  \geq a+b- 2ab(a+b)/3ab= 1/3*(a+b)
So we have:
 \sum (a^3+b^3)/(a 2 +ab+ b 2 )  \geq 2/3*(a+b+c) \geq 2*(abc)^1/3 =2


abstract algebra
group theory
induction
superior algebra
superior algebra unsolved

I just wanted to show it (it leads to be very easy prove of Cauchy's theorem). Let $ G$ be a finite abelian group. Let $ p$ be a prime dividing the order of $ G$. Using elementary group theory prove there is a subgroup of order $ p$.

We can first of all assume that no proper subgroup has order divisible by $ p$, or else we are done by induction. Now there exists an element of the group whose order is not $ 1$, or else the group is the trivial group. Let that order be $ n$. Note that $ f(x) \equal{} x^n$ is a homomorphism with non-trivial kernel. Both the kernel and the image are a proper subgroup of $ G$ in this case. Furthermore, $ |Ker(f)\parallel{}Im(f)| \equal{} |G|$. Thus either $ |Ker(f)|$ or $ |Im(f)|$ is divisible by $ p$, so we have exhibited a proper subgropu whose order is divisible by $ p$, and we are done.

Even more elementary:
Use the ideas [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=27505]here[/url] to find a subgroup $ H$ of order $ p^k$, where $ p^k$ is the largest power of $ p$ dividing the order of $ G$. Choose an element $ a$ other than the identity in $ H$: if it has order $ p^r$, $ \{e,a^{p^{r\minus{}1}},a^{2p^{r\minus{}1}},\dots,a^{(p\minus{}1)p^{r\minus{}1}}\}$ is a subgroup of order $ p$.

Here another approach, that I really like.
[hide="Is using factor groups considered elementrary"]
Argue by induction on $ n$. Let $ G$ be a finite group and $ p$ divide the $ n\equal{}|G|$. Pick $ a\not \equal{} 1$ and construct $ H \equal{} \left< a \right>$. If $ H \equal{} G$ then the group is cyclic and the proof is complete (because it certainly holds for cyclic groups). If $ H\subset G$ then there are two cases: (i) $ p$ divides $ H$ then by the inductive step it has a subgroup of order $ p$, (ii) $ p$ does not divide $ m\equal{}|H|$. It is safe to assume (ii). Construct $ G/H$, now $ |G/H|<n$ and $ p$ divides its order. Thus, by inductive step there is a coset $ bH$ with order $ p$. The claim is if we let $ b^m$ then it has order $ p$.
[/hide]


algebra polynomial calculus quadratics algorithm search integration

I don't know if this is the right place to ask this but I am learning Partial Fraction Decomposition in my Calculus Class and my teacher told me that all polynomials with real coefficients can be factored into linear and quadratic terms. Like 1 / ( x^3 - x^2 + x - 1 ) = 1 / ( ( x^2 + 1 ) * ( x - 1 ) ) = ( -.5x-.5) / ( x^2 +1 ) + ( .5 ) / (x-1) but is this really true? I can't seem to decompose 1 / ( x^4 - x^3 + x^2 - x +1) . Any help would be greatly appreciated!

Any real polynomial can be factored that way, but the coefficients don't have to be nice. Your example $ \frac{x^5\plus{}1}{x\plus{}1}$ is nicer than most, as its roots (four of the fifth roots of $ \minus{}1$) can be written in terms of rational numbers using nothing worse than nested square roots. $ x^4\minus{}x^3\plus{}x^2\minus{}x\plus{}1\equal{}\left(x^2\minus{}\frac{1\minus{}\sqrt{5}}{2}x\plus{}1\right)\left(x^2\minus{}\frac{1\plus{}\sqrt{5}}{2}x\plus{}1\right)$

In the exercises you'll be given on this, don't expect factorization to be a serious difficulty- because if it is, you should really be doing things a different way.

[quote="LightningT"]... my teacher told me that all polynomials with real coefficients can be factored into linear and quadratic terms. [/quote]
This is one of several equivalent formulations of the theorem known as the Fundamental Theorem of Algebra. But the Fundamental Theorem of Algebra is a "pure existence" theorem: it asserts that the roots (or factors) exist, but does not give an algorithm for finding those roots. And in fact, it is also known (this is the prime subject of Galois Theory) that there [i]can't[/i] be such an algorithm. There exist some polynomials of every degree at least 5 whose roots cannot be expressed as roots (or combinations of nested roots) of rational numbers. And even in degrees 3 and 4, for which the roots (factors) of such a polynomial can be algorithmically constructed by the Tartaglia-Cardano-Ferrari formulas ... let's just say about using those formulas that the cure may be worse than the disease.

You can systematically search for rational roots (which is to say integer-coefficient linear factors) using the Rational Root Theorem; there are related results for searching for rational factors of other degrees. These methods can tell you what to try to see [b]if[/b] anything works nicely, but that's all it is. If no such factor works, then you're out of luck.

The pure existence theorem does give us the confidence to try numerical methods - if the factors exist, then there must be ways to reasonably approximate them. Of course, once you start down that path ... if the original problem was to compute an integral, then you might as well directly attack the integral with a numerical method.

As jmerry implied, the world of elementary textbooks is a world of rigged problems in which the factors can be found, or are already given.


Ok here are the modified rules:

1) Exactly eight people start the game.
2) Each one starts with thirty points.
3) The person who is hurt loses 3 points.
4) The person who is healed gains 2 points.
5) You may NOT heal the person just healed, though you may hurt him.
6) You may NOT hurt the person just hurt, though you may heal him.
7) A post is valid only if it hurts exactly one person and heals exactly one other person, neither of which is the poster himself. In all other cases the post is devoid of value, and the next poster should ignore it.
8) No person is allowed to withdraw from the game once it starts.

and the rules that carry over:

9) No consecutive posts (which means that there should be at least one valid post in between two of your posts).
10) Anyone can post - not just the participants.

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

In this trial version, we start the game with five (somewhat) senior players with eighteen points each.

bubka - 18
Klebian - 18
randomdragoon - 18
Ignite168 - 18
junggi - 18

the game begins from the next post.

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

In the full version of the game that is coming soon, participation will be by registration. Keep looking forward to it ...

im a senior player? uhhhh
alright....

hurt ignite heal bubka

bubka - 20
Klebian - 18
randomdragoon - 18
Ignite168 - 15
junggi - 18

you are somewhat senior compared to some of the other people  :)

hurt Ignite168 heal junggi

bubka - 20
Klebian - 18
randomdragoon - 18
Ignite168 - 12
junggi - 20

hurt bubka heal klebian
bubka - 17
Klebian - 20
randomdragoon - 18
Ignite168 - 12
junggi - 20

i have this feeling people aren't very interested

hurt ignite 168 heal juggi
bubka - 17
Klebian - 20
randomdragoon - 18
Ignite168 - 9
junggi - 22

heal ignite 168 hurt juggi
bubka - 17
Klebian - 20
randomdragoon - 18
Ignite168 - 11
junggi - 19


A number is 18 less than its additive inverse. What is the
number?


MATHCOUNTS calculus integration geometry ARML AoPSwiki AMC

People from the Missouri Mathcounts team should meet here to prepare and strategize. Missouri will win!  :rotfl:

Yeah, if they have no life like xpmath :P

Don't use this forum.  As you can see, it's pretty deserted and useless.  Use your team meetings to discuss stuff, which there isn't much of.

These are the people on the MO MATHCOUNTS team(In order):

Skyler Wright(einsteinjr35)

Max Schindler（mewto55555)

Me(Tiny Tim)

Katie Zhang- no username

and Victor Wang(math154) won Countdown and got 5th individual

xpmath what are you even doing in the missouri forum? R u a lifeless noob?

Mewto55555 is the integration of e^ x times y

[quote="mewto55555"] R u a lifeless noob?

 [/quote]

I take offense at that. I have some life and I'm not [i]completely[/i] a noob.

Well...I kind of did, for like a few months when we first moved to the US (I was really small).
But I don't now.

[quote="pianoforte"]Any Mccalla students?[/quote]

What do you mean?

Charles McCalla? nvm, only applies to KC area students.

You're right she is from KC. Any idea what school?

hmm, thats our motivation. Along with this:http://www.artofproblemsolving.com/Forum/viewtopic.php?t=200297&start=20 (posts #27 and #28).

Post #27 just happened to be from someone who has done much better than you at nats.

haha...ZzZzZzZzZzZz

wow, i'm kinda suprised that from someone who made 10th individual made a very false statement.

[quote="tinytim"]10th individual? That's not MUCH better than I did... :P[/quote]

And considering that you'll probably be better than 10th individual this year...

rofl, I can't believe this account still works!!! The Missouri multiple accounts still live on...

mewto, I'll pm you the pass to this account so you can post.

Acutally it was me.  :P

I made the account last year and since mewto55555 can't post anymore, I am giving it to him so he can post...

[quote="ZzZzZzZzZzZz"]Again, Missouri is not a winning team.  For example, try solving this:

For each integer $ n > 1$ determine which of the two positive real numbers $ a$ or $ b$ satisfying
\[ a^n \equal{} a \plus{} 1,\ \ \ \ \ \ b^{2n} \equal{} b \plus{} 3a
\]
is larger.[/quote]

Just manipulate the equations a bit.

$ a^{2n}\minus{}a\equal{}a^2\plus{}a\plus{}1$

$ b^{2n}\minus{}b\equal{}3a$

Blah.

$ (a\minus{}1)^2>0 \implies a^2\plus{}a\plus{}1>3a$

and the rest follows.

nice job not looking at the official solution, max. copying the aopswiki solution shows so much more integrity.

[quote="xpmath"]Eh you can't go assuming that people look at solutions like that. Just wait and see how they do at competitions  :wink: .

Where was that problem from anyway?[/quote]

USAMO 1993 #1 :wink: Let's look at the AoPSWiki solution now, shall we?

OMG, Mewto's a guy!

Ok, sorry, to interupt, but you guys have the meanest, most un-civilized people ever. Oh, and I hope you don't mind me intruding but I lived in Springfield for like 3 years.

I cannot believe you guys have spam wars...California is so organized and peaceful.


hi so results have been declared so how have you al fared :?:

Got 169  92.9 percentile :(  :(
Chappli got 208!!
And gaurav got 223!!!  :o 99.1 percentile!
Saw d others but wont post

was getting 255
stupid fiitjee has given me 235
my 20 marks dissappeared
got 99.3

160 90.5 Percentile.

I got -1 in 1 chem paper.. This is ridiculous. But the pathetic fallacy is that -1 is 25.9 percentile..... Where has my 12 marks gone????????

And funnily I got 73 in physics but still 92.5 Percentile. :((

Anit this is ridiculous  :mad: .
I AM MAD! I THOUGHT I EXPECTED 150 BUT SHRIDS REMINDED ME THAT I EXPECTED 180  :mad:  SO 11 MARKS LESSER  :mad:

159..i have improved my chem...got 37 in paper 1 :D ...still less but i am concentrating :D

yes i know...i compare myself with myself dA :D

got 175.. expected mark.. 94.2 pcile

physics as usual flop with a total of 37  :rotfl: im happy only wid my math performance this time: 39 + 47.. got 99.7 pcile in that :)

and the shock of the year: got 51 in chem paper 1  :rotfl:

where r these jee crazy guys who get 298 in these fiitjee tests

Me too just realised that i lost 18 marks in subjectives  :(   :mad:  :o

[quote="Chapplikka"]I got 208,98.4 percentile and I have discovered where I lost the 18 marks I expected..
Yes, you guessed it right.. SUBJECTIVE..

Lost 2 subjectives in Chem and one in math, all decimal subjectives.. Only then did I realise that Prasad had told me a few days ago that the way I wrote decimal subjectives was wrong.. Thats y I was doomed...  :D
 :D[/quote]

how did u rite it like??

I got 156 well atleast 6 marks more than expected :roll:
But I can't blame FIITJEE anyway because I flopped in everything and expected this apalling total
[quote="madness"]Actually paper 1 was easy da, chemrock got 64!!!  :o  :o   :D[/quote]
Jothi got 72 and prasad got 68 so shut up da rohit :mad:

So wat, its still a gr8 mark and btw can you tell me jothi's total and his overall split up?  :maybe:

I think Jothi got 90 in Chem.

And Yes, I think I didn't get marks in subj as well. The mark I lost in Phy & Maths paper is correctly a multiple of 6. :(

The shocking thing was that a 51 in Phy Paper II got me a lousy 95.4 percentile only. This is ridiculous.

no jothi is surely less than equal to 74 :D

Why da he got 72 in paper 1 itself  :huh:

yes he did get that in paper-1 but in paper-2 both jothi and me were those unfortunate ones who lost all the subjectives. :(  :(  My gain in mark in also a multiple of 6 so probably he would have reconsidered one of them :rotfl:

wth????

everyone is online...good....

are any of u able to access the BIts server....

It's too slow

My marks were a similar flop thanks to the subjective neck cutting..(kazhutharuppu)
@sori: I wrote my subjectives as 000.5 da.. and I heard decimal subj shud be written as 0.500 and only the non decimal ones should be written as 00005.

[quote="Chapplikka"]
@sori.[/quote]
ru s ure about what u wrote.. :rotfl:  :rotfl:  :D

Sori==hash_include==shridhar :rotfl:  :rotfl:


Prove for nonegative reals:
$ {a}^{3}\plus{}{b}^{3}\plus{}{c}^{3} \geq a\sqrt{b\plus{}c}\plus{}b\sqrt{a\plus{}c}\plus{}c\sqrt{a\plus{}b}$

No, it doesn't... And it's wrong! For $ a \equal{} b \equal{} c \equal{} 1$ we get $ 3 \ge 3\sqrt2$.

URBAN1990, you forgot to say that:

this probleme is proposed on JBMO.

[hide]

with Chebychev:

Cauchy-chwarz:

it's easy to prove that:  (shur or AM-GM)

therefore:

 [/hide]

See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=248080
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=248093


Consider a $ 3\times3\times3$ cube made out of $ 27$ hollow subcubes. The subcubes are connected by doors on their faces (so every subcube has six doors, although of course some cubes have doors that open to the "outside," where you can't go). From which subcubes is it possible to start such that you can visit every other cube exactly once?

[quote="mewto55555"]Are we to assume that you cannot visit a cube twice?[/quote]Yeah, problem edited.

[hide]This is a rather common problem. Alternately color the subcubes black and white like a 3-D checkerboard. Without loss of generality, let the 8 corners and the 6 face centers be colored black (14 black faces) and the rest be colored white (13 white faces). Now, when traveling from one subcube to another, the cube you travel to must have a different color than the cube you started in. Therefore, the possible sequences of COLORS you can travel to are BWBWBWBWBWBWBWBWBWBWBWBWBWB and WBWBWBWBWBWBWBWBWBWBWBWBWB... but there are no more whites to go to and one black remains that is impossible to reach at this point. Thus the valid starting cubes are the 14 originally colored black.[/hide]

You have to show that the 14 subcubes that you said are possible actually work, but this isn't too hard (there are only two cases).

[quote="aidan"]From any corner cube, we can traverse the cubes on one plane through an S pattern and then do the same in the next two planes. For the center cube of a face, we go outwards in a spiral, traversing all the cubes on one plane and then perform S-traversals in the next 2 planes. I don't know how to explain in more concrete terms though  :P[/quote]If you were writing a proof of this for a competition, I'd suggest drawing the cube as a graph and drawing out the path WLOG in each case.


Not very-very difficult, but beautiful :)!

Several boxes are arranged on a circle. Each box contains several balls (or no ball at all). One [i]move[/i] consists in selecting a non-empty box $ B$, taking all its balls and placing one ball in each box following $ B$ in clockwise direction, until we exhaust all balls. Prove that:

[i]A)[/i]  if at each next move we select the box where the last ball has been placed at the previous move, then after a finite number of moves, we will obtain the initial distribution again.

[i]B)[/i]  if, unlike at [i]A)[/i], we are free to chose any non-empty box for the next move, then we can reach from any initial distribution of balls any other distribution (provided that the total number of balls is the same in these $ 2$ distributions).

[hide]Actually the problem IS rather difficult but the idea has been quite often used in (also) difficult problems in the last years' competitions :)[/hide]
Hope it hasn't been posted before...

can you post the solution freemind
thank you very much

Because I've been asked, here's the solution. But I suggest you think a bit more about this beautiful problem before reading the solution :)
[hide="Solution"]
Let $ n$ denote the number of boxes and $ S$ the number of balls.
[b]a)[/b]
Call an [i]object[/i] a sequence of $ n+1$ numbers - first $ n$ numbers are the number of balls in boxes $ 1,2,\ldots,n$ and $ n+1$th - the index of the [i]marked[/i] box, where the last ball has been placed at the last move (and where the next [i]move[/i] starts). Note that this box is never empty. Because $ S$ is finite, the number of [i]objects[/i] is finite as well. Note that each [i]object[/i] generates a single new [i]object[/i], and conversely, each object $ X$ can be obtained from a unique [i]inverse object[/i] $ X'$.

The moves determine a sequence of objects: $ X_1\rightarrow X_2\rightarrow\ldots$. Because the number of objects is finite, there is a minimal $ k$, such that $ X_k\in\{X_1,X_2,\ldots,X_{k-1}\}$, that is, the object $ X_k$ has been obtained before. Suppose $ X_1\neq X_k$. Then $ X_k=X_i$ for some $ 1<i<k$. Then $ X_{i-1}=X_k'$ (from the uniqueness of the inverse object), so $ X_{i-1}=X_{k-1}$, contradicting the minimality of $ k$. So $ X_k=X_1$ and we're done.

[b]b)[/b]
Define now an [i]object[/i] as a sequence of $ n$ numbers, denoting the number of balls in the respective boxes. Consider an object $ X$ with $ k$ non-empty boxes, and $ n-k$ empty.
[b]Lemma 1[/b] Pick a box $ Z$. We show that in a sequence of moves we can obtain an object $ F$ with all balls gathered in $ Z$. Indeed: start each move with the closest non-empty box $ B$ to $ Z$ in counterclockwise direction. After this move, either the number of balls in $ Z$ increases, or the closest nonempty box to $ Z$ in counterclockwise direction gets even closer to $ Z$.

[b]Lemma 2[/b] We prove that if we can obtain with some sequence of moves the object $ B$ from object $ A$ then we can obtain $ A$ from $ B$. Firstly we show that if object $ X$, after a single move, creates object $ Y$, then there is a sequence of moves, so that from $ Y$ we can get $ X$. For this, apply [b]a)[/b] to the initial situation $ X$ creating at the first move $ Y$. But at [b]a)[/b] we have shown that after some moves, we obtain again $ X$.
Now suppose we have obtained object $ B$ from $ A$ like this: $ A\rightarrow X_1\rightarrow\ldots\rightarrow X_s\rightarrow B$. According to the above we can obtain from $ B$ object $ X_t$, from $ X_t$ object $ X_{t-1},\ldots$ from $ X_2$ - $ X_1$ and finally, $ A$.

Now take any two objects $ X$ and $ Y$. From both we can get $ F$. According to the above Lemma 2 we can get from $ F$ object $ Y$, therefore the following transformations are possible: $ X\rightarrow F\rightarrow Y$
[/hide]
___
A problem using a somewhat similar idea is [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=155696]ISL 2006 C4[/url]


That's clearly not true -- either you've left something out, or it's just wrong.  Let a = b = 4 to provide a counterexample.

Try [hide]AM-GM, implemented somewhat cleverly.[/hide]

Scrambled wrote:Is there a way without calculus to prove that 3ab :le: a :^3: +2b :sqrt: b?

If Joel's hint not enough :

[hide]

Write ab as :  a * b^(2/3) * b^(1/3) and  ... use Joel's hint ![/hide]

[hide]We have from AM-GM that (a+b+c)/3 >= (abc)^(1/3)

So 2*bsqrt(b) = b^(3/2) + b(3/2)

Now apply AM-GM to get that RHS >= 3*(a^3*b^3)^(1/3) = 3ab, which is the desired result.[/hide]

[quote="Scrambled"]Is there a way without calculus to prove that 3ab :le: a :^3: +2b :sqrt: b?[/quote]

I assume both a and b are non-negative.

Here is a proof [b]without using AM-GM[/b]:

It's trivial if b=0. So assume b>0, let x = a/(:sqrt: b), then the inequality is equivalent to

x :^3:-3x+2 :ge: 0 ------------------ (*)

The LHS of (*) can factored into  (x-1) :^2: (x+2) which  :ge: 0.


What is the value of [i]n[/i] when (3!)(5!)(7!)=[i]n[/i]!

* I figured out the answer by just doing them..

But I think there's a faster way.

If you guys know it, I would appreciate if you kindly post it.

Thanks.

[hide]Answer: 10[/hide]

[hide="this works"]u know n has to be greater than 7. now using the 3! and 5! given, try to see if u can combine the products to make an extra factors. e.g. 8, 9, 10. for example, u have $ 7!\cdot(3\cdot 2\cdot\ 5\cdot 4\cdot 3\cdot 2)$ combine these to get other factors that would lead to greater factorials. [/hide]

[quote="DonkeyKong"]So what you are saying is that since we know that it has to be greater than 7,

you then know the range of an answer.

Then you simplify the equation.

Right??  :maybe:[/quote]
[hide]Right.  First of all, you know the number has to be at least 7.  So, you split up the other factors as follows.
$ 3!5!7!\equal{}7!\cdot 1\cdot 2\cdot 3\cdot 1\cdot 2\cdot 3\cdot 4\cdot 5$
Now, we first try to make an $ 8$ out of some of the factors, so we can get $ 8!$.  We can pull out a $ 4$ and a $ 2$:
$ \equal{}8!\cdot 2\cdot 3\cdot 3\cdot 5$
Next, we should find a $ 9$ by pulling out a $ 3\cdot 3$:
$ \equal{}9!\cdot 2\cdot 5$
And, we can pull out the $ 2\cdot 5\equal{}10$ to get $ 10!$[/hide]


[quote="FOURRIER"]some one post a solution  :|  :([/quote]

You have not to wait so long, FURRIER, Mathlinks is a village of genius inequality-solvers. They will give you the solution soon.  :)

I really wait for a nice solution.

Does anybody find out a solution now? The solution is actually just a simple application of Schur for 3 numbers.

I am waiting!

Can you tell me the minimum of above expression?  $ \ge$ not $ >$ (clearly value)

Great inequality hungtkhn . it reminds me the also great ineq of vasc
if a,b,c non negative real numbers and
$ (a\plus{}b\minus{}c)(\frac{1}{a}\plus{}\frac{1}{b}\minus{}\frac{1}{c})\equal{}4$  then prove that

$ (a^4\plus{}b^4\plus{}c^4)(\frac{1}{a^4}\plus{}\frac{1}{b^4}\plus{}\frac{1}{c^4})\geq 2304$

Now my solution
Put $ a\plus{}b\plus{}c\equal{}s$ , $ ab\plus{}bc\plus{}ca\equal{}q$ , $ abc\equal{}p$

Solving the given equality for $ p^2$ and subtituting in the one we want to prove we have to show that

$ \frac{36q^2(s^2\minus{}2q)}{s^2(q^2\minus{}2ps)}>34$  which is equivalent to

$ q^2s^2\plus{}34ps^3>36q^3$

But from the given one we have that $ q^2s^2\equal{}36p^2\plus{}2ps^3$ so we have to prove that
$ p^2\plus{}ps^3>q^3$  which is equivalent to

$ abc(a^3\plus{}b^3\plus{}c^3)\plus{}a^2b^2c^2\geq a^3b^3\plus{}b^3c^3\plus{}c^3a^3$  which is clearly true

Hope it is right...

It was very legal, Siluan, but It is actually incomplete. If it is true, it was so amazing since your solution is very nice and simple.

P.S. Siluan. You are totally right when referencing VASC's problem, since it is actually my inspiration for my problem now.

(but of course, they have nothing in common, both the title and the solution)

$ (a^2 \plus{} b^2 \plus{} c^2)\left(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\right)^2 \minus{} 34$

$ \equal{} (a^2 \plus{} b^2 \plus{} c^2)\left(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\right)^2 \minus{} \left\{(a \plus{} b \plus{} c)^2\left(\frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2}\right) \minus{} 2\right\}$

$ \equal{} (a^2 \plus{} b^2 \plus{} c^2)\left(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\right)^2 \minus{} \{a^2 \plus{} b^2 \plus{} c^2 \plus{} 2(ab \plus{} bc \plus{} ca)\}\left\{\left(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\right)^2 \minus{} 2\left(\frac {1}{ab} \plus{} \frac {1}{bc} \plus{} \frac {1}{ca}\right)\right\}$

$ \plus{} 2$

$ \equal{} 2(a^2 \plus{} b^2 \plus{} c^2)\left(\frac {1}{ab} \plus{} \frac {1}{bc} \plus{} \frac {1}{ca}\right) \minus{} 2(ab \plus{} bc \plus{} ca)\left(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\right)^2$

$ \plus{} 4(ab \plus{} bc \plus{} ca)\left(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\right) \plus{} 2$

$ \equal{} 2\{(a \plus{} b \plus{} c)^2 \minus{} 2(ab \plus{} bc \plus{} ca)\}\left(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\right) \minus{} 2(ab \plus{} bc \plus{} ca)\left(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\right)^2$

$ \plus{} 4(ab \plus{} bc \plus{} ca)\left(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\right) \plus{} 2$

$ \equal{} 2(a \plus{} b \plus{} c)^2\left(\frac {1}{ab} \plus{} \frac {1}{bc} \plus{} \frac {1}{ca}\right) \minus{} 2(ab \plus{} bc \plus{} ca)\left(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\right)^2 \plus{} 2$

$ \equal{} \frac {2(a \plus{} b \plus{} c)^3}{abc} \minus{} \frac {2(ab \plus{} bc \plus{} ca)^3}{(abc)^2} \plus{} 2$

$ \equal{} \frac {2}{(abc)^2}\{abc(a \plus{} b \plus{} c)^3 \minus{} (ab \plus{} bc \plus{} ca)^3 \plus{} (abc)^2\}$,

We are to prove $ abc(a \plus{} b \plus{} c)^3 \minus{} (ab \plus{} bc \plus{} ca)^3 \plus{} (abc)^2 > 0.$

Nice transformation, too , kunny (the final result of you and siluan are the same). However, just notice that following inequality is wrong

\[ (a\plus{}b\plus{}c)^2\left(\frac{1}{a^2}\plus{}\frac{1}{b^2}\plus{}\frac{1}{c^2}\right) \le (a^2\plus{}b^2\plus{}c^2)\left(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\right)^2\plus{}2.\]

[quote="FOURRIER"][quote="silouan"]
$ abc(a^3 \plus{} b^3 \plus{} c^3) \plus{} a^2b^2c^2\geq a^3b^3 \plus{} b^3c^3 \plus{} c^3a^3$  which is clearly true
[/quote]

How? give me a hint please  :yankchain:[/quote]

No, Fourrier. It is not clear, even it is very hard.

[quote="FOURRIER"][quote="hungkhtn"]
No, Fourrier. It is not clear, even it is very hard.[/quote]

Ok, so how do you prove it ?     :D
I am still waiting... :what?:

Thanks  :wink:[/quote]

It is better if we wait for Siluan or Kunny finish their solution first. Their idea seems to be very nice,

[quote="hungkhtn"]Let $ a,b,c$ be positive real numbers such that
\[ (a \plus{} b \plus{} c)^2\left(\frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2}\right) \equal{} 36.
\]
Prove that
\[ (a^2 \plus{} b^2 \plus{} c^2)\left(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\right)^2 > 34.
\]
[/quote]

Let $ a \plus{} b \plus{} c \equal{} x,\ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \equal{} y\Longrightarrow ab \plus{} bc \plus{} ca \equal{} yabc$.

$ \therefore \frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2} \equal{} \left(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\right)^2 \minus{} \frac {2(a \plus{} b \plus{} c)}{abc} \equal{} y^2 \minus{} \frac {2x}{abc}$.

Thus $ (a \plus{} b \plus{} c)^2\left(\frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2}\right) \equal{} 36\Longleftrightarrow x^2\left(y^2 \minus{} \frac {2x}{abc}\right) \equal{} 36$,yielding $ abc \equal{} \frac {2x^3}{x^2y^2 \minus{} 36}\ \because xy\geq 9$.

$ \therefore (a^2 \plus{} b^2 \plus{} c^2)\left(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\right)^2 \minus{} 34 \equal{} (x^2 \minus{} 2yabc)y^2 \minus{} 34$

$ \equal{} \left(x^2 \minus{} 2y\cdot \frac {2x^3}{x^2y^2 \minus{} 36}\right)y^2 \minus{} 34$

$ \equal{} x^2y^2 \minus{} \frac {4x^3y^3}{x^2y^2 \minus{} 36} \minus{} 34$

$ \equal{} \frac {t^4 \minus{} 4t^3 \minus{} 70t^2 \plus{} 34\cdot 36}{t^2 \minus{} 36}$, where $ t \equal{} xy\Longrightarrow t\geq 9$.

Let $ f(t) \equal{} t^4 \minus{} 4t^3\minus{}70t^2 \plus{} 34\cdot 36 \equal{} t^2\{(t \minus{} 2)^2 \minus{} 74\} \plus{} 34\cdot 36$, fot $ t\geq 9$ we have

$ f'(t) \equal{} 4t^3 \minus{} 12t^2 \minus{} 140t \equal{} 4t\{t(t \minus{} 3) \minus{} 35\} > 4\cdot 9(9\cdot 6 \minus{} 35) > 0$

$ \therefore f(t)\geq f(9) \equal{} 9^2\{(9 \minus{} 2)^2 \minus{} 74\} \plus{} 34\cdot 36$

$ \equal{} 9(25\cdot 9 \minus{} 34\cdot 4) > 0$, which completes the proof.

[quote="kunny"][quote="hungkhtn"]Let $ a,b,c$ be positive real numbers such that
\[ (a \plus{} b \plus{} c)^2\left(\frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2}\right) \equal{} 36.
\]
Prove that
\[ (a^2 \plus{} b^2 \plus{} c^2)\left(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\right)^2 > 34.
\]
[/quote]

Let $ a \plus{} b \plus{} c \equal{} x,\ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \equal{} y\Longrightarrow ab \plus{} bc \plus{} ca \equal{} yabc$.

$ \therefore \frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2} \equal{} \left(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\right)^2 \minus{} \frac {2(a \plus{} b \plus{} c)}{abc} \equal{} y^2 \minus{} \frac {2x}{abc}$.

Thus $ (a \plus{} b \plus{} c)^2\left(\frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2}\right) \equal{} 36\Longleftrightarrow x^2\left(y^2 \minus{} \frac {2x}{abc}\right) \equal{} 36$,yielding $ abc \equal{} \frac {2x^3}{x^2y^2 \minus{} 36}\ \because xy\geq 9$.

$ \therefore (a^2 \plus{} b^2 \plus{} c^2)\left(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\right)^2 \minus{} 34 \equal{} (x^2 \minus{} 2yabc)y^2 \minus{} 34$

$ \equal{} \left(x^2 \minus{} 2y\cdot \frac {2x^3}{x^2y^2 \minus{} 36}\right)y^2 \minus{} 34$

$ \equal{} x^2y^2 \minus{} \frac {4x^3y^3}{x^2y^2 \minus{} 36} \minus{} 34$

$ \equal{} \frac {t^4 \minus{} 4t^3 \minus{} 70t^2 \plus{} 34\cdot 36}{t^2 \minus{} 36}$, where $ t \equal{} xy\Longrightarrow t\geq 9$.

Let $ f(t) \equal{} t^4 \minus{} 4t^3 \minus{} 70t^2 \plus{} 34\cdot 36 \equal{} t^2\{(t \minus{} 2)^2 \minus{} 74\} \plus{} 34\cdot 36$, fot $ t\geq 9$ we have

$ f'(t) \equal{} 4t^3 \minus{} 12t^2 \minus{} 140t \equal{} 4t\{t(t \minus{} 3) \minus{} 35\} > 4\cdot 9(9\cdot 6 \minus{} 35) > 0$

$ \therefore f(t)\geq f(9) \equal{} 9^2\{(9 \minus{} 2)^2 \minus{} 74\} \plus{} 34\cdot 36$

$ \equal{} 9(25\cdot 9 \minus{} 34\cdot 4) > 0$, which completes the proof.[/quote]

By the way, we can prove similarly that

Let a,b,c be positive real numbers such that
\[ (a \plus{} b \plus{} c)^2\left(\frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2}\right) \equal{} 36.\]
Prove that
\[ (a^2 \plus{} b^2 \plus{} c^2)\left(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\right)^2 > 36.\]

?

[quote="kunny"][quote="hungkhtn"][quote="kunny"]I will tackle your second problem.
.[/quote]

It is not what I want, kunny, since the problem I've just posted is obviously wrong.[/quote]

 :blush:[/quote]

Don't be so seriously. Think again, kunny. There is a mistake somewhere, and believe that your solution is approaching the right.

I'm confusing now, but Is my solution wrong :?:

[quote="silouan"][quote="silouan"]$ abc(a^3 \plus{} b^3 \plus{} c^3) \plus{} a^2b^2c^2\geq a^3b^3 \plus{} b^3c^3 \plus{} c^3a^3$  which is clearly true
[/quote]

In fact it is not clear . :oops:

But here is a try of a  solution . Due to the homogenity assume that $ abc \equal{} 1$
Then it suffices to prove that $ a^3 \plus{} b^3 \plus{} c^3 \plus{} 1\geq\frac {1}{a^3} \plus{} \frac {1}{b^3} \plus{} \frac {1}{c^3}$

So the problem becomes
If $ a,b,c$ positive reals such that $ abc \equal{} 1$ then prove that $ a \plus{} b \plus{} c \plus{} 1\geq\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}$

Put $ a \equal{} \frac {x}{y}$ etc

Then we have to show that $ \frac {x}{y} \plus{} \frac {y}{z} \plus{} \frac {z}{x} \plus{} 1\geq\frac {y}{x} \plus{} \frac {z}{y} \plus{} \frac {x}{z}$

or equivalently $ 1\geq \frac {(x \minus{} y)(x \minus{} z)(y \minus{} z)}{xyz}$  or $ xyz\geq (x \minus{} y)(x \minus{} z)(y \minus{} z)$[/quote]

I hope someone can figure out his mistake  :(  Seem to be obvious but it is wrong.

[quote="silouan"]$ xyz\geq (x \minus{} y)(x \minus{} z)(y \minus{} z)$[/quote]

Wrong one.Try $ x\equal{}5,y\equal{}2,z\equal{}1$
$ LHS\equal{}10,RHS\equal{}12$. :wink:

[quote="FOURRIER"][quote="stephen38"][quote="silouan"]$ xyz\geq (x \minus{} y)(x \minus{} z)(y \minus{} z)$[/quote]

Wrong one.Try $ x \equal{} 5,y \equal{} 2,z \equal{} 1$
$ LHS \equal{} 10,RHS \equal{} 12$. :wink:[/quote]

Then the Original Inequalitie is [b][color=red]WRONG[/color][/b]  :mad:

Let $ a,b,c$ be positive real numbers such that
\[ (a \plus{} b \plus{} c)^2\left(\frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2}\right) \equal{} 36.
\]
Prove that
\[ (a^2 \plus{} b^2 \plus{} c^2)\left(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\right)^2 > 34.
\]
:roll:[/quote]

Actually, the last inequality of Siluan and Kunny hold if $ a,b,c$ such that

\[ (a \plus{} b \plus{} c)^2\left(\frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2}\right) \equal{} 36.
\]
but not trufor all positive number (there exits a difference here).

SOme days pasted and no solution is found. Here is my hint:

Consider the following expressions, build their relationships first then using Schur inequality lastly.

\[ x\equal{}\frac{a}{b}\plus{} \frac{b}{c}\plus{}\frac{c}{a}.\]

\[ y\equal{}\frac{a}{b}\plus{} \frac{b}{c}\plus{}\frac{c}{a}.\]

\[ m\equal{}\frac{a^2}{bc}\plus{}\frac{b^2}{ca}\plus{}\frac{c^2}{ab}.\]

\[ n\equal{}\frac{bc}{a^2}\plus{}\frac{ca}{b^2}\plus{}\frac{ab}{c^2}.\]

We have, for examples, the following relationship

\[ xy\equal{}3\plus{}m\plus{}n, x^3\plus{}y^3\equal{}mn\plus{}6(m\plus{}n)\plus{}12.\]

Sorry for reviving something almost 9 years old, but could someone post a solution? Thanks!

[quote=hungkhtn]Let $ a,b,c$ be positive real numbers such that

\[ (a\plus{}b\plus{}c)^2\left(\frac{1}{a^2}\plus{}\frac{1}{b^2}\plus{}\frac{1}{c^2}\right)\equal{}36.\]

Prove that

\[ (a^2\plus{}b^2\plus{}c^2)\left(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\right)^2 >34.\][/quote]
Let $a+b+c=3u$, $ab+ac+bc=3v^2$, $abc=w^3$ and $u^2=tv^2$.
Hence, the condition gives $u^2(9v^4-6uw^3)=4w^6$ or $w^3=\frac{3u}{4}\left(\sqrt{u^4+4v^4}-u^2\right)$.
$(a-b)^2(b-c)^2(c-a)^2=27(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)\geq0$, which gives $1.037...\leq t\leq\frac{16}{15}$.
We need to prove that $(9u^2-6v^2)9v^4>34\left(\frac{3u}{4}\left(\sqrt{u^4+4v^4}-u^2\right)\right)^2$
or $9t-6>\frac{17}{8}t(\sqrt{t^2+4}-t)^2$ or $t>1.028...$, which is obvious. Done!
We can prove that$ (a^2+b^2+c^2)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2 >34.7$ is also true.

Nice job arqady!

Is there a solution using Schur's, like in hungkhtn's hint? Thanks!


geometry
geometric transformation
reflection
analytic geometry
graphing lines
slope

ok. this just suddenly popped into my head. this is "the Poem game". basically, one person starts by making a line in a poem, then every poster adds one line to the made-up poem. follow any ryme sceme u want. i dont care what its about. just TRY to make it nice. (i know its a really lame game but oh well. dont post if u dont want to)

i'll start:

[i]I heard the Wind,[/i][/i]

i hear the wind,
[b]dancing through the trees,[/b]

i hear the wind,
dancing through the trees,
a calm, soothing scene.

I heard the wind,
dancing through the trees,
a calm, soothing scene.
[b]I glanced at the sunset so serene.[/b]

I heard the wind,
dancing through the trees,
a calm, soothing scene.
I glanced at the sunset so serene.

[b]I felt the rain,[/b]

I heard the wind,
dancing through the trees,
a calm, soothing scene.
I glanced at the sunset so serene.

I felt the rain,
[b]beating like a drum,[/b]

This is the worst rhyme scheme ever.

Rhymes don't really matter; they're there to evoke a past line.

Lines 2-3 actually match really well.

Line 4 doesn't really work as it's just too long; the rhyme comes too late, and it's not really related to the rest of the stanza. if you cut out the "I glanced at" and started a new stanza it would work better.

Substituting "rain" for "storm" might work better as drum has a hard abrupt ending which is characterized by "storm" more than "rain."
I like the change from the flowing of the first stanza to the abruptness of the second stanza.

OK, since it seems that this topic won't be getting anywhere if we don't follow ignite's advice....

I heard the wind
dancing through the trees,
a calm, soothing scene.

[b]I saw the hills
fading in the distance[/b]
and the sunset so serene.

I felt the storm,
beating like a drum,

I heard the wind
dancing through the trees,
a calm, soothing scene.

I saw the hills
fading in the distance
and the sunset so serene.

I felt the storm,
beating like a drum,
pounding on the earth with more than a hum.

i know, its REALLY lame. somebody change it if they want.

I heard the wind
dancing through the trees,
a calm, soothing scene.

I saw the hills
fading in the distance
and the sunset so serene.

I felt the storm,
beating like a drum,
pounding on the earth, with more than a hum.

in poems, each stanza should have an equal amount of lines

btw, this is how I would edit this poem to keep as much of it as possible.

I heard the wind
dancin' through the trees,
I heard a calm
a soothing scene.

I saw the hills
fadin' far away
the sunset gleaming
like a light show play.

I felt the storm
beatin' like a drum
pounding on the earth
with more than a hum.

so continued?

And then I noticed

(I think we should finish it on this stanza)

[quote="Ihatepie"]in poems, each stanza should have an equal amount of lines with a difference of the last stanza, which should have a different amount.[/quote]

......what?

dam bro,  :mad:  edits it when I go to the b-room. Ok I'll edit it.

well here is how it is

I heard the wind
dancin' through the trees,
I heard a calm
a soothing scene.

I saw the hills
fadin' far away
the sunset gleaming
like a light show play.

I felt the storm
beatin' like a drum
pounding on the earth
with more than a hum.

so continued?

And then I noticed

(make this second word rhymeable)

with all the cutoffs like "dancin'' and "fadin" , it kinda ruins the peaceful, serious, serene air that i had intended for it. now it sounds all carefree and sway-y. but o well. it's good. sadly, my terrible poet skills prevent me from adding a new line.  :(

theres no rule about the number of lines a stanza should have...
in fact breaking the pattern draws special attention

I heard the wind
dancin' through the trees,
I heard a calm
a soothing scene.

I saw the hills
fadin' far away
the sunset gleaming
like a light show play.

I felt the storm
beatin' like a drum
pounding on the earth
with more than a hum.

And then I noticed
That nature was

i know, it doesnt ryme. but oh well.  :mad:

A very lonely number is infinity,
That reveals its name without divinity,
and yet the other numbers do so yearn,
to be like her, or just to from her learn.

A number neither even nor odd
Neither rational or irrational
Which yields a shimmering facade
Over values international.

Her towers rise forever yonder
That all who gaze above might ponder,
What is such a number as infinity,
who reveals her hame without divinity?

And then in the skies above
And the things or lack thereof,
infinity penetrates through and beyond,
as if from some wrongdoing to abscond...

...she finds another solitary creature,
infinitesimal be its only feature,
A number that equals itself negated,
Negative infinity has long been hated.

As now we see with our own eyes,
The infinities are beyond the skies,
In this world you can never hide,
Even if you add, subtract, or divide.
Infinity is to heaven a door,
nothing less, and nothing more.

And now we reflect on our happy days,
those long, forgotten ways,

----------------------
the longest poem it is, then.

Ok, that'll make for an interesting ending. Let's watch Yongyi's reaction  :lol: .

Next poem:

Plsplsplsplzplzplz use same meter and rhyme scheme as http://thinks.com/words/nonsense/william.htm

[b]"You are young, my dear son," the old man said,
and your dreams still have yet to come true;[/b]

Okay here is the first few lines of another poem.

  There is a small little number that is often forgotten,
  Don't divide by this number, because no value is gotten
  I am going to inform you that this number is zero.
  This number is no hero.
  It is the slope of a line with no rise,
  When using this number you must be wise.

  Okay, that is the first stanza. I don't know if you guys can continue on from there.

Okay here is the first few lines of another poem.

There is a small little number that is often forgotten,
Don't divide by this number, because no value is gotten
I am going to inform you that this number is zero.
This number is no hero.
It is the slope of a line with no rise,
When using this number you must be wise.

And now we shall travel to the other end of the line,
This number is one that is not so divine,
This number's value shall also never be divided by,
For if you doth so, all your homework shall die.

A very lonely number is infinity,
That reveals its name without divinity,
and yet the other numbers do so yearn,
to be like her, or just to from her learn.

A number neither even nor odd
Neither rational or irrational
Which yields a shimmering facade
Over values international.

Her towers rise forever yonder
That all who gaze above might ponder,
What is such a number as infinity,
who reveals her hame without divinity?

And then in the skies above
And the things or lack thereof,
infinity penetrates through and beyond,
as if from some wrongdoing to abscond...

...she finds another solitary creature,
infinitesimal be its only feature,
A number that equals itself negated,
Negative infinity has long been hated.

As now we see with our own eyes,
The infinities are beyond the skies,
In this world you can never hide,
Even if you add, subtract, or divide.
Infinity is to heaven a door,
nothing less, and nothing more.

And now we reflect on our happy days,
those long, forgotten ways,

There is a small little number that is often forgotten,
Don't divide by this number, because no value is gotten
I am going to inform you that this number is zero.
This number is no hero.
It is the slope of a line with no rise,
When using this number you must be wise.

And now we shall travel to the other end of the line,
This number is one that is not so divine,
This number's value shall also never be divided by,
For if you doth so, all your homework shall die.

A very lonely number is infinity,
That reveals its name without divinity,
and yet the other numbers do so yearn,
to be like her, or just to from her learn.

A number neither even nor odd
Neither rational or irrational
Which yields a shimmering facade
Over values international.

Her towers rise forever yonder
That all who gaze above might ponder,
What is such a number as infinity,
who reveals her hame without divinity?

And then in the skies above
And the things or lack thereof,
infinity penetrates through and beyond,
as if from some wrongdoing to abscond...

...she finds another solitary creature,
infinitesimal be its only feature,
A number that equals itself negated,
Negative infinity has long been hated.

As now we see with our own eyes,
The infinities are beyond the skies,
In this world you can never hide,
Even if you add, subtract, or divide.
Infinity is to heaven a door,
nothing less, and nothing more.

And now we reflect on our happy days,
those long, forgotten ways,
[b]This poem is boring - let's have better fun -
by changing the topic to a new one:

"You are young, my dear son," the old man said,
and your dreams still have yet to come true;[/b]

HEY! Did you see what Yongyi781 and Wickedestjr did? Everything is so messed up and it's all being blamed on me  :(  *sob*

[hide]
 :D
[/hide]

what did they do? i like the new infinity/zero poem! the two fit together well! if u think its boring then wait till its DONE, leoxnlin!  :lol:  :P

[quote="dreambold22"]new rule: maximum 2 lines per post. this is getting out of hand. seriously. and dont start a new poem when the current one isnt done yet. so now we are continuing this new "infinity/zero" poem. (i'm actually TRYING to make some good poems here, *cough* *cough* leoxnlin)

There is a small little number that is often forgotten,
Don't divide by this number, because no value is gotten
I am going to inform you that this number is zero.
This number is no hero.
It is the slope of a line with no rise,
When using this number you must be wise.

And now we shall travel to the other end of the line,
This number is one that is not so divine,
This number's value shall also never be divided by,
For if you doth so, all your homework shall die.

A very lonely number is infinity,
That reveals its name without divinity,
and yet the other numbers do so yearn,
to be like her, or just to from her learn.

A number neither even nor odd
Neither rational or irrational
Which yields a shimmering facade
Over values international.

Her towers rise forever yonder
That all who gaze above might ponder,
What is such a number as infinity,
who reveals her hame without divinity?

And then in the skies above
And the things or lack thereof,
infinity penetrates through and beyond,
as if from some wrongdoing to abscond...

...she finds another solitary creature,
infinitesimal be its only feature,
A number that equals itself negated,
Negative infinity has long been hated.

As now we see with our own eyes,
The infinities are beyond the skies,
In this world you can never hide,
Even if you add, subtract, or divide.
Infinity is to heaven a door,
nothing less, and nothing more.

And now we reflect on our happy days,
those long, forgotten ways,
to the times of infinite glad,
and negative infinate sad.[/quote]

Well I think we can agree this is done.

New poem: about fractions
Fractions are a lonely bunch
never complete, never whole
Part of pie, a piece to munch,
always finite, like a foal. (finite would pronounced fineet)

[quote]
Fractions are a lonely bunch
never complete, never whole
Part of pie, a piece to munch,
always finite, like a foal. (finite would pronounced fineet)
[/quote]

I think we can agree that this is done  :D . Ok fine. Can I add a whole stanza?

Fractions are a lonely bunch
never complete, never whole
Part of pie, a piece to munch,
always finite, like a foal. (finite would pronounced fineet)

Fractions dwell simply everywhere
Between two, always another
and yet they are in fact quite rare
it's very true - the space is ether.

ok fine. new rule: maximum 1 stanza per post. unless u'r finishing someone else's.

Fractions are a lonely bunch
never complete, never whole
Part of pie, a piece to munch,
always finite, like a foal. (finite would pronounced fineet)

Fractions dwell simply everywhere
Between two, always another
and yet they are in fact quite rare
it's very true - the space is ether.

fractions aren't irrational,
they're quite real, never fake,
but neither are they natural,
with parts and wholes to make.

You messed up the rhyme scheme  :(

That's ok though.

We should start a new one. I can't think about a topic though, so feel free.

[quote]3. Don't revive old threads. If you want to continue a pre-existing game, just make a new thread, saying it's a revival in the subject description, as well as a link of the original thread and credit to the original poster in the post[/quote]

Nuff said.

But would about the newer users that have never seen it before?


So. Discuss any interesting sprint round problems or the homework here. If you want to talk about a sprint problem, please post the entire problem because I don't know which test you did and I probably don't have the test anyway. Week 2 answers will be posted eventually--most likely not today because I've been procrastinating on USAMTS and IT'S DUE TONIGHT!!! AAAHHHH!!! Yeah. So who else is doing USAMTS?

WHOOOOOOOOOOOOOOOOOOOOOOOOOOOOOAAAAAAAAAA
THERE
you just spammed!  :o

COOL YOU DIDNT DO USAMTS EITHER  :rotfl:  :rotfl:
me neither

[quote="bookaholic"]Whoa, that was surprisingly quick. I just finished off my last USAMTS problem. Some of my answers are making me uncomfortable, though, so I'll have to go back and check.

I'm getting a really, really bad feeling about the difficulty level of round 2...[/quote]
OMG! YOU ACTUALLY DID IT?
WHOA I SHOULDNT PROCRASTINATE ANYMORE
ps hows w00t

scores for sprint round since u guys wanna kno:
bluestalk: 27/30
blondazn:27
they were the best scores
me-10 haha im so lame

Do you know what year/level it was?

I'm sort of behind on WOOT--still haven't gotten around to doing the Building Bridges B problems. There's no class this Thursday, right?

did you look at the p set yet? at first glance, they look kinda hard. Didn't attempt any yet

AHH... I havn't really started any of the problems before today, so I'm hoping I can finish (mostly) everything in this "cram session" (basicly I'm doing usamts because I need a good reason to procrastinate math homework). :|

i think sylvia and evelyn (dont think i spelled that rite) got 10 also....but im not sure u kno em
and i think it was from 1991...old one anyway
and sum other guy i dont kno got 23...so yea

Yup, it was the chapter sprint for 1991. The top scores were 27, 27, and 24...so that would be me, Bluestalk, and Andrew, (this guy in my math class.) Other than that, I don't know...

cool. chapterrrrr......... thats pretty good though 27, thats 90%  :)

27 actually is really good--now if you guys can do the same thing in the real competition and get a perfect on the Target Round, which isn't hard if you don't make stupid mistakes, you'll be almost certainly in the top ten and quite possibly top five, which means we would probably make state. See? Told you it could be done.

WRONG I got like a 26 or something. I'm not counting right. I know thought, that benign_blondeazn is waaaaay better than me though. DON'T KILL ME.

Plus I'm stressing over my poor little Analysis grade.....BLAH

lame. my trig grade is like lame. at least i don't drop the class -_-

[quote="bookaholic"]Don't be so negative...what are you people doing in analysis now? I'm beginning to wonder if Paly doesn't put some Gunn analysis material in trig and Gunn trig material in analysis. I mean we're STILL REVIEWING CONICS. It's so so boring.[/quote]
we're REVIEWING FUNCTIONS

SO IM LEARNING FUNCTIONS AND THE REMAINDER THEOREM. AND DESCARTES RULE OF SIGNS. ITS LAME. ok i had to rant. math is getting lame
plus last year math was lame too.
slack was like good, and she gave us free poptarts ($\$$ $\$$). other than that it was lame

[quote="benign_blondazn"]I think she said it had to do with money(she said she spent too much money on food last year or something)[/quote]
ROFLROFL
DAVID THATS PROBABLY YOUR FAULT
lol you used to take 3 poptarts everyday and sell it to kids at school

[quote="benign_blondazn"]How many degrees did you raise for your parabolic cooker from last year? I just tested mine...2 hours ago... and roasted a marshmellow.[/quote]
adljgasldkgjasdlkgjadsklg I failed that project. probably b/c i was with ccy(lol jk).

[quote="benign_blondazn"]Mrs. Slack gave you POPTARTS?? I don't think she's brought in much for us so far...we haven't had any food in the morning for a looooooooong time. I think she said it had to do with money(she said she spent too much money on food last year or something). Well.

How many degrees did you raise for your parabolic cooker from last year? I just tested mine...2 hours ago... and roasted a marshmellow. :P[/quote]

I don't know how we got a passing grade on ours last year.  I think it was because our thermometer gave us inaccurate (to our advantage) readings.

[quote="benign_blondazn"]HAHAAAAA. Kimberly and I raised it to around 80 degrees Celsius... but the temperature kept on going down after that because of the wind...[/quote]
kimberly as in my sister?
OMG HOW IS SHE ACING THAT PROJECT ASDLGKASHDLGKASGDKLJASDKLGJASKL

[quote="junggi"][quote="benign_blondazn"]I think she said it had to do with money(she said she spent too much money on food last year or something)[/quote]
ROFLROFL
DAVID THATS PROBABLY YOUR FAULT
lol you used to take 3 poptarts everyday and sell it to kids at school[/quote]
YAY POPTARTS YAY POPTARTS :D  :D  :D  :D

what about the people who drank 3 chocolate milks (evan, neil)? and what about john xia and his poptart cookings. Still I paid like $\$$20 and made more like $\$$100. pretty good profits there... omer zach was a good customer. Boris and I just were all poptart smugglers. We smuggled out of jordan and sent them to jls, and sold them in the library as the black market. I don't think Slakc ever figured it out. And we also failed the parabolic cooker thing. giveme ur meauserments i might need them this year (we are doing it again in the spring)

-edit-
I use d to take 3 poptart PACKS, not 3 poptarts. 3 poptarts is impossible. 3 pax=6 poptarts=3 dollars=snacks, and more money


Let S denote the set of all 6-tuples $(a,b,c,d,e,f)$ of positive integers such that
$a^2 + b^2 + c^2 + d^2 + e^2 =f^2$
 Find the G.C.D of all the possible products $abcdef$. :)

$24$:
We have $1^2+1^2+1^2+2^2+3^2=4^2$, so it's at most $24$.
Now look $\mod 3$: if all numbers wouldn't be divisible by $3$, their squares would be $\equiv 1\mod 3$ causing the contradiction $5 \equiv 1\mod 3$.
Thus, the $\gcd$ is divisible by $3$.
Now look $\mod 8$ (the squares are all $\equiv 0,1,4 \mod 8$):
If $4|f$, then not all other numbers can be odd since that would give $odd=even$, so in this case $8|\gcd$.
If $2|f , 4 \nmid f$: then the RHS is $\equiv 4 \mod 8$ and either one on the RHS is divisible by $4$ or two of them are even again.
If $2 \nmid f$, the RHS is $\equiv 1 \mod 8$. Now checking shows that at four of the numbers on the LHS are even.
So in all cases we have $8|\gcd$ and the result follows.


For question #1, "What are your long-range goals?"

Should I interpret this as "What do you want to do for a career, where do you see yourself in 20 years?" or "How do you think RSI will enrich you, why do you want to go to RSI?" I'd hate to have to answer it in the former way, it's just like putting down a major on your college app, except there isn't an "undecided" box to choose.

I think they want the first way.  And it's probably okay to say, "I don't know."  Just, say it in several paragraphs.

Although, you might want to listen to answers from people who actually applied to RSI ...

I took it to mean what I would do after college. I think it's to see how serious you really are about science and your interests. RSI 03

follow up: for the teacher recommendations, if my primary research interest is math, should i get recs from two math teachers or one math, one other science?

I got both recs from math teachers. 1 was a professor who i did a research project with and the other was just a teacher I had in high school. If you're doing math you probably should get at least one from a math teacher, but both isn't neccessarily better. Just make sure whoever you get it from will write really good stuff.


function
integration
calculus
real analysis
real analysis unsolved

I'm wondering if we can write for a $ C$ function $ f$ : $ \int_F^x f(k)(x \minus{} k)dk \equal{} \int_F ^{\infty} 1_{[x \ge k]}f(k)(x \minus{} k)dk$ ? using that i get a contradiction while developing in the Taylor with rest integral formulas !

You certainly can write it this way if $ x>F$ but not if $ x<F$ when the first integral means $ \minus{}\int_x^F$ and the second is just $ 0$.

thank you fedja, I realized after a while that the transformation I've made ( which is very useful for what I do) led me to a contradiction, It seems stupid but I was tring to have $ 1_{[x\ge k]}f(k)(x \minus{} k) \equal{} f(k)(x \minus{} k)^{ \plus{} }$ for a decomposition on the basis $ (x\rightarrow(x \minus{} k)^{ \plus{} } , k)$ ! so can we do it in the general case without separate the two cases ?


topology
real analysis
real analysis unsolved

Hello!

Consider $ \mathcal D'(\mathbb R)$.

Does there exist a linear continuous operator $ A\colon \mathcal D'(\mathbb R) \to \mathcal D'(\mathbb R)$, which is not closed?

Please explain the notation and give the definition of "closed" in this context. Thanks.

Here [i]closed[/i] means that operator $ A$ maps closed sets to closed sets (in sense of $ \mathcal D'(\mathbb R)$ topology)


Let n and k be positive integers such that $ n\geq \frac {k(k\plus{}1)}{2}$ .Find the number of solutions $ (x_1,x_2,x_3....,x_k),x_1\geq 1,x_2\geq 2,...x_k\geq k$,for all integers satisfying the condition $ x_1\plus{}x_2\plus{}x_3\plus{}.....\plus{}x_k\equal{}n$

Assume $ n \equal{} \frac{k(k\plus{}1)}{2}$. If $ i$ is the biggest number where $ x_i \neq i$, then we have $ < \frac{i(i\minus{}1)}{2}$ pigeons for the smaller holes but we need to fill up $ \frac{i(i\minus{}1)}{2}$ holes. Thus there can be no such $ i$.

Then, for $ n > \frac{k(k\plus{}1)}{2}$ we have $ n \minus{} \frac{k(k\plus{}1)}{2}$ extra units to place in $ k$ possible bins. This gives us $ {n \minus{} \frac{k(k\minus{}1)}{2} \minus{} 1 \choose k \minus{} 1}$ possibilities.


Eh. *points to topic*

I'm not particularly excited about getting started on this.

I suppose it's because I don't really know that much about the test--but could anyone point me to any resources on the topic? Also, if someone could just give a general idea of what books to read, how to study, etc., that would be good. :lol:

Thanks,
~The_Scintillator

Hi Scintillator,

The first thing I would do is to go to the bookstore (or online) and buy a few SAT study books with 10-12 practice tests each and *solutions* (not just answers). They generally come under brand names like Princeton Review, Kaplan, and McGraw-Hill, etc. and are about 1.5 - 2in thick. They’re about \$20-30 each, but it’s well worth the investment and saves *a lot* of energy searching for resources. They are really good instruction and practice; some of them even have sample essays for each score 0-6 for each prompt! (And don’t worry about them being “official” or not; they are mostly the same.)

Next, needless to say, practice, practice, practice. Assuming solid math skills since you’re an AoPSer, do not waste energy practicing math (if you can do AMC problems, SAT math is no sweat). If fact, if you are trying to majorly improve skills, don’t practice it like a real SAT, i.e. do not do it section-by-section like CR-M-W-CR, etc. Focus on one aspect at a time; for example, practice only grammar for two weeks, then critical reading for two weeks, etc. Make sure to read the solutions on the problems you are not sure/got wrong and make sure you know *why* you got them wrong!

One other thing—if your vocab is solid, do some studying, but don’t kill yourself over memorization  :maybe: . Vocab is just a small portion of the CR sections, and it is much more productive and interesting to practice reading/grammar/essay skills than to spend countless hours memorizing vocab that you, more likely than not, won’t need.

After this, you should begin the section-by-section real SAT conditions practicing. I wouldn’t recommend doing too many of this kind, actually; just do enough sets before the test to get “in the flow.”

If you practice consistently, you’ll be surprised at how fast you improve. (I practiced as I described for two months and ended up improving by (an estimated) 350+ points on CR/W on the first test.) Don’t be afraid of boredom either, as you have hinted, because some of the critical reading passages are actually pretty interesting.

The bottom line is this: don’t let practicing interfere with your life  :lol: . IMO, the SAT is, at the end of the day, just a minor portion of a college app.

Hope this helps, and good luck :) !

Emton


the sides $ AC$ and $ BC$ of triangle $ ABC$ habe equal lengths.  let $ P$ be a point inside triangle $ ABC$ such that $ <PAB\equal{}<PBC$ and let $ M$ be the midpoint of $ AB$ prove that $ <APM\plus{}<BPC\equal{}180$

let CP meet the circumcircle $ \Gamma$ of APB again in P'. By angle BC is tangent to $ \Gamma$ and so also CA, then P'P is the P-symmedian on CP'B, then CPBP' is and harmonic quad, then also PP' is the P symmedian in PAB so $ \angle APM \equal{} \angle P'PB \equal{} 180 \minus{} \angle BPC$.

can you say what is symmedian point???is there any elementary solution???

Another solution. Apply an inversion with center C and radius AC. We get an inscribed quadrilateral P'BPA. M' is center of circumscribed circle of P'BPA. We see that 2<APP'=<AM'P', <P'PM=<180-PM'P' and <CPB=<APP'+AM'C => <APM+<CPB=<AM'P'+<AM'C+180-(<P'M'C) QED


superior algebra
superior algebra unsolved

Let $K$ be a field of characteristic $p$ and let $u,v$ be algebraically independent over $k$.Prove the following:

a). $k(t,u)$ has degree $p^{2}$ over $k(t^{p},u^{p})$.
b). There exist infinitely many extensions between $k(t,u)$ and $k(t^{p},u^{p})$.

a) $[k(t,u^{p}): k(t^{p},u^{p})] = p$ because $X^{p}-t^{p}\in k(t^{p},u^{p})[X]$ is irreducible because $t \not \in k(t^{p},u^{p})$ (if $a$ has no $p$th root, $X^{p}-a$  is irreducible in general, here even more evident because it defines an extension of degree $p$ prime), the same for $[k(t,u): k(t,u^{p})] = p$.

$k$ should be infinite, I think.

b) $k(u^{p},t^{p}, u+at)$ for $a \in k$ gives infinitely many different extensions of degree p: $X^{p}-(u^{p}+a^{p}t^{p}) \in k(u^{p},t^{p})[X]$ is obviously irreducible and if for $a \neq b \in k$ both $u+at,u+bt \in intermediate field$, then also $(a-b)t$ and $t$ in intermediate field, hence also $u$, hence equal $k(t,u)$, contradiction


Find the product of all the roots of $x^{2\log_2 x}=8x$.

[hide]Doing $2^k=x$:

$x^{2\log_2 x}=8x \implies 2^{2.k^2}=2^{k+3} \implies 2k^2=k+3$

By Viete's the sum of the roots of the equation $2k^2-k-3=0$ is $\frac{1}{2}$ in consequence the product of the root of the primitive equation is $2^{\frac{1}{2}}=\sqrt2$[/hide]

[hide="Solution"]The equation is the same as
\[ 2\log_2 x \log_2 x = \log_2 8x \]
\[ 2(\log_2 x)^2=\log_2 x + 3 \]
\[ 2(\log_2 x)^2-\log_2 x - 3=0 \]
\[ \log_2 x =\frac32,\text{ or} \log_2 x=-1 \]
Therefore, $x_1=2\sqrt 2,\ x_2=\frac12$.
Thus the product of the roots $=x_1x_2=\sqrt 2$[/hide]


Hi,

We can write the inverse of $ f(x)$ as $ f^{\minus{}1}(x).$ But why can't the inverse of  $ f(2x)$ be written as $ f^{\minus{}1}(2x)?$ I mean, can't we just substitute in a $ y$ for the $ 2x$ like so: $ 2x \equal{} y,$ thus the inverse of $ f(y)$ is $ f^{\minus{}1}(y),$ which is essentially the same thing as the inverse of $ f(x)$ being $ f^{\minus{}1}(x),$ isn't it?

Thank you for the help.  :)

I'm pretty sure you can, unless I'm missing something.

Ihatepie, that's what I thought at first, but the Solution Manual's answer makes more sense...but it is different. Here's how the SM did it:

Let $ g(x) \equal{} f(2x).$ Thus, if $ y \equal{} g^{\minus{}1}(x),$ then we have $ g(y) \equal{} x.$ Or, $ g(y) \equal{} f(2y).$ Thus, we have $ f(2y)\equal{}  x.$ So, since $ f$ is invertible, we have $ f^{\minus{}1}(f(2y)) \equal{} f^{\minus{}1}(x) \implies 2y \equal{} f^{\minus{}1}(x) \implies y \equal{} \boxed{\frac{1}{2}f^{\minus{}1}(x)}.$

The definition of $ f^{\minus{}1}(x)$ is the inverse of that function, so it should work either way.

Yeah, but at the end of the solution, the Solutions Manual specifically said: "... thus the inverse of $ f(2x)$ is $ \frac{1}{2}f^{\minus{}1}(x)$ [b]not[/b] $ f^{\minus{}1}(2x).$"

Oh, I forgot to realize that the inverse gives back what you inputted, which in this case is $ x,$ not $ 2x.$ Thank you very much Bictor :)

Avoid learning mathematics through notation, you'll learn nothing but literal adventures in La-La Land.

Pay attention to the single quotations:
What does $ f(2x)$ mean? It means that you 'multiply' $ x$ by 2 and then 'function' it to get a corresponding $ y$. You want to find the inverse function, right? Then, 'play backwards' your steps: 'Inverse function' $ y$ to get $ f^{\minus{}1}(y)$ and then 'inverse multiply' by 2 to get $ \frac12f^{\minus{}1}(y)$.

$ 2[g(x)^{3}] \equal{} g(2x^{3} \minus{} 5)$
by looking i can only get $ g(x) \equal{} x$

[quote="mik3"]I need some help on functions. I'm not sure when to tell if something is even or odd. Like this problem F(x)= g(x)- g(-x). I don't know if this is even or odd. I'm so confused @_@. I also need some help with problems that are f of g of x or stuff like that. If f(x)= 2x^3-5, then find g(x), such that f o g(x)= g o f(x). Thanks Mr. Green[/quote]A function is even when $ f(\minus{}x)\equal{}f(x)$ and odd when $ f(\minus{}x)\equal{}\minus{}f(x)$. Now, try $ F(\minus{}x)$. Is that equal to $ F(x)$, $ \minus{}F(x)$, or none?

For your next question, $ f(x)\equal{} 2x^3\minus{}5$ which is a bijective function. Then, if $ f\circ g\equal{}g\circ f$, this means that $ f$ commutes with $ g$ and the only two functions that can commute is either the identity function or its inverse. So, $ g(x)\equal{}x$ or $ g(x)\equal{}f^{\minus{}1}(x)\equal{}\sqrt[3]{\frac{x\plus{}5}{2}}$.


Let ABCD be a cyclic quadrilateral and let E, F be the midpts of AC and BD respectively. Show that <AEB=<AED => BFA=<BFC.

grobber I expected that you knew the rule about not posting problems from magazines not older than 1 year at the moment of the post :) just wait a couple of more months :)

[color=red][Moderator edit: Now it's almost one year since the publication of the problem. A solution can be found at http://www.mathlinks.ro/Forum/viewtopic.php?p=22938 now.][/color]


algebra
polynomial
Vieta
number theory
relatively prime
number theory unsolved

Prove that for any positive integer $ m$ there exist an infinite number of pairs of integers $(x,y)$ such that
$(\text{i})$ $x$ and $y$ are relatively prime;
$(\text{ii})$ $x$ divides $y^2+m;$
$(\text{iii})$ $y$ divides $x^2+m.$

Solve Pell's equation $ x^2\plus{}y^2\plus{}m\equal{}kxy$.

Your conditions equavalent to $ xy|x^2\plus{}y^2\plus{}m$, therefore we can solve $ x^2\plus{}y^2\plus{}m\equal{}kxy$ and show, that for one k it had infinetely mane solutions.

[quote="Rust"]therefore we can solve $ x^2 \plus{} y^2 \plus{} m \equal{} kxy$ and show, that for one k it had infinetely mane solutions.[/quote]

I hate to say this, but can you show it?

Fix $ k\equal{}m\plus{}2$. Then you can generate infinitely many solutions $ (x_n,y_n)$ using the following recursion:

i. $ x_0\equal{}y_0\equal{}1$

ii. $ x_{n\plus{}1}\equal{}y_n$, $ y_{n\plus{}1}\equal{}(m\plus{}2)y_n\minus{}x_n$

[hide="Idea"]Here I used Vieta Jumping for the equation $ x^2\minus{}(m\plus{}2)xy\plus{}y^2\plus{}m\equal{}0$. All we need to see is that $ (1,1)$ is a solution, and if $ (x,y)$ is a solution then so are $ ((m\plus{}2)y\minus{}x,y)$ and $ (y,x)$.[/hide]

$ x\equal{}1,y\equal{}m\plus{}1,k\equal{}m\plus{}2$ always solution, and $ D\equal{}4\minus{}k^2<0$, therefore for $ k\equal{}m\plus{}2$ we had infinetely mane solutions.


inequalities
calculus
integration
search
inequalities unsolved

[quote="RDeepMath91"]$x,y,z$ positive reals. Prove that
$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{27}{x+y+z}\geq \frac{36}{4(x+y+z)-3x}+\frac{36}{4(x+y+z)-3y}+\frac{36}{4(x+y+z)-3z}$
[/quote]
It's integral from vasc's inequality:
If $a,$ $b$ and $c$ are positive numbers such that $abc=1$ then
$a^{3}+b^{3}+c^{3}+9\geq4(ab+ac+bc).$

Still don't get it... Can u give more explanation, please ??  :o

Is it integral with 3 variables? if it is, how we can do that? [I just know integral and differenciate 1 variable]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=165582#p165582]This[/url] was the history of this method :lol:

EDITED:
@arqady: Imitating this method, it means that we have to prove this inequality: \[{a_{1}}^{n}+{a_{2}}^{n}+\ldots+{a_{n}}^{n}+n^{2}a_{1}a_{2}\ldots a_{n}\geq (n+1)\sum_{cyc}{a_{1}}^{\frac{n+1}{n}}{a_{2}}^{\frac{n+1}{n}}\ldots{a_{n-1}}^{\frac{n+1}{n}}{a_{n}}^{\frac{1}{n}}\] But how? Is it a well-known theorem or sumthing? :blush:  :?:

It's vasc's inequality too. If I am not wrong, Mildorf proved it.

Well, I don't spend too much time to learn how to integrate the inequality in such ways...  :oops:
So, can u please tell me about the trick of integrating some inequalities, please ??   :)

arqady, can you give the link for the proof of that inequality? I've search and found too much vasc's :D

[quote="dyd"] it means that we have to prove this inequality: \[{a_{1}}^{n}+{a_{2}}^{n}+\ldots+{a_{n}}^{n}+n^{2}a_{1}a_{2}\ldots a_{n}\geq (n+1)\sum_{cyc}{a_{1}}^{\frac{n+1}{n}}{a_{2}}^{\frac{n+1}{n}}\ldots{a_{n-1}}^{\frac{n+1}{n}}{a_{n}}^{\frac{1}{n}}\] But how? Is it a well-known theorem or sumthing? :blush:  :?:[/quote]

Don't know if it's well known but it is not hard.
If $p_{k}$ denotes the $k$th power mean, we can write it as $\frac{p_{n}^{n}}{p_{0}^{n}}+n\ge (n+1)\frac{p_{0}}{p_{-1}}$.
In view of ineq(3)  of http://www.mathlinks.ro/Forum/viewtopic.php?t=110149 by Vasc, which states $\frac{p_{n}^{n}}{p_{0}^{n}}+(n-1)\ge n\frac{p_{1}}{p_{-1}}$, it is sufficient to prove $np_{1}+p_{-1}\ge(n+1)p_{0}$. Write that as $nA+H\ge(n+1)G$ or $n+\frac{H}{A}\ge (n+1)\frac{G}{A}$. But we have $n+\frac{H}{A}\ge n+(\frac{G}{A})^{n}$ by MacLaurin. Putting $x=\frac{G}{A}\le1$, we get by AM-GM $n+x^{n}\ge(n+1)x^{\frac n{n+1}}\ge(n+1)x$. Done.


I want to know what you all think of this stuff:

http://www.informationliberation.com/index.php?id=15397

Now even if you descredit this actual article ( you may, I've skimmed it, and this is pretty early in the morning for me ), I still want your oppinions on this whole idea of "USA building a scary tommorow".

Reguardless of what the states DO do, they do have a massive propaganda machine spilling pretty constantly down there... and I'm just a little intimidated.

More proactivly: what can I do, to keep this world safe? I'm no superhero, but I like this planet a lot.

but I think thats what the black helecopter use to find my lawn furnature.

Man, idea stuf is expensive these days...

Duh... obviously they need Lebanon vaporized so they can get to Syria and Iran.  The Nazis had to take Poland before they could invade Russia.

Really they are just delaying the inevitable though... obviously the Chinese would catch up a lot faster if they could actually get all the oil they needed, but they'll catch up inevitably.  America's days are pretty numbered; essentially their only way of staying in place as world superpower is to obtain control of all the oil supply in the middle east by occupying the countries Nazi-style and imposing their own western-style government on other countries.

They've got Israel on their side (who have been successfully occupying Palestine and Gaza for decades; Palestine is just one big brown version of the WWII jew-ghetto), and Saudi Arabia/Bahrain/UAE/Yemen/Kuwait/etc are pretty western friendly (ie... they're just in it for the money; they don't have loyalties).  So they just need to destroy all the resistance in the northern countries (Iraq/Iran/Syria/Lebanon/Afganistan) to secure their middle-eastern stronghold.  (btw, "resistance fighters" are called "terrorists" in the US, and therefore the US won't negotiate with them).

The best thing you guys can do is send money/weapons/etc to Hamas, Hezbollah, and various other resistance organizations who are doing their best to stop the American push into the middle east.  :roll:  :roll:

I actually am kinda surprised that US hasn't invaded Iran already... Iran just bought a bunch (28 or 29) anti-aircraft missile sites that will make US air strikes a lot more difficult... and by sometime in September, they'll be set up and fully operational.  But maybe the US forces were stretched too thin to attack this soon, since Iraq is going so badly (didn't those idiots learn back in Vietnam that guerilla wars are IMPOSSIBLE to win?!?!?  Maybe they thought it would be easier in the desert than in the jungle... hah...)

Really, I think that America is likely to get its ass kicked very soon.  Not so bad for Americans, cuz they're not gonna get nukes dropped on their faces (although they'll be facing lots of other bad economic problems).  But Iran wants Israel wiped off the map (and why the hell wouldn't they after what Israel has done...).

So well... don't go to the middle east for the next little while.  If you know anybody who lives there, tell them to get out.  LOL

But seriously, it's a gamble whether or not the US actually invades Syria... because it's a huge risk, considering their chances of success and the amount of loss of life and economic loss that they will incur.  I wouldn't entirely bet on it, although they've made enough stupid decisions in the past that they might just go ahead and take that risk.

What bothers me, is how quickly these things can deteriorate if the US invades Iran.

Reasoning:

Iran is 70 million people, mostly nationalisists. While they may not be crazy, they will defend themselves to the death. The united states knows they can't win a ground war. You can't just 'occupy' a 70 million person nation, you have to destroy it.

My bet, is that the US government is trying to leverage enough of the popular oppinion, to have everyone convinced that Iran has nukes, and is giving them to 'terrorists'. With this in hand, they will stage/fake/prod untill they can say they were 'forced' to nuke Tehran.

Once that happens, things get really bad. Not only will the mildeast compleatly desend into a chaotic wasteland ( remember the fallout ), but all of the security councile nations will be damn pissed that the united states has just killed millions in cold blood.

I honestly I think if they nuke tehran, we will eventualy see a nuclear exchange between some combination of the security council nations.

Stupid America. I like this planet.

[quote="rzsolt"]What bothers me, is how quickly these things can deteriorate if the US invades Iran.

Reasoning:

Iran is 70 million people, mostly nationalisists. While they may not be crazy, they will defend themselves to the death. The united states knows they can't win a ground war. You can't just 'occupy' a 70 million person nation, you have to destroy it.

My bet, is that the US government is trying to leverage enough of the popular oppinion, to have everyone convinced that Iran has nukes, and is giving them to 'terrorists'. With this in hand, they will stage/fake/prod untill they can say they were 'forced' to nuke Tehran.

Once that happens, things get really bad. Not only will the mildeast compleatly desend into a chaotic wasteland ( remember the fallout ), but all of the security councile nations will be - pissed that the united states has just killed millions in cold blood.

I honestly I think if they nuke tehran, we will eventualy see a nuclear exchange between some combination of the security council nations.

Stupid America. I like this planet.[/quote]

That makes no sense at all... nuking Tehran won't win a Guerilla war.  They'll bomb the crap outta the place if they can afford to, but I don't think what you're saying makes any sense.

The only thing that's different this time is that people KNOW that Iran is pursuing nuclear capabilities, whereas Iraq NEVER HAD ANY WMDs.  The Americans' biggest concern is Israel getting wiped off the map (obviously America itself could never get nuked) because then their entire outpost in the middle east is gone.  And Iran's current leader has said he wants Israel wiped off the map.

Israel alone has an arsenal big enough to wipe out most of the middle east.  America really has no need to launch any nukes.

This upcoming stage of the game is likely to be a long, costly, drawn out war.  The Americans aren't going to simply decide that they'd be happier if Iran was a giant parking lot.  They'll push in at a pace that their economy can sustain, and they'll secure whatever oil fields they can (as they did in Afghanistan and Iraq).  All this bull about the "war on terror", "liberating Iraq", "bringing democracy to the middle east", etc. is just a lot of rhetoric to enhance public support for the war effort.

9-11 was staged by the US gov't as a public relations stunt to build support for an oil crusade... there's really nothing more to it.

Already about 20% of Americans believe that the WTC towers were brought down with the aid of a controlled demolition... considering the number of rednecks and idiots in the states, that's quite a lot.  Really if anyone here hasn't watched Loose Change and the 9-11 eyewitness reports, they should.

The only thing that got me about loose change, was: wouldn't 'they' try and stop the info from getting out if it was true?

Maybe they see it as to much of a give away if they kill the producer of it?

Also, why has no one steped out and said " I was part of this, and have a real moral delima? ".

[quote="rzsolt"]The only thing that got me about loose change, was: wouldn't 'they' try and stop the info from getting out if it was true?

Maybe they see it as to much of a give away if they kill the producer of it?
[/quote]

 :rotfl:  :rotfl:

WTF man... America is a pretty messed up place, but it's not frigging North Korea.


inequalities
floor function
logarithms
limit
number theory
prime numbers

Let $p_{n}$ again denote the $n$th prime number. Show that the infinite series \[\sum^{\infty}_{n=1}\frac{1}{p_{n}}\] diverges.

Copy from a script I've written a year ago:

Every $k \in \mathbb{N}$ can be written as $k=t \cdot s^{2}$ with $t$ not divisible by a square $>1$. This gives the inequality
\[\sum_{k=1}^{n}\frac{1}{k}\leq \prod_{ \substack{ p \in \mathbb{P}\\ p \leq n}}\left(1+\frac{1}{p}\right) \sum_{s=1}^{n}{\frac{1}{s^{2}}}.\]
Since $\frac{1}{s^{2}}\leq \frac{1}{s(s-1)}= \frac{1}{s-1}-\frac{1}{s}$ for $s \geq 2$ we get $\sum_{s=1}^{n}{\frac{1}{s^{2}}}\leq 1+\sum_{s=2}^{n}\frac{1}{s-1}-\frac{1}{s}= 2-\frac{1}{n}\leq 2$.
Together with the easy to verify property $1+x \leq e^{x}$ for all $x \in \mathbb{R}$ this yields
\[\sum_{k=1}^{n}\frac{1}{k}\leq \prod_{ \substack{ p \in \mathbb{P}\\ p \leq n}}\left(1+\frac{1}{p}\right) \sum_{s=1}^{n}{\frac{1}{s^{2}}}\leq 2 \prod_{ \substack{ p \in \mathbb{P}\\ p \leq n}}\left(1+\frac{1}{p}\right) \leq 2 \cdot \prod_{ \substack{ p \in \mathbb{P}\\ p \leq n}}e^\frac{1}{p}= 2 \cdot e^{\sum_{}\frac{1}{p}}\]
where the last sum also runs over all primes $p \leq n$.
To show that $\sum_{ p \in \mathbb{P}}\frac{1}{p}$ diverges, it now suffices to show that $\sum_{k=1}^\infty \frac{1}{k}$ diverges. But the latter one is a well known property, shown by the following since for all $n \geq 2^{m}$ we have:
\[\sum_{k=1}^{n}\frac{1}{k}\geq \sum_{k=1}^{2^{m}-1}\frac{1}{k}= \sum_{s=0}^{m-1}\sum_{k=2^{s}}^{2^{s+1}-1}\frac{1}{k}\geq \sum_{s=0}^{m-1}\sum_{k=2^{s}}^{2^{s+1}-1}\frac{1}{2^{s+1}}= \sum_{s=0}^{m-1}\frac{1}{2}= \frac{m}{2},\]
giving the divergence because $m$ can be chosen arbitrary large.

[quote="Peter"]Let $ p_{n}$ again denote the $ n$th prime number. Show that the infinite series
\[ \sum^{\infty}_{n=1}\frac{1}{p_{n}}\]
diverges.[/quote]

[b]Reference[/b]: A Classical Introduction to Modern Number Theory by K. Ireland and M. Rosen

[b] Second Solution [/b] We first prepare a lemma. Let $ \varrho(n)$ denote the set of prime divisors of $ n$. Let $ {\mathcal{S}}_{n}(N)$ denote the set of positive integers $ i\leq N$ satisfying that $ \varrho(i)\subset\{ p_{1},\cdots, p_{n}\}$.

 [b]Lemma.[/b]  We have $ \left\vert{\mathcal{S}}_{n}(N)\right\vert\leq 2^{n}\sqrt{N}$.

 [b] Proof [/b] It is because  every positive integer $ i\in{\mathcal{S}}_{n}(N)$ has a unique factorization $ i = s t^{2}$, where $ s$ is a divisor of $ p_{1}\cdots p_{n}$ and $ t\leq\sqrt{N}$. In other words, $ i\mapsto (s,t)$ is an injective map from $ {\mathcal{S}}_{n}(N)$ to $ \mathcal{T}_{n}(N) =\{\, (s,t)\;\vert\; s\,\vert\, p_{1}\cdots p_{n},\, t\leq\sqrt{N}\}$, which means that
\[ \left\vert{\mathcal{S}}_{n}(N)\right\vert\leq\left\vert\mathcal{T}_{n}(N)\right\vert\leq 2^{n}\sqrt{N}.\]
Now, assume to the contrary that the infinite series $ \frac{1}{p_{1}}+\frac{1}{p_{2}}+\cdots$ converges. Then we can take a sufficiently large positive integer $ n$ satisfying that
\[ \frac{1}{2}\geq\sum^{\infty}_{i>n}\frac{1}{p_{i}}=\frac{1}{p_{n+1}}+\frac{1}{p_{n+2}}+\cdots .\]
Take a sufficiently large positive integer $ N$ so that $ N > 4^{n+1}$. By its definition of $ {\mathcal{S}}_{n}(N)$, we see that each element $ i$ in $ \{1,\cdots, N\}-{\mathcal{S}}_{n}(N)$ is divisible by at least one prime $ p_{j}$ for some $ j > n$.  Since the number of multiples of $ p_{j}$ not exceeding $ N$ is $ \left\lfloor\frac{N}{p_{j}}\right\rfloor$, we have
\[ \left\vert\{1,\cdots, N\}-{\mathcal{S}}_{n}(N)\right\vert\leq\sum_{j>n}\left\lfloor\frac{N}{p_{j}}\right\rfloor\]
or
\[ N-\left\vert{\mathcal{S}}_{n}(N)\right\vert\leq\sum_{j>n}\left\lfloor\frac{N}{p_{j}}\right\rfloor\leq\sum_{j>n}\frac{N}{p_{j}}\leq\frac{N}{2}\]
or
\[ \frac{N}{2}\leq\left\vert{\mathcal{S}}_{n}(N)\right\vert.\]
It follows from this and from \textsc{Lemma 3.2} that
\[ \frac{N}{2}\leq 2^{n}\sqrt{N}\]
so that $ N\leq 4^{n+1}$. However, this is a contradiction for the choice of $ N$.

 [b]Third Solution[/b] We use an auxiliary inequality without a proof.

 [b]Lemma.[/b]  We have $ \frac{1}{1-t}\leq e^{t+t^{2}}$ for all $ t\leq\left[0,\frac{1}{2}\right]$.

 Let $ l\in\mathbb{N}$. By  [b]Fundamental Theorem of Arithmetic[/b], each positive integer $ i\leq p_{l}$ has a unique factorization $ i ={p_{1}}^{e_{1}}\cdots{p_{1}}^{e_{l}}$ for some $ e_{1},\cdots, e_{l}\in\mathbb{Z}_{\geq 0}$. It follows that
\[ \sum_{i=1}^{p_{l}}\frac{1}{i}\leq\sum_{ e_{1},\cdots, e_{l}\in\mathbb{Z}_{\geq 0}}\frac{1}{{p_{1}}^{e_{1}}\cdots{p_{l}}^{e_{l}}}=\prod_{j=1}^{l}\left(\sum_{k=0}^{\infty}\frac{1}{{p_{j}}^{k}}\right)=\prod_{j=1}^{l}\frac{1}{ 1-\frac{1}{p_{j}}}\leq\prod_{j=1}^{l}e^{\frac{1}{p_{j}}+\frac{1}{{p_{j}}^{2}}}\]
so that
\[ \sum_{j=1}^{l}\left(\frac{1}{p_{j}}+\frac{1}{{p_{j}}^{2}}\right)\geq\ln\left(\sum_{i=1}^{p_{l}}\frac{1}{i}\right).\]
Together with the estimation
\[ \sum_{j=1}^{\infty}\frac{1}{{p_{j}}^{2}}\leq\sum_{j=1}^{l}\frac{1}{(j+1)^{2}}\leq\sum_{j=1}^{l}\frac{1}{(j+1)j}=\sum_{j=1}^{l}\left(\frac{1}{j}-\frac{1}{j+1}\right)=1,\]
we conclude that
\[ \sum_{j=1}^{l}\frac{1}{p_{j}}\geq\ln\left(\sum_{i=1}^{p_{l}}\frac{1}{i}\right)-1 .\]
Since the divergence of the harmonic series $ 1+\frac{1}{2}+\frac{1}{3}+\cdots$ is well-known, now we get the result.

 [b]Note[/b]  The fact that the sum of the reciprocals of all prime numbers diverges was first proved by Euler. Using the well-known result $ 1+\frac{1}{2}+\cdots+\frac{1}{N}\geq\ln (N+1)$ and modifying the method in the second  solution, we get the following result.

     For all real numbers $ x\geq 2$, we obtain
\[ \sum_{p: prime,\; p\leq x}\frac{1}{p}\geq\ln\left(\ln x\right)-1.\]
It turns out that this lower bound inequality is sharp. In fact, it is known that
\[ \mathcal{M}=\lim_{n\rightarrow\infty }\left(\sum_{p: prime,\; p\leq n}\frac{1}{p}-\ln\left(\ln n\right)\right)\]
converges. Here, $ \mathcal{M}$ is called Meissel-Mertens constant. It is known that $ \mathcal{M}= 0.2614972128476427837554268386086958590516\cdots$

[quote="Peter"]Let $ p_{n}$ again denote the $ n$th prime number. Show that the infinite series
\[ \sum^{\infty}_{n\equal{}1}\frac{1}{p_{n}}\]
diverges.[/quote]

 [b] Reference. [/b]  Elementary Number Theory by   David M. Burton

 It can be deduced from [b]Prime Number Theorem[/b] Let $ \pi(x)$ denote the prime counting function. Since $ pi(x)\rightarrow\frac{ x }{\ln x}$ as $ x\rightarrow\infty$,  we can  find a constant $ \lambda > 0$ satisfying that
\[ \pi(x) >\lambda\frac{ x }{\ln x}\]
for all sufficiently large positive real numbers $ x$. This means that $ n >\lambda\frac{ p_{n}}{\ln p_{n}}$ when $ n$ is sufficiently large.  Since $ \lambda\frac{ x }{\ln x}>\sqrt{x}$ for all sufficiently large $ x > 0$, we also have $ n >\lambda\frac{ p_{n}}{\ln p_{n}}>\sqrt{ p_{n}}$  or $ n^{2}> p_{n}$ for all sufficiently large $ n$. We conclude that
\[ n >\lambda\frac{ p_{n}}{\ln p_{n}}>\lambda\frac{ p_{n}}{\ln\left( n^{2}\right)}\]
or
\[ \frac{1}{ p_{n}}>\frac{\lambda }{2n\ln n}.\]
when $ n$ is sufficiently large. Since
\[ \sum_{n\equal{}2}^{\infty}\frac{1 }{n\ln n}\equal{}\infty\]
Comparison Test yields that
\[ \sum_{n\equal{}1}^{\infty}\frac{1}{ p_{n}}\equal{}\infty.\]

[quote="Peter"]Let $ p_{n}$ again denote the $ n$th prime number. Show that the infinite series
\[ \sum^{\infty}_{n\equal{}1}\frac{1}{p_{n}}\]
diverges.[/quote]

 On the other hand, Viggo Brun established that the sum of the reciprocals of the [b]twin primes[/b] converges. In other words,
 \[ \mathcal{B}\equal{}\sum_{p, p\plus{}2: prime}\left(\frac{1}{p}\plus{}\frac{1}{p\plus{}2}\right) \equal{}\left(\frac{1}{3}\plus{}\frac{1}{5}\right)\plus{}\left(\frac{1}{5}\plus{}\frac{1}{7}\right)\plus{}\left(\frac{1}{11}\plus{}\frac{1}{13}\right)\plus{}\cdots\]
 converges. The constant $ \mathcal{B}\equal{}1.90216\cdots$ is called [b]Brun's Constant[/b].

Other solution:
[b]Lemma [/b]
Consider  $ a_i\in (0,1)$
Then if $ \prod_{i \equal{} 1}^{ \plus{} \infty}(1\minus{}a_i) \equal{} 0$ then
$ \sum_{i \equal{} 1}^{ \plus{} \infty} a_i \equal{} \plus{} \infty$
Now we has :
$ \prod_{k \equal{} 1}^{ \plus{} \infty}(\sum_{i \equal{} 0}^n\frac {1}{p_k^i}) \equal{} \sum_{i \equal{} 1}^n\frac {1}{i}$
because $ \sum_{k \equal{} 1}^{ \plus{} \infty}\frac {1}{k} \equal{} \plus{} \infty$
so $ \prod _{i \equal{} 1}^{ \plus{} \infty}(1 \minus{} \frac {1}{p_i}) \equal{} 0$
Imply that $ \sum_{k \equal{} 1} \frac {1}{p_k} \equal{} \plus{} \infty$

How do you prove the Lemma? I only see $ \sum a_i\ge1$ by Bernouilli...
And why is $ \prod_{k \equal{} 1}^{ \plus{} \infty}(\sum_{i \equal{} 0}^n\frac {1}{p_k^i}) \equal{} \sum_{i \equal{} 1}^n\frac {1}{i}$?

And why is $ \prod_{k \equal{} 1}^{ \plus{} \infty}(\sum_{i \equal{} 0}^n\frac {1}{p_k^i}) \equal{} \sum_{i \equal{} 1}^n\frac {1}{i}$?[/quote]
Because every number can represent in the form $ n\equal{}\prod_{i\equal{}1}^{\plus{}\infty}p_i^{r_i}$
Where $ r_i\geq 0$

That means that $ \sum_{i \equal{} 0}^n\prod_{k \equal{} 1}^{ \plus{} \infty}\frac {1}{p_k^i} \equal{} \sum_{i \equal{} 1}^n\frac {1}{i}$, but how are sum and product commutative here?

Aah, now I get it.  :oops:  Sorry, it seems I'm not very sharp today.


modular arithmetic
number theory unsolved
number theory

Note that $ 561\equal{}3\cdot 11\cdot 17$
Given that $ a^{561}\equiv a\pmod{3}$, $ a^{561}\equiv a\pmod{11}$, and $ a^{561}\equiv a\pmod{17}$,
prove that

$ a^{561}\equiv a\pmod{561}$


How can i simplify tan(arctan(a)+arctan(b)+arctan(c)...)

use the fact that arctan(a) + arctan(b) = arctan($ \frac {a\plus{}b}{1\minus{}ab}$)  :)