FTW ARML trigonometry AMC AIME USA(J)MO USAMO

We probably are, as even though we are slightly understaffed for states due to an orchestra conflict, we will probably still win there (by the way, I'm one of the people who can't go).

But yeah, unless a HUGE SURPRISE arises, we'll have our whole team for the New England Meet, and we will be ready to try to beat all of you again.

haha...lexington is crippled

Acton FTW  :P

We're going to States tomorrow (Medium schools).

We might make it to NEAML. We've made it the last two years, but it's always been close. Hopefully our team shows up strong.

[quote="ZhangPeijin"]Hmm, but it says you're in China, meaning that you can't come to MA to join the Acton Team and beat Lexington anyways...
What is NEAML anyways?
Sounds like NEML but it can't be...[/quote]NEAML (New England Association of Mathematics Leagues) is the conclusion (if you don't count ARML, which is totally different) of the chain of competitions which begin with local mathematics leagues throughout New England which have meets throughout the first half of the academic year. These feed into state competitions, and then qualification for NEAML is dependent either on the state meet in some states or the league performance in others.

We're going to pwn Small Schools this year. Hopefully NEAML will be better than CSAML, which was horrendously run. (as in, they forgot all the perfect scorers when giving out prizes, the competition went an hour later than it needed to, they ran out of packets when announcing the ARML team, etc.) The problems weren't great either. (got a 17 because I forgot how to do determinants, which I didn't even think were fair game)

[quote="Pineappleperson"]haha...lexington is crippled[/quote]

lol at looking at things in retrospect :lol:

We made it to New Englands!

We'll be there

I shall be going--All-State conflicted with CSAML, but thankfully, nothing conflicts with NEAML. It's a terrible date for juniors, though, with the SAT administered the next day and AP exams looming.

Pingree will be there (small school), first time ever. :D

[quote="Pineappleperson"]haha...lexington is crippled[/quote]

...which makes our state win even more pwnage

I'm going too!   :lol:

BTW, does anybody know the approximate schedule of regionals?  I need to know when the team round will be over, because there's somewhere else in MA I sorta need to be. :P

[quote="Harmonikdiskorde"]Question: I've been doing some of the old NEAML contests. Does it seem that the individual rounds are about the same difficulty as local (GNHML, e.g.) rounds, whereas the team round is significantly harder?

Also, I second the question on approximate schedule. (start time, end time)[/quote]Both are significantly harder. GNHML was harder this year than usual.

Yeah, the first five rounds were pretty triv, then I went into round 6 and was like "ok this will probably be triv as well," and it pretty much was, but i made a couple of stupid mistakes. Whatever.

[quote="worthawholebean"]We didn't stay for large schools. I was happy with my 18. :)

Yet again, round 6 was epic fail. They just made it ugly, not requiring skill. Numbers one and three were just understanding polar form and unit-circle trig, and 2 was just a trick. I was hoping it would be more difficult after what I heard about MAML; sadly, not the case.[/quote]

Epic fail indeed. It took me 5 minutes to do number 1, and CatalystOfNostalgia and I both got solutions for number 3 that had $ \frac{\pi}{2}<\theta<\pi$. It was really computationally-heavy -- I ran out of time to try number 2.

:spam:

If I took either rounds 1 or 2 and not round 6 I definitely would have been a high scorer, maybe a 17 and definitely not a 12. arg...

Considering we only had 1 senior this meet, Noah, Stephen and Sway not competing, and a few new kids coming through Diamond and Clarke, I feel pretty good about our chances next year.

Anyways, congrats ABRHS on um...scoring twice as much as we did on round 6 and tying with us for first.

We had 0 seniors this year... though continuing to pwn Small Schools is not really a huge achievement.

You can always move up a division...I believe Canton is a Medium Schol in terms of population but competes as a Large School in most/all of the stuff it does.

Bedford tied for 2nd in Medium

I got a 15/18. I missed #3 in round 6.

I was about 30 seconds away from getting it. I just had to add 3 fractional radian values, but I ran out of time. The first problem took me too long.

NEAML was quite easy this year.

I don't like how they banned frisbee in the gym :(

I did 6/3 in degrees because I saw it would be ugly in radians...

I definitely should have

$ 15\minus{}63$ is much better than $ \frac{\pi}{12} \minus{} \left(\frac{\minus{}\pi}{20} \plus{} \frac{2\pi}{5}\right)$

(I think that's what I was trying to do in the last 15 seconds)

[quote="chickendude"]I definitely should have

$ 15 \minus{} 63$ is much better than $ \frac {\pi}{12} \minus{} \left(\frac { \minus{} \pi}{20} \plus{} \frac {2\pi}{5}\right)$

(I think that's what I was trying to do in the last 15 seconds)[/quote]Except it was $ 135\minus{}63$.

oh yeah, I see it now.

Well then I just got it wrong =\.

[quote="Ignite168"]Why did they ask for grades?[/quote]Very good question.

NEAML results are online:

Team:  http://www.wocomal.org/statistics/neaml/2008/2008n_team.html
Individual:  http://www.wocomal.org/statistics/neaml/2008/2008n_individual.html

(WOCOMAL schools/students are highlighted.)

Round difficulty, based on averge score:

Round 4 -- 2.917 points (easiest)
Round 3 -- 2.672 points
Round 1 -- 2.525 points
Round 2 -- 2.450 points
Round 5 -- 2.183 points
Round 6 -- 0.842 points (hardest)

I mentioned last year that WOCOMAL has a tiebreaker based on the differential between a student's score and a student's expected score (based on the average score in each round).  e.g., A student who scored 18 in rounds 2, 5, 6 on Friday would have scored 18-(2.450+2.183+0.842)=12.525 points more than expected, the best possible.

If NEAML had that tiebreaker in place, the eight students with 18's would be ranked this way:

1. Patrick Hulin (Hopkins) -- 11.568 points above expected score
(only student with an 18 who was in round 6)
2. Jeremy Rose (Sharon, MA) -- 10.619 points
3. Joshua Scripture (Bangor, ME) -- 10.375 points
4. Misha Kushner (Darien, CT) -- 10.353 points
4. Luyi Zhang (Hopkins School) -- 10.353 points
6. Peter Baratta (Belmont, MA) -- 10.227 points
7. Michael Lai (Choate) -- 9.961 points
7. Wang, Kai (Acton-Boxborough, MA) -- 9.961 points

There was a lot of partity between the divisions.  Multiply the ABRHS/crippled Lexington team scores by 3/4 and you get 92.25, just a hair above the Hopkins 92 and Choate 92.

Rick Yanco
Worcester Academy

[quote="worthawholebean"][quote="rjyanco"]1. Patrick Hulin (Hopkins) -- 11.568 points above expected score
(only student with an 18 who was in round 6)[/quote]Ahh. I like this. :)

[quote="rjyanco"]There was a lot of partity between the divisions.  Multiply the ABRHS/crippled Lexington team scores by 3/4 and you get 92.25, just a hair above the Hopkins 92 and Choate 92.[/quote]How was Lexington crippled?

Hopkins would have scored a 95 if I had seen that stupid $ \le$ on team round 6... whatever. I'm happy.[/quote]

Our top three people (Noah Arbesfeld, Stephen Xu, Sway Chen) were all at DOE Nationals. With them one could reasonably have expected our score to be bettered by somewhere between 5 and 15 points.

...And (maybe a bit selfishly) I probably wouldn't have needed to take round 6.  :wink:

My school went....Our two AIME qualifying seniors, and usamo qualifying juniors couldnt come, so only I was there from our usual A team. Including myself everyone was freshman, i got an 11 on the two freshman rounds; and then had to take a suicide round i hadnt learned yet and got a 0. The second highest score on our team was a 4.

Since this will be my first time going to any of these competitions (i'm in middle school), how long are each of the rounds.


Would you do as Andrew Wiles and lost all contact with humans for about 7 years just to prove something??? Why?? Why not???????.

But its not true that he lost contact with all humans for the 7 years that he was working on the proof. He did work on the problem in secret for 7 years, but he worked on other things at the same time. He had several very good papers in other areas of number theory. He also had lots of contact with other humans, including his family, colleagues, and students. I had him as a teacher shortly after he had just completed the proof. He was affable, friendly, and a good teacher.

Wow!!! Andrew Wiles as your teacher? You are lucky.

Google loves you <3

Anyways, he's the guy who proved Fermat's Last Theorem.

Wouldn't me not using Google imply that Google hates me?  Lol.  I love Google, I'm just incredibly lazy, and don't want to wade through a bio to find out that one little fact.   :D

[quote="G-UNIT"][quote="mysmartmouth"]I'm sorry, but fiil me in.  Who in the world is Wiles?[/quote]

  You do not deserve to know.  :D[/quote]

Cuac .........        Andrew Wiles proved Fermat Last Theorem...


calculus
integration
calculus computations

How in the world do i find the integral of e^(x^(1/3))

hello, do you mean $ \int e^{x^{\frac{1}{3}}}dx$?
Sonnhard.

yes! sorry but yeah that is  what i meant.

hello, by substituting $ z\equal{}x^{\frac{1}{3}}$ we get $ x\equal{}z^3$ and from here $ dx\equal{}3z^2dz$ and our integral is right now $ 3\int z^2e^zdz$. This can be solved with integration by parts.
Sonnhard.


Let us all reveal each others real names and help each other to the next level.
My name Is Diptarka Hait  from west bengal.
also anyone got a letter from hbcse?

Well, I'm[i] S Krishna Savant[/i] from Warangal AP.
No I didn't yet get any letter from HBCSE.
Also the results are not put on the hbcse site.

Hi dear fellow AoPSians and MAthlinkers [hide=" :D "]including me ... :blush: [/hide], please use this thread for furthur discussions regarding INOs 2010

[quote="savy2020"]Is Syllabus of IN(C/P)hO same as I([url=http://www.icho2009.co.uk/files/pdf/IChOsyllabus.pdf]C[/url]/[url=http://www.jyu.fi/ipho/syllabus.html]P[/url])hO?
Or is it just JEE/CBSE syllabus?
Any idea?[/quote]
No one seens to bother to answer me  :huh:  :maybe:

[quote="diptarka"]Please post on the INO 2010 forum regarding this. I mean NSEs are history by now and so we should let the dead past bury its dead and move to the present INO's.
p.s:any advice regarding INCHO . I am doing past papers but i am sinking into panic mode a bit too fast and ending up ruining all of it.Plus i change answers a bit too fast and the small box plus banning pencil written answers gets on my nerves.i feel the time too is limited.
do you think i should invest in a pen eraser?[/quote]
OK!!  It's a good idea I think to get a pen eraser.
U'd feel time to be limited bcos you're changing answers and panicking. Be cool adn time would definitely be enough. :)

But the instructions say that we can't change our answer once we've marked it and even if we use pen eraser its going to show, isn't it??
As per the syllabus I really have no idea and the HBCSE site only says that INOs syllabus are broadly similar to the NSEs which are based on class XI and XII CBSE

Has anyone got incho hall ticket. I haven't and am getting extremely nervous.
@abhitayal
pen eraser is not whitener. it really removes ink rather than covering it. but i haven't got a good one so far.

@Abhitayal
Where did you see the "instructions"?

@Diptarka
I've received my INChO hall-ticket today. But there's more time.... Don't worry... You'll get yours soon[hide=" :roll: "]
[/hide]
btw how's your preparation going on..

Even I have got my INAO card
but don't worry, you'll get urs soon

@Savy
I read it on the instructions page on NSE
Hers's an attatched copy on NSEC-2008(Sorry, couldn't find my 2009 paper) instruction page
See point 5,6
It says sctratching not allowed
and the pen-eraser scratches off the page where you rub it

Well Abhitayal, we were talking of INOs :P
BTW, the pattern of INChO was subjective i.e, you need to "write" the answer. So I think pen erasers are allowed as the answer sheets are NOT machine-graded.
May be for MCQs scratching etc isn't allowed
Last years INPhO pattern was mostly obj. What was pattern of INAO?

DOES ANYBODY HERE HAS INPHO 2007 (OR EARLIER) PAPER?
ANYBODY GOT INPHO 2008 SOLUTIONS?
what was cutoff in 2008?

I thought that even INOs are objective + subjective
And so I thought that the instructions for MCQ remain the same as in NSEs
Forgive me for my ignorance :blush:

@Abhitayal
nvm..
ofc Physics and Astronomy were Obj+Subj.
But chemistry was fully subjective...

But who knows what may be the pattern this year :maybe:

@revengeofST
You can but the book containing papers(with solutions) from HBCSE . Pay a visit to HBCSE if you are near Mumbai.
http://olympiads.hbcse.tifr.res.in/subjects/physics/references
But by post you may get the book after the examination [hide="because.."]
That's the speciality of indian postal service.. :rotfl:  [/hide]

Thanks , i just got my hall ticket. I think the INCho will be subjective+ a little objective like the icho papers. I wonder why they made physics mcq? Sounds impractical as IPhO is not mcq. I think pen erasers are permissible as answers wont be machine graded.

I think it only means that we can't solve equations using the calculator

[quote="savy2020"]Well INChO instructions say that "[i]non-programmable calculators[/i]" are allowed. What does that mean?
Does a scientific calculator qualify as non-programmable?[/quote]
Yes.
And the book thing- there's one from MtG as well.
I got my admit card yesterday, lost by the morning, and my mom found it again in the evening. All set to lose it again <_<
Ok, any suggestions where to study all of class 11-12 physics in 10 days? I prefer doing examples. Is DC Pandey any good for that? OR should I stick to Halliday Resnick and HCV?
I'm thinking of actually studying physics for once :P

Thanks nuke and abhi.
I've never used a calculator :oops:  So I didn't knew...

@nuclear
If you prefer to do "examples", D.C Pandey's books may be good.
But if you want to strengthen your concepts or revise, then it is NOT the one. :D

[quote="savy2020"]
But if you want to strengthen your concepts or revise :rotfl: then it is NOT the one. :D[/quote]
All right, so I guess I'll study from Halliday and do extensive examples from DC

[quote="nuclear_alchemist"][quote="savy2020"]Well I need a suggestion...
From where shall I study "Spectroscopy"?
Some concepts are being asked in INChO..[/quote]
From a spectrometer, duh![/quote]
Lol, what do you mean... :o
I was asking from which "book" to study the concepts of spectroscopy..

[quote="savy2020"][quote="nuclear_alchemist"][quote="savy2020"]Well I need a suggestion...
From where shall I study "Spectroscopy"?
Some concepts are being asked in INChO..[/quote]
From a spectrometer, duh![/quote]
Lol, what do you mean... :o
[/quote]
I was joking, idiot!

[quote="savy2020"][quote="nuclear_alchemist"][quote="savy2020"]Well I need a suggestion...
From where shall I study "Spectroscopy"?
Some concepts are being asked in INChO..[/quote]
From a spectrometer, duh![/quote]
Lol, what do you mean... :o
[/quote]
joking  :|

[quote="nuclear_alchemist"]joking  :|[/quote]
nvm.....btw your suggestion?

and why did you double post :rotfl:

I said 'no post mode specified' when I first tried.
And sorry, no idea about chem :P

[i]Morrison & Boyd[/i] has an entire chapter devoted to spectroscopy [hide="but"]I'm guessing you'd have already read that  :) [/hide]

ofc but that's just 70-80pgs..... bit superficial .....

Do you think INPhO will be objective this year also??

It's really bad that it was objective last year.. :(

I hope it is subjective :roll:   ... Any chance for that?

If you ask me, i will say inpho has very low chance of being subjective. The Indian team did very well last year and so they wouldn't change pattern. If however the chem guys get inspired, it spells disaster. I dont see why they were made objectives as no INO is objective except IJSO ( test section only, there is a seperate subjective part). But here we are all puppets in the hands of the system and so anything can happen

ok!! SO NO HOPE...
At least letz hope that Chemistry wouldn't be objective..

 :(
Very bad news for me.... :(  Pre-boards from Wednesday :(

What is there to worry about pre boards? I have hell as i have history but you are free from the shackles of social science. Fortunately mine are over.

Hey!
I've never touched the board text-books till now  :oops:
Anyways nvm... Even if I weren't to write INOs I wouldn't study :P

And more importantly I could bunk an exam on Saturday! Hurray! :

[quote="savy2020"]Hey!
I've never touched the board text-books till now
Anyways nvm... Even if I weren't to write INOs I wouldn't study

And more importantly I could bunk an exam on Saturday! Hurray! :[/quote]

i can understand your delight :D   :rotfl: :D
i wish i also had an opportunity to miss some, :( (but mine preboards r over)

@spacetime
 :( No bunking...
My Principal told me to write the exam on Monday... So  :(
This is really bad..

That's hard luck
But AAL IZZ WELL!!!!! :P

[quote="savy2020"]@spacetime
No bunking...
My Principal told me to write the exam on Monday... So
This is really bad..[/quote]
oh!what a bad news for u.  :(
your principal has same line of toughts as our headmaster :police: .
probably they are some sort of relatives... :wow:

[quote="spacetime"]
oh!what a bad news for u.
your principal has same line of toughts as our headmaster :police: .
probably they are some sort of relatives... [/quote]
Probably :P

Any way as Abhi says  $ AAL$ $ IZZ$ $ WELL$!!!! :)


calculus
integration
logarithms
function
calculus computations

Evaluate $\int_{0}^{\infty} \frac{x}{e^x - 1} \, dx$.

My friends and I encountered this on the BC Calculus Class Test and I don't think any of us were able to solve it  :huh:

[quote="pleurestique"]$e^x - 1 = u$; $e^x dx = du$;

$\int \frac{x}{e^x-1}dx = \int \frac{\ln (u+1)}{u(u+1)} du = \int \frac{\ln(u+1)}{u}du - \int \frac{\ln(u+1)}{u+1} du$

and the rest isn't too hard[/quote]

it's interesting... what about $\int \frac{\ln(u+1)}{u}du$?? :?:

well, i was thinking power series for that part... though it gets a little complicated. there's probably a better way. if i think of one, i'll post.

[quote="jmerry"]That substitution is trouble- once you take the difference, both improper integrals diverge here. The antiderivative isn't an elementary function anyway.

Here's an idea: $\frac1{e^x-1}=e^{-x}+e^{-2x}+e^{-3x}+\dots$; it's a geometric series. Now find $\int_0^\infty xe^{-nx}\,dx$ and add them up.[/quote]

Hey, that works.  Thanks.  :)

[quote="anik_andrew"][quote="pleurestique"]$ e^x \minus{} 1 \equal{} u$; $ e^x dx \equal{} du$;

$ \int \frac {x}{e^x \minus{} 1}dx \equal{} \int \frac {\ln (u \plus{} 1)}{u(u \plus{} 1)} du \equal{} \int \frac {\ln(u \plus{} 1)}{u}du \minus{} \int \frac {\ln(u \plus{} 1)}{u \plus{} 1} du$

and the rest isn't too hard[/quote]

it's interesting... what about $ \int \frac {\ln(u \plus{} 1)}{u}du$?? :?:[/quote]

See [url=http://mateforum.ro/viewtopic.php?t=30]here[/url]

[quote="t0rajir0u"]Evaluate $ \int_{0}^{\infty} \frac {x}{e^x \minus{} 1} \, dx$.

[/quote]
Substitute $ x\equal{}\minus{}\ln u.$

$ \int_0^\infty  {\frac{x}
{{e^x  \minus{} 1}}\,dx}  \equal{}  \minus{} \int_0^1 {\frac{{\ln u}}
{{1 \minus{} u}}\,du} .$

One more substitution defined by $ v\equal{}1\minus{}u.$ After series expansion yields $ \zeta(2).$

We encounter a very similar integral when deriving Planck's law of Black Body radiation. We can also calculate the following generalisation:

\[ \displaystyle\int_{0}^{\infty} \frac{x^n}{e^x\minus{}1}\equal{}\zeta(n\plus{}1)\Gamma(n\plus{}1)\]

The derivation of this can be found on wikipedia here:

[url]http://en.wikipedia.org/wiki/Planck_law#Appendix[/url]


I am opened to sugestions about a main page poll - the screen resolution one is obsolete, and I will have to take it away ...

Well I have always wondered whether I'm right or not about this : most mathematicians ENJOY (to say the least) rock music. So a good topic for a poll would be (I think) : what's your favourite music genre ? Of course do not forget to include 'manele' as well :D I know a few amateurs myself :) (who are into maths as well)

Yeah I think the idea with the music genre is very good. Also the poll can be called: What sport do you prefer? Football,Basketball,Tennis.....etc

okay. anyother ideas for a poll. I will leave the music poll running for a couple more days (to see if the classical music can beat the rock :D) and would like some more suggestions ... :D

so lagrangia I see you want to introduce an infinite list...  :D

I feel it is very important to train his mind. But don't forget about the sport matter. It is really important to keep you fit. But it is also a social event in our culture. So my question is:

what is your favourite sport ?

table tennis
tennis
basketball
soccer
ice hockey
handball
volleyball

I hope you all like sports games. Otherwise there has to be a new topic including sports as athletics, gymnastics,...

this polls' next. I will leave the current poll for another week or so, since the Classical music has got real close to Rock :)

Another poll suggestion:

Where did you get to know about mathlinks ?

- yahoo IMO group
- another web site giving a link to mathlinks: Which site provided the link ?
- browsing the web (Which search engine did you use ?)
- from a teacher, friend,...

Hmmm.. I think that the last pool Orlando suggested should actually be a question when you register for the forum.. anyway this is a thought! :)

cheers! :)

The question would be: how useful is that information, or is it just for the fun of knowing the statistics?

It is not just for fun. If you make a survey concerning a certain product, you will also ask about the preferences of the common population regarding it. One question might be: How did they get to know about the product ? Collecting information is very important in economy and commerce matters. You will always make this step before thinking about your strageties promoting a special product. You see I try to make the forum very well-known and therefore I am gathering information...


inequalities
function
logarithms
inequalities unsolved

Your inequality can be written as
\[ \frac{1}{3}\sum \frac{1-3a}{3^a}\geq 0. \]
Let's consider function $f(t)=\frac{1-3t}{3^t}$ for $0<t<1$. Then
\[ f'(t)=-3\cdot 3^{-t}+(3t-1)\cdot\ln{3}\cdot 3^{-t} \]
and
\[ f''(t)=-3\cdot (-\ln{3}) \cdot 3^{-t}+\ln{3}(3\cdot 3^{-t}+(3t-1)(-\ln{3}\cdot 3^{-t})= \]
\[ =3^{-t}\ln{3}(6+\ln{3}-3t\ln{3}). \]
Function $f$ is convex for $0<t<1$, because $f''(t)>0$ is equivalent to
\[ 6+\ln{3}>3t\ln{3}\iff t<\frac{6+\ln{3}}{3\ln{3}}. \]
It will be proved if
\[ 1<\frac{6+\ln{3}}{3\ln{3}}\iff 3\ln{3}<6+\ln{3}\iff \ln{3}<3 \]
wich is obvious because $3<e^3$. Applying Jensen inequality we get
\[ \frac{1}{3}\sum \frac{1-3a}{3^a}\geq \frac{1-3\cdot\frac{a+b+c}{3}}{3^\frac{a+b+c}{3}}=0 \]
which ends the proof.

[hide]
Let's work on $LHS=\frac{1}{3} \cdot \frac{1}{3^a}+\frac{1}{3} \cdot \frac{1}{3^b}+\frac{1}{3} \cdot \frac{1}{3^c}$

Since $f(x)=\frac{1}{3^x}$ is convex, $LHS \ge \frac{1}{3^{\frac{a+b+c}{3}}}=\frac{1}{\sqrt[3]{3}}$ by Jensen.

Now look at $RHS=\frac{1}{3} \cdot \frac{a}{3^{a-1}} + \frac{1}{3} \cdot \frac{b}{3^{b-1}}+\frac{1}{3} \cdot \frac{c}{3^{c-1}}$.

Function $f(x)=\frac{x}{3^{x-1}}$ is concave in $[-1,0]$,

so again by Jensen, $RHS \le \frac{1}{3^{\frac{a+b+c}{3}}}=\frac{1}{\sqrt[3]{3}}$, and we have the proof[/hide]

well , isnt this a direct application of chebserve since

$\{a,b,c\}$ and $\{3^{-a},3^{-b},3^{-c}\}$ is oppositely sorted . We just apply chebserve and done .


If $ f(m\plus{}1)\equal{}m(\minus{}1)^{m\plus{}1}\minus{}2f(m)$ for integers $ m>\equal{}1$, and $ f(2001)\equal{}1$, compute $ f(1)\plus{}f(2)\plus{}f(3)\plus{}...\plus{}f(2000)$

$ f(2001)\equal{}1 \implies 1\equal{}\minus{}2000\minus{}2f(2000) \implies f(2000)\equal{}\minus{}2001/2$

In general, $ f(m) \equal{} m/2(\minus{}1)^{m\plus{}1}\minus{}f(m\plus{}1)/2$ so we have $ \sum_{m\equal{}1}^{2000} f(m) \equal{} \sum_{m\equal{}1}^{2000} m/2(\minus{}1)^{m\plus{}1} \minus{} 1/2 \sum_{m\equal{}2}^{2001}f(m)$

Are you sure it isn't $ f(2001)\equal{}f(1)$?  Looks very much like a former ARML problem.  Otherwise, this problem needs brute force...  If it is supposed to be $ f(2001)\equal{}f(1)$ then we have $ 3\sum_{m\equal{}1}^{2000}f(m) \equal{} \sum_{m\equal{}1}^{2000} m(\minus{}1)^{m\plus{}1} \equal{} (1\minus{}2)\plus{}(3\minus{}4)\plus{}(5\minus{}6) ... \plus{}(1999\minus{}2000) \equal{} \minus{}1000$ so then our answer is $ \minus{}1000/3$.


Did anybody  from NC get a perfect score (amc 8) this year

[quote="chess64"][quote="MysticTerminator"]yeah I beasted it dude[/quote]

Good job! What did you get?[/quote]
how can you question arnav's beastliness? >=0

[quote="BFG"][quote="MysticTerminator"]yeah I beasted it dude[/quote]

AMC history book shows that you did not get a perfect score in AMC 8 dude.[/quote]
and everyone/thing else says "no one cares" dude

Wait, so what did Mystic Terminator get? He's what a 6th grader this year? I mean I know he got a perfect score but like... what did he get? You know what I'm saying? :) I just want to know what Mystic Terminator got on the AMC 8 this year. :) :) :) :) :) :) :) :)

where did you guys see who made honor roll?

[quote="b-flat"][quote="BFG"][quote="b-flat"]I imagine there will be several perfects this year because NC had 9 schools make national honor roll.[/quote]

There were 96 schools on the list last year but only 61 this year.

By the way, did you get a perfect score?[/quote]

Yes, I did  :)[/quote]

heh obviously... this is B-FLAT here!

I got soooooo close!!!!! I got a 24! (again :maybe: )

Wow, nice.
My best friend, Jessie Duan [Ligon], got perfect. So did Marshall Lochbaum [Martin]. And Christine Hong and Jenny Chen [both from Arendell Parrott, I believe]. And Bryce, of course, obviously. I don't know the others:
[url]http://www.unl.edu/amc/e-exams/e4-amc08/e4-1-8archive/2006-8a/2006-amc8Perfect.shtml[/url]
scroll down to NC

Dude... I am pretty proud to have a perfect score on AMC 8... from NC... go me I have a lot of potential huh... like how many other people can say they got a perfect score... oh man I'm definitely like IMO perfect score material now

I don't think the AMC8 is that hard...most of my class finished it in about half an hour, and the material is mostly below Algebra II. Making careless mistakes was probably what stopped a lot of people from getting perfect scores. I mean, I got a perfect score on the AMC8, but I got an 85 on my last Geometry test and I'm having trouble in Algebra II. >.<

I failed. Period.

[hide="this is what I got"]
(insert random equation/formula for my score)=24
[/hide]


calculus
integration
logarithms
trigonometry
function
absolute value
calculus computations

can you solve :D

$ \int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\;\;\ln\;\bigg|\;\frac{\tan y+\sqrt 7}{\tan y-\sqrt 7}\;\bigg|\;\;\textbf dy$

[ PS: i have a feeling didilica, jmerry, sylow, perfect radio
and sos440 will spend sleepless nights on this one !  :rotfl: ]

1) Let $ f(x)=ln(\frac{tan(x)+\sqrt{7}}{tan(x)-\sqrt{7}})$, without the absolute value in the $ log$ (obviously we use “the” complex $ log$). The analytic function $ dilog$ is defined as follows : let $ u$ be a complex number; a) if $ \parallel 1-u\parallel \leq1$ then $ dilog(u)=\sum_{n=1}^\infty\frac{(1-u)^{n}}{n^{2}}$; b) else let $ v=\frac{-u}{1-u}$; then $ \parallel 1-v\parallel \leq1$ and $ dilog(v)=-dilog(u)-\frac{\pi^{2}}{6}-\frac{1}{2}(ln(u-1))^{2}$ (it's a particular case of the polylog function).
Let $ G(x)$ be an antiderivative (continuous in $ arctan(\sqrt{7})$) of $ ln(|\frac{tan(x)+\sqrt{7}}{tan(x)-\sqrt{7}}|)$.
 Maple can calculate $ F(x)$, an antiderivative of $ f(x)$; we can choose $ G$ s.t. if $ \frac{\pi}{2}>x>arctan(\sqrt{7})$ then $ F(x)=G(x)$; if $ \frac{\pi}{3}\leq{x}<arctan(\sqrt{7})$ then $ Re(F(x))=G(x)$. Thus $ G(\frac{\pi}{2}-0)-G(\frac{\pi}{3})= F(\frac{\pi}{2}-0)-Re(F(\frac{\pi}{3}))$.
2) With Maple we obtain $ F(\frac{\pi}{2}-0)=-Im(dilog(\alpha))\approx{0.485}$ where $ \alpha=\frac{-7+\sqrt{7}I}{-3+\sqrt{7}I}$.
$ Re(F(\frac{\pi}{3}))=(arctan(\frac{\sqrt{3}+\sqrt{7}}{1-\sqrt{21}})+\pi)ln(\frac{5+\sqrt{21}}{2})-Im(dilog(\beta))\approx{-0.404}$ where $ \beta=\frac{7\sqrt{3}+\sqrt{7}+(7-\sqrt{21})I}{(\sqrt{7}-\sqrt{3})(-3+\sqrt{7}I)}$.
We deduce easily the asked result ($ \approx{0.889}$).


I liked Japanese theorem so I got some anothers results:

Let ABCD be a inscribed quadrilateral, led (X, x) be the A-excircle of ABD, (Y,y) the B-excircle of ABC, (Z,z) the C-excircle of BCD and W the D-excircle of ACD. Then:

i. x+z=y+w
ii. XYWZ is a rectangle.

Are They known? I'd like those were Chavez's theorems  :blush: (I'm José Carlos Chávez)

I'm looking for another ones (I mean I'm looking for the result wich says that the sum of the inradiis of the triangles made by a partition of a inscribed poligon doesn't depend of the diagonalization)

Could anybody tell me?

thanks a lot

José Carlos

Unfortunately this is a well know fact. It is a corollary of  Japanes Theorem.
I did read this in a book proposed  as problem, but i don't remember the author. By the way,  you can see my short proof of Japanes Theorem at:
http://www.campus-oei.org/oim/revistaoim/numero13/TeoremasPorJCS.pdf


For any natural numbers $ n\ge2$, prove that

$ n^{\frac n 2} < n!\le 2^{\frac {n(n\minus{}1)} 2}$

 :w00t:

We will  prove that $ n^{\frac n 2} \le n!$ for all $ n\in \mathbb N$.

1.If $ n\equal{}2k$, $ k\in \mathbb N$ then

$ n!\equal{}1\cdot 2...k(k\plus{}1) ...2k\equal{}(1\cdot 2k)\cdot (2(2k\minus{}1))...(k(k\plus{}1))\geq (2k)^k\equal{}n^{\frac n 2}$

2.If $ n\equal{}2k\plus{}1$, $ k \in \mathbb N_0$ then

$ n!\equal{}1\cdot 2...k(k\plus{}1)(k\plus{}2) ...(2k\plus{}1)\equal{}(1\cdot (2k\plus{}1))\cdot (2(2k))...(k(k\plus{}2))(k\plus{}1)\geq (2k\plus{}1)^k(k\plus{}1)\geq(2k\plus{}1)^{k\plus{}\frac{1}{2}}\equal{}n^{\frac n 2}$


function
complex analysis
real analysis
real analysis unsolved

Help me solve the problem below:
Does there exist a function $ f$ :$ N$  $ \rightarrow$$ N$ satisfies the following three conditions:
1/ $ f(m) = 1\Longleftrightarrow m = 1$
2/Let $ d = gcd(m,n)$,we have  $ f(mn)f(d) = f(m)f(n);m,n\in N$
3/ $ f(f(f(f...f(m))...) = m;\all m\in N$
(2010 times)

Well, this was a marginally more appropriate place to post this problem than the complex analysis forum  :roll:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=306684


modular arithmetic
number theory unsolved
number theory

Prove that if $\left[n\sqrt{2}\right]\equiv a_{n}\pmod{10}$ and $-1<a_{n}<10,$ then the sequence $a_{n}$ is not periodic.

If $a_{n}$ is periodic then there exists $T\in\mathbb{N}$ such that for any $n>n_{0}$
\[[(n+T)\sqrt2]\equiv[n\sqrt2]\pmod{10}\]
But $[(n+T)\sqrt2]-[n\sqrt2]=[T\sqrt2+\{n\sqrt2\}]$, hence $[T\sqrt2+\{n\sqrt2\}]$ is constant for $n>n_{0}$.
But $\varliminf_{n\to\infty}[T\sqrt2+\{n\sqrt2\}]=[T\sqrt2]$  and  $\varlimsup_{n\to\infty}[T\sqrt2+\{n\sqrt2\}]=[T\sqrt2]+1$


conics
parabola
analytic geometry
function
calculus
integration
calculus computations

Consider a lamina that occupies a region bounded by the parabola x=1-y^2 and the coordinate axes in the first quadrant with density function p(x,y)=y. Find the mass and the center of mass of the lamina.

I am not sure where the region actually bounded, any one can help me out? Thanks alot.

Since the density is a function of $y$ alone, it would be more convenient to slice the figure liyi drew into horizontal strips and set up the integrals you need with respect to $dy.$


vector
integration
LaTeX
calculus
calculus computations

A vector field $ \vec{G}$ in 3-space is defined outside the cylinder $ x^2 + y^2 = 4$

$ \vec{G} = \frac {6y\vec{i} - 6x\vec{j}}{x^2 + y^2}$

Find $ \int\limits_S \vec{G} \cdot d\vec{A}$ where S is the open cylinder $ x^2 + y^2 = 16 , 0 \leq z \leq 7$ oriented outward.

[color=green]Moderator edit: fixed a bad character in $ \color{green}{\text{\LaTeX}}$.[/color]

If what you're looking for is the flux of $ \vec{F}$ trhough the surface, it's null, cause the vector is tangent to the surface.


In
\[ \Delta ABC
\]

, show that
\[ s \ge \frac {3}{2}\sqrt {6Rr}
\]
with usual notations

hello, from the AM-GM inequality we get:
$ \frac{a\plus{}b\plus{}c}{3}\geq \sqrt[3]{abc}$, this is equivalent with
$ \frac{(a\plus{}b\plus{}c)^3}{27}\geq abc$
$ \left(\frac{a\plus{}b\plus{}c}{3}\right)^2\geq \frac{6abc}{4s}$
$ \frac{a\plus{}b\plus{}c}{3}\geq \sqrt{6\frac{abc}{4A}\frac{A}{s}}$
$ \frac{a\plus{}b\plus{}c}{3}\geq \sqrt{6Rr}$
$ \frac{a\plus{}b\plus{}c}{2}\geq \frac{3}{2}\sqrt{6Rr}$
Note that the following equations are true :$ A\equal{}\frac{abc}{4R}$ and $ A\equal{}sr$
Sonnhard.


Ring Theory
superior algebra
superior algebra unsolved

Prove that if 5|d and  d is 2 or 3 mod 4, then Q(sqrt(d)) is not a UFD.

I showed that it's not always Euclidean. eg. when d<-11.


A positive integer k is given. Define a sequence (an) by

a1 = k + 1, an+1 = an2 - kan + k.

Show that if m, n are distinct, then am and an are relatively prime.

...

It is easy to show
an+1=a1a2...an+k.
Thus we must prove (an,k)=1. It is obvious (by induction).


Hello Is there anyone who guesses which kind of problem will come in IMO this year in NT and others

It is not the correct place to discuss this type of questions ...

Davron


I think you makes mistakes in finding $S_{a}$  :o  !


inequalities
geometry
3D geometry
sphere
LaTeX
inequalities unsolved

The problem say let p_1 , p_2, p_3,..., p_n be a point on the sphere of radius then if dij is the distance between point i and j then

:Sigma:d2ij < = n2

where i< j .

And how to enable the latex posting. i wanted to post using latex command but it did not work.
regards,
Amit

The problem say let $p_1 , p_2, p_3,..., p_n$ be a point on the sphere of radius then if $d_{ij}$ is the distance between point $i$ and $j$ then $\sum_{i<j} d^2_{ij} \leq n^2$.

You can use "Quote" to watch reply source and learn Latex from examples :)

There is a beautiful proof via moment of inertia and Steiner's theorem.

I got it :). I know latex but did not knew how to use it here.  Anyway.. MOre information about the solution will be gratefully acknowledged. can you tell me any sourse where i can get any hint. :) thanx myth..

Amit

[quote="Myth"]Rrrr... [b]M[/b]yth.

BTW, math is feminine in Russian...  :lol:[/quote]

Thank god. I am hetrosexual :) . I searched net on  what is Steiner's theorem and still have not got a clue how to sovle this prob. :( i spent my whole of last night trying to solve this . I can't live like this :((.

Amimt

Its a sphere.

hence

$P_i=(x_i,y_i,z_i)$.  Anyway now i know how to solve it for the 3D prob. thanx Myth. You be the way you are.

Regards,
Amit

Ops... Really, it is a sphere. But the method exactly the same.

how to change it to a physics problem?! (you mentioned moment of inertia..)

Consider our points as mass-points of mass 1. There is a Lagrange formula for moment of inertia wrt to center of gravity $M$. In this particular case it gives $\frac{1}{n}\sum_{i<j} d_{ij}^2$. On the other hand there is Steiner's theorem: $I_O=I_M+m\cdot OM^2$. Let $O$ be center of sphere. Then $I_O=n$, so $n=I_O\geq I_M=\frac{1}{n}\sum_{i<j} d_{ij}^2$.

thx!
unluckily.... i don't know that formula :(


function
topology
real analysis
real analysis unsolved

Let f:[0,1]->[0,1] be a continuous function s.t. for each x in [0,1] the sequence f(x),f(f(x),... contains 0. Does it follow that for n sufficiently large f<sup>(n)</sup>(x)=0 ofr all x in [0,1]?

It's been giving me a headache for quite some time now. I think I'm very close to solving it, but some details keep not working.

Uf. I think, it follows, but I have just discover something monstrous.

1. f(0)=0 and f(x)<x for x>0 (elsewhere there exists positive x suxh that f(x)=x, that immediately lead to contradiction).

2.  So, f^n(x)<=f^k(x)  for n>k.

3. Assume that for some a>0, f(y)=0 for all y<a. Then for any x, consider the open neighborhood d(x) of x and positive integer k(x) such that
f^{k(x)}(y)<a for all y from d(x) (take k(x) such that f^{k(x)}(x)=0 and then we may find d(x) by continuity).

4. There is a finite number of d(x_i), i=1..n, which cover the segment [0,1] (Borel lemma, or how is it called?) Put k:=max(k(x_i)). By (2),
f^k(x)<a for any x from [0,1], so f^{k+1}=0 identically.

5. So, for every a>0 there exists x<a: f(x)>0. But clearly, for every a>0 there exists positive x<a: f(x)=0 (else start with some x<a, then f(x), f(f(x)) etc is decreasing- by (2)- sequnece of positive numbers).

6. Then, the open set X={x:f(x)>0} is union of countable many disjoint intervals, and it has intervals arbitrarily close to zero.

7. Now we construct the sequence of those intervals. Start with interval
D1:=[a1,b1], let c1:=max_{D_1} f. Then take D2=[a2,b2] such that b2<c1; c2:=max_{D_2} f and so on.

8. Consider the set S_n of points x in D_1 such that f(x) in D_2, f(f(x)) in D_3, ..., f^n(x) in D_{n+1}.

9. I claim that S_n is non-empty. Indeed, take arbitrary c_{n+1} from D_{n+1}, then we find c_n in D_n such that f(c_n)=c_{n+1}, then c_{n-1} in D_{n-1} such that f(c_{n-1})=c_n and so on, we find c_1 from S_n.

10. Each S_n is a closed set (if D_k are closed intervals). So, their intersection is non-empty.

11. But for every point a from their intersection, the sequence a, f(a), f(f(a)) and clearly does not vanish.

Hmmm.. I was trying to prove that it doesn't follow, so maybe that was my mistake. I didn't check everything you wrote here, but I will.

Anyway, thanks for solving it!

Hello!

I read this proof and it seemes to be correct.
I tried to solve that problem yestoday, but I thought there exists function with "strange" property.

[quote="Fedor Petrov"]
6. Then, the open set X={x:f(x)>0} is union of countable many disjoint intervals, and it has intervals arbitrarily close to zero.

7. Now we construct the sequence of those intervals. Start with interval
D1:=[a1,b1], let c1:=max_{D_1} f. Then take D2=[a2,b2] such that b2<c1; c2:=max_{D_2} f and so on.
[/quote]
I was browsing through the old post and found this. I liked the problem but had trouble understanding Fedor's solution: what exactly are $D_1$, $D_2,\dots$? My first impression was that they were the constituting intervals of the open set $f>0$, but then they should be open, not closed. Another possibility would be to have them closures of the constituting intervals, but then the Step 11 in the proof fails. One more possibility is to have them compact subintervals of the constituting intervals, but then I do not understand step 9. Can anybody clear this mistery?  :?

Oops, stupid me  :blush: ... I didn't realize that $f(x)\neq 0$ because $f(x)\in D_2$, not just because $x\in D_1$! The proof is correct, of course. Sorry for the false alarm...


function
calculus
derivative
analytic geometry
graphing lines
slope

$ y\equal{}(x\plus{}3)(x\minus{}2)^2$

[img]http://i22.tinypic.com/xdz7tf.jpg[/img]

When you are trying to sketch $ y^2\equal{}f(x)$ how do you know if at $ x\equal{}2$ the line pass straight through or bounces and goes back up

the graph you sketched doesn't seem to be right.
e.g. x=0 then y=12

It will probably bounce back up.
All values of a squared real function is positive,  So it can't be negative.

That's what I think.

[quote="bos1234"]$ y = (x + 3)(x - 2)^2$

(Image)

When you are trying to sketch $ y^2 = f(x)$ how do you know if at $ x = 2$ the line pass straight through or bounces and goes back up[/quote]

Please clarify what you are asking. A function such that:
$ f(x)=y$
$ f(x)=y^2$

Is not a function since it returns [b]more than one[/b] value of $ f(x)$. And in general $ f(x)=y$ is just to show that the output of the function $ f$ is represented on the y-axis in a graph, so I really don't see what $ y^2=f(x)$ is supposed to mean. And your graph, which looks like a Jesus fish, makes no sense either.  :huh:

Though if instead are using $ y$ to represent $ (x+3)(x-2)^2$ (next time use a different variable that is not so common in functions), and $ f(x)=((x+3)(x-2)^2)^2$, then I think the graph would go straight upwards. This is because the leading coefficient of the expansion will be positive, and that the function has a degree that is even. Thus as $ x\to\infty$, $ f(x)\to\infty$.

anazerzia seems to be right.
I made a mistake.  Accidently thought of the square on the other side.
IT will be good if the question could be clearer.

The graph seems to be correct.

It is not a function but a relation. (does not need to pass vertical line test)

I dont know what you mean by this. (line pass straight through or bounces and goes back up)
Please clarify.

I can tell you this that there is a discontinuity point in the derivative of that point, since the slope has two different values there.

[quote="t0rajir0u"]Do you mean the relation $ y^2 \equal{} (x \plus{} 3)(x \minus{} 2)^2$?

The answer to your question is that it doesn't matter.  At such a point (known as a [b]crunode[/b]), the graph can be drawn either way, as long as the end result is that there are two intersecting branches.  The idea of graphs as lines that are drawn from beginning to end loses meaning when crunodes and other singular points are introduced.[/quote]

Is a singular point only located on the x axis?

If so then in three dimensions would the singular points only be located on the x-y plane?

[quote="t0rajir0u"]Absolutely not (why would you think so?).  Graphs can be translated arbitrarily, and so can singular points.  For example, the graph $ (y \minus{} b)^2 \equal{} (x \minus{} a)^2$ is simply two lines intersecting at the point $ (a, b)$, which is a crunode.

In three dimensions, singular points become [b]very complicated.[/b][/quote]

lol. thank you.

The reason I thought this was because I solely looked at $ (y)^2 \equal{} (x \minus{} a)^2$ and not $ (y \minus{} b)^2 \equal{} (x \minus{} a)^2$

[quote="t0rajir0u"]Do you mean the relation $ y^2 \equal{} (x \plus{} 3)(x \minus{} 2)^2$?

The answer to your question is that it doesn't matter.  At such a point (known as a [b]crunode[/b]), the graph can be drawn either way, as long as the end result is that there are two intersecting branches.  The idea of graphs as lines that are drawn from beginning to end loses meaning when crunodes and other singular points are introduced.[/quote]

ok I understand now

thanks for the helps everyone


If 4 horses eat 4 bales of hay in 4 days, how many days will it take 20 horses to eat 30 bales of hay at the same rate?

If $ h$ horses eat $ b$ bales of hay in $ d$ days, then the values $ hd$, $ \frac{h}{b}$, and $ \frac{d}{b}$ are all constants, though not necessarily equal to each other.  The square root of their product, which is $ \frac{h\cdot d}{b}$, must also be a constant then.

From this, we can say that \[ \frac{4\cdot 4}{4}\equal{}\frac{20\cdot d}{30} \Rightarrow d\equal{}\boxed{6}\]


number theory
least common multiple
algebra unsolved
algebra

Let $ ^n\!C_r \equal{} C_r$ and $ (a_1, \ a_2,\ a_3, \ldots, a_n)$ denote the l.c.m of the numbers $ a_i$'s.
Prove the following:
$ (C_0,\, C_1,\,C_2,\,\ldots,\,C_n) \equal{} \dfrac{1}{n\plus{}1}\,(1,\,2,\,3,\ldots,\, n\plus{}1)$

It deeply saddens me to have to say this once again, after I thought the entire issue had died out. This is an Indian Postal problem and it is still 'open for submission'. So, please do not post any solutions.


Prove that:$ x^2\plus{}4y^2<1$ With x,y be positive real numbers such that :$ x^3\plus{}y^3\equal{}x\minus{}y$.

I am not sure about the validity of the following solution.

If $ x\equal{}y$ then from the relation $ x^3\plus{}y^3\equal{}x\minus{}y$ we can receive that $ x\equal{}y\equal{}0$.
Since $ x,y$ are positive numbers this can't happen.However if we consider $ x,y$ to be $ 0$ as
well then $ 0<1$ so the inequality is still satisfied.But anyways,

Let's assume that $ y>x$.Then we get:

 $ 1\equal{}\frac{x^3\plus{}y^3}{x\minus{}y}$. Hence it suffices to prove that:

  $ \frac{x^3\plus{}y^3}{x\minus{}y}>x^2\plus{}4y^2$ or $ x^3\plus{}y^3>(x\minus{}y)(x^2\plus{}4y^2)\equal{}x^3\plus{}4xy^2\minus{}x^2y\minus{}4y^3$ or

  $ 5y^2>4xy\minus{}x^2$. But $ 5y^2\equal{}4y^2\plus{}y^2>4xy\plus{}x^2>4xy\minus{}x^2$.

[quote="new_member"]
Let's assume that $ y > x$.Then we get:[/quote]

Actually it is obvious that $ x>y$


If you get a warning, go to internet security settings and go to the advanced tab, then click "Active content"

Oh, you will also need the source files. Save each file in the same folder as the source file.


calculus
integration
algebra
function
domain
Ring Theory
superior algebra

let R be commutative ring with identity s.t.
 x^t=x for some natural no. t depending on x,foer all x in R.then show any prime ideal of R is maximal. :)

Let $\mathfrak{p}$ be an ideal.

$\mathfrak{p}$ prime $\iff$ $R/\mathfrak{p}$ is a integral domain.
$\mathfrak{p}$ maximal $\iff$ $R/\mathfrak{p}$ is a field.

But every nonzero element $x$ of $R/\mathfrak{p}$ is invertible: by assumption, we find $t$ such that $x^{t}\equiv x \mod \mathfrak{p}\iff x(x^{t-1}-1) \equiv 0 \mod \mathfrak{p}$, thus (remember that we are in an integral domain) $x=0$ (excluded) or $x^{t-1}\equiv 1 \mod \mathfrak{p}$, giving that $x^{t-2}$ is the desired inverse and we get a field. This means that our prime ideal was maximal.


I was reading about the massage tool in ACoPS. I didnt really understand it. Can someone give me an example of it.(the example they gave in the book was find $ \sum_{n\equal{}1}^{10000} \frac{1} {\sqrt{n}},$ and i didnt really understand that)

I have no idea what you could possibly mean by "massage tool."  Could you elaborate?

I think he means [url=http://www.1800wheelchair.com/asp/view-product.asp?product_id=1994]this[/url]:

[img]http://www.1800wheelchair.com/siteimages/large/pressure-jackII.jpg[/img]

very funny.

I dont know. I dont understand it either.

The book says to massage the terms so they become easier to work with. like it says $ 2(\sqrt{n+1}-\sqrt{n}) < \frac{1} {\sqrt{n}} < 2(\sqrt{n+1}+\sqrt{n})$. so $ \sum_{n=1}^{10000} {2(\sqrt{n+1}-\sqrt{n}) < \sum_{n=1}^{10000} {\frac{1} {\sqrt{n}} <\sum_{n=1}^{10000} {2(\sqrt{n+1}+\sqrt{n})}}}$
$ 2\sqrt{100000}-\sqrt{0}< \sum_{n=1}^{10000} {\frac{1} {\sqrt{n}} < 2\sqrt{10001}-2}$
and its easy to work from there.

I just dont understand how to use it in problems other than this one.

That technique is essentially approximating an integral by left and right [url=http://en.wikipedia.org/wiki/Riemann_sum]Riemann sums[/url].  It has application in other problems of this type, such as bounding partial sums of the harmonic series by logarithms, but general estimating tricks usually aren't things that can be written down in the form of hard lemmas.

There isn't any "tool" here per se. What they mean by "massaging" is that you tweak, play with, modify and coax the expression until it does what you want it to do. This is an art, not a tool or a recipe. In this case, it's bounding an expression between telescoping terms. Great technique but it's not necessarily obvious how to find the right expression.


hi,
can anyone help me to find a book about  olympiad combinatorics (free)
please , it's lifeblood to me !!
thnx

102 Combinatorial Problems by Titu Andreescu,Zuming Feng. It is free, i think so.

thnx
but where can i find it ? at google i think  :oops: ?!
can i find in it some solutions to those problems ? there is some theorems in it ?
please answer me  :maybe: !!


Find the number of positive integers $ a$ of $ n$ digits s.t. $ a\plus{}b$ has only odd digits, where $ b$ is obtained from $ a$ by reversing the order of digits.

[quote="ovidius"]Find the number of positive integers $ a$ of $ n$ digits s.t. $ a \plus{} b$ has only odd digits, where $ b$ is obtained from $ a$ by reversing the order of digits.[/quote]

http://www.mathlinks.ro/viewtopic.php?t=258711

No need to repost it.


Let us assume that the sun radiates like a blackbody with a surface temperature of $T = 6000 K$ and the earth absorbs radiation coming only from the sun.If both the earth and the sun are considered as perfect spheres with distance between the centre of the earth and the centre of the sun to be $200 times$ the radius of the sun.Find the temperature in Kelvin of the surface of the earth in steady state.

( in this question there is a small thingwhich cannot be ignored Well first give it a try you may possibly get it rightl ...........

I think I had already posted this or a very similar problem. Let's assume both the Sun and the Earth are absolutely black bodies. The power of the Sun's radiation is $P = \sigma T^{4}\cdot 4R^{2}\pi$, where $R$ is the radius of the Sun. Considering the Sun as a particle homogenously radiating the power $P$ in the space, we can find the portion of this power falling on the Earth. This is the crucial point. Since the Sun radiates uniformly in all directions, the Earth will receive the power $\omega P/4\pi$, where $\omega$ is the spatial angle, in this case approximately $r^{2}\pi/d^{2}$, where $r$ is the radius of the Earth and $d$ is the distance between the Sun and the Earth (we have neglected the Sun's radius here).

Since the Earth is in equilibrium, it emits all the power it absorbs, so we have that
\[4 \sigma\pi R^{2}T^{4}\frac{r^{2}\pi}{4d^{2}\pi}= 4\sigma\pi r^{2}T_{0}^{4},\]
so we can obtai that
\[T_{0}= T \sqrt{\frac{R}{2d}}= 300 \text K.\]

oh no i did exactly the same thing......but the stupidest of all calculation errors....should slap myself for that.....never mind ...thanks


Bueno, el problema 2 no lo posteo, xq sinceramente es un problema muuuuuuy facil, y no vale la pena. Aki el problema 3.

Sea n un entero mayor que 1. ¿De cuantas formas se pueden acomodar todos los numeros de $1, 2, ..., 2n$ en las casillas de una cuadricula de 2xn, uno en cada casilla de manera que cualesquiera dos numeros consecutivos se encuentren en casillas que comparten un lado de la cuadricula?

Espero soluciones... :lol:

Una pregunta:
¿1 y 2n tienen que estar pegados?

Para cada acomodación, sea $A$ el conjunto de los $1<i<n$ tales que en la $i$-ésima columna se ubican $2$ números consecutivos.
Separemos las acomodaciones posibles en $2$ tipos.
Si $A=\emptyset$; en este caso, la acomodación queda determinada por la ubicación del número $1$ y la dirección (de las $2$ posibles) en la que se ubica el $2$.
Entonces tenemos $4n$ acomodaciones.
Si $A\neq\emptyset$; en este caso, $A$ consiste de $k$ números consecutivos ($1\leq k\leq n-2$) y entonces, cada acomodación queda determinada por $A$, por la fila en la que se ubica el menor número de todos los números ubicados en las columnas correspondientes a $A$ y, hacia que lado (izquierda o derecha) de las columnas correspondientes a $A$ se ubica el número $1$ .
Entonces tenemos $4$ acomodaciones, para cada conjunto $A$.
En total, tenemos
$4n+4((n-2)+(n-3)+\dots+1)=2(n^{2}-n+2)$
posibles acomodaciones.

Ola jonfv, sinceramente no entendi tu solucion muy bien, aunque esa es la respuesta correcta.
Mi solucion fue fijandome cuantos arreglos existen para cada ubicacion del 1. Para esto separe en 2 casos, cuando el 1 se encuentra en una esquina y cuando no se encuentra en la esquina. Dejo como tarea que prueben que el numero de arreglos en el primer caso es n y en el segundo caso es (n-1).
Ya con eso concluyes que hay $(n-1)(2n-4)+4n=2n^{2}-2n+4$

Notemos que lo que hay que hacer es un camino que vaya de $1$ hasta $2n$. Si coloreamos como ajedrez nos damos cuenta que el $1$ y el $2n$ deben estar en dos casillas de distinto color. Hay $2n^{2}$ formas de poner el $1$ y el $2n$ en casillas que cumplan. Sin embergo para todas las casillas que no sean esquinas no es posible terminar en la misma columna (si hay $n$ columnas), pues si empieza y acaba ahi corta en dos el tablero, y no se puede pasar de un lado al otro. Hay $2n-4$ casillas que no son esquinas. Entonces las restamos. Es facil ver que el primero y el ultimo termino determinan los demas numeros del tablero (¿cómo pongo una imagen para explicar?) y que todos esos se pueden.
Entonces hay $2n^{2}-2n+4$ caminos.

Disculpa jasamper88, kreo k la parte k dice: "Es facil ver que el primero y el ultimo termino determinan los demas numeros del tablero" no es tan trivial y requiere la demostracion de la prueba!!
 :D

Por eso dije que necesitaba poner una imagen..  :wink:
Se basa en que despues del 1(si no esta en la esquina) hay que arrancar hacia uno de los lados hasta llegar al borde y devolverse pasar por debajo del 1 (hay que ir hacia el lado en que no esta 2n, pues en el momento en que pase por debajo de 2n, deben estar cubiertos todas las casillas que esten del mismo lado de 1 comparado con 2n, pues el tablero no se puede separar en dos pedazos.).  Y eso es mas o menos.. (pero seria mejor con una imagen por eso insisto).. no se en que formato ponerlas para poder pegarlas.
Si el 1 esta en una esquina tenemos que llenar todas las casillas que estan del mismo lado de 2n antes que pasar por la columna de 2n.

PS. Podrias poner los otros problemas de la olimpiada?


Right for $x\in [0,\pi]$
To expand on $x\in [-\pi, 0]$ just replace $x$ by $|x|$ in RHS.

[quote="digger"]Right for $x\in [0,\pi]$ [/quote]

How would one prove that? I tried taking derivatives, but I found nothing...

This is done with Fourier series and i dont think it's place is here

LHS is serie Fourie for RHS.
Expanding by period $2\pi$ we can conclude that RHS is $\frac{\pi}{4}\cdot \arcsin (\cos x)$. This is true for $x\in R$.

Taking second derivative of $f(x)=\sum_{n=1}^\infty\frac{\cos(2n-1)x}{(2n-1)^{2}}$ we get $f''(x)=-\sum_{n=1}^\infty\cos(2n-1)x=-\Re\left(\sum_{n=1}^\infty\text{cis}^{2n-1}x\right)=-\Re\left(\frac{\text{cis}\,x}{1-\text{cis}^{2}x}\right)$. After simplification, we find $-\Re\left(\frac{\text{cis}\,x}{1-\text{cis}^{2}x}\right)=-\Re\left({i\over 2\sin x}\right)=0$. Hence $f(x)=ax+b$. Putting $x=0$ we get $b=\sum_{n=1}^\infty{1\over (2n-1)^{2}}={\pi^{2}\over 8}$, and putting $x={\pi\over 2}$ we get $a{\pi\over 2}+b=0\iff a=-{\pi\over 4}$, giving $f(x)=-{\pi\over 4}x+{\pi^{2}\over 8}={\pi\over 4}\left({\pi\over 2}-x\right)$

On the other hand, the last series shouldn't be treated as convergent, since $|\text{cis}^{2}x|=1$...  :(


linear algebra
matrix
probability
inequalities
limit
probability and stats

Suppose $ \{X_n\}$ is a finite state Markov chain such that a state $ i$ is reachable from every other state $ j$. Can we conclude from this that $ \exists n$ s.t.$ P(X_n\equal{}i)\equal{}1$?

no, e.g., consider the transition matrix $ T\equal{}\left(\begin{array}{cc}0.5 & 0.5\\0.5 & 0.5\end{array}\right)$. Each state is reachable, but $ T^2\equal{}T$

Assume that states $ \{0,1,2,3,4\}$ are related as follows: $ P(0\to1) \equal{} \frac {1}{3}$, $ P(0\to 2) \equal{} \frac {1}{3}$, $ P(0\to3) \equal{} \frac {1}{3}$, $ P(1\to2) \equal{} \frac {1}{2}$, $ P(2\to1) \equal{} 1$, $ P(1\to3) \equal{} \frac {1}{2}$, $ P(3\to4) \equal{} 1$, and $ P(4\to1) \equal{} 1$, with initial state $ X_0 \equal{} 0$.  Note that since $ P(X_i \equal{} 1)$ means that $ P(X_{i \plus{} 2} \equal{} 1) \equal{} P(X_{i \plus{} 3} \equal{} 1) \equal{} \frac {1}{2}$, and $ P(X_1 \equal{} 1) \equal{} P(X_2 \equal{} 1) \equal{} \frac {1}{2}$, then $ P(X_i \equal{} 1)\neq1$.  Note also that, except for $ 0$, all other states are reachable from every other state.

Edit: Just posted my solution and saw hjbrasch's clearly much simpler...   :oops:

But, if my state $ i$ is further an absorbing state then my conclusion follows, isn't it?

[b]no[/b] to your question about absorption, e.g., consider for $ 0<a<1$ the transition matrix $ T\equal{}\left(\begin{array}{cc}a & 1\minus{}a\\ 0 & 1\end{array}\right)$, i.e. state 2 is reachable from state 1 and it is absorbing, but $ T^n\equal{}\left(\begin{array}{cc}a^n & 1\minus{}a^n\\ 0 & 1\end{array}\right)$

Can I atleast conclude that $ lim_{n\to \infty}P(X_n\equal{}i)\equal{}1$ if $ i$ is the only absorbing state and is reachable from all other states?

[quote="limsup"]Can I atleast conclude that $ lim_{n\to \infty}P(X_n \equal{} i) \equal{} 1$ if $ i$ is the only absorbing state and is reachable from all other states?[/quote]

As far as I know, that would be a yes, but I cannot think right now of a proof, I'll see if I can cook something up.  In this last case that you mention, the condition that the number of states is finite is actually necessary, since otherwise you could have an infinite loop leading to the absorbing state...

I think you meant  $ P((\exists n)(X_n \equal{} i)) \equal{} 1$.
If $ i$ is the only absorbing state, and can be reached directly from everywhere, all other states must be transient.
Hence each one is visited atmost finitely many times.
Hence with probability $ 1$ we end up in $ i$.

[quote="bubka"]I think you meant  $ P((\exists n)(X_n \equal{} i)) \equal{} 1$.[/quote]
Yes I meant this only.

Moreover, if I have only two absorbing states, say, $ i$ and $ j$, and both are reachable from all other states, can I conclude that  $ P((\exists n)(X_n \equal{} i) \cup (\exists m)(X_m \equal{} j) ) \equal{} 1$

We have $ \mathbb{P}(\exists n: X_n\equal{}i)\equal{}1$ already by definition, since state $ i$ was assumed to be reachable from every starting point.

@ hjbrasch: I understand that must be true for a finite chain, but I am curious to know why according to you it that follows 'by definition'.

[b]Bubka[/b], for the proof I only use linear algebra.

Perform the block decomposition $ T \equal{} \left(\begin{array}{cc}A & (I \minus{} A)\bar{e} \\
\bar{0}^t & 1\end{array}\right)$ for the transition matrix, where $ \bar{e}\in\mathbb{R}^{n \minus{} 1}$ consists only of $ 1$-entries and let the last row "correspond" to the absorbing state. Direct calculation yields $ T^m \equal{} \left(\begin{array}{cc}A^m & (I \minus{} A^m)\bar{e} \\
\bar{0}^t & 1\end{array}\right)$.
Recall that $ \mathbb{P}(\exists k\leq m: X_k \equal{} i|X_0 \equal{} j) \equal{} 1 \minus{} \mathbb{P}(\forall k\leq m: X_k\not \equal{} i|X_0 \equal{} j) \equal{} 1 \minus{} \sum_{r \equal{} 1}^{n \minus{} 1}(A^m)_{jr}$ holds.

Since $ A$ is a non-negative matrix, we can apply Perron-Frobenius and get for all eigenvalues $ \lambda$ of $ A$ the inequality $ |\lambda |\leq \rho(A)\leq\max_{k \equal{} 1,\ldots,n \minus{} 1}\left\{\sum_{j \equal{} 1}^{n \minus{} 1}A_{kj}\right\}\leq 1$, where $ \rho(A)$ is the spectral radius. We now perform a case distinction:

[b]Case 1:[/b] For $ \rho(A) \equal{} 1$ let $ \bar{v}\in\mathbb{R}^{n \minus{} 1}$ denote an eigenvector of $ A$ corresponding to the eigenvalue $ 1$. After reordering we can assume without restriction $ 0\leq v_1\leq\ldots\leq v_{n \minus{} 1}$ and $ v_{n \minus{} 1} > 0$. Choose $ r \equal{} \min_{k \equal{} 1,\ldots,n \minus{} 1}\{v_k \equal{} v_{n \minus{} 1}\}\leq n \minus{} 1$. This yields $ \forall k \equal{} r,\ldots,n \minus{} 1$ the inequality:

$ v_k \equal{} \sum_{j \equal{} 1}^{n \minus{} 1}A_{kj}v_j\leq\sum_{j \equal{} 1}^{n \minus{} 1}A_{kj}v_{k}\leq v_{k}$, i.e. we must have equality in each step implying $ \sum_{j \equal{} 1}^{n \minus{} 1}A_{kj} \equal{} 1$ and $ A_{kj} \equal{} 0$ for all $ j < r$. From this we conclude that $ A$ has the form $ \left(\begin{array}{cc}* & * \\
0 & A'\end{array}\right)$ with $ A'\in M_{n \minus{} r}(\mathbb{R})$ a stochastic matrix. But then the state set $ r,\ldots,n \minus{} 1$ is closed in the sense that we can not leave it, in particular, we can not reach the state $ n$ from it, a contradiction to the assumption that this state is reachable.

[b]Case 2:[/b] $ \rho(A) < 1$ yields $ \lim_{m\rightarrow\infty}A^m\equal{}0$ and therefore $ \mathbb{P}(\exists n: X_n \equal{} i|X_0 \equal{} j) \equal{} 1 \minus{} \lim_{m\rightarrow\infty}\sum_{r \equal{} 1}^{n \minus{} 1}(A^m)_{jr} \equal{} 1$


geometry
inequalities
analytic geometry
vector
circumcircle
function
trigonometry

Let $C_1$ and $C_2$ be two concentric circles in the plane with radii $R$ and $3R$ respectively. Show that the orthocenter of any triangle inscribed in circle $C_1$ lies in the interior of circle $C_2$. Conversely, show that every point in the interior of $C_2$ is the orthocenter of some triangle inscribed in $C_1$.

Let $O$, the center of $C_1$ and $C_2$, be the origin - I am going to use vectors. Let $ABC$ be any triangle inscribed in $C_1$ and write $\mathbf{a} = \overline{OA}$, $\mathbf{b} = \overline{OB}$ and $\mathbf{c} = \overline{OC}$. Then $|\mathbf{a}| = |\mathbf{b}| = |\mathbf{c}| = R$. Also, the orthocentre $H$ of $\Delta ABC$ is given by $\mathbf{h} = \overline{OH} = \mathbf{a} + \mathbf{b} + \mathbf{c}$. By the triangle inequality, we have \[ |\mathbf{h}| = |\mathbf{a} + \mathbf{b} + \mathbf{c}| \leq |\mathbf{a}| + |\mathbf{b}| + |\mathbf{c}|= 3R, \] and it follows that $H$ lies inside the circle with centre $O$ and radius $3R$.

Anyone for pure geometry or coordinate geometry?
Seems quite tough .....
The problem with of is that i havent yet learnt vector use in geometry.........

After the last post , I realised that a very short and trivial solution exists using trigonometry.....

It is a well known relation that

$ OH \equal{} R \sqrt{1\minus{}8cosAcosBcosC}$

where O is the circumcentre of a triangle, H is the orthocentre, R is the circumradius, and A,B,C the angles of the triangle.

Since $ cosA \cdot cosB \cdot cosC\equal{} cos(180\minus{}(B\plus{}C)) \cdot cos B \cdot  \equal{}\minus{}cos(B\plus{}C) \cdot cosB \cdot cosC < \minus{}1$ since the maximumum value of the cos function is $ 1$ which exists at angle $ 0$ not possible....

Therefore
$ OH \equal{} R \sqrt{1\minus{}8cosAcosBcosC} < R \sqrt{1\plus{}8} \equal{} 3R$

This proves the first part.....

For the second part....
Since $ cosAcosBcosC$ assumes every value between $ \minus{}1$ and $ \frac{1}{8}$ with equality existing only in the latter.....
Therefore every point in the interior of $ C_2$ is the orthocenter of some triangle inscribed in $ C_1$.

It is a well known relation that

$ OH \equal{} R \sqrt {1 \minus{} 8cosAcosBcosC}$

where O is the circumcentre of a triangle, H is the orthocentre, R is the circumradius, and A,B,C the angles of the triangle.

$ cosA \cdot cosB \cdot cosC \equal{} cos(180 \minus{} (B \plus{} C)) \cdot cos B \cdot cos C \\
\equal{} \minus{} cos(B \plus{} C) \cdot cosB \cdot cosC > \minus{} 1$
since the maximum value of the cos function is $ 1$ which exists at angle $ 0$ not possible....

Therefore
$ OH \equal{} R \sqrt {1 \minus{} 8cosAcosBcosC} < R \sqrt {1 \plus{} 8} \equal{} 3R$

This proves the first part.....

For the second part....
Since $ cosAcosBcosC$ assumes every value between $ \minus{} 1$ and $ \frac {1}{8}$ with equality existing only in the latter.....
Therefore every point on a radius of the circle is the orthocentre of some triangle in $ C_1$...So rotating those triangles we get the required statement of the proof...
Therefore every point in the interior of $ C_2$ is the orthocenter of some triangle inscribed in $ C_1$.

Here's a solution using pure geometry for the first part:
Use the relation OH = 3OG, and the fact that G (the centroid) must lie inside the triangle, and hence inside the circle. It follows that H must lie inside the circle of radius 3R

My first part is same as Pascal96's solution.

For the second part (without trigo), let $O$ be the center of both circles, and $HO$ intersects $C_1$ at $A$ and $X$, where $AH\leq AX$. Note that the midpoint of $HX$ lies in $C_1$, so the perpendicular bisector of $HX$ intersects $C_1$ at 2 points $B,C$. Then $H$ is the orthocenter of $\triangle ABC$, not hard to show.

another proof for the second part:
Let $C_1$ be the unit circle in Argand Plane and the let the positive x-axis be along $OH$ where $O$ is the origin and $H$ is an arbitrary point inside $C_2$.We have $h=|OH|=k$(say).Using the fact $h=a+b+c$ it suffices to prove that k can be written as $e^{ix}+e^{iy}+e^{iz}$.This is equivalent to \[\cos x+\cos y+\cos z=k\] and \[\sin x+\sin y+\sin z=0\]Choosing $\cos x=\cos y=\frac{k-1}{2}$(this is possible because $k<3$),$\cos z=1$ and $x=-y$ we get a triangle.
$QED$

[hide=Solution]
For the first part just notice that in any triangle we have
\[OH < AH + AO = 2R\cos A + R \leq 3R\]

Now let \(H\) be any point in \(C_2\). Then construct any line through \(H\) intersecting \(C_1\) at \(A,P\). Construct the perpendicular bisector of \(HP\), and suppose it intersects \(C_1\) at \(B,C\). Then since \(AH\perp BC\) and the reflection of \(H\) in side \(BC\) lies on \(C_1\), it is clear that \(H\) is the orthocenter of \(ABC\).
[/hide]


how many squares of any size appear in a 3 by 5 grid of unit squares?

i think i know how to do this, but im not sure.

First, the number of 1x1 squares: $15$
Next, the number of 2x2 squares: $8$
Finally, the number of 3x3 squares: $3$

Adding it all up,
$15+8+3=26$

Can you generalize?

How many squares of any size appear in a $m\times n$ grid of unit squares where $m<n$

How many squares appear in an $n\times n$ grid?  :D

And [b]vishalarul[/b]'s sum with [b]SplashD[/b]'s correction can be simplified into $K(m,n)=\frac{n(n+1)(3m-n+1)}{6}$, and $K(n,n)$ is of course equal to what [b]lotrgreengrapes7926[/b] posted.


LaTeX
limit
integration
calculus
calculus computations

what is the
                                         [b] lim[(1^p+2^p+...+n^p)/(n^(p+1))][/b]                       :?:
                                                n :arrow: infinety

Please use, or learn to use, $\LaTeX$.

Anyways,

$\lim_{n\to\infty} \sum_{i=1}^{n} \frac{i^p}{n^{p+1}}=\lim_{n\to\infty} \frac{1}{n}\sum^n_{i=1} \left(\frac{i}{n}\right)^p=\int^1_0 x^pdx = \frac 1{p+1}$ for $p\geq 0$.

And please don't start new problems out in "solved problems."

I dont know latex ans dusing it is not popular here and i should instell it in the internet cafe fromwhere i am writing
i am sorry

You write it in much the same way as you would write normal mathematical language, except that instead of writing something like

f(x) = 1+x+x^2

you would enclose the exact same thing in dollar signs to obtain

$f(x)=1+x+x^2$

You don't need to install any software - the forum software takes care of it for you.  You can also click on other people's latex to find the commands they used to produce certain symbols.


function
number theory unsolved
number theory

Find a function $f(n)$ on the positive integers with positive integer values such that $f( f(n) ) = 1993 n^{1945}$ for all $n$.

[quote="Rust"]There are many functions, satisfyed your condition.
For example $f(n)=1993n$ if exist m, suth that $n=m^{1945}$,
else $f(n)=n^{1945}$.[/quote]
Are you sure? From what you wrote , then $f(1)=1993$ and if we choose $n=1$ then $f(1993)=1993$ , but $f(1993)=1993^{1945}$.

We start with the positive integers $1, 2, \ldots$ and sieve through the numbers by taking the smallest number $n$ still in the list, and remove $n, f^2(n), f^4(n), \ldots $ from the list. That list of numbers well will call a "chain".
First, I will prove that any two chains are disjoint. Indeed, if two chains shared a value, then applying $f^{-1}\circ f^{-1}$ to that value will output the value of the previous number in each chain; repeating this process until we get to one of the chains' beginnings shows that that chain's beginning was already used in a previous chain via the sieve, contradiction to the construction.
Now we pair up the chains arbitrarily. Let $a_1, a_2, \ldots $ be one chain and $b_1, b_2, \ldots $ be the chain it is paired with. We define $f(a_i)=b_i$ and $f(b_i)=a_{i+1}$ for all $i\ge 1$. Note that this definition is consistent with the problem statement; in addition, since every number shows up in some chain, all positive integers are successfully mapped. Thus, this function $f$ works, and we are done.


Hi! I haven't been on AOPS for a whiles.

I'm doing some stuff about parabolas, foci, and Directices (hopefully i got those plurals right ;))
Can you infer the equation of a Parabola from knowing the Vertex and one point? And how?

Also, does anyone know of a proof that the distant from any point on a parabola to the focus is the same distance as the shortest distance from that point to the Directix? Just a link would be great

Also, to find the Focus, you have to put the equation in the form $ x^2\equal{}4py$. Where does the 4p come from?

I hope I didn't write too many questions in one thread :wink:

For number 1
We know the vertex $ (k,h)$, so write an equation in the form
$ y\equal{} a(x\minus{}k)^2 \plus{} h$
Now sub in the other point that you know, $ (x_1,y_1)$, to find $ a$, and presto we have our equation

What is a "directix"?

[quote="ashpotter"]
Can you infer the equation of a Parabola from knowing the Vertex and one point? And how?
[/quote]
[hide="Click Here!"]
Yes, first You have to find the value of $ p$ from the standard form of the equation(where the equation depends according to the direction where your parabola opens)... by simply substituting the values of your $ Vertex(h,k)$ and a $ Point(x,y)$ being contained by the parabola. Then, Set it later to its Standard form....

$ e.g.$ A Parabola opening upward has its Vertex at the origin and it passes through $ (4,2)$. Find for the equation of this parabola.....

$ (x\minus{}h)^2\equal{}4p(y\minus{}k)$
$ (4)^2\equal{}4p(2)$

Solving this gives $ p\equal{}2$

Therefore.... The eq'n of the parabola is $ x^2\equal{}8y$
So here's a list of the standard forms of a parabola according to its opening:
a.)upward: $ (x\minus{}h)^2\equal{}4p(y\minus{}k)$
b.)downward: $ (x\minus{}h)^2\equal{}\minus{}4p(y\minus{}k)$
c.)to the left: $ (y\minus{}k)^2\equal{}\minus{}4p(x\minus{}h)$
d.)to the right:$ (y\minus{}k)^2\equal{}4p(x\minus{}h)$
[/hide]
[quote="ashpotter"]Also, to find the Focus, you have to put the equation in the form $ x^2 \equal{} 4py$. Where does the 4p come from?[/quote]
[hide="Click Here!"]
$ 4p$ is the length of the parabola's Focal Chord(Latus Rectum)..... and the value of $ p$ helps you know how far should be a vertex to its Focus and to its directrix, Thus, helping you to know where is the exact location of your focus and what eq'n of directrix you have.....  :)

$ Let:$
$ VF$= be the distance from the vertex to its focus
$ Vd$= be the distance from the vertex to its directrix

Such that your Vertex is at $ (0,0)$:
$ x^2\equal{} 2(VF\plus{}Vd)y$
Since $ VF \equal{} Vd\equal{}p$
$ x^2\equal{}2(2p)y$
Thus, gives you the eq'n $ x^2\equal{}4py$

[color=red]This is only true when your vertex is at $ (0,0)$ and the parabola opens above or upward.[/color]
[/hide]

[quote="ashpotter"]Also, does anyone know of a proof that the distant from any point on a parabola to the focus is the same distance as the shortest distance from that point to the Directix? Just a link would be great[/quote]
Hmm... Do you mean a point  contained by the parabola which is also contained by the line passing through the focus???  If that's what you mean, a proof is possible that its distance is equal to that of the distance from that focus to the directrix... Otherwise, then there might be no proof....  :|
[quote="ocha"]What is a "directix"?[/quote]
A directrix is a line perpendicular to the axis of symmetry and it has the same distance($ p$) from a Vertex to its Focus.....  :lol:

[quote="maximos"][quote="ocha"]What is a "directix"?[/quote]
A directrix is a line perpendicular to the axis of symmetry and it has the same distance($ p$) from a Vertex to its Focus.....  :lol:[/quote]

i.e. When the equation of a parabola is $ y\minus{}y_1\equal{}a(x\minus{}x_1)^2$, directrix is $ x\equal{}x_1$

When the equation of a parabola is $ x\minus{}x_1\equal{}a(y\minus{}y_1)^2$, directrix is $ y\equal{}y_1$

For help on this problem and the definition, I think you should refer to my blog.
http://www.artofproblemsolving.com/Forum/weblog.php?w=1373
Look at "$ conic$ $ sections$"

If you need general help on this kind of stuff, then I would recommend [url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=300]Intermediate Algebra[/url]


Does anyone know of any good books/websites with practice problems and answers for the AMC10? I want to let my son work on some of them so he can have a shot at it next year. Thanks :)

You can purchase previous year's exams for AMC12 and AMC10 on their website [url=http://www.unl.edu/amc]here[/url].  The AMC10 is relatively new, so there are only a few year's practice tests.

Few things I did this year for my son:

1. Ordered the old tests on AMC website
2. Enrolled him in the AMC-10 classes offered through this website (started in Sept. '04)
3. Bought him "the art of problem solving" books for him
4. Had him work on the relavent forums to practice problem solving skills and ask questions.

There is a Parents forum at the bottom of the forum list.  You can ask other parents on things they do to advance their children's math interest.


calculus
integration
function
geometry
trigonometry
logarithms
calculus computations

Hi

I have the function below which I have to find the area under the curve, but dont seem to be getting anywhere. I have the answer using Mathcad, but I need to be able to do it by hand.

$ \int_{\minus{}\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{2\sec x}{2\plus{}\tan x}dx$

I can simplify it to

$ \int_{\minus{}\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{2}{2\cos x\plus{}\sin x}$

But I dont know if this is any better to integrate. I have been told to use the Weierstrass substitution, but it is something we haven't covered in the course.

Any help would be appreciated.

Thanks

Hi bobred.

The Weierstrass substitution is simply $ t \equal{} \tan(\frac {x}{2})$,so that $ \sin x \equal{} \frac {2t}{1 \plus{} t^{2}}$, $ \cos x \equal{} \frac {1 \minus{} t^{2}}{1 \plus{} t^{2}}$ and $ dx \equal{} \frac {2}{1 \plus{} t^{2}}\,dt$.

However, looking at the denominator, I tried

$ 2\cos x \plus{} \sin x \equal{} \sqrt {5}(\frac {2}{\sqrt {5}}\cos x \plus{} \frac {1}{\sqrt {5}}) \equal{} \sqrt {5}\cos (x \minus{} \alpha)$, where $ \tan\alpha \equal{} \frac {1}{2}$

Therefore your integral becomes

$ \frac {2}{\sqrt {5}}\int\, \sec(x \minus{} \alpha) d(x \minus{} \alpha)$

$ \equal{} \frac {2}{\sqrt {5}}\ln(\sec(x \minus{} \alpha)\tan(x \minus{} \alpha))$

Now just put in the limits.

If you want I can probably work through the Weierstrass substitution which begins with

$ \int\, \frac{2\,dt}{1\plus{}t\minus{}t^{2}}$

=$ \int\, \frac{2\,dt}{\frac{5}{4}\minus{}(t\minus{}\frac{1}{2})^{2}}$

We can then use partial fractions. The integrand is of the form

$ \frac{2}{a^{2}\minus{}y^{2}} \equal{} \frac{1}{a}(\frac{1}{a\plus{}y} \plus{} \frac{1}{a\minus{}y})$

where $ a\equal{}\frac{\sqrt{5}}{2}$ and $ y\equal{}(t\minus{}\frac{1}{2})$

This gives

$ \frac{2}{\sqrt{5}}\ln(\frac{\sqrt\frac{5}{2} \plus{}(t\minus{}\frac{1}{2})}{\sqrt\frac{5}{2} \minus{} (t\minus{}\frac{1}{2})})$

so now we just put in the limits.

Hope this is of some help.

the best way to integrate such functions as Andrewtom writely said is to substitute $ t \equal{}\tan(\frac{x}{2})$ .

Though i din know it was called weistrass substitution ,thats the standard procedure i follow. Maybe we have a shorter method .
Anyways thanks ANDRETOM  for ur information (i mean the name of t :) he substitution )

hello, after the hint above i have got
$ 4/5\,{\it artanh} \left( 1/5\,\sqrt {5}\plus{}2/5\,\sqrt {5}\cot \left( 3/8
\,\pi  \right)  \right) \sqrt {5}\plus{}4/5\,{\it artanh} \left( \minus{}1/5\,
\sqrt {5}\plus{}2/15\,\sqrt {5}\sqrt {3} \right) \sqrt {5}$
Sonnhard.


geometry
rectangle
circumcircle
geometry open

Hi
Solve this difficult (I think) problem :
Two points A , B are fixed and M is variable on the circle C(O,R) . Find locus foot into the middle AM perpendicular to the line BM .
[url=http://www.iimmgg.com/image/532c900764ac32338d37ee9d57c322f2][img]http://i.iimmgg.com/images/th/0a2a5d99b7082697c31427b49a70cdb1.png[/img][/url]
I'v tried to solve it for a long time , but I couldn't solve it in any way .
Please help me as soon as you can !  :help:
Thanks

[quote="livetolove212"]I have some hints for you:
Let $ N, J$ be the midpoints of $ AB, AM$. Construct a rectangle $ ANOF$. Then prove that $ FJ\perp BM$. You can use angle chasing and properties of cyclic quadrilateral. Note that $ A,F,J,O,N$ are concyclic.[/quote]

Thanks . I'm thinking on your hint ...

Let $ (O')$ be, the circle with diameter $ AB$ and we denote as $ C,$ the point of intersection of the circle $ (O),$ from the line through $ A$ and perpendicular to $ AB.$

Let be the point $ D\equiv (O')\cap BM$ and we denote as $ E,\ F,\ Z,$ the midpoints of the segments $ AC,\ AM,\ MD,$ respectively.

From the similar right triangles $ \bigtriangleup ABC,\ \bigtriangleup DAM,$ we conclude that $ \angle AEB \equal{} \angle AZB$ $ ,(1)$ ( the angles formed by the homologous line segments of that triangles, are equal ).

From $ (1)$ we conclude that the point $ Z,$ where $ FZ\perp BM,$ lies on a circle so be it $ (K),$ as the circumcircle of the constant right triangle $ \bigtriangleup ABE$ and the solution og the proposed problem is completed.

Kostas Vittas.

Midpoints $ C, N$ of $ AB, AM$ are on a circle $ (O')$ with diameter $ AO.$ Circles $ (O') \sim (O)$ are similar with center $ A$ and coefficient $ \frac{_1}{^2}.$ Let $ NH$ cut $ (O')$ again at $ F.$ Angle $ \angle FNA \equal{} \angle HNM \equal{} 90^\circ \minus{} \angle AMB$ is fixed and $ A \in (O')$ $ \Longrightarrow$ $ F \in (O')$ is also fixed. Angle $ \angle FHB$ is right $ \longrightarrow$ $ H$ is on a circle $ (L)$ with diameter $ BF.$

When $ M \in (O)$ is diametrically opposite to $ B,$ $ N \in (O')$ is diametrically opposite to $ C$ and the line $ NH \perp BM$ is tangent of $ (O')$ at $ F$ $ \Longrightarrow$ $ N \equiv F$ are identical $ \Longrightarrow$ $ F \in (O')$ is diametrically opposite to $ C.$ In this case, $ BF$ is the B-median of the right $ \triangle ABM$ with hypotenuse $ BM$ $ \Longrightarrow$ $ BF^2 \equal{} \frac{_1}{^4}(2AB^2 \plus{} 2BM^2 \minus{} AM^2),$ $ BF \equal{} \frac{_1}{^2}\sqrt{3AB^2 \plus{} 4R^2}.$


calculus geometry trigonometry FTW parameterization probability invariant

Each subject starts at 40. Reach 0 = doomed.

No reviving doomed subjects, hurt = -1, heal = 1
---------------------
ALIVE SUBJECTS

Algebra: 40
Calculus: 40
Combinatorics: 40
Geometry: 39
Graph Theory: 40
Number Theory: 41
Simple Arithmetic: 40
Trigonometry: 40

DEAD SUBJECTS

---------------------
Ok, start. Post as much as you want so long as it isn't consecutive doublepost/triple/etc.

Algebra: 40
Calculus: 40
Combinatorics: 41
Geometry: 39
Graph Theory: 40
Number Theory: 40
Simple Arithmetic: 40
Trigonometry: 40

Algebra: 40
Calculus: 40
Combinatorics: 42
Geometry: 39
Graph Theory: 40
Number Theory: 40
Simple Arithmetic: 39
Trigonometry: 40

[quote="james4l"]Algebra: 40
Calculus: 40
Combinatorics: 43
Geometry: 39
Graph Theory: 40
Number Theory: 39
Simple Arithmetic: 39
Trigonometry: 40[/quote]

[quote="TZF"][b]Proposition:[/b] Heal = +1, Hurt = -1, 2 hurts and 1 heal per turn (hurts can be on same subject)[/quote]

Accept proposal.

Algebra: 40
Calculus: 40
Combinatorics: 42
Geometry: 39
Graph Theory: 40
Number Theory: 41
Simple Arithmetic: 38
Trigonometry: 40

Proposal: Let's all kill simple arithmetic.

Algebra: 40
Calculus: 40
Combinatorics: 43
Geometry: 39
Graph Theory: 40
Number Theory: 41
Simple Arithmetic: 36
Trigonometry: 40

[color=green]Algebra: 41 [/color]
[color=red]Calculus: 39[/color]
Combinatorics: 43
Geometry: 39
Graph Theory: 40
Number Theory: 41
Simple Arithmetic: 36
Trigonometry: 40

start using the colors

Algebra: 42
Calculus: 38
Combinatorics: 43
Geometry: 39
Graph Theory: 40
Number Theory: 41
Simple Arithmetic: 36
Trigonometry: 40

Algebra: 42
Calculus: 38
Combinatorics: 44
Geometry: 39
Graph Theory: 40
Number Theory: 39
Simple Arithmetic: 36
Trigonometry: 40

Algebra: 43
Calculus: 38
Combinatorics: 44
Geometry: 37
Graph Theory: 40
Number Theory: 39
Simple Arithmetic: 36
Trigonometry: 40

Let's kill geometry because I hate writing double column proofs.  :P

Algebra: 43
Calculus: 38
Combinatorics: 45
Geometry: 37
Graph Theory: 39
Number Theory: 39
Simple Arithmetic: 36
Trigonometry: 39

Algebra: 43 - 1 = 42
Calculus: 38
Combinatorics: 45
Geometry: 37
Graph Theory: 39
Number Theory: 39 +1 = 40
Simple Arithmetic: 36
Trigonometry: 39

Proposal: Let's kill Algebra because I suck at it.

Hm... that's not going to be accepted.

Nominated for kills:

Simple Arithmetic (good idea)
Algebra
Geometry

Algebra: 42
Calculus: 38
Combinatorics: 46
Geometry: 37
Graph Theory: 38
Number Theory: 39
Simple Arithmetic: 36
Trigonometry: 39

SIMPLE ARITHMETIC FTW!!!

Algebra: 42
Calculus: 38
Combinatorics: 45 <--- a little too high???
Geometry: 37
Graph Theory: 38
Number Theory: 39
Simple Arithmetic: 37
Trigonometry: 39

Arithmetic: 109
Algebra: 58
Number Theory: 39

PWNT: Counting/Combinatorics
PWNT: Geometry
PWNT: Calculus
PWNT: Invariants
PWNT: Graph Theory
PWNT: Probability

Arithmetic: 109
Algebra: 56
Number Theory: 40

PWNT: Counting/Combinatorics
PWNT: Geometry
PWNT: Calculus
PWNT: Invariants
PWNT: Graph Theory
PWNT: Probability

Arithmetic: 110
Algebra: 56
Number Theory: 38

PWNT: Counting/Combinatorics
PWNT: Geometry
PWNT: Calculus
PWNT: Invariants
PWNT: Graph Theory
PWNT: Probability

Arithmetic: 111
Algebra: 56
Number Theory: 36

PWNT: Counting/Combinatorics
PWNT: Geometry
PWNT: Calculus
PWNT: Invariants
PWNT: Graph Theory
PWNT: Probability

Arithmetic: 112
Algebra: 56
Number Theory: 34

PWNT: Counting/Combinatorics
PWNT: Geometry
PWNT: Calculus
PWNT: Invariants
PWNT: Graph Theory
PWNT: Probability

Arithmetic: 113
Algebra: 56
Number Theory: 32

PWNT: Counting/Combinatorics
PWNT: Geometry
PWNT: Calculus
PWNT: Invariants
PWNT: Graph Theory
PWNT: Probability

Arithmetic: 113
Algebra: 57
Number Theory: 30

PWNT: Counting/Combinatorics
PWNT: Geometry
PWNT: Calculus
PWNT: Invariants
PWNT: Graph Theory
PWNT: Probability

Originally I was increasing Number Theory, but I have very few reasons as to why. I have millions of reasons to like Algebra. So now I am increasing Algebra.

Arithmetic: 114
Algebra: 57
Number Theory: 28

PWNT: Counting/Combinatorics
PWNT: Geometry
PWNT: Calculus
PWNT: Invariants
PWNT: Graph Theory
PWNT: Probability

Arithmetic: 115
Algebra: 57
Number Theory: 26

PWNT: Counting/Combinatorics
PWNT: Geometry
PWNT: Calculus
PWNT: Invariants
PWNT: Graph Theory
PWNT: Probability

Arithmetic: 115
Algebra: 55
Number Theory: 27

PWNT: Counting/Combinatorics
PWNT: Geometry
PWNT: Calculus
PWNT: Invariants
PWNT: Graph Theory
PWNT: Probability

Arithmetic: 116
Algebra: 55
Number Theory: 25

PWNT: Counting/Combinatorics
PWNT: Geometry
PWNT: Calculus
PWNT: Invariants
PWNT: Graph Theory
PWNT: Probability

Arithmetic: 114
Algebra: 55
Number Theory: 26

PWNT: Counting/Combinatorics
PWNT: Geometry
PWNT: Calculus
PWNT: Invariants
PWNT: Graph Theory
PWNT: Probability

Arithmetic: 115
Algebra: 55
Number Theory: 24

PWNT: Counting/Combinatorics
PWNT: Geometry
PWNT: Calculus
PWNT: Invariants
PWNT: Graph Theory
PWNT: Probability

Arithmetic: 113
Algebra: 55
Number Theory: 25

PWNT: Counting/Combinatorics
PWNT: Geometry
PWNT: Calculus
PWNT: Invariants
PWNT: Graph Theory
PWNT: Probability

Arithmetic: 113
Algebra: 56
Number Theory: 23

PWNT: Counting/Combinatorics
PWNT: Geometry
PWNT: Calculus
PWNT: Invariants
PWNT: Graph Theory
PWNT: Probability


algebra
polynomial
inequalities
ratio
real analysis
real analysis unsolved

Let $P(z)$-polynomial of degree n, such that $\displaystyle \max_{|z|=1}|P(z)|\leq1$. Prove that if $P(\alpha)=0$, then
$\displaystyle \max_{|z|=1}|\frac{P(z)}{z-\alpha}|\leq \frac{n+1}{2}$, and $\displaystyle \max_{|z|\leq1}|\frac{P(z)}{z-\alpha}|\leq \frac{n}{1+|\alpha|}$

[quote="eugene"]maybe any ideas???[/quote]
Both inequalities would follow if one can prove the following:
Let $P$ be a polynomial of degree $n$, $|\alpha|=1$, $P(\alpha)=0$. Then for every $z$ on the unit circumference, $\left|\frac{P(z)}{z-\alpha}\right|\leq \frac n2$. This reminds the famous Bernstein inequality and, indeed, applying Bernstein's inequality to the trigonometric polynomial $z^{-n}P(z^2)$, one can show that a similar, but somewhat smaller ratio  $\left|\frac{P(z)}{\arg z/\alpha}\right|$ is bounded by $n/2$. So my idea would be to take the proof of Bernstein's inequality and try to modify it to get the result.

thanks for you reply, fedja.
Berstein inequality: $||f'||\leq n||f||$, where $f$-polynomial of degree $n$ and norm is taken for example on $|z|=1$. I think that you meant this (first) Berstein inequality
Could you please explain how you've proved that $\left| \frac{P(z)}{\arg(\frac{z}{\alpha})}\right|\leq \frac{n}{2}$ I must be missing something or too stupid, but i can't understand
(I've proved our problem only for upper bound $n+1$(it's almost evident) not for $\frac{n+1}{2}$)

[quote="eugene"]
Berstein inequality: $||f'||\leq n||f||$, where $f$-polynomial of degree $n$ and norm is taken for example on $|z|=1$. I think that you meant this (first) Berstein inequality [/quote]
Yes, that's what I meant except I need it for trigonometric rather than algebraic polynomials, i.e., for the sums $f(z)=\sum_{-n\leq k\leq n}a_k z^k$
[quote="eugene"]
Could you please explain how you've proved that $\left| \frac{P(z)}{\arg(\frac{z}{\alpha})}\right|\leq \frac{n}{2}$ [/quote]
WLOG, $\alpha=1$. Consider $f(z)=z^{-n}P(z^2)$. This is a trigonometric polynomial of degree $n$ (note that it includes both negative and positive powers of $z$ up to $n$). Clearly, $\|f\|=\|P\|\leq 1$. Using the Bernstein inequality, we see that $\|f'\|\leq n$. Also $f(1)=0$, so $|f(z)|=|P(z^2)|\leq n|\arg z|$, which is equivalent to the statement to prove. :)


inequalities
integration
inequalities theorems

It's equivalent to Cauchy-Shwarz, to obtain it just square both sides...

They are different (although they are related).

[size=150][b]Cauchy-Schwarz Inequality[/b][/size]
Let $a_i, b_i$ be real numbers, $i=1, 2, \cdots , n$, then
$(a_1^2+a_2^2+\cdots +a_n^2)(b_1^2+b_2^2+\cdots +b_n^2)\geq (a_1b_1+a_2b_2+\cdots +a_nb_n)^2$.
Equality holds when $\frac{a_1}{b_1}=\frac{a_2}{b_2}=\cdots =\frac{a_n}{b_n}$

[b]Generalizations[/b]
1. [b]Holder Inequality[/b]
If $p$ and $q$ are real numbers such that $\frac{1}{p}+\frac{1}{q}=1$ and $a_i, b_i$ are positive real numbers, $i=1, 2, \cdots , n$, then
$(a_1^p+a_2^p+\cdots +a_n^p)^{\frac{1}{p}}(b_1^q+b_2^q+\cdots +b_n^q)^{\frac{1}{q}}\geq a_1b_1+a_2b_2+\cdots +a_nb_n$
Equality holds when $\frac{a_1}{b_1}=\frac{a_2}{b_2}=\cdots =\frac{a_n}{b_n}$

2. [b]Generalized Holder Inequality[/b]
Let $a_{ij}$ be positive real numbers, $i,j=1, 2, \cdots n$, then
$(a_{11}^n+a_{12}^n+\cdots +a_{1n}^n)(a_{21}^n+a_{22}^n+\cdots +a_{2n}^n)\cdots (a_{n1}^n+a_{n2}^n+\cdots +a_{nn}^n)\geq (a_{11}a_{21}\cdots a_{n1}+a_{12}a_{22}\cdots a_{n2}+\cdots +a_{1n}a_{2n}\cdots a_{nn})^n$

[size=150][b]Minkowski Inequality[/b][/size]
Let $p>1$ and $a_i, b_i$ be real numbers, $i=1, 2, \cdots , n$, then
$\sqrt[p]{a_1^p+b_1^p}+\sqrt[p]{a_2^p+b_2^p}+\cdots +\sqrt[p]{a_n^p+b_n^p}\geq \sqrt[p]{(a_1+a_2+\cdots +a_n)^p+(b_1+b_2+\cdots +b_n)^p}$.
Equality holds when $\frac{a_1}{b_1}=\frac{a_2}{b_2}=\cdots =\frac{a_n}{b_n}$

[b]Generalization[/b]
Let $p>1$ and $a_{ij}$ be positive real numbers, $i=1, 2, \cdots m, j=1, 2, \cdots n$, then
$\sqrt[p]{a_{11}^p+a_{12}^p+\cdots +a_{1n}^p}+\sqrt[p]{a_{21}^p+a_{22}^p+\cdots +a_{2n}^p}+\cdots \sqrt[p]{a_{m1}^p+a_{m2}^p+\cdots +a_{mn}^p}\geq \sqrt[p]{(a_{11}+a_{21}+\cdots +a_{m1})^p+(a_{12}+a_{22}+\cdots +a_{m2})^p+\cdots +(a_{1n}+a_{2n}+\cdots +a_{mn})^p}$

Well, I thought that original Minkowski is for $p=2$, and then of course it's C-S in equivalent form.

thank you.. But how how can i prove that?  :|

OK, this proof was presented in my university course. I hope you know basic properties of $L^p$ functions.

Minkowsky inequality: Let $p\geq1$ and $f,g\in L^p$. Then $f+g\in L^p$ and
\[ ||f+g||_p\leq ||f||_p+||g||_p. \]

Proof (case $p>1$). Let $q=\frac{p}{p-1}$. Then $p$ and $q$ are Hölder conjugates of each other. If $a,b\geq 0$ we have
\[ (a+p)^p\leq (2\max(a,b))^p=2^p(\max (a,b))^p\leq 2^p(a^p+b^p).  \]
We may assume that $f(x),g(x)\in \mathbb{R}$ for all $x\in \mathbb{R}$ and therefore we have
\[ |f+g|^p\leq (|f|+|g|)^p\leq 2^p(|f|^p+|q|^p).  \]
On the other hand
\[ |f+g|^p=|f+g||f+g|^{p-1}\leq |f||f+g|^{p-1}+|g||f+g|^{p-1}.  \]
and
\[ (|f+g|^{p-1})^q=|f+g|^p\implies |f+g|^{p-1}\in L^p.  \]
From Hölder inequality
\begin{eqnarray*} ||f+g||_p^p&=&\int |f+q|^p\leq|f||f+g|^{p-1}+\int |g||f+g|^{p-1}\\ &\leq&||f||_p\left (\int (|f+g|^{p-1})^{q}\right )^{1/q}+||g||_p\left (\int (|f+g|)^{p-1}^q\right )^{1/q}\\ &=&||f||_p\left (\int |f+g|)^{p}\right )^{1/q}+||g||_p\left (\int |f+g|)^{p}\right )^{1/q}\\ &=&(||f||_p+||g||_p)||f+g||_p^{p/q}\\ \end{eqnarray*}

Since $p/q=p-1$ we have $||f+g||_p\leq ||f||_p+||q||_p$ as was to be proved.

Here is another proof in the case of sums and $p>1$. $p<1$ similarly.

By Hölder's inequality,
\[\left (\sum_{i=1}^{n}(a_{i}+b_{i})^{p}\right )^{1/p}=\max_{c\in L}\sum_{i=1}^{n}(a_{i}+b_{i})c_{i},\]
where $L=\{c|\sum_{i=1}^{n}c_{i}^{q}=1\}$, we have
\begin{eqnarray*}\left (\sum_{i=1}^{n}(a_{i}+b_{i})^{p}\right )^{1/p}\leq \max_{L}\sum_{i=1}^{n}a_{i}c_{i}+\max_{L}\sum_{i=1}^{n}b_{i}c_{i}\\ \leq \left (\sum_{i=1}^{n}a_{i}^{p}\right )^{1/p}+\left (\sum_{i=1}^{n}b_{i}^{q}\right )^{1/q}\end{eqnarray*}


Find the all pairs of integers $ (x,y)$ such that $ 54x^3\plus{}1\equal{}y^3$

We let 3x=a, the equation becomes $ 2a^3\plus{}1\equal{}y^3$
You should the simple property: The product of two squares is a square.
After some days, I will post my solution if nobody give ones.

[quote="mr.danh"]We let 3x=a, the equation becomes $ 2a^3 + 1 = y^3$
You should the simple property: The product of two squares is a square.
After some days, I will post my solution if nobody give ones.[/quote]
[hide="My solution"]Let a=3x. we have $ 2a^3 + 1 = y^3$
so y is the odd number. Let $ y = 2k + 1 (k\in Z)$, then $ 2a^3 + 1 = 8k^3 + 12k^2 + 6k + 1$
$ \Leftrightarrow a^3 = 4k^3 + 6k^2 + 3k (1)$
So $ a^3$ is divisible by $ k \Rightarrow a\vdots k$
Let a=tk with $ t\in Z$, (1) becomes: $ t^3k^3 = 4k^3 + 6k^2 + 3k$
$ \Leftrightarrow t^3k^3 = k^3 + 3k(k + 1)^2$
$ \Leftrightarrow k[t^3k^2 - k^2 - 3(k + 1)^2] = 0$
$ \Leftrightarrow \left[\begin{k = 0} \\
{t^3k^2 = k^2 + 3(k + 1)^2}$
We have two cases:
1. If $ k = 0 \Rightarrow y = 1,x = 0$
2. If $ t^3k^2 = k^2 + 3(k + 1)^2$ then $ (t^3 - 1)k^2 = 3(k + 1)^2$
We resume having two cases:
1. If $ t = 1 \Rightarrow k = - 1 \Rightarrow y = - 1,x = \frac {1}{3}$ (weed out this case because $ x\not\in Z$)
2. If $ t\neq 1 \Rightarrow k\neq - 1 \Rightarrow 3(k + 1)^2\vdots k^2$.
But $ (k,k + 1) = 1$ hence $ 3\vdots k^2 \Rightarrow k = 1 \Rightarrow y = 3 \Rightarrow 54x^3 = 26 \Rightarrow x\not\in Z$ (weed out this case)
Therefore only 1 solution $ \boxed{x = 0,y = 1}$  :wink: [/hide]

[quote="natural"]
So $ a^3$ is divisible by $ k \Rightarrow a\vdots k$
[/quote]
I'm sure that your statement made a mistake.
$ 3^3\vdots 9$, but $ 3\not\vdots 9$!!!
So, your solution is incorrect.
This problem is very nice and its solution is very excellent.

Hmm, I'd like to see that easy solution, because I got quite far, but I sstopped at one moment.
This is where I got to:

$ a^{3} \equal{} 4k^{3} \plus{} 6k^{2} \plus{} 3k$ (basing on the natural's beginning of the proof)
Let us denote $ g \equal{} GCD(a,k)$
So:
$ a \equal{} a'g \ k \equal{} k'g \ GCD(a',k') \equal{} 1$
Assume $ 3\not \vdots g$
We have another equation:
$ a'^{3}g^{3} \equal{} 4k'^{3}g^{3} \plus{} 6k'^{2}g^{2} \plus{} 3k'g$
$ a'^{3}g^{2} \equal{} 4k'^{3}g^{2} \plus{} 6k'^{2}g \plus{} 3k'$ thus
$ (g|k')\Rightarrow (k \equal{} k''g^{2})$
$ a'^{3}g^{3} \equal{} 4k''^{3}g^{6} \plus{} 6k''^{4}g^{4} \plus{} 3k''g^{2}$
$ a'^{3}g \equal{} 4k''^{3}g^{4} \plus{} 6k''^{2}g^{2} \plus{} 3k''$, thus $ k \equal{} k'''g^{3}$
$ a'^{3}g^{3} \equal{} 4k'''^{3}g^{9} \plus{} 6k'''^{2}g^{6} \plus{} 3k'''g^{3}$
$ a'^{3} \equal{} 4k'''^{3}g^{6} \plus{} 6k'''^{3}g^{2} \plus{} 3k'''$ which is impossible, beacause
$ (k'''|RHS)\Rightarrow (k'''|LHS)$, but $ GCD(k''',a') \equal{} 1$
Now, what if $ 3|g$
Let us define $ s\in \mathbb{N}$ such as:
$ (g \equal{} 3^{s}g') \ (3\not \vdots g')$
Now:
$ a \equal{} a'g'3^{s} \ k \equal{} k'g'3^{s} \ GCD(a',k') \equal{} 1 \ GCD(g',3) \equal{} 1$, thus we land in an equation:
$ a'^{3}g'^{3}3^{3s} \equal{} 4k'^{3}g'^{3}3^{3s} \plus{} 6k'^{2}g'^{2}3^{2s} \plus{} 3k'g'3^{s}$
Now proceeding the same way as before we land in $ k'''|a'^{3}3^{s}$, thus
$ k''' \equal{} 3^{k} \ k\in \mathbb{N} \ k\leq s$
Finally what is last to be considered is:
$ a \equal{} a'g'3^{s} \ k \equal{} g'^{3}3^{s \plus{} k}$
Putting this in our original equation gives us
$ a'^{3}3^{2s} \equal{} 4g'^{6}3^{2s \plus{} 3k} \plus{} 6g'^{3}3^{s \plus{} 2k} \plus{} 3^{k \plus{} 1}$, but that is impossible, because
$ GCD(g',3) \equal{} 1$, thus $ k \plus{} 1 \equal{} max_{n\in \mathbb{N}} \ 3^{n}|RHS$ and $ k\leq s$, and $ 2s \equal{} min_{l\in \mathbb{N}} \ 3^{l}|LHS$
Thus we aquire $ (k \plus{} 1 \equal{} 2s)\Rightarrow s \equal{} k \equal{} 1$, and to suffice s has its minimum $ GCD(a',3) \equal{} 1$
$ a'^{3} \equal{} 108g'^{6} \plus{} 18g'^{3} \plus{} 1$
Setting $ t \equal{} g'^{3}*3$ we have a new equation:
$ a'^{3} \equal{} 12t^{2} \plus{} 6t \plus{} 1$ with $ GCD(a',t,3) \equal{} 1$
From $ mod3$ we conclude that
$ a' \equal{} 3n \plus{} 1$
So:
$ 27n^{3} \plus{} 27n^{2} \plus{} 9n \equal{} 12t^{2} \plus{} 6t$
$ 9n^{3} \plus{} 9n^{2} \plus{} 3n \equal{} 2(2t^{2} \plus{} t)$
From $ mod2$ we have
$ n \equal{} 2n_{1}$
$ 72n_{1}^{3} \plus{} 36n_{1}^{2} \plus{} 6n_{1} \equal{} 2t^{2} \plus{} t$
Now, observe that $ 6|LHS$, thus looking mod6 we acquire $ t \equal{} 6t_{1} \plus{} 4$
$ 72n_{1}^{3} \plus{} 36n_{1}^{2} \plus{} 6n_{1} \equal{} 72t_{1}^{2} \plus{} 102t_{1} \plus{} 36$
$ 12n_{1}^{3} \plus{} 6n_{1}^{2} \plus{} n_{1} \equal{} 12t_{1}^{2} \plus{} 17t_{1} \plus{} 6$
And I'm stuck here, any ideas?  :blush:  :huh:  :wink:

[quote="mr.danh"]Find the all pairs of integers $ (x,y)$ such that $ 54x^3 \plus{} 1 \equal{} y^3 (1)$[/quote]
This problem is interesting, isn't it?
Here is my solution:
Let $ a \equal{} 54x^3\Rightarrow (6xy)^3 \plus{} 1 \equal{} 4y^3a \plus{} 1 \equal{} 4a(a \plus{} 1) \plus{} 1 \equal{} (2a \plus{} 1)^2$
We get $ k \equal{} 6xy$ satisfying $ (2a \plus{} 1)^2 \equal{} k^3 \plus{} 1 \equal{} (k \plus{} 1)(k^2 \minus{} k \plus{} 1)$
It is easy to show that $ (k \plus{} 1,k^2 \minus{} k \plus{} 1) \equal{} 1$. The product of $ k \plus{} 1$ and $ k^2 \minus{} k \plus{} 1$ is a square number
Hence, they are two square numbers, which means that $ k \plus{} 1 \equal{} m^2,k^2 \minus{} k \plus{} 1 \equal{} n^2$ ($ m,n$ are integers)
If $ k > 1$, $ k^2 > n^2 > (k \minus{} 1)^2$, we made a contradiction.
If$ k < 0$, $ k^2 < n^2 < (k \minus{} 1)^2$, we also made a contradiction.
If $ k \equal{} 1, k \plus{} 1 \equal{} 2$ is not a square.
So $ k \equal{} 0$, the statement leads the problem's answer $ (x,y) \equal{} (0,1)$

hmm ,
$ 54x^3\plus{}1\equal{}y^3$
or $ 54x^3\equal{}(y\minus{}1)(y^2\plus{}y\plus{}1)$
we call $ d\equal{}(y\minus{}1,y^2\plus{}y\plus{}1)$ since then $ y^2\plus{}y\plus{}1\minus{}y^2\plus{}y \equal{}2y\plus{}1$ is divisible by d , furthermore , $ 2y\plus{}1\minus{}2y\plus{}2\equal{}3$ is divisible by d . On the other hand  ,  if d=3 then $ y\minus{}1\equal{}6x^{k} ; y^2\plus{}y\plus{}1\equal{}3x^{p}$ where $ p\plus{}k\equal{}3$ and k <p
if k=0 then y=7 in which we find x is not interger
if k=1 then $ y\minus{}1\equal{}6x$ and $ y^2\plus{}y\plus{}1\equal{}3x^2$ , leads to $ 11x^2\plus{}6x\plus{}1\equal{}0$ gives non-interger root
if d=1 then $ y\minus{}1\equal{}54$ in which we find y=55 and x is not an interger ..
final solution is (0,1)

[quote="ali666"][quote="HTA"]
 if d=3 then $ y \minus{} 1 \equal{} 6x^{k} ; y^2 \plus{} y \plus{} 1 \equal{} 3x^{p}$
[/quote]
I think this step is not true...can you say it in more details plesase?[/quote]
yeah you are right  :blush: , that step is not true  :blush:

Mr danh wrote: It is easy to show that $ (k\plus{}1,k^2\minus{}k\plus{}1)\equal{}1$. How? Let's say k was 5, you'd have 6 and 21, with 3 in common.

In fact, $ \gcd(k \plus{} 1, k^2 \minus{} k \plus{} 1) \equal{} \gcd(k \plus{} 1, 3k) \equal{} 1, 3$.

[quote="dgreenb801"]Mr danh wrote: It is easy to show that $ (k \plus{} 1,k^2 \minus{} k \plus{} 1) \equal{} 1$. How? Let's say k was 5, you'd have 6 and 21, with 3 in common.[/quote]
Note: k=6xy, so k is even  :mad:


I think we could really use a forum dedicated to posts that are meant to linked to the resources section. That way, we won't have to worry about flooding a forum that is meant to have people answer other people's questions. This would make it easier for the problem posters to most more problems faster (you generally have to wait before posting more problems so as to not flood the entire forum you're posting in). It gets really annoying when you try to post an entire contest such as the AHSME which has a whopping 50 problems and have a lot of people complaining.

EDIT: It also doesn't have to be disorganized either. It can have a section for AMC, Olympiads, etc.

I agree. Yesterday, it took up a whole page (High School Basics).

National Olympiads... Post links to problems there and one can easily find them.

But, there's still a problem of flooding forums. Maybe, in Olympiad Section, make another subforum ina every category like Algebra Competition Problems (as there is Algebra Unsolved Problems)...


I think guess and check works the best on this problem.

[hide]Berris is 14 years old and fueler is 12 years old.[/hide]

[quote="Olga1146B"]i challenge u all to accomplish this problem.

berris and his brother feuler are between 10 and 20 years old. the sum of thier ages has a units digit of 6 and the difference between thier ages is 2. if feuler's age is an even number how old is berris???????
 :rotfl:[/quote]
Well if they add up to 6 in the units digit, and have a difference of 2...
Then it's either.
7 and 9
or
2 and 4.
But since the brother's age is even,
They have to be 12 and 14.


In a company of $n$ persons each person has no more than $d$ acquaintances and in that company there exist a group of $k$ persons, $k \geq d$, who are not
acquainted to each other. Prove that the number of acquainted couples is not greater than $ \left[ \dfrac{n^2}{4} \right].$

eum... isn't your other question about lines in the plane is sort of the same? :)