wow...andy murray owned rafa nadal 6-0, 6-0!

At the Rotterdam open in Holland ( Netherlands )

Is that surface clay? :huh:

Err 6-3, 4-6, 6-0? And it's only because of Rafa's knee problem.

And it's hard court I think?

yeah math's score is right. Justin. Where did you get 6-0 6-0? made it up?

lol, sorry!

i forgot!

and math154, i like nadal, but murray is a complete beast!

Hes better than djokovic and fed.

Um, I'm not from Missouri, but I just have to comment on this.  Andy Murray is definitely not better than both Federer and Djokovic.  He's never won a Grand Slam, for one, and he doesn't have as many ATP singles titles as either of the two people.  He's not ranked as highly as the two and went out in the fourth round of the Australian Open this year to Fernando Verdasco.  True, Murray is very good compared to other players, but don't even try to compare him to the top three players, let alone assert his superiority over them.  He's not nearly as elegant and talented as Federer and not nearly as good at groundstrokes as Djokovic is.

OH WOW! You just got me pissed!

Okay, so Murray doesn't have the same amount of titles as djokovic and federer.  But,  Murray is younger than both, and if you have looked at games between them, you will see that murray leads federer head to head 5-2!! And with Djokovic, Murray has won BOTH meetings on Djokovic's most favorable surface: Hard Court.  And you saying that Federer is all great, well he is, but not the all-time best.  For example, Sampras had beaten Federer in Korea/Japan in straight sets after retirement!  Bjorn Borg, Rod Laver, Sampras, McEnroe - these are the best there are ever.  Also, Laver has technically won the most grand slams because he had won the same tournaments back then ( Wimbledon, aus....) but were considered amateur tournaments with all the best players such as laver and jean lacoste.  And you and your Djokovic's groundstrokes!  They are good, but he needs to offer more to the game such as sportsmanship which Djokovic lacks, SIR!  Also, John McEnroe says that Murray has tons of potential and has the ability to be in the top two on a ESPN 2 interview.  As you can see, Murray will and I Mean Will rush past by Federer and Djokovic and will be a strong competetor with Rafael Nadal.  Oh yeah, and get your facts right. Andy murray lost in the quarterfinal not the 4th round.  Young players like verdasco will race past fed and djoko- it wont take very long.

Um, not to be mean, but I think you need to get your facts straight.  Verdasco beat Murray in the round of 16, not the quarterfinals.  Federer is definitely one of the best, and the only reason people doubt his true mastery of the sport is his almost debilitating mental defeat at the hands of Nadal.  Anyways, Murray is far from winning a Grand Slam any time soon, both physically and mentally, but who knows?  He may become as skilled as if not more so than Nole in the future, but [b]as of now[/b], he does not deserve to be compared to Federer, Nadal, Djokovic.  I'm not saying he will not grow to become a fearsome player on the tour, but at the moment, he has yet to win a Slam and yet to rise any higher than world No. 4.  Sorry, but that's the truth.

No, not just his ( Federer's) demoralizing losses against Nadal have been given him the character of not deserving to be in the top 5.  Federer, in fact, has lost to players ranked in the 20's and 30's such as Giles Simon who has recently moved up to number 6.  Federer is neck and neck with Djokovic, and Murray is not far behind.  Going at the rate that Nadal and Murray are going, they will both surpass federer and djokovcome later this year or next year.  I've already given straightforward facts to you:  Murray has a better record then Federer overall and against Djokovic on djokovic's favorable hard courts.  If that is not enough proof to say that Murray is much comparable to Federer and Djokovic, then I don't know what will.  It's the psychological aspect of being a tennis fan.  You have these favorite players, and when they start to lose, you think that is nothing.

Also on the topic of Federer's downfall:  He has had mononucleosis which basically "threw him off his horse", resulting in numerous defeats to lower ranked players, not just Nadal.  I hope we can agree that Federer is not the best player of all time because the only competition he had when he was in top form was Andy Roddick, whom i do admire much, but is simply not that great.  Again look back to the 20th century and research players like Agassi, McEnroe, Sampras, Connors, Bjorg, Lendl, Laver, Becker...  Those are some of the greatest players there were, and " That's a Fact, Jack."

Uh...We're getting nowhere in this argument, so I'll drop it.  I still and will always like Federer as my favorite player, but I respect your opinions.

And one last thing: if Federer had continued at the rate he was going racking up Slams and other titles years ago, he would have gotten very far, too.  The best argument is what will come later in the year, so you'll see!!!!  The Fedexpress is unbeatable!!!!!! :)

Lol, we'll see! As they say, Only time can tell!  And to those Ms.Meyer students, as Ponyboy would say: "nothing gold can stay." - means federer, jk!

Heh, Federer will have at least a short comeback like Sampras did.

I think Federer's comeback was basically the U.S. open last year.  Indeed, he will be victorious with a few more slams, and will pass Sampras's record, but he won't be as good as he was before.  Nadal, Djokovic ( unfortunately), and Murray will surpass him along with a few other players...come a couple of years.  Agreed?

The US Open was not really unexpected at all, so I don't think that was part of the comeback. Obviously Federer will go out of his prime eventually, so yeah, I agree.

Federer is nearing his 30s, while players like Nadal, Djokovic, and Murray are in their prime.  I agree they will surpass him, but not yet, at least.

Justin Ahamann:  Going at the rate that Nadal and Murray are going, they will both surpass federer and djokovcome later this year or next year

What do you mean by that Nadal part. That he will suprass djokovic and federer. If you mean ranking, Nadal's numbr one!!!. so he has already suprassed both in that. And if its grand slams won he has suprassed djokovic in that too!. what do you mean by he will suprass djokovic.

Sorry, I think I didn't make that clear enough.... :oops:

I mean that [b]Nadal[/b] [b]and[/b] [b]Murray[/b] will both surpass Federer in grand slams.  I'm sure that in

some time in the near future, Murray will achieve the number 2 ranking. ( It will take some time, though)..

but currently, Murray is more consistent than Djokovic and Federer.  And Nadal ? Well, he's just Nadal. :D

[quote="justinahmann"]I mean that [b]Nadal[/b] [b]and[/b] [b]Murray[/b] will both surpass Federer in grand slams.[/quote]

I have to disagree with this for obvious reasons...

Ha-ha very funny, math. You know what I mean.  And sadly, this may be my last post on AOPS, because our

great friend jjx1 has banned me. : (.  I will miss everybody except you know who ( sniff sniff ).

Good bye math, i may never talk to u again... : (

I can't ban. levans did for waht you said after you got muted.

You know what?

[b]Shut up[/b], jjx1.

Being a prominent FTW citizen, posting in the Missouri forum (eww, dirty!) is not considered setting a good example. It would be [b]highly advisable[/b] for you [i]not[/i] to reply to this.

Oh, and don't worry, Shoujit...I'm banned until 2011 for saying "shut up." And I'm proud of it. :)  :D  :D

MATH, YOU ARE THE MAN!

And really jjx1, this is completely and outrageously crazy.  You can't keep doing this, by muting people just

because I said your name.  By the way, I [b]whispered[/b] your name, so you can't have that type of

ridiculous excuse against people. If you hadn't done that,  I wouldn't have said what I had said.  I am

appalled that you have so much pleasure in muting people, and causing them to be banned.  This is why most

users don't respect you as much.  [b]Something to think about.[/b]


After I applied to early decision on the commonapplicataion website, the whole application locks and i can't change it anymore.  I just want to make sure i'm not doing something wrong.  I mean why wouldn't they allow you to make some changes and add new things for the regular decision colleges?

thanks

They tell you in the instructions that you can't change the application after you submit it.  What you can do is clone a new version of the application that you can send to different schools.  You can find the link in the instructions.

Check again. You should have a means of modifying your earlier essays, etc. for your other colleges. This was a software change made after the beginning of this school year. At first (this year) everything locked after the first submission, but now it is possible to make updated versions for your regular round applications.

t0rajir0u, where is the link that you are talking about? the instructions as the commonapp website.

It's under "Application Versions."  The link is http://app.commonapp.org/application/applicantlogin.aspx?allowcopy=true (you have to log out first).


The circle has center (-2,-4) and passes through point (6,14). How would I find the equation of the circle described and sketch a graph about it?

Don't double post, I've already answered this at http://www.mathlinks.ro/Forum/viewtopic.php?p=829051#829051


calculus
integration
inequalities
logarithms
real analysis
real analysis unsolved

Prove the following inequality.

\[\int^e_1 (\ln x)^{2n} dx<\frac{3\cdot5\cdots (2n-1)}{4\cdot6\cdots (2n)}(e-2)\ \ (n\geqq 2)\]

[b]kunny[/b] are you sure that your inequality is right?

For $n\geq 3$ the trivial estimate $\int_1^e(\ln x)^{2n}\,dx=\int_0^1 y^{2n}e^y\,dy<e\int_0^1 y^{2n}\,dy=\frac e{2n+1}$ is stronger. The case $n=2$ can be considered separately (after all, the integral can be computed and the resulting inequality for $e$ can be verified). But this solution is too ugly...


Revival of the uber image pwning game. You must post an image that pwns the one before it.

I'll start.

[img]http://mashable.com/wp-content/uploads/2009/03/o_rly.jpg[/img]

[img]http://dcist.com/attachments/dcist_sommer/i-is-sorry-i-has-rabies.jpg[/img]
BWAHAHA!! The rabid kitty pwns OWL! HAHA!

[img]http://www.blavish.com/wp-content/uploads/2006/12/viking-48-oven-12-56-2006.jpg[/img]

FULL POWER

:P

The message is too small. Please make the message longer before submitting.

jjfun1's image is really popular look at how many times the attachment was veiwed compared to the other pictures!!

The message is too small. Please make the message longer before submitting.

it's a desert btw lolz

[img]http://firsttube.com/uploads/lolcats(7).jpg[/img]

 :rotfl:

I don't know...
I'm pwning my own image.
Sad.
:P
[img]http://1.bp.blogspot.com/_8tZf9lm-smo/SHIcmCVMJEI/AAAAAAAAAVY/L8tJhSPl2tU/s400/invisiblestuckpicklejar.jpg[/img]

Apparently my image is so un-pwnable it has won the game :)

LET'S START WITH A NEW ONE!
I'm starting out with something easy because I don't want this thread to like... die. When we get farther in the game MEEWHEE WILL PWN YOU! :)
[img]http://www.cricketfizz.com/cricketblog/wp-content/uploads/2008/12/pineapple.jpg[/img]

[hide="Oh no! :o "]
[hide="Oh wait... :maybe: "]
[hide="I got it: :idea: "]

[img]http://trinities.org/blog/wp-content/uploads/hammer.jpg[/img] :P
[size=59]squashed[/size][/hide][/hide][/hide]

[hide="What could it be?"]
[hide="let us find out!"]
[hide="ready?"]

[img]http://static.howstuffworks.com/gif/hammer-1.jpg[/img]

[/hide]
[/hide]
[/hide]


There is a  regular tetrahedron ABCD with side t.
V is a point inside ABCD such that VA = a, VB = b, VC = c and VD = d. How to determine t in terms of a, b, c and d?
Please help.

Please pay attention to the difficulty level of each forum.  This is certainly not a problem for Middle School/Classroom math.


A certain bacteria doubles in size every ten minutes. At noon a small amount of this bacteria is placed in a container, and by 6 p.m. of the same day the container was just full. At what time that afternoon was the container one-fourth full?

You notice that at $ 5: 50$, the container is $ 1/2$ full, so at $ \boxed{5: 40}$, the container is $ 1/4$ full.


Why dont you integrate a latex extension in the phpBB forum?

everyone expects that feature. :)

But it depends on Valentin

unfortunately as I have said in the previous posts concerning this issue, I am trying for months now to find another provider: the current server is located in Targu Mures, and it is actually the server of the firm Reea which was kind enough to suport the site (and its 5Gb / month bandwidth).

latex environment is not that secure. by now I have seen a phpBB-latex inforced forum cracked within minutes (ie the guy I was watching gained access to the phpBB admin panel - that is to all posts, e-mails, everything) through latex. that is a serious security threat, and it is hard to find someone that can deal with it, without banning all Chinese IPs, or other drastic measures, like it is done now.

But tomorrow I am going to have a talk, and hopefully if all goes well this Monday, MathLinks will have a 1Gbit/s transfer with outside, space on the server and latex installed, also all the ip-s that are currently banned will be unbanned.

Please understand that all these efforts cost time and money for me, and these are all for making your stay on the forum better. I could have put up nasty pop-ups, or adds of sorts (there are numerous boards which do that), but I prefere it clean.


This is not the right forum to ask this question.To Latex your code: Enclose it with the dollar signs.

Visit this link for more info: http://www.mathlinks.ro/LaTeX/AoPS_L_About.php


How to prove that if $n>m$ then
$ \sum_{k=0}^n {n \choose k} {n-k \choose m-k}  = 2^m { n \choose m }$

Consider a group of $n$ persons. From them, you want to form a group of $m$ persons with some chiefs (possibily none).
The first way is to select $m$ persons ($C_n^m$ choices), and for each of these $m$ decide if this person is a chief or not ($2$ choices for each so a total of $2^m$ choices). Thus, the RHS is the number of such groups.

In another hand, if $k$ is the number of chiefs, you can first select these $k$ chiefs in the initial group ($C_n^k$ choices) and them complete the group with $m-k$ among the $n-k$ remaining ones ($C_{n-k}^{m-k}$ choices). Summing over $k$ will leads to the total number of groups with $m$ persons with some chiefs, and you will obtain the LHS.

Pierre.

Thanx alot!!
Is there a way to show it with algebra? or is it too complicated?

if you want an algebraic way(usually the combinatorial ways are better and also more difficult) here is what you can do:

we want to prove that $\displaystyle \sum_{k\geq 0} \binom{n}{k} \binom{n-k}{m-k} = 2^m \binom{n}{m}$.what we can do instead is prove that $\displaystyle\sum_{m\geq 0} \sum_{k\geq 0} \binom{n}{k} \binom{n-k}{m-k} x^m = \sum_{m\geq 0} 2^m \binom{n}{m}x^m$.the RHS clearly is $(1+2x)^{n}$.on the other hand the LHS =
$\displaystyle\sum_{k\geq 0} \binom{n}{k} \sum_{m\geq 0}\binom{n-k}{m-k} x^m=\sum_{k\geq 0}\binom{n}{k}x^k \sum_{m\geq 0} \binom{n-k}{m-k}x^{m-k}=(1+x)^n \sum_{k\geq 0}\binom{n}{k} \Big(\frac{x}{1+x}\Big)^{k}=(1+x)^n\Big(1+\frac{x}{1+x}\Big)^n=(1+2x)^n$

But how did you get this:
$ \sum_{m\geq 0} \binom{n-k}{m-k}x^{m-k}= \sum_{k\geq 0}\binom{n}{k} \Big(\frac{x}{1+x}\Big)^{k} $

there is a step in between-
$\displaystyle \sum_{k\geq 0}\binom{n}{k}x^k \sum_{m\geq 0} \binom{n-k}{m-k}x^{m-k} =
 \sum_{k\geq 0}\binom{n}{k}x^k \big(1+x\big)^{n-k}= \big(1+x\big)^n \sum_{k\geq 0} \binom{n}{k} \Big(\frac{x}{1+x}\Big)^k$
  and this is what i had there.


The MathZoom Summer Camp 2008 will be held at the University of California, Los Angeles, from July 13 through August 2, 2008.

Please visit the [url=http://www.artofproblemsolving.com/Forum/index.php?f=453]MathZoom[/url] forum or [url]http://www.mathzoom.org[/url] for details.

That is quite an ironic name, am I right?


sorry,I think we can't discuss this problem.admin please lack it...


function
calculus
derivative
calculus computations

I am trying to find the generating function, say $G(x)$, for a number series defined as follows:
$a_{n+1}= a_{n}+(n+1)!$ and $a_{0}= 1, n \ge 0$.
Using the power series on the left hand side we have:
$\sum_{n=0}x^{n}a_{n+1}= a_{1}+xa_{2}+x^{2}a_{3}+\cdots+x^{n}a_{n+1}$ Manipulating the series a little we have:
$\frac{a_{0}+(xa_{1}+x^{2}a_{2}+\cdots+x^{n+1}a_{n+1})-a_{0}}{x}= \frac{G(x)}{x}$ (Since we want $G(x)$ to have matching $n$'s.)
Applying the power series to the RHS we have:
$\sum_{n=0}x^{n}(a_{n}+(n+1)!) = (a_{0}+1!)+x(a_{1}+2!)+\cdots+x^{n}(a_{n}+(n+1)!)$. We can break this into two summations so that we have:
$\sum_{n=0}x^{n}a_{n}= G(x)+1$ (I think?)
And
$\sum_{n=0}x^{n}(n+1)! = 1!+x2!+x^{2}3! \cdots$.
So we have (not sure if I'm doing this right):
$\frac{G(x)}{x}= G(x)+1+\sum_{n=0}x^{n}(n+1)!$. At this point can we use the identity of $e^{x}$ somehow to eliminate the $x^{n}(n+1)!$ summation? Can we multiply $e^{x}$ into the summation so that $\sum x^{n}e^{x}(n+1)! = \frac{(n+1)!}{n!}= n+1$?

I got the impresion that all your work was for nothing  :blush: . (there are also couple of things with the indexes that are not clear to me)
It is clear that $a_{n}=\sum_{k=0}^{n}n!$. I dont believe that there is a (nice) closed form for $a_{n}$

You'll probably want to go with an exponential generating function for this.
\[E(x) = \sum_{n = 0}^\infty \frac{a_{n}x^{n}}{n!}.\]
Then \[E(x)-1 = x\sum_{n = 0}^\infty\frac{a_{n+1}x^{n}}{(n+1)!},\]

etc.

Or  $a_{n}=a_{n-1}+n! \backslash \sum_{n\geq 0}$ ,and get
$\sum_{n\geq 0}a_{n}x^{n}=x^{0}a_{0}+x\sum_{n\geq 1}a_{n-1}x^{n-1}+\sum_{n\geq 0}n!x^{n}$ , that is if we donate $G(x): =\sum_{n\geq 0}a_{n}x^{n}$ , we have $G(x)=1+xG(x)+\sum_{n \geq 0}n! x^{n}$ , and try to deal with the last term.

Thats why I'm lost - I don't know what to do with that last term. I'm trying JBL's suggestion, but I'm confused, maybe you could steer me in the right direction.  We define $E(x) = \sum_{n=0}^\infty \frac{a_{n}x^{n}}{n!}$. So then if we divide out by $(n+1)!$ and multipy the $x^{n}$ we have:
$\sum_{n=0}\frac{a_{n+1}x^{n}}{(n+1)!}= \sum_{n=0}\frac{{a_{n}}x^{n}}{(n+1)!}+\sum_{n=0}x^{n}$.  So then $\sum_{n=0}\frac{a_{n+1}x^{n}}{(n+1)!}= \frac{E(x)-1}{x}$.  How do we define$\sum_{n=0}\frac{{a_{n}}x^{n}}{(n+1)!}$? From here I don't know what to do.

Do what I did but instead of using (ordinary) generating function use exponential,you will get $G(x)=1+xG(x)+\sum_{n \geq 0}x^{n}$, but this time $G(x)=\sum_{n \geq 0}\frac{a_{n}}{n!}x^{n}$. I miss lead you in the last post. (I hope I am right this time).

P.S. From what I understand you have some problems with indexes (or indices),you keep on considering generating funct. for $a_{n+1}$, and thats what confusing you.

[quote="illcrowflu"]So then if we divide out by $(n+1)!$ and multipy the $x^{n}$ we have:
$\sum_{n=0}\frac{a_{n+1}x^{n}}{(n+1)!}=$...[/quote]

The first sentence I've quoted doesn't make any sense.  I'm not sure (as Mr. Doe also noted) why you switched from my perfectly nice $E(x)$ to the generating function you have here, which is the generating function for $a_{n+1}$.

...

although it's possible that to finish this will require you to take a derivative somewhere, in which case you'll likely end up being able to express your answer in terms of a non-elementary integral.


A late passenger, sprinting at 8 meters per second, is 30 meters away from the rear of a train when the train starts at rest and accelerates away from the station at 1 meter per second per second.  Can the passenger catch the train if the platform is long enough.  Show work please.

$30+8t=\frac{t^{2}}{2}\Longrightarrow t^{2}-16t-60=0$

$\Delta>0$, so he can catch the train. :P

I dont really understand how you got your answer...Can you explain more?

well, the $v_{i}$ of the train is $0$ because it starts from rest.

The guy has a constant velocity, so the distance he travels is $8t$ plus the $30$ b/c that's the distance away originally.

The distance the train will travel is $(0)(t)+\frac{1}{2}(1)(t)^{2}$

set em equal


------------------------------
 Let $x, y, z >0$ real numbers, such that: $xyz = 1$

Prove that:

\[ \frac{1}{x+2} + \frac{1}{y+2} + \frac{1}{z+2} \leq 1. \]

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

Does anybody know the reference (in this forum or another NET resource) about this ineq??

I can't find it....

Carlos Bravo :)
Lima-PERU

About this problem, I think it has been proposed by Arthur Engel for an IMO  german test in 2005. Anyway, the idea
 is quite simple: just take $x=\frac{a}{b}, y=\frac{b}{c}, z=\frac{c}{a}$ and the inequality reduces to the well-known, $\sum\frac{b}{a+2b}\leq 1$ and from here it is clear how to proceed. ;)

[quote="cezar lupu"]About this problem, I think it has been proposed by Arthur Engel for an IMO  german test in 2005.[/quote]

???

Anyway, the inequality was posted at http://www.mathlinks.ro/Forum/viewtopic.php?t=19330 and http://www.mathlinks.ro/Forum/viewtopic.php?t=45439 .

By the way, nice solution, Cezar!

  darij

[quote=carlosbr]------------------------------
 Let $x, y, z >0$ real numbers, such that: $xyz = 1$

Prove that:

\[ \frac{1}{x+2} + \frac{1}{y+2} + \frac{1}{z+2} \leq 1. \]
[hide=*]Does anybody know the reference (in this forum or another NET resource) about this ineq??

I can't find it....

Carlos Bravo :)
Lima-PERU[/hide][/quote]
http://www.artofproblemsolving.com/community/c6h1163695p5551265


[quote="now a ranger"]Find the sum of the next $3$ odd numbers in this sequence.

$1,4,6,8,9,10,12,14,15......$[/quote]
[hide]note that this is all numbers without the primes.
So the next three would be 21, 25, 27.
So our answer is 73[/hide]


LaTeX
AMC
AIME
HMMT
trigonometry
algebra
partial fractions

O.K., I don't really take the time to learn Latex, so just bear with me.

Can someone please post tough problems involving series notation, and then explain them? :whistling:  :stretcher:  :bomb:

We have no idea what your motivation is, or what you need, from your wording.

Since you (hopefully) have a good idea of what you're looking for, I suggest you look through past AIME problems or HMMT problems and get a feel of how series are used in problems and in solutions.

Problems involving notation?
You mean like, "Put $ 1\plus{}2\plus{}...\plus{}n$ in sigma notation"?

Grn turtle uses series notation alot. It's the greek E thing with a number on top, on the bottom, and to the right.

Sort of like this?

$ \sum_{n \equal{} 1}^{50} n \equal{} \dfrac{50\cdot 51}{2} \equal{} 1 \plus{} 2 \plus{} \cdots \plus{} 49 \plus{} 50$

If so, here are some problems:

1) Find the value of $ \sum_{n \equal{} 1}^{20} 2n \minus{} 1$

2) Find the value of $ \sum_{n \equal{} 1}^{10} n^2$

3) Find the value of $ \sum_{n \equal{} 1}^{50} \dfrac{1}{(2n \minus{} 1)\cdot (2n \plus{} 1)}$

4) Find the value of $ \sum_{n \equal{} 1}^k \dfrac{1}{(2n \minus{} 1)\cdot (2n \plus{} 1)}$

5) Find the value of $ \sum_{n \equal{} 0}^k \binom{k}{n}$

6) Find the value of $ \sum_{n \equal{} 0}^{90} \sin^2 n$

[hide="Answer 1"]
$ 20^2 \equal{} 400$
[/hide]

[hide="Answer 2"]
$ \dfrac{(10)(11)(21)}{6} \equal{} 385$
[/hide]

[hide="Answer 3"]
$ \dfrac{1}{2} \cdot \left(1 \minus{} \dfrac{1}{101}\right) \equal{} \dfrac{50}{101}$

You get this by noticing that each term can be written as $ \left(\dfrac{1}{2n\minus{}1}\minus{}\dfrac{1}{2n\plus{}1}\right)\cdot \dfrac{1}{2}$ and then in the sum you take out a factor of $ \dfrac{1}{2}$ and all the terms cancel out except for the first term and the last, which in this case are $ 1$ and $ \dfrac{1}{101}$
[/hide]

[hide="Answer 4"]
$ \dfrac{1}{2}\cdot \left(1 \minus{} \dfrac{1}{2n \plus{} 1}\right) \equal{} \dfrac{n}{2n \plus{} 1}$

Read the last solution
[/hide]

[hide="Answer 5"]
$ 2^n$

Consider $ (1\plus{}1)^n$ and note that this expands to $ \sum_{i\equal{}0}^{n} \binom{n}{i}$ and this equals $ (1\plus{}1)^n$ or just $ 2^n$
[/hide]

[hide="Answer 6"]
$ 45\cdot 1 \plus{} \dfrac{1}{2} \equal{} 45.5$

Since $ \sin^2 x \equal{} \cos^2 90\minus{}x$, and for values of $ x>45$ you use this substitution to get $ (\sin^2 1 \plus{}\cos^2\plus{}1)\plus{}(\sin^2 2\plus{}\cos^2 2)\plus{}\cdots\plus{}(\sin^2 44\plus{}\cos^2 44) \plus{} \sin^2 45\equal{} 1 \plus{}1\plus{}1\plus{}\cdots 1 \plus{} \dfrac{1}{2}\equal{}45\plus{}\dfrac{1}{2}\equal{}45.5$
[/hide]

Thanks so much tonypr(did I spell your name right?) your help was highly appreciated.   :D

Heh, weird, I always use $ k$ for the dummy variable and $ n$ for the upper bound.

batteredbutnotdefeated, do the series chapter (chapter 13 or so) from AoPS 2 - it's very helpful here, and it'll show you the derivations of the formulas you need for problems like #2 and #5, as well as how to handle weird denominator things like #3 and #4 with partial fractions and telescoping sums.

Thanks Grn Turtle! Ever since I started posting on this forum you've been helping me! :lol:  :)


analytic geometry
complex numbers
imaginary numbers

Find the value of x:
x, x:^2:, x:^3:, x:^4:,x:^5:...=x, -1, -x, 1, x... (the solution repeats itself every four terms)

For people who knows the answer, please don't answer this, not even in spoiler, because people learn this in school at some point.

then what are we supposed to do?

P.S.
the answer is somewhere in aops v1

I think we're supposed to do nothing. I guess the more advanced people should just stay out of this one completely, and let the "Getting Started" people post on this one.

ah. but they wont be able to answer either though.

so im guessing the only posts can be questions on how to do it. and then we can post solutions. right?

Well...I was kind of hoping a beginner guessing the right answer...I'll give it more time.

there arent exactly that many beginners here. but sure. ill wait. i can be very impatient at times.

People who haven't had very many math classes won't be able to solve this question. So if you are in 8th or 9th grade, why don't you try to post an answer, at least in the spoiler. It is an important topic for all to know.

[quote="Syntax Error"]ill wait. i can be very impatient at times.[/quote]
Isn't that a contradiction?

Um, ComplexZeta, I wrote the question...

Yeah..the answer is pretty much whether you've learned the concept yet or you haven't.

well i havent learned it in school. i got it from aops v1. but im gonna leave it to someone who is beginning to solve this problem

Hmm...a bad question then...

The answer is [hide]"i", the imaginary number which equals radical -1. Usually square roots of negative numbers doesn't exist in the "real" plane, but by using the concept of i not only can you write imagenary solutions easier, (usually in the form ai+b, whereas a and b are real number) but it opens up a whole new coordinate system for us[/hide]

I still hope that someone gets the answer without looking at the above...

Actually that isn't quite right. -i also satisfies the conditions of the problem.

yeah, but there was a radical in his thingie. but it doesnt matter. most likely there wont be a person posting the answer

[quote="Syntax Error"]well it kinda gives it up by putting a radical, but hey, whatever floats your boat[/quote]
I fixed that.

To themathguy:
i is a term used when expressing imaginary numbers. (Imaginary numbers are the square roots of negative numbers) i=square root of -1.
It is primarily used in the form ai+b whereas a and b are real numbers (numbers like 2 and -5.623)

I hope I explained it alright...

[quote="Syntax Error"]who's the mathguy?[/quote]
He's the one who said "i?"
FYI (and for others) "i?" is tied with my post "No" for the shortest post ever...Just in case you wanted to know (or cared to know) :)
And yes, I'll delete this soon.

ah. i seemed to have skipped over his post

[quote]-i also satisfies the conditions of the problem.[/quote]

huh? (-i)^3 = -1 * i^3 != i^3.

Who said
[quote]-i also satisfies the conditions of the problem.[/quote]
?

[quote="ComplexZeta"]I said that, and it is true. (-i)^3=i=-(-i).[/quote]
 :?: Can you double check the answer in case of x:^2: and x:^3:?

[quote="Fierytycoon"][quote]I said that, and it is true. (-i)^3=i=-(-i).[/quote]

Yes, but i^3 = -i, not i[/quote]
Also, I think -i:^2:=1 which does not equal -1, as in the problem.

[quote="Fierytycoon"][quote]Also, I think -i=1 which does not equal -1, as in the problem.[/quote]

Actually, i^2 = (-i)^2.

(-i)^2 = (-1)^2 * i^2 = i^2[/quote]
I don't think exponents work that way...

aargh, im too confused in what im saying.

[quote="Fierytycoon"]Yes, but we are not looking at -i^2, we are looking at (-i)^2[/quote]
...you're right. Still, -i doesn't work in case of x:^3:

[quote="Fierytycoon"]Ok, a summary:

i^1 = i
i^2 = -1
i^3 = -i
i^4 = 1

(-i)^1 = -i
(-i)^2 = -1
(-i)^3 = i
(-i)^4 = 1

Therefore, as i^3 != (-i)^3, only i satisfies the requirements of the problem.[/quote]
*clap* nicely done. very neat and organized, something I should do :)

[quote="Tare"]Find the value of x:
x, x:^2:, x:^3:, x:^4:,x:^5:...=x, -1, -x, 1, x... (the solution repeats itself every four terms)[/quote]

Let us test the hypotheses for -i. Note: this is -i EVERYWHERE, not i when it feels like being i.

x^1: -i
x^2: (-i)^2=-1
x^3: (-i)^3=i=-(-i)
x^4: (-i)^4=1

This satisfies every single condition. It is just as good of a solution as i.

wooooooot. weve finally come to a conclusion

All this for such a simple problem... :)

It is i right? Well it looks like it has to be... And I am blind and did not see the other 2 pages...It has already been answered, sorry.

Ah, my mistake. Sorry guys. ComplexZeta is correct.


trigonometry
geometry
circumcircle
perpendicular bisector
geometry proposed

[color=darkblue]If, in a $\triangle ABC$, median $\overline{BD}$ is such that $\measuredangle\ A=\measuredangle\ CBD=x$, and $\measuredangle\ ADB=45^\circ$, then prove that $x=30^\circ$[/color]

[hide="About a powerful tool"]
[color=indigo]Trigo. sometimes provides quickly solution, but not interesting one[/color][/hide]

Maybe you are right, but this problem follows too obvious trigonometrically:
From $\angle{CBD}=\angle{BAC}$, we have $CD \cdot CA = CB^{2}\Leftrightarrow a^{2}=\frac{b^{2}}{2}$. Now just compute $BD^{2}=\frac{c^{2}}{2}$, from the median theorem, and now use the law of sin : $\frac{BD^{2}}{\sin^{2}{x}}=\frac{c^{2}}{\frac{1}{2}}$, therefore $\sin{x}=\frac{1}{2}$, therefore $x=30^\circ$, because it's acute.

Yep, but I hope somebody try to find an elementary solution  :)

Here's the [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=825947#825947]Problem 2[/url]

Through $A,$ we draw a line perpendicular to $BD,$ intersects the segment line $BD,$ at a point so be it $E$ and the line through $C$ and parallel to $BD,$ at a point , so be it $F.$

Because of $\angle CBD = \angle BAD,$ we conclude that the sideline $BC$ of $\bigtriangleup ABC,$ tangents to the circumcircle of the triangle $\bigtriangleup ABD,$ at the point $B.$

So, we have that $(CB)^{2}= (CD)\cdot (CA) = \frac{(CA)^{2}}{2}= (CF)^{2}$ $\Longrightarrow$ $CB = CF$ $,(1)$

From $(1)$ and $AF\perp CF,$ we conclude that the segment line $AF,$ tangents at the point $F,$  to the circle $(C),$ centered at $C,$ with radius $CB = CF.$

Hence, we have $\angle AFB = \frac{\angle BCF}{2}$ $,(2)$

But $\angle AFB = \angle FAB$ $,(3)$ $($ because $BA = BF$ $)$ and $\angle BCF = \angle CBD = \angle A = \angle x$ $,(4)$

From $(2),$ $(3),$ $(4)$ $\Longrightarrow$ $\angle FAB = \frac{\angle x}{2}$ $,(5)$

From $(5)$ and because of $\angle FAB+\angle x = 45^{o},$ we conclude that $\angle x = 30^{o}$ and the proof is completed.

Kostas Vittas.

Nice proof vittasko

[color=indigo]Here's my solution[/color]

[color=darkred]Let $M$ be the ortogonal projection from $C$ to the extension of $\overline{BD}$. Denote $\overline{AD}=\overline{CD}=m\implies\overline{CM}= \frac{1}{{\sqrt 2 }}m$. Note that $\triangle ABC\sim\triangle BDC\implies\overline{BC}= m\sqrt 2\ \therefore\ \triangle CMB$ is a half of an equilateral triangle with side $m\sqrt{2}\implies x=30^\circ$[/color]

I like better your approach, because of:

$(CB)^{2}= (CD)\cdot (CA) = 2(CD)^{2}= 4(CM)^{2}$ $\Longrightarrow$ $CB = 2(CM)$ $\Longrightarrow$ $\angle x = 30^{o}.$

Best regards, Kostas Vittas.

[color=darkblue][b][u]Method I (trigonometric).[/u][/b] Prove easily that $m(\widehat{DBA})=135^{\circ}-x$, $m(\widehat{BCA})=45^{\circ}-x$. From the well-known relation

$\frac{DA}{DC}=\frac{\sin\widehat{BCA}}{\sin\widehat{BAC}}\cdot\frac{\sin\widehat{DBA}}{\sin\widehat{DBC}}$ obtain $\sin(45^{\circ}-x)\sin (135^{\circ}-x)=\sin^{2}x$ $\implies$ $\cos 2x=1-\cos 2x$ $\implies$ $\boxed{x=30^{\circ}}$.

[b][u]Method II (metric).[/u][/b] $\{\begin{array}{ccccccccc}1\blacktriangleright & \widehat{CAB}\equiv\widehat{CBD}& \Longrightarrow & \triangle CAB\sim\triangle CBD & \Longrightarrow & \frac{b}{a}=\frac{c}{m_{b}}=\sqrt 2\\\\ 2\blacktriangleright & m(\widehat{ADB})=45^{\circ}& \Longrightarrow & c^{2}=\frac{b^{2}}{4}+m^{2}_{b}-bm_{b}\cdot\frac{\sqrt 2}{2}& \Longrightarrow & c^{2}+ac\sqrt 2-a^{2}=0 & \Longrightarrow & \frac{c}{\sqrt 3-1}=\frac{a}{\sqrt 2}\end{array}\|$ $\implies$
\[\boxed{\frac{a}{\sqrt 2}=\frac{b}{2}=\frac{c}{\sqrt 3-1}}=\frac{m_{b}\sqrt 2}{\sqrt 3-1}\]
$3.1\blacktriangleright$ From the Cosinus' theorem $a^{2}=b^{2}+c^{2}-2bc\cos A$ obtain :

$2=4+(\sqrt 3-1)^{2}-4(\sqrt 3-1)\cos x$ $\implies$ $2\sqrt 3(\sqrt 3-1)=4(\sqrt 3-1)\cos x$ $\implies$ $\cos x=\frac{\sqrt 3}{2}$ $\implies$ $\boxed{x=30^{\circ}}$.

$3.2\blacktriangleright$ Otherwise, from the Sinus' theorem $\frac{AB}{\sin 45^{\circ}}=\frac{BD}{\sin x}$ obtain :

$c\sqrt 2=\frac{m_{b}}{\sin x}$ $\implies$ $\sqrt 2\sin x=\frac{m_{b}}{c}=\frac{1}{\sqrt 2}$ $\implies$ $\sin x=\frac{1}{2}$ $\implies$ $\boxed{x=30^{\circ}}$.

[b][u]Method III (synthetic).[/u][/b] The circumcircle $C(O)$ of $\triangle ABC$ cuts again the line $BD$ in the point $E$. Therefore,

$\{\begin{array}{c}\widehat{BAC}\equiv\widehat{EAC}\\\ OD\perp AC\\\ m(\widehat{CDE})=45\end{array}\|$ $\implies$ the line $\overline{BDE}$ is the bisector of the segment $[OC]$ $\implies$ $m(\widehat{BAE})=60^{\circ}$ $\Longrightarrow$ $\boxed{x=30^{\circ}}$.

[b][u]Method IV (synthetic).[/u][/b] Let $R$- the circumradius of $\triangle ABD$. Thus, $\{\begin{array}{ccc}\triangle CAB\sim\triangle CBD & \implies & BD\sqrt 2=AB\\\\ m(\widehat{ADB})=45^{\circ}& \implies & AB=R\sqrt 2\end{array}\|$ $\implies$ $BD=R$ $\implies$ $x=30^{\circ}$.[/color]

[size=150][color=darkred][b][u]Extension.[/u][/b][/color][/size] $\{\begin{array}{c}\triangle ABC\\\\D\in (AC)\ ,\ DA=DC\\\\ m(\widehat{ADB})=\phi\\\\ m(\widehat{BAD})=m(\widehat{CBD})=x\end{array}\|\Longrightarrow \{\begin{array}{c}\cos 2x=\cos^{2}\phi\mathrm{\ or\ }\boxed{\ \sin\phi=\sqrt 2\sin x\ }\\\\ \frac{a}{1}=\frac{b}{\sqrt 2}=\frac{c}{\sqrt{1+\cos^{2}\phi}-\cos\phi}\end{array}$

[size=134][color=darkblue][b]Remark.[/b] Very nice the poster's and Vittasko's methods ![/color][/size]

Hahahaha, Virgil Nicula always looking for new ideas  :wow:

[size=150] [color=red]See my solution to an older problem. Change a little the wording to take your problem.[/color][/size]
[quote="stergiu"][quote="stergiu"][quote="senouy"] $ABC$  is a triangle ,
$M$ is midpoint of  side $BC$ such that: $\angle AMB = 45$ and $\angle B = \angle MAC$ Show that $\angle B = 30$[/quote]

[b] HINT(A first approach )[/b]

 Triangles $ABC , AMC$ are similar. From this fact  we take $\angle A = 135$ and $CA^{2}= CM.CB$. Now take sinus law in triangle $ABC$ to find $\sin B = \frac{1}{2}$ ect.(Use  : $CA = \sqrt 2 BM , CB = 2 BM$ in the sinus law , so you can simplify  !!!)

[u] Babis[/u][/quote]

 [b]Synthetic Solution(HINT) ![/b]
[color=blue]
During a pause in the school I found this solution. I hope it is not bad ![/color]

 If we draw $AD \perp BC , CE \perp BA$ then we easily find that $ADCE$ is inscribed ,   $EA = EC$ and $DE$ is the perpendicular bisector of $AM$ , so $EM= EA$ .Thus $EC=EA=EM= \frac{BC}{2}= MC$.This gives $\angle BCE = 60$ , so $\angle B= 30$.

[u]Babis[/u][/quote]


How many three-digit numbers are not multiples of 5 nor 7?

but the question asks "NEITHER 5 NOR 7", so not 5 AND not 7?
Is there something wrong?

[quote=nicolahcm][quote="ernie"]Using PIE, we get $ 900\minus{}\lfloor\frac{900}{5}\rfloor\minus{}\lfloor\frac{900}{7}\rfloor \plus{} \lfloor\frac{900}{35}\rfloor \implies 900 \minus{} 180 \minus{} 128 \plus{} 25 \implies \boxed{617}$.[/quote]

The multiplies of 35 from 1 to 900 are different from the multiples of 35 from 100 to 999.[/quote]

So...... PIE only apply to the ones that starts with one, right??

[hide =Solution] $\frac{4}{5}$ of integers from $1-35$ are not multiples of $5$ and  $\frac{6}{7}$ of integers from $1-35$ are not multiples of $7$. $900\equiv25\pmod{35}$ and $105$ is one of the first $25$ three digit integers. Since $35$ is a multiple of both $5$ and $7$ the answer is $$\lceil{\frac{4}{5}*\frac{6}{7}*900}\rceil=618$$[/hide]

[quote=Mrdavid445][quote="ernie"]Using PIE, we get $ 900\minus{}\lfloor\frac{900}{5}\rfloor\minus{}\lfloor\frac{900}{7}\rfloor \plus{} \lfloor\frac{900}{35}\rfloor \implies 900 \minus{} 180 \minus{} 128 \plus{} 25 \implies \boxed{617}$.[/quote]
[hide="Except.."]You have add $1$ at the end, or else you would be missing 1.

So our answer is actually $\boxed{618}$[/hide]

Edit: Thanks for abusing my rating.[/quote]

Why did you add one?

PIE is the Principle of Inclusion-Exclusion, which is a method of finding the union of multiple sets. See http://artofproblemsolving.com/wiki/index.php?title=Principle_of_Inclusion-Exclusion

Simply put, you add them up, then subtract repeats, then add repeats of repeats, on and on

You don't have to blame people for getting it wrong. You should tell them politely, and not restate what  Mrdavid445 ALREADY said before you

I know PIE is the inclusion exclusion method but how can it help you solve problems
            Thank you.

[hide = Hint1]We use complementary counting here![/hide]
[hide = Hint2]A multiple of 5 and 7 is a multiple of 35![/hide]
[hide = Solution][b]There are 900 total three digit numbers : 100 to 999[/b]
We want to count :
[b]1. "3 digit" multiples of 5[/b]
5 * 20 = 100 & 5 * 199 = 995 | So we get that there are 180 "3 digit" multiples of 5.
[b]2. "3 digit" multiples of 7[/b]
Similarly, 7 * 15 = 105 & 7 * 142 = 994 | So we get that there are 128 "3 digit" multiples of 7.
[b]3. "3 digit" multiples of 5 and 7 (or 35) [/b]
Lastly, 35 * 3 = 105 & 35 * 28 = 980 | So we get that there are 26 "3 digit" multiples of 5/7 (or 35)

Using PIE we get, $900 - 180 - 128 + 26$ $\implies \boxed{618}$
[/hide]

But why does PIE not work in this case? It gives 617 and the correct answer is 618.

It does work. Double check your arithmetic maybe?

[hide = My solution]
We have sequences:
$5,10...1000$
and
$7,14...994$
When we divide the first sequence by $5$ we get $1,2...200$ With $200$ terms.
When we divide the second sequence by $7$ we get $1,2...142$ With $142$ terms.
We also have to subtract the multiples of $35$. Using the same process as above, we get $28$

So, $ 200 + 142 - 28 = 314$
We know that there are $900$ three digit numbers. Now, we subtract $314$ from this. We get $586$. That is our answer. [/hide]

Concepts from Intro to Counting & Probability.

@above

[quote=GameBot]How many [b]three-digit[/b] numbers are not multiples of 5 nor 7?[/quote]

You subtracted one and two digit multiples of 5 and 7 too.

I'm pretty sure you also included 1000 into your total.

I think PIE doesn’t work because doing 900/35 means we’re looking for multiples of 35 from 0 to 900. But in this case, we’re looking for multiples of 35 from 100 to 999. I used PIE too and got 617...
I don’t know, just a theory.

[hide=my sol]3-digit numbers divis. by $5=180$, 3-digit numbers divis by $7=128$, 3-digit numbers divis by $35=26$. $180+128-26=\boxed{618}$[/hide]

[hide=Sol]There are $900$ 3-digit numbers from $100$ to $999$. There are $180$ multiples of $5,$ $128$ multiples of $7$, and $26$ multiples of $35.$ To remove the overcounts: $900-(180+128-26) = 900-282=\boxed{618}.$ (I used PIE.)[/hide]

[quote=Shadowlight]This is probably one of the oldest and most popular GameBot threads[/quote]

There are some with over 100 posts, and others from 2008. But yeah, been active.

[quote=Shadowlight]This is probably one of the oldest and most popular GameBot threads[/quote]

Ya, gamebot threads usually only have solutions.

Hello!
NOTE: This is meant to be a detailed solution for people who may not understand the other solutions, or may not understand why post #2 was incorrect.
We want to find the number of 3 digit numbers not divisible by $5$ or $7$. Obviously, it would be tedious to find each one, so instead, let find the ones that ARE divisible by $5$ and $7$, and subtract them from the total.

We know that there are $900$ 3-digit numbers, from $100$ to $999$. Let's first find how many are divisible by $5$. We can use this expression to find out: $\lfloor{\frac{999}{5}}\rfloor-\lfloor{\frac{99}{5}}\rfloor$. [hide=explanation: read if you don't understand exactly what I wrote to clear things up :)]First, before using it, let's understand what it means. It is finding how many numbers less than $999$ are divisible by $5$ (term 1), and subtracting the number of numbers less than $99$ that are divisible by $5$ (term 2). If you don't understand what the bars around the fractions mean, I suggest you look up "floor function" online. The subtraction yields the total number of numbers between $100$ and $999$ that are divisible by $5$. You might ask [quote=You]HIA2020, why not just clear up the expression and write $\lfloor{\frac{999-99}{5}}\rfloor=\lfloor{\frac{900}{5}}\rfloor$?[/quote]
Good question! Although with normal subtraction and division, this would be correct; with the floor function, it isn't. Just because the two expressions are equivalent with this example doesn't mean it always will be. With the original expression, you are finding how many numbers between $100$ and $999$ are divisible by $5$. However, with the second expression, which is incorrect, you are only finding how many numbers are divisible by $5$ from $1$ to $900$. You will see how this will have an impact later in the solution.[/hide]
This would be $199-19=180$.

Now, let's do this for $7$: $\lfloor{\frac{999}{7}}\rfloor-\lfloor{\frac{99}{7}}\rfloor=142-14=128$.

You may be tempted, now that we've found the total number of numbers divisible by $5$ and $7$, to subtract this from the 900 3-digit numbers. However, you must realize that we counted the numbers divisible by $5$ AND $7$ TWICE! We must subtract them off. Let's keep track of what we are doing:

[hide=What we are doing]$\text{Number of 3 digit numbers not divisible by 5 or 7 = Total number of 3 digit numbers - }$ $\text{(3 digit numbers divisible by 5 + 3 digit numbers divisible by 7 - 3 digit numbers divisible by 5 and 7, or 35)} $ [/hide]

Now, let's find the number of 3 digit numbers divisible by $35$.
 $\lfloor{\frac{999}{35}}\rfloor-\lfloor{\frac{99}{35}}\rfloor=28-2=26$

[hide=Common error]Like I said, doing $\lfloor{\frac{999-99}{35}}\rfloor=\lfloor{\frac{900}{35}}\rfloor$ would've thrown you off! This time, it gives an incorrect answer. [/hide]

Now, let's plug into the equation in the hide tag "what we are doing" above.
$900-(180+128-26)=\boxed{618}$

Unfortunately, due to the unclear wording, I put 284 instead of 618, the problem should be revised to make it easier to understand; by rewording it like, "How many three digit numbers are not multiples of 5 or 7?"


inequalities
ratio
function
calculus
derivative
inequalities proposed

For $ a,b,c>0$ prove

$ \prod_{sym}(a\plus{}2b)\ge27(ab\plus{}bc\plus{}ca)^3.$ ;)

[hide="remark"]Putting $ a\equal{}\frac1x$ etc, we can write it equivalently as
$ \prod_{sym}( x\plus{}2y)\ge27xyz(x\plus{}y\plus{}z)^3$ or

\[ \sum_{sym}\log(x\plus{}2y) \ge \log(3x)\plus{} \log(3y)\plus{} \log(3z)\plus{}3 \log(x\plus{}y\plus{}z) .\]
Compare this with Harazi’s one
\[ \sum_{sym}\frac1{x\plus{}2y}\le \frac1{3x}\plus{} \frac1{3y}\plus{} \frac1{3z}\plus{} \frac3{x\plus{}y\plus{}z}.\]

Rather funny, isn’t it? :lol: I don’t think though that mine can be ‘derived’ (so to say :) ) [i]directly[/i] from Schur – although my proof uses Schur. [/hide]

This follows from a more general version of Schur's inequality:

If $ f: (a,b)\to R$ such that $ f''(x) \ge 0$ and the sign of $ f'''(x)$ is constant on $ (a,b)$, then
$ f(3x) \plus{} f(3y) \plus{} f(3z) \plus{} 3f(x\plus{}y\plus{}z) \ge \sum_{sym} f(2x\plus{}y)$.

Just plug in $ f(x) \equal{} \minus{}log(x)$ on the interval $ (0,\infty)$ to solve it now.

I think one might have an elementary solution thanks to the identity
\[ (x \plus{} 2y)(y \plus{} 2z)(z \plus{} 2x) \minus{} 3(x \plus{} y \plus{} z)(xy \plus{} yz \plus{} zx) \equal{} (x \minus{} y)(y \minus{} z)(z \minus{} x)
\]
So \[ \prod_{sym}(x \plus{} 2y)\equal{}9(x \plus{} y \plus{} z)^2(xy \plus{} yz \plus{} zx)^2\minus{}(x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2\] and your inequality gives an improvement of a specific case of Newton's inequality. But I think from here it is easy to conclude

[quote="spanferkel"]For $ a,b,c > 0$ prove

$ \prod_{sym}(a \plus{} 2b)\ge27(ab \plus{} bc \plus{} ca)^3.$ ;)

[hide="remark"]Putting $ a \equal{} \frac1x$ etc, we can write it equivalently as
$ \prod_{sym}( x \plus{} 2y)\ge27xyz(x \plus{} y \plus{} z)^3$ or
\[ \sum_{sym}\log(x \plus{} 2y) \ge \log(3x) \plus{} \log(3y) \plus{} \log(3z) \plus{} 3 \log(x \plus{} y \plus{} z) .
\]
Compare this with Harazi’s one
\[ \sum_{sym}\frac1{x \plus{} 2y}\le \frac1{3x} \plus{} \frac1{3y} \plus{} \frac1{3z} \plus{} \frac3{x \plus{} y \plus{} z}.
\]
Rather funny, isn’t it? :lol: I don’t think though that mine can be ‘derived’ (so to say :) ) [i]directly[/i] from Schur – although my proof uses Schur. [/hide][/quote]
$ 9(a \plus{} b \plus{} c)^2(ab \plus{} bc \plus{} ca)^2 \minus{} (a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2\ge 27(ab\plus{}bc\plus{}ac)^2$
$ \frac{9}{2}(ab\plus{}bc\plus{}ac)^2((a\minus{}b)^2\plus{}(b\minus{}c)^2\plus{}(c\minus{}a)^2)\ge (a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2$
its obvious that if ve mix var like this ($ c\equal{}min(a,b,c)$), $ a\equal{}a\minus{}c,b\equal{}b\minus{}c,c\equal{}0$ LHS will increase RHS - const so we must prove it only when $ c\equal{}0$ so
$ \frac{9}{2}a^2b^2((a\minus{}b)^2\plus{}b^2\plus{}a^2)\ge (a\minus{}b) ^2b ^2a^2$ if $ ab\equal{}0$ its obvious so
$ \frac{9}{2}((a\minus{}b)^2\plus{}b^2\plus{}a^2)\ge (a\minus{}b) ^2$ and its obvious to because$ SOS\ge0$

[quote="spanferkel"]For $ a,b,c > 0$ prove

$ \prod_{sym}(a \plus{} 2b)\ge27(ab \plus{} bc \plus{} ca)^3.$
[/quote]

Let $ a^2 \plus{} 2bc \equal{} x,b^2 \plus{} 2ca \equal{} y,c^2 \plus{} 2ab \equal{} z,ab \plus{} bc \plus{} ca \equal{} s$,
We can prove this inequality :
$ xyz\ge s^3$

Then,
$ \prod_{sym}(a \plus{} 2b) \equal{} \prod_{cyc}(a \plus{} 2b)(a \plus{} 2c) \equal{} \prod_{cyc}(x \plus{} 2s)$
$ \equal{} 8s^3 \plus{} 4s^2(x \plus{} y \plus{} z) \plus{} 2s(xy \plus{} yz \plus{} zx) \plus{} xyz\ge 8s^3 \plus{} 12s^3 \plus{} 6s^3 \plus{} s^3 \equal{} 27s^3$
$ \equal{}27(ab\plus{}bc\plus{}ca)^3$, done!

I remember Arqady posted in 2007 on this forum that:

If $ a,b,c$ are nonnegative real numbers such that $ ab\plus{}bc\plus{}ca\equal{}3$, than

$ (a\plus{}2b)(b\plus{}2c)(c\plus{}2a) \ge 27$.

@ zeb:
[quote="zeb"]This follows from a more general version of Schur's inequality:

If $ f: (a,b)\to R$ such that $ f''(x) \ge 0$ and the sign of $ f'''(x)$ is constant on $ (a,b)$, then
$ f(3x) \plus{} f(3y) \plus{} f(3z) \plus{} 3f(x \plus{} y \plus{} z) \ge \sum_{sym} f(2x \plus{} y)$.
[/quote]
Nice. This is stronger than Popoviciu. :roll: Can you give a reference for this? In the proof, how does the $ f'''$ condition come in?

@Vasc (& Arqady):
[quote="Vasc"]I remember Arqady posted in 2007 on this forum that:

If $ a,b,c$ are nonnegative real numbers such that $ ab \plus{} bc \plus{} ca \equal{} 3$, than

$ (a \plus{} 2b)(b \plus{} 2c)(c \plus{} 2a) \ge 27$.[/quote]

This is cool :cool:  :!: Sooooo...
Here is something.

 For $ x,y,z>0$, prove
$ \frac1{3x}\plus{} \frac1{3y}\plus{} \frac1{3z}\plus{} \frac3{x\plus{}y\plus{}z} \ge 2\sum_{cyc}\frac1{x\plus{}2y}.$     :whistling:

[quote="spanferkel"]@ zeb:
[quote="zeb"]This follows from a more general version of Schur's inequality:

If $ f: (a,b)\to R$ such that $ f''(x) \ge 0$ and the sign of $ f'''(x)$ is constant on $ (a,b)$, then
$ f(3x) \plus{} f(3y) \plus{} f(3z) \plus{} 3f(x \plus{} y \plus{} z) \ge \sum_{sym} f(2x \plus{} y)$.
[/quote]
Nice. This is stronger than Popoviciu. :roll: Can you give a reference for this? In the proof, how does the $ f'''$ condition come in?[/quote]

Well, I originally heard it from Gjergji (aka Albanian Eagle), so I decided to try to prove it myself. But my proof is tooo complicated... first I got rid of the condition that $ f'' \ge 0$ by assuming $ f'''\ge 0$ and $ x\ge y\ge z$ and replacing $ f (x \plus{} 2z) \plus{} f (y \plus{} 2z)$ with $ f (3z) \plus{} f (x \plus{} y \plus{} z)$ and $ f (2x \plus{} y) \plus{} f (x \plus{} 2y)$ with $ f ((2 \plus{} \phi)x \plus{} (1 \minus{} \phi)y) \plus{} f ((1 \minus{} \phi)x \plus{} (2 \plus{} \phi)y)$ on the right hand side ($ \phi$ is golden ratio between $ 0$ and $ 1$). After I did that, the sum and the sum of squares on both sides were the same, and it followed from an inequality I had already proved for functions with third derivative positive (namely, that if we have $ x_1,...,x_4$, $ y_1,...,y_4$ with equal sum and sum of squares, and if the biggest $ x_i$ is bigger than the biggest $ y_i$ and the smallest $ y_i$ is smaller than the smallest $ x_i$, and if $ f'''\ge 0$ then $ \sum f(x_i)\ge\sum f(y_i)$).

Anyway there should be an easier way to prove it (i.e. the method that the person who originally told it to Gjergji proved it with).

[quote="zeb"][quote="spanferkel"]@ zeb:
[quote="zeb"]This follows from a more general version of Schur's inequality:

If $ f: (a,b)\to R$ such that $ f''(x) \ge 0$ and the sign of $ f'''(x)$ is constant on $ (a,b)$, then
$ f(3x) \plus{} f(3y) \plus{} f(3z) \plus{} 3f(x \plus{} y \plus{} z) \ge \sum_{sym} f(2x \plus{} y)$.
[/quote]
Nice. This is stronger than Popoviciu. :roll: Can you give a reference for this? In the proof, how does the $ f'''$ condition come in?[/quote]

Well, I originally heard it from Gjergji (aka Albanian Eagle), so I decided to try to prove it myself. But my proof is tooo complicated...

Anyway there should be an easier way to prove it (i.e. the method that the person who originally told it to Gjergji proved it with).[/quote]Thank you. Yes indeed, it is a bit technical - as expected. I had barely hoped for something generalizable :huh:

I asked him about it the other day, he says there's a much easier way to prove it if we look at $ h(x) \equal{} f(3ax) \plus{} f(3bx) \plus{} f(3cx) \plus{} 3f(ax \plus{} bx \plus{} cx) \minus{} \sum_{sym}f(2ax \plus{} bx)$, take its derivative and use standard methods on increasing convex functions to prove that $ h'(x) \ge 0$ for $ x \ge 0$.

It also isn't too hard to prove $ \sum_{cyc} f(4a) \plus{} f(2a\plus{}b\plus{}c) \ge \sum_{cyc} f(3a \plus{} b) \plus{} f(3b\plus{}a)$ under the same conditions, so it can definitely be generalized.

Indeed the "function" version of Suranyi's inequality (general of Schur's) can be proven to be true under the conditions $ f,f'$ are convex.
It's a result by M.Bencze

I've proven that under this conditions even http://www.mathlinks.ro/Forum/viewtopic.php?&t=89757 post #5 holds in its "function" version.

[quote="spanferkel"]
 For $ x,y,z > 0$, prove
$ \frac1{3x} \plus{} \frac1{3y} \plus{} \frac1{3z} \plus{} \frac3{x \plus{} y \plus{} z} \ge 2\sum_{cyc}\frac1{x \plus{} 2y}.$     :whistling:[/quote]
This inequality is equivalent to the following inequality
Let $ a,b,c >0$, then
\[ a\plus{}b\plus{}c\plus{}\frac{9abc}{ab\plus{}bc\plus{}ca} \ge 6\left( \frac{ab}{a\plus{}2b}\plus{}\frac{bc}{b\plus{}2c}\plus{}\frac{ca}{c\plus{}2a}\right) \qquad (\bigstar)\]
Unfortunately, $ (\bigstar)$ does not always hold, for example with $ c\rightarrow 0, a\equal{}1.5,b\equal{}1$. But we can force it to become true and the condition is: $ a,b,c$ are the sidelengths of a triangle, then we can show that $ (\bigstar)$ is valid. :)

I think this is true.

If $ x,y,z \ge 0$, then

$ \sqrt x\plus{}\sqrt y\plus{}\sqrt z \plus{}3\sqrt \frac{x\plus{}y\plus{}z}{3} \le 2(\sqrt \frac {x\plus{}2y}{3}\plus{}\sqrt \frac{y\plus{}2z}{3}\plus{}\sqrt \frac {z\plus{}2x}{3}\ )$.

I think,
$ \sum x^k\plus{}3 \left(\frac{x\plus{}y\plus{}z}{3}\right)^k \le 2\sum_{cyc}\left(\frac{x\plus{}2y}{3}\right)^k$ for $ 0\le k\le 1$
and
$ \sum x^k\plus{}3 \left(\frac{x\plus{}y\plus{}z}{3}\right)^k \ge 2\sum_{cyc}\left(\frac{x\plus{}2y}{3}\right)^k$ for $ k\ge1$. ;)

For $ k<0$, both are wrong.

hmm. any progress with these last inequalities?

[quote="spanferkel"]I think,
$ \sum x^k \plus{} 3 \left(\frac {x \plus{} y \plus{} z}{3}\right)^k \ge 2\sum_{cyc}\left(\frac {x \plus{} 2y}{3}\right)^k$ for $ k\ge1$. ;)
[/quote]
This is not true for $ k > 5.8913.....$ (see case $ x \equal{} 0$ and $ y\equal{}1$).

[quote="Vasc"][quote="spanferkel"]I think,
$ \sum x^k \plus{} 3 \left(\frac {x \plus{} y \plus{} z}{3}\right)^k \ge 2\sum_{cyc}\left(\frac {x \plus{} 2y}{3}\right)^k$ for $ k\ge1$. ;)
[/quote]
This is not true for $ k > 5.8913.....$ (see case $ x \equal{} 0$ and $ y \equal{} 1$).[/quote]
Yes I have realized in between (but wasn't online) that this must be wrong. Because otherwise, we could sum them up for all $ k$ taking $ \sum \frac {x^k}{k!}$etc.  to obtain a "skew" form of Schur (i.e. cyclic RHS), which is obviously wrong  :blush:

[quote="Vasc"]I remember Arqady posted in 2007 on this forum that:

If $ a,b,c$ are nonnegative real numbers such that $ ab \plus{} bc \plus{} ca \equal{} 3$, than

$ (a \plus{} 2b)(b \plus{} 2c)(c \plus{} 2a) \ge 27$.[/quote]
Thank you, Vasc! It's here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=151210
There is a nice proof for this inequality.  :wink:
By the way, what is the best $ k$ for which an inequality
\[ (a \plus{} kb)(b \plus{} kc)(c \plus{} ka)\geq(1 \plus{} k)^3
\]
holds for all non-negative $ a,$ $ b$ and $ c$ such that $ ab \plus{} ac \plus{} bc \equal{} 3$ $ ?$
For $ k \equal{} 1$ we obtain the known Carlson's inequality.

[quote=can_hang2007]
Let $ a,b,c >0$, then
\[ a\plus{}b\plus{}c\plus{}\frac{9abc}{ab\plus{}bc\plus{}ca} \ge 6\left( \frac{ab}{a\plus{}2b}\plus{}\frac{bc}{b\plus{}2c}\plus{}\frac{ca}{c\plus{}2a}\right) \qquad (\bigstar)\]
Unfortunately, $ (\bigstar)$ does not always hold, for example with $ c\rightarrow 0, a\equal{}1.5,b\equal{}1$. But we can force it to become true and the condition is: $ a,b,c$ are the sidelengths of a triangle, then we can show that $ (\bigstar)$ is valid. :)[/quote]
The Ravi's substitution gives
$\sum_{cyc}(8x^5y+4x^5z+33x^4y^2+3x^4z+12x^3y^3+24x^4yz-20x^3y^2z-28x^2z^2y-36x^2y^2z^2)\geq0$,
which is obviously true.


Find all pairs of integers $(m,n)$ such that an $m\times n$ board can be totally covered with $1\times 3$ and $2 \times 5$ pieces.

one can work this out in a straightforward manner-first note that any $n\geq 8$ can be written in the form $3a+5b$ with both $a,b\geq 0$(and that $7$ can't be writen so) and also that any integer $n\geq 2$ can be written as $2a+3b$ with $a,b\geq 0$.thus if $m$(or $n)\geq 8$ then writing $m=3a+5b$ we note that $2\times m$ can be tiled with tiles of the specified types.also since $3\times m$ can always be tiled with $1\times 3$ tiles, we see that writing $n=2c+3d$ gives a tiling of the $m\times n$ square.
what now remains is seeing which of the ones that have been left out($m,n\leq 7$) can be tiled and which can't.wlog, assume $m\leq n$.clearly $3\times n,6\times n$ can be tiled for all $n\geq 1$.if $m=2$ we have the following list: $(2,3),(2,5),(2,6)$.
for $m=4$ we have $(4,5),(4,6)$((4,7) can't be tiled and that is probably the only slightly nontrivial one in this listing,but one can argue that out rather easily since any potential tiling of it has to contain precisley one $2\times 5$ tile and $6\ 1\times 3$ tiles).
finally one can check(!) that $5\times 5$ and $7\times 7$ can also be tiled(for the $5\times 5$ note that this can be decomposed into a $2\times 5$ and $3\times 5$).
the last case is that of $7\times 7$ and this can also be tiled(form an outer layer of the $2\times 5$ tiles using $4$ of them;what remains inside is a $3\times 3$ grid).

Well, you just did de 4x7 wrongly: there is a covering, exactly with the 5x2 centered and the 3x1 surrounding it.

Note that every $3\times n$ can be covered by just the $1\times 3$ piece.

Now we prove that every $4\times n$ board and every $5\times n$ board can be covered for $n\ge 5$ by induction.
By a simple check, we find that the $4\times5$, $4\times6$, and $4\times7$ boards can be covered, and that a $4\times n$ board can be covered iff a $4\times (n+3)$ board can be covered by adding or removing four $1\times 3$ pieces. Hence by induction, we are done.
Similarly, note that the $5\times4$, $5\times5$, and $5\times6$ boards can be covered, and a $5\times n$ board can be covered iff a $5\times(n+3)$ board can be covered by adding or removing five $1\times 3$ pieces. Hence by induction, we are done.

So, if $n\ge 5$, then every $k\times n$ board can be covered for $k=3,4,5$.

Fix $n\ge 5$. Suppose all the $k\times n$ boards can be covered for $k=3,4,5,...,m$ where $m\ge 5$. Then we may choose a $k_1$ and $k_2$ in the set $\{3,4,5,\dots, m\}$ such that $k_1+k_2=m+1$. For such integers, we note that the $k_1\times n$ and $k_2\times n$ boards can be covered. By joining the tilings for these two boards along the common edge of length $n$, we obtain a tiling for the $(m+1)\times n$ board. Thus by strong induction, we have proven that for any $m,n\ge 5$, the $m\times n$ board can be tiled.

It now remains to determine which boards cannot be tiled. We know that every $4\times n$ board can be tiled if $n\ge 5$ (and thus the $5\times 4$ board as well). By simply checking, we find that the $4\times 1$, $4\times 2$, and $4\times 4$ boards cannot be tiled since an $2\times 5$ tile cannot fit in these boards and the number of squares is not a multiple of $3$ which would be required for it to be able to be tiled solely by $1\times 3$ pieces.
Every $3\times n$ board can be tiled trivially.
It is easy to see that a $2\times n$ board can be tiled iff it can be tiled with $2\times 5$ and $2\times 3$ pieces in which case this is possible if and only if $n=3a+5b$ for nonnegative integers $a$ and $b$. By the Chicken-McNugget theorem, every integer $n$ beyond $7$ can be written in this way. It is easy to check, then, that the $2\times 1$, $2\times 2$, $2\times 4$, and $2\times 7$ boards cannot be tiled.
Finally, any $1\times n$ board can be tiled iff $n$ is a multiple of $3$ since the $5\times 2$ piece cannot fit in this board.

Thus the boards that cannot be covered are the $4\times 4$, $2\times 2$, $2\times 4$, $2\times 7$, $1\times(3k+1)$, and $1\times (3k+2)$ boards. And every other board can be covered.


[quote="GameBot"]The $ 22$ members of Troop $ 71$ and their leader, Dave, are planning an overnight camping trip.  Dave is planning to make hamburgers, and his recipe calls for $ 1$ pound of beef per $ 4$ burgers.  Assuming each person eats one burger, how many pounds of beef should he purchase to serve the troop, himself, and the $ 3$ chaperones? Express your answer as a decimal to the nearest tenth.[/quote]

There are 26 people, so 26 burgers will be eaten.

So, $ \frac{26}{4}\equal{}6.5$.

[quote=davidaops]what are you saying? you get two tries per problem no matter what[/quote]

This is FTW, not Alcumus, davidaops. :P

ohhh sorry

the headers look the same :blush:


geometry
geometric transformation
reflection
circumcircle
trigonometry
symmetry
incenter

[color=darkred]Quadrilateral $ ABCD$ is inscribed in the circle of diameter $ BD$. Point $ A_1$ is reflection of point $ A$ wrt $ BD$ and $ B_1$ is reflection of $ B$ wrt $ AC$. Denote $ \{P\}\equal{}CA_1 \cap BD$ and $ \{Q\}\equal{}DB_1\cap AC$. Prove that $ AC\perp PQ$.[/color]

Denote $(O,\varrho)$ the circumcircle of $ABCD$ and $R \equiv AC \cap DB.$ Clearly $\angle BPC \equal{} \angle BAC \minus{} \angle DAA_1$ $\Longrightarrow$ $ \angle BPC \equal{} \angle BAC \minus{} \angle BAO \equal{} \angle OAC$ $\Longrightarrow$  $AOCP$ is cyclic. Thus, $AR \cdot RC \equal{} PR \cdot OR=\varrho^2 \minus{} OR^2$ $\Longrightarrow$

$(PO \minus{} OR) \cdot OR \equal{} \varrho^2 \minus{} OR^2 \Longrightarrow PO \cdot OR \equal{} \varrho^2 \equal{} OD^2 \ \ (1)$

If $ H$ is the intersection of lines $ BB_1$ and $ AC,$ then $OH$ is a midline of $\triangle  BDB_1$

$\Longrightarrow OH \parallel DQ \Longrightarrow \triangle DQR \sim \triangle OHR \Longrightarrow \frac {DR}{OR} \equal{} \frac {DQ}{OH} \ \ (2)$

From $(1)$ and $(2)$ we get $\frac{PD}{OD} \equal{} \frac {DQ}{OH}$ $\Longrightarrow \frac {PD}{OB} \equal{} \frac {DQ}{OH},$

which implies that $\triangle QDP \sim  \triangle BHO$ $\Longrightarrow$ $ PQ \parallel BH \perp AC.$

I think the formulation of the problem it's OK and easy to prove.

Kostas Vittas.

[color=darkblue]Here is my solution from TST:
There are lot of cases (I mean drawings) and because of this I thnk I will lose some points.
Suppose that everything is as in the figure (other cases are similliar).

[img]http://pic.ipicture.ru/uploads/090315/Ds17Pjs60C.png[/img]

We wanna prove that $ PQ \parallel BB'$ or

$ \dfrac{DQ}{QB_1} \equal{} \dfrac{DP}{PB}$

Let $ \angle{ABD} \equal{} \alpha$, $ \angle{DBC} \equal{} \beta$.

Then

$ \angle{BPC} \equal{} 90^\circ \minus{} \angle{PA_1A} \equal{} 90^\circ \minus{} \angle{CDA} \equal{} 90^\circ \minus{} (180^\circ \minus{} \angle{ABC}) \equal{} \alpha \plus{} \beta \minus{} 90^\circ$.
$ \angle{PCB} \equal{} \angle{A_1AB} \equal{} 90^\circ \minus{} \angle{ABD} \equal{} 90^\circ \minus{} \alpha$.

$ BC \equal{} BD\sin\angle{BDC} \equal{} BD\cos{\beta}$.

So in $ \triangle BPC$:

$ \dfrac{PB}{BC} \equal{} \dfrac{\sin \angle{PCB}}{\sin{\angle{BPC}}} \equal{} \dfrac{\cos \alpha}{\sin{(\alpha \plus{} \beta \minus{} 90^\circ})}$

$ \dfrac{PB}{BD} \equal{} \dfrac{\cos \alpha \cos \beta}{ \minus{} \cos{(\alpha \plus{} \beta)}} \Longleftrightarrow \dfrac{BD}{PB} \equal{} \dfrac{ \minus{} \cos{(\alpha \plus{} \beta)}}{\cos \alpha \cos \beta}$

$ \dfrac{BD \plus{} PB}{PB} \equal{} \dfrac{ \minus{} \cos{(\alpha \plus{} \beta)} \plus{} \cos \alpha \cos \beta}{\cos \alpha \cos \beta}$

$ \boxed{\dfrac{DP}{PB} \equal{} \dfrac{\sin \alpha \sin \beta}{\cos \alpha \cos \beta}}$

Let $ N$ be proiection of $ D$ onto $ AC$. Then $ ND \parallel B_1M$, so

$ \dfrac{QB_1}{QD} \equal{} \dfrac{B_1M}{DN}$

But

$ B_1M \equal{} BM \equal{} AB \sin \angle{MAB} \equal{} AB \sin {\angle{CDB}} \equal{} AB\cos \beta \equal{} BD \cos \alpha \cos \beta$

$ DN \equal{} CD \sin \angle{ACD} \equal{} CD \sin \angle{ABD} \equal{} BD \sin \alpha \sin \beta$, so

$ \boxed{\dfrac{QB_1}{CD} \equal{} \dfrac{\sin \alpha \sin \beta}{\cos \alpha \cos \beta}}$

And we are done.[/color]

Here is another solution:

Since $B_1$ is reflection of $B$ on $AC,$ then $QC$ bisects $\angle BQC$ internally. Tangents of $\odot(ABCD)$ at $A,A_1$ meet on $BD$ due to obvious symmetry, hence $ABA_1D$ is harmonic $\Longrightarrow$ $C(A_1,A,B,C)=-1$ $\Longrightarrow$ $(P,E,B,D)=-1,$ where $E \equiv AC \cap BD.$ Therefore $QP$ is the external bisector of $\angle BQD$ $\Longrightarrow$ $PQ \perp QC \equiv AC.$

Let be the point $ C'\equiv (O)\cap PA$ and then, because of $ AE \equal{} EA_{1},$ where $ E\equiv BD\cap AA_{1},$ we have $ \angle C'PB \equal{} \angle BPC$ ( because of symmetry ) $ \Longrightarrow$ $ \angle C'AB \equal{} \angle BAC$ $ ,(1)$

So, because of $ AD\perp AB,$ we conclude that the pencil $ A.PDRB,$ where $ R\equiv BD\cap AC,$ is in harmonic conjugation.

Because of now, the pencil $ Q.PDRB,$ is also in harmonic conjugation and $ \angle DQR \equal{} \angle RQB,$ from the isosceles triangle $ \bigtriangleup QBB_{1}$ with $ QB \equal{} QB_{1},$ we conclude that $ PQ\perp QR\equiv AC$ and the proof is completed.

Kostas Vittas.

Cool...
That's the nicest way to solve this nice problem (it was also the official solution).

[quote="Ahiles"][color=darkred]$ ABCD$ is inscribed in the circle with diameter $ [BD]$ . Point $ A_1$ is reflection of  $ A$ w.r.t. $ BD$ and $ B_1$ is

reflection of $ B$ w.r.t. $ AC$ . Denote $ P\in  CA_1 \cap BD$ and $ Q\in DB_1\cap AC$ . Prove that $ AC\perp PQ$ .[/color][/quote]

[color=darkblue][b][u]Proof[/u].[/b] Denote $ S\in AC\cap BD$ . Observe that $ \left\|\begin{array}{c}
 CB\perp CD\\\\
 \widehat {BCP}\equiv\widehat {BCS}\end{array}\right\|$ $ \Longrightarrow\ \{P,S\}$ are conjugate w.r.t. $ \{B,D\}$ .

Since $ \widehat {SQB}\equiv\widehat {SQD}$ and $ \{P,S\}$ are conjugate w.r.t. $ \{B,D\}$ obtain $ QS\perp QP$ , i.e. $ AC\perp PQ$ .[/color]

We can also prove that $ B$ is incenter of $ \triangle ACP$ and $ D$ is $ P$-excenter in this triangle. Then the result is quite easy to prove.

Stragely , it is the same problem for Mongolian TST 2008, day3 problem 2
. See http://www.mathlinks.ro/viewtopic.php?p=1135126#1135126


[quote="236factorial"]Two 2-digit numbers, one twice as big as the other, multiply up to be 1250. What is the sum of the two numbers?[/quote]

[hide]
? x 2? = 1250
2?^2 = 1250
?^2= 625

the smaller number is 25.
the larget number is 50.

the sum is 75
[/hide]

[quote="buzzer11"][hide]25 and 50[/hide][/quote]

They wanted the sum of th two  numbers, and show your work  ;)


In how many ways can we place 5 green, 5 blue, and 5 red balls in a row?

In how many ways can we do it if no 2 blue balls are adjacent?

jen7 wrote:[hide]For the first one:
15!/(5!5!5!)=756,756

For the second one:
(10!/(5!5!))*C(11,5)=116,424.[/hide]

I'm also getting the same answers..are they correct ??

[quote="blahblahblah"]I agree with both of the previous answers (well, I didn't multiply them out).[/quote]

Neither did I do the actual multiplications..my answers are in terms of C(n,r).


For any positive integer n, prove the identity

$ (sinx)(1+2cos2x+2cos4x+...+2cos2nx)=sin((2n+1)x)$

I should've put this in the intermediate section.

Using the formula $ 2\sin a\cos b=\sin (a+b)+\sin (a-b)$ we have:
$ \sin x=\sin x$
$ 2\sin x\cos 2x=\sin 3x-\sin x$
$ 2\sin x\cos 4x=\sin 5x-\sin 3x$
...............................................
$ \sin x\cos 2nx=\sin (2n+1)x-\sin (2n-1)x$
Summing these equalities yields
$ \sin x+2\sin x\cos 2x+\ldots 2\sin x\cos 2nx=\sin (2n+1)x$

[hide="mkay"]Let $ z=\text{cis}\, x$

Then $ \sin x=\frac{z-z^{-1}}{2i}$ and $ 2\cos kx=z^{k}+z^{-k}$. Make these substitutions and the equality becomes obvious by the geometric series formula.[/hide]

INDUCTION!!!... after already figuring out that n=1 works:

[hide]Assume n works.  Then, for n=1, we have:
$ \sin((2n+1)x)+2\sin(x)\cos((2n+2)x)=\sin((2n+3)x)$
$ \sin((2n+2)x-x)=\sin((2n+2)x+x)-2\sin(x)\cos((2n+2)x)$

$ \sin((2n+2)x)\cos(x)-\sin(x)\cos((2n+2)x)$
$ =\sin((2n+2)x)\cos(x)+\sin(x)\cos((2n+2)x)-2\sin(x)\cos((2n+2)x)$

Sorry for the last line being on two lines...not too familiar with LaTeX...

Everything cancels and we're done[/hide]


Here's a cool problem that I generalized (probably nothing new, though):

A plastic rod having a uniformly distributed charge of $ Q$ has been bent in a circular arc of radius $ r$ and angle $ \alpha$. $ P$ is the center of curvature of the rod.  In terms of $ Q$, $ r$, $ \alpha$, and fundamental constants, what is the electric field $ \vec{E}$ due to the rod at point $ P$?  Be sure to indicate the coordinate system.

[hide="Small Hint"]
Place $ P$ at the origin of a coordinate system, and let an $ x$-axis be the rod's axis of symmetry.  Resolve the electric field into its components.[/hide]

$ \vec{E} \equal{} \cfrac{2kQ\sin\cfrac{\alpha}{2}}{\alpha r^2}\vec{e_x}$

What do you mean by $ \vec{e_x}$?

My answer, when an $ x$-axis is placed through the rod's axis of symmetry and $ P$ is positioned at the origin, is:

$ \vec{E} \equal{} \frac {2kQ\sin{\frac {\alpha}{2}}}{\alpha r^2} \hat{i}$

where $ k$ is Coulomb's Law constant, which is $ \frac {1}{4\pi \epsilon_0}$, where $ \epsilon_0$ is the permitivity constant of free space.

The above expression can also be written as:

$ \vec{E} \equal{} \frac {Q\sin{\frac {\alpha}{2}}}{2\pi \epsilon_0 \alpha r^2} \hat{i}$

My $ \vec{e_x}$ is your $ \hat{i}$ because I use the coordinate system in your hint :)


function
integration
limit
calculus
logarithms
inequalities
geometric series

Consider the sequence of numbers $(a_{n})$ defined by:
1) $0<a_{1}<1$
2) $a_{i+1}=a_{i}+ \frac{a_{i}^2}{i^2}$ $\forall i \ge 1$
Prove that:
exist  $k>0$ satisfying $a_{i} \leq k$ $\forall i \ge 1$

Nice problem! But I just solve it if we take $0 < a_1 < \frac{1}{2}$. Here is my solution!

Let $f: R^*\to R$ such that $f(x)=\frac{1}{x^2}$. It´s easily prove that $\{a_n\}$ is an increase sequency, and we can considerate a partition on this form $\{a_1,a_2,...,a_n\}$. But $f$ is a decreasing function, so we have:

$\sum_{k=1}^{n-1}f(a_k)(a_{k+1}-a_k) > \int_{a_1}^{a_n}\frac{1}{t^2}dt$. But we have $a_{k+1}-a_k = \frac{{a_k}^2}{k^2}$ and so we can write $\sum_{k=1}^{n-1}\frac{1}{k^2} \geq -\frac{1}{a_n}+\frac{1}{a_1}$ -> $\frac{1}{a_n} \geq \frac{1}{a_1}-\sum_{k=1}^{n-1}\frac{1}{k^2}$. But we can prove that $\sum_{k=1}^{n-1}\frac{1}{k^2} \leq 2-\frac{1}{n-1}$ -> $\frac{1}{a_n} \geq \frac{1}{a_1}+\frac{1}{n-1}-2 \geq \frac{1}{a_1}-2$. And now it´s cleary that $a_n \leq \frac{a_1}{1-2a_1}$, and this is true if $0 < a_1 < \frac{1}{2}$.

I´ll try extend this prove.

Consider the sequence $b_{i+1} = (1 + \frac{1}{i^2}) b_{i}, b_1 = a_1$.  We have $b_i > a_i$, with near-equality if $a_1$ is very close to $1$.

Hence, if we can show $\lim_{i \to \infty} b_i = b_1 \prod_{k=1}^{\infty} (1 + \frac{1}{k^2})$ is finite, then the original sequence is also bounded.  In fact, $\prod_{k=1}^{\infty} (1 + \frac{1}{k^2}) = \frac{ \sinh \pi}{\pi}$, so for any integral $i$, we have $a_i < b_i < a_1 \frac{ \sinh \pi}{\pi} < \frac{ \sinh \pi}{\pi} \approx 3.676...$.  Q.E.D.

Alternately, take the natural log and we get $\sum_{k=1}^{\infty} \left( \ln(1 + \frac{1}{k^2}) \right)$.  Using the Taylor Series for $\ln x$, we have $\ln (x+1) = x - \frac{x^2}{2} + \frac{x^3}{3} -+...$, from which we derive $\ln (1 + \frac{1}{k^2}) = \frac{1}{k^2} - \frac{1}{2 k^4} + \frac{1}{3 k^6} -+...$

This means, for all $k > 1$, we have $\ln (1 + \frac{1}{k^2}) < \frac{1}{k^2}$, so our sum $\sum_{k=1}^{\infty} \left( \ln(1 + \frac{1}{k^2}) \right) < \sum_{k=1}^{\infty} \frac{1}{k^2} = \frac{\pi^2}{6}$.  Hence we have, for any integral $i$, the inequality $a_i < b_i < \prod_{k=1}^{\infty} (1 + \frac{1}{k^2}) < e^{ \frac{\pi^2}{6} } \approx 5.18067...$, a worse upper bound, but a valid one nonetheless.

[quote="t0rajir0u"]Consider the sequence $b_{i+1} = (1 + \frac{1}{i^2}) b_{i}, b_1 = a_1$.  We have $b_i > a_i$, with near-equality if $a_1$ is very close to $1$. [/quote]

That's false because if $a_1$ is very close to $1$  then $b_i < a_i$  for $i > 2$...

[quote="t0rajir0u"]
 $a_i < b_i < \prod_{k=1}^{\infty} (1 + \frac{1}{k^2}) < e^{ \frac{\pi^2}{6} } \approx 5.18067...$, a worse upper bound, but a valid one nonetheless.[/quote]
Still false ! Actually $\lim_{a_1\rightarrow 1, a_1 < 1} (a_i) = i$....
 Try again :cool:

[quote="Diogene"]
That's false because if $a_1$ is very close to $1$  then $b_i < a_i$  for $i > 2$...
[/quote]

Oops, sorry, didn't catch that  :(

I'll try again, then...

$a_2 = a_1 + a_1^2$
$a_3 = (a_1 + a_1^2) + ( \frac{a_1 + a_1^2}{2} )^2 = a_1 + a_1^2 + \frac{a_1^2 + 2a_1^3 + a_1^4}{4} = a_1 + \frac{5}{4} a_1^2 + \frac{1}{2} a_1^3 + \frac{1}{4} a_1^4$
...
$a_i = \sum_{k=1}^{2^{i-1}} c_k a_1^k$ for some set of coefficients $c_k$ such that $\sum_{k=1}^{2^{i-1}} c_k = i$.

Thus, if it can be shown that each individual value of $c_k$ stays finite as $i \to \infty$, then $a_i$ can be bounded by the (finite) sum of a geometric series with $a_1$ common ratio.

I'm too tired to think tonight, so somebody finish this off, it shouldn't be that hard, there's almost definitely a combinatoric formula for $c_k$ that I don't have the energy to find  :help:

I have a complete proof but I had some troubles with Latex. Later I´ll post here.

In fact I didn´t could obtain a better aproximately. I proved that if $0<a_1<0.74912138391$ then $a_n$ is limited. I don´t know if it´s worse post this proof. I´m still trying aprimorate.


function
trigonometry
topology
real analysis
real analysis unsolved

Can you find a continuous bijection $ f: [0,1]\rightarrow [0,1]\times[0,1]$ ?

You can't, because you can't find a continuous injective function  $ g : [0,1]\times[0,1] \rightarrow [0,1]$ .
Take for exemple the circle $ x(t)\equal{}\frac {1\plus{}\cos t}2,\ y(t)\equal{}\frac {1\plus{}\sin t}2$ . I define a continuous function $ h(t)\equal{} g(x(t),y(t))\minus{}g(x(\pi\plus{}t),y(\pi\plus{}t))$
Manifestly $ h(0)\equal{}\minus{} h(\pi)\Longrightarrow h(0)h(\pi) \leq 0$ so ther is $ t_0\in[0,\pi ]$ such that
$ h(t_0)\equal{}0 \Longrightarrow g(x(t_0),y(t_0))\equal{}g(x(\pi\plus{}t_0),y(\pi\plus{}t_0))$ , so $ g$ is not injective.
....
An other one  :
$ g$ is continuous and $ [0,1]\times[0,1]$ is connected, so the image-set $ g([0,1]\times[0,1])$ is connected : $ g([0,1]\times[0,1])\equal{}[a,b]$
For all  $ c\in ]a,b[$ there is a point $ P_c\in [0,1]\times[0,1]$ such that $ g(P_c)\equal{}c$
$ g$ is injective so $ g([0,1]\times[0,1]\minus{}\{P_c\})\equal{}[a,c[\  \cup\  ]c,b]$,
and that's impossible because $ [0,1]\times[0,1]\minus{}\{P_c\}$ is connected and $ [a,c[\ \cup \ ]c,b]$ is not connected !
....
 :cool:

It's not clear to me that that's relevant: the question asks for a function going the other direction, and there's no particular reason for the inverse of a continuous function to be continuous.

[quote="JBL"]and there's no particular reason for the inverse of a continuous function to be continuous.[/quote]
Except that a continuous bijection from a compact space to a Hausdorff space is always a homeomorphism ;).

Ok, that's true. Is there a more elementary proof? Continuity and bijections are more elementary notions than Hausdorff spaces and topology topics.


Find the largest odd natural number that is a factor of 5!.

5x4x3x2x1=5x3x2^3. Obviously if it has any powers of 2 it isn't odd, so 5x3=15.

Note: $ n!\equal{}n\cdot (n\minus{}1)\cdot ...\cdot 2\cdot 1$


Find a closed form for:

$1-\binom{n}{2}+2\binom{n}{3}-3\binom{n}{4}+...$

It was posted very many times.
[hide]$\sum_{k=0}^{n}(-1)^{k}\cdot k\cdot \binom{n}{k}=$
$\sum_{k=1}^{n}(-1)^{k}\cdot k\cdot \binom{n}{k}=$
$\sum_{k=1}^{n}(-1)^{k}\cdot n\cdot \binom{n-1}{k-1}=$
$-n\cdot \sum_{k=0}^{n}(-1)^{k-1}\cdot \binom{n-1}{k-1}=$
$-n\cdot (1-1)^{n}= 0$. [/hide]
For generalisation see [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=90529[/url].

i agree that the answer is 0, but the problem is asking to find $\sum (-1)^{k}k\binom{n}{k+1}$ not $\sum (-1^{k}k\binom{n}{k}$

[quote="m4467"]i agree that the answer is 0, but .... the problem is asking to find $\sum (-1)^{k}k\binom{n}{k+1}$ not $\sum (-1^{k}k\binom{n}{k}$[/quote]
But the difference is $\sum_{k=0}^{n-1}(-1)^{k}\binom{n}{k+1}=1$, which gives your reqired extra 1 in your former expression.


Expand out and it is same as

$\sum a^3(b^2+c^2)+\sum a^3bc-3\sum a^2b^2c\geq 0$

Since $\sum a^3(b^2+c^2)\geq 2\sum a^3bc$ so it is equivalent to

$3abc(\sum a^2-\sum ab)\geq 0$ $\Longleftrightarrow 3abc(\sum (a-b)^2)\geq 0$


Let $ P(x) \in \mathbb{Z}[x]$ be a polynomial of degree $ \text{deg} P \equal{} n > 1$. Determine the largest number of consecutive integers to be found in $ P(\mathbb{Z})$.

[i]B. Berceanu[/i]

Consider the polynomial $ P(x\plus{}1)\minus{}P(x)\minus{}1$ it has degree $ \le 1$ so we get that we can not have more than $ n$ consecutive elements in the range of $ P$. This is reached when for example $ P(x)\equal{}x\plus{}\prod_{i\equal{}1}^n (x\minus{}i)$

Hi [b] Albanian Eagle [/b],

What was your point to consider $ P(x\plus{}1) \minus{} P(x) \minus{} 1$ ? Why does the fact that its degree $ \leq 1$ deduce there are at most $ n$ consecutive elements?

I think that we are to find the maximum number of integers $ a_1, a_2, ... a_k$ such that $ P(a_{i\plus{}1}) \equal{} P(a_i) \plus{} 1$ without the restriction of $ a_i$ to be consecutive integers.

Oh, right sorry for not explaining that part.
Can you see why $ a_{i\plus{}1}\minus{}a_i|P(a_{i\plus{}1})\minus{}P(a_i)\equal{}1$  :wink:


I've decided to make a Problem Solving Marathon. Like all other marathons, here are the rules (slightly changed to suit my flavor :lol: )

(1) First, I will post 1 problem. Be the first to answer it!
(2) The person who answers it first may post 1 to 2 problems. The first person to answer BOTH questions correctly may post 1 to 2 problems. This process is repeated until we reach 100 problems.
(3) If whoever answers it first doesn't want to post a problem, please say so. I will post a problem for them.
(4) If you've posted a problem, and say 2 people answer it (the max is 4 people can answer it), YOU have to tell me who got the right answers first. So, wait until 4 people answer it, and then tell me the correct solution.
(5) Please hide your solutions :P

Ready? Let's get started!

Problem 1:

Rationalize $\frac{1}{\sqrt{3}+\sqrt{7}+2}$

What do you mean? :?:  This is my first marathon.

He means we already have a marathon in Getting Started. You could start a competition if you want. :wink:

What is a competition? (Sorry, I am kind of new here)

a competition is where you get a bunch of people to sign up to do some problems that you made up and you can organize it however you want to determine a winner

[quote="vishalarul"]So, do I have to get a moderator to delete this topic for me? I'll post a new topic with a competition.[/quote]

ill just lock it


inequalities
absolute value
inequalities unsolved

Let x_1,x_2,...x_n and y_1,y_2,...,y_n be real numbers satisfying x_1^2+x_2^2+...+x_n^2=y_1^2+y_2^2+...+y_n^2=1.Prove that (x_1.y_2-x_2.y_1)^2 <= 2|1-(x_1.y_1+x_2.y_2+...+x_n.y_n)|   and determine when equality holds. ( |...| is absolute value :P )

First,This one should be posted in Inequality section.
Second ,It has been posted many times before.
Third,It can be solved by Legrange Identity.( the method to prove Chauchy inequality)


[quote="furia"]man these questions are hard..

given forumlae are these:
-should be familiar

[code]Terms forumlae

tn= ar^n-1

sum sequence formulae

sn = a ( r^n  -  1 )
     ----------------
           r   -  1[/code]

a = first term         r = ratio        n = term number

pleases show work, thanks

1. Given the expression of Sn[size=100]n[/size]= n^2 +2n for the sum of the first n terms of a series, determine an expresssion, in simplified form, for the n^th term
t[size=100]n[/size] in terms of n

2. x-9, x-1, 3x-3 are the second, third, and fourth terms of a geometric sequence of non-zero terms. determine the value of the first term

3. The sum of the first three terms of a geometric sequence is -14, while the seecond term is 12. Find all such sequences

4. The sum of the first two terms of a geometric sequence is 16. The sum of the reciprocals of the first two terms is 1/3. Find the first two terms of the sequence[/quote]

[hide="4"]$a+ar=16$
$\frac{1}{a}+\frac{1}{ar}=\frac{1}{3}$
solving, $r=\frac{1}{3}$, $r=3$
$a=12$, $a=4$
first two terms are 12 and 4, no real order [/hide]
[hide="2"]
$a(x-9)=x-1$
$a(x-1)=3x-3$
dividing, $\frac{x-9}{x-1}=\frac{1}{3}$
$x=13$
second, third, and fourth terms are $4$, $12$, and $36$
the ratio is $\frac{12}{4}=3$, so the first term is $\frac{4}{3}$ [/hide]

[hide="3"]The second term of the sequence, $ar$, is 12.

The sum of the first three is -14, so the sum of the first and the third, $a(1+r^2)$, is -26.

Multiply both sides of the second equation by $r$ and divide by 2. $6(1+r^2)=-13r$.

Then $6r^2+13r+6=(3r+2)(2r+3)=0\Rightarrow r=-\frac{3}{2},\ -\frac{2}{3}$.

Since $a=\frac{12}{r}$ from first equation, the sequence with $r=-\frac{3}{2}$ has $a=-8$ and the sequence with $r=-\frac{2}{3}$ has $a=-18$.[/hide]

For #4, they don't really need to be terms of a geometric sequence.  Two terms given already are, so they can just be $a$ and $b$...


algebra
difference of squares
special factorizations

1. For how many minutes in a 12-hour period will a 4-digit, 12-hour, digital clock have digits showing that have a sum greater than or equal to 17?

[hide]84[/hide]

2. Jose invested 50,000 dollars for 2 years at an annual interest rate of 4% compounded yearly. Patricia invested 50,000 dollars for the same period of time, at the same interest rate, but the interest was compounded quarterly.  To the nearest dollar, how much more money did Patricia's investment earn than that of Jose?

[hide]63[/hide]

[quote="abacadaea"]well whats your way? i dont know if mine is faster tahn yours or not if i dont know your solution.

#2. use the compound intrest formula and do difference of squares[/quote]

abacadaea- What's the compound interest formula, and how does the difference of squares become important?

For problem 1, I listed out [u]all[/u] of the possible combinations.

compound intrest formula:
$ (p*r)^t$ in this case t=2 and p=50000

$ (50000*1.04)^2\minus{}(50000*1.04)^8$
$ (52000)^2\minus{}(52000)^2$
difference of squares just helps if you dont have a calculator.

[quote="abacadaea"]compound intrest formula:
$ (p*r)^t$ in this case t=2 and p=50000

$ (50000*1.04)^2 \minus{} (50000*1.04)^8$
$ (52000)^2 \minus{} (52000)^2$
difference of squares just helps if you dont have a calculator.[/quote]

abacadaea- I still don't get it.  Can you give me more details.

[quote="abacadaea"]alright:

$ (p*r)^t$ is the compound intrest formula.
p=pricipal
r=rate
t=time
for both cases, the pricipal and the rate is 50000 and 1.04 repspectivley

the time is still 2 years, but the period of time for which the pricipal is compounded is different.

jose gets intrest once every year. so he gets intrest twice. for joses equation, t=2. patricia gets intrest once every quarter of a year or 3 months. she gets intrest 8 times in 2 years. so in patricias equation, t=8. then you can subtract.[/quote]

abacadaea- When I subtract them and take the positive difference, I get the wrong answer.  Have you tried it and got the right answer?  [hide="Correct Answer"]The correct answer is 63[/hide]

[quote="CircleSquared"][hide="1"]
The lowest digit-sum that works for the hour will be 12 or 3.  That needs to go with 59.
Then comes 4.  That can go with 49, 58, or 59.
Then 5.  That can go with 39, 48, 57, 49, 58, or 59.
Then 6.  That goes with 38, 29, 39, 48, 57, 49, 58, or 59.
Well you get the idea.
[/hide][/quote]

So for #1, there's no algebraric expression or easier way to solve it except plugging?

i think so. if it were something like a die or something, theres a formula.


This problem is truely beautiful and really strong :flex:
No ideas from me  :(

[quote]This problem is truely beautiful and really strong[/quote]
Also wrong :). The $r_i$'s should be the inradii of triangles $OAB,OBC,OCD,ODA$, where $O$ is the intersection of the quad's diagonals.

Enescu, I was thinking about that. But i didn`t dare to spit it out.
Can you show the solution with this condition, becouse i don`t want to loose another hour to think about it.Thank you

[quote="letiendat"]Let be given a quadrilateral ABCD.$r_{1}, r_{2}, r_{3},r_{4}$ are the incircleradius of triangles BCD,CDA,DAB,ABC .Prove that : there is a circle inscribed in the quadrilateral ABCD if and only if:
$\frac{1}{r_{1}}+ \frac{1}{r_{3}}= \frac{1}{r_{2}}+ \frac{1}{r_{4}}$[/quote]
For this inradis we have that $r_1+r_3=r_2+r_4$ :) ,but not$\frac{1}{r_{1}}+ \frac{1}{r_{3}}= \frac{1}{r_{2}}+ \frac{1}{r_{4}}$ :(  :D

[quote="Tiks"]For this inradis we have that $r_1+r_3=r_2+r_4$ :) ,but not$\frac{1}{r_{1}}+ \frac{1}{r_{3}}= \frac{1}{r_{2}}+ \frac{1}{r_{4}}$ :(  :D[/quote]
Not exactly. This equality holds in a [b]cyclic[/b] quadrilateral.

NO, the correct problem is: the same but with ABCD cyclic and if and only if $r_1+r_3=r_2+r_4$

Actually, there are two problems here. Read carefully my posts above.


When you say inscribed angle, the angle's vertex is behind the central angle that it inscribes correct?

[img]http://mathworld.wolfram.com/images/eps-gif/InscribedAngle_1000.gif[/img]

In the above picture, angle b is inscribed in the circle, so angle b is an inscribed angle. Is that what you were looking for?

Thanks.

I've seen in AoPS vol 1 that it is true in the case of 2 secants, but I am confused about for 2 tangents and a tangent and a secant. Cases in which the bottom arc can be greater than 180 degrees. In such a case, one should treat the situation as the 2 secant case right? Basically the measure for the bottom arc (distinguishing between major or minor arc) is the arc measure that opens away from the vertex of the 2 segments intersecting the circle and the arc measure for the top arc is the one opening toward the vertex?

[quote="nebula42"][quote="breez"]I am wondering which arc measure to use when applying the identity of the angle between 2 segments intersecting a circle, $\theta = \frac{\beta-\alpha}{2}$
Because there is a major and minor, and I am not sure which to use.

Is $\beta$ always arc measure opening away from $\theta$? And I think $\alpha$ is always the arc measure opening towards $\theta$.[/quote]
See the diagram I've attached.
In the diagram, $\angle APC = \frac{\angle BOD-\angle AOC}{2}=\frac{\widehat{BD}-\widehat{AC}}{2}$.
(Is the symbol for an arc "$\widehat{AB}$"? Or is there something better?)[/quote]

This is in AoPS Vol 1, Chapter 10...\ :wink:  :D  :)  :)  :!:


function
algebra
domain
induction
limit
functional equation

Can anyone tell me or recommend a good website for learning to solve  functional equations?
I have no previous knowledge about this.

Thanks

[quote="Extra"]say if

$x^2f\left( x\right) +f\left(1-x\right)=2x-x^4$

how would one go about solving it?[/quote]

Plug in $x=t$ and $x=1-t$ into the equation. Since $1-x=t$, you will have a system of two equations in two functions, $f(t)$ and $f(1-t)$, in terms of $t$. Solve for $f(t)$ in terms of $t$.

i think i sort of get it now, but how would you prove that this is the only solution?

It already does prove that the function is the only solution since all values in the domain must satisfy the identity. Else, there is a contradiction.

In scorpius119's example, suppose $f(y)\neq y$. But the identity must still hold true. Namely, \[ 2f(y)+f(1-y)=y+1, \] \[ 2f(1-y)+f(y)=(1-y)+1=2-y, \]from which we conclude $f(y)=y$ and $f(1-y)=1-y$. Contradiction. (Note that the assumption $f(y)\neq y$ doesn't actually do anything.)

In some other functional equations uniqueness may be proved through conditions such as continuity, boundedness, etc. [hide="Click here to see one of the more common functional equations"]\[ f(x+y)=f(x)+f(y). \] Plugging in $y=x$, we see that \[ f(2x)=2f(x). \] Plugging in $y=2x$, we see that \[ f(3x)=f(x)+f(2x)=f(x)+2f(x)=3f(x). \]Through induction, we see that \[ f(nx)=nf(x) \] for all natural numbers $n$. For a rational $x=\frac{p}{q}$ with natural numbers $p,q$, we have $qx=1p$. (Note that if $x$ is negative, then we can just set a dummy variable $z=-x$ and proceed.) Then \[ f(qx)=f(1p), \] \[ qf(x)=pf(1), \] \[ f(x)=\frac{p}{q}\left(f(1)\right), \] \[ f(x)=xf(1). \] Set $f(1)$ to be the constant $c$ so that $f(x)=cx$, for all rational $x$.

[hide="Click here to see more advanced proof of uniqueness under continuity"]Suppose $f(x)$ is continuous. If $t$ is irrational, then by the density of rationals, there exists a sequence of rational numbers $t_n$ converging to $t$. Since $f(t_n)=ct_n$ for all of these $t_n$, then \[ \lim_{x\rightarrow t}f(x)=ct, \] from the sequential characterization of limits. From the definition of continuity, $f(t)=ct$. If that last part didn't make sense to you, just imagine a bunch of rational points on the $x$-axis approaching $x=t$ from both sides getting arbitrarily close. As they get closer to $x=t$, their $f(x)$ values get closer to $ct$. Because $f(x)$ is continuous, then the value $f(t)$ is "squeezed" to be $ct$. Okay, if that did not help or even worsened your understanding, then just ignore this last paragraph altogether. :)[/hide][/hide]

[quote="Extra"]i'm sorta lost again...

if we set x=t and x=1-t and use them to solve the functional equation, doesn't that imply that t=1-t?[/quote]

No, you're just evaluating the function at two different values of $x$. If $f(x)=x^2+1$ and you let $x=0$ and then $x=1$ to get $f(0)=1$ and $f(1)=2$, then this does not mean that 0 has to equal 1 (for obvious reasons).

The only difference here is that instead of a specific number, a variable is used.

[quote="Virgil Nicula"][hide="Exercises for you from the my book with the functional equations."]
$\blacksquare\ 1^{\circ}.\ \ f(x)+f\left(\frac{1}{1-x}\right)=x,\ x\not\in \{0,1\}.$
$\blacksquare\ 2^{\circ}.\ \ 2f(x)-f\left(\frac{1-x}{1+x}\right)=x-1,\ x\not\in \{-1,0,1\}.$
$\blacksquare\ 3^{\circ}.\ \ f(x)+f\left(\ -\frac 1x\right)+f\left(\frac{x-1}{x+1}\right)=x,\ x\not\in \{-1,0,1\}.$
$\blacksquare\ 4^{\circ}.\ \ f(x)+f\left(\frac{x-1}{x}\right)=\frac 1x-x+1,\ x\not\in \{0,1\}.$
$\blacksquare\ 5^{\circ}.\ \ f,g\ : \ R$$\rightarrow R,\ g(0)=0,\ 2f(x)+f(1-y)+g(x)-g(y)=3(x+1)^2-6y,\ \left(\forall\right)x,y\in R.$
$\blacksquare\ 6^{\circ}.\ \ f\ : \ Q\rightarrow Q,\ f(x+y)+f(x-y)=2\left[ f(x)+f(y)+1\right],\ \left(\forall\right)x,y\in Q.$
$\blacksquare\ 7^{\circ}.\ \ f\ : \ N\rightarrow Q,\ f(x+1)=f(x)+\frac 12,\ \left(\forall\right) x\in N.$
$\blacksquare\ 8^{\circ}.\ \ f\ : \ R\rightarrow R,\ f\nearrow ,\ f\left[ x+f(y)\right]=f(x+y)+1,\ \left(\forall\right) x,y\in R.$
$\boxed { \mathrm { \ Wish\ you\ and\ another\ exercises\ ?\ }}$ [/hide][/quote]Nice. It's getting kinda late here today, so I'll do them tomorrow.  :)

This seems really really wrong, but for number 1:
[hide]
let x=t

$f(t) + f(\frac{1}{1-t}) =t$

let $x= \frac{1}{1-t}$

$f(\frac 1{(1-t)}) + f(t) = \frac 1{(1-t)}$

so $t-f(t) +f(x) = \frac 1{(1-t)}$
giving t=1/(1-t)

subbing into question gives:

$2f(x)=x$
hence, $f(x)=0.5x$
[/hide]

With functional equations, you should always plug your answer back into the identity to see whether or not it is indeed valid. Trying your answer $f(x)=0.5x$ yields \[ f(x)+f\left(\frac{1}{1-x}\right)=x\rightarrow 0.5x+\frac{1}{2-2x}=x,  \] which is not always true. Thus, your answer is not correct.

In your solution, when you let $x=\frac{1}{1-t}$, you should get \[ f\left(\frac{1}{1-t}\right)+f\left(\frac{1}{1-\frac{1}{1-x}}\right)=\frac{1}{1-t}  \] instead of \[ f\left(\frac{1}{1-t}\right)+f(t)=\frac{1}{1-t}.  \] Keep at it though! :)

[quote="scorpius119"]simpler example in case you couldn't quite understand: $2f(x)+f(1-x)=x+1$

[hide="solution"]the equations from $x=t$ and $x=1-t$ are $2f(t)+f(1-t)=t+1$ and $2f(1-t)+f(t)=2-t$, respectively.

Add the equations to get $3(f(t)+f(1-t))=3\Rightarrow f(t)+f(1-t)=1$.

Substitute into first equation to get $f(t)=t$.[/hide][/quote]

how does knowing $f(t)=t$ help you solve for x in the original equation?

[quote="Tecnogram888"]how does knowing $f(t)=t$ help you solve for x in the original equation?[/quote]

Because the letter by which you denote the argument of the function is totally irrelevant, as long as you use the same letter on the RHS. Hence, these are all the same: $f(x)=x, f(t)=t, f(a)=a, f(math)=math, f(technogram)=technogram$. And in the functional equations we don't solve for $x$ but for $f(x)$ - a huge difference.

Think of it this way: You know how to solve the equation $x-2=5$. Now, if we write $y-2=5$, would you think it's any different from the first one coz it's not in $x$?

[hide="8"]Setting $x=0$, $f(f(y))=f(y)+1$ for real $y$.

Then $f(x+f(y))=f(x+y)+1=f(f(x+y))$; since $f$ is one-to-one, $x+f(y)=f(x+y)$.

Set $y=0$ to get $f(x)=x+f(0)$. Then $f(f(x))=x+2f(0)=x+f(0)+1\Rightarrow f(0)=1\Rightarrow f(x)=x+1$ as only possible solution.

If $f(x)=x+1$, $f(x+f(y))=f(x+y+1)=x+y+2=f(x+y)+1$, which checks. So the only function $f$ is $\boxed{x+1}$.[/hide]

[quote="Extra"][hide]
i'm stuck on number two.

$2f(x)- f(\frac{1-x}{1+x}) = x-1$

let x=1-x/1+x

then $2f(\frac{1-x}{1+x}) - f(x) = \frac{1-x}{1+x}-1=\frac{-2x}{1+x}$

so $2[2f(x)-x+1] -f(x)=\frac{-2x}{1+x}$
$=3f(x) = \frac{-2x}{1+x}+2x-2= \frac{2x^2-2x-2}{1+x}$
thus, $f(x)=\frac{2(x^2-x-1)}{3(1+x)}$

but substituting into question doesn't give the correct answer.
what have i done wrong?[/hide][/quote]

Nothing wrong - you must have made an error in the final checkup.

[quote="Extra"][hide]
i'm stuck on number two.

$2f(x)- f(\frac{1-x}{1+x}) = x-1$

let x=1-x/1+x

then $2f(\frac{1-x}{1+x}) - f(x) = \frac{1-x}{1+x}-1=\frac{-2x}{1+x}$

so $2[2f(x)-x+1] -f(x)=\frac{-2x}{1+x}$
$=3f(x) = \frac{-2x}{1+x}+2x-2= \frac{2x^2-2x-2}{1+x}$
thus, $f(x)=\frac{2(x^2-x-1)}{3(1+x)}$

but substituting into question doesn't give the correct answer.
what have i done wrong?[/hide]

edit: my bad.[/quote]

i think i did do something wrong. because if i substitute x=1-x/1+x into eq., shouldn't i get $2f(\frac{1-x}{1+x}) - f(\frac{1-\frac{1-x}{1+x}}{1+\frac{1-x}{1+x}}) = \frac{1-x}{1+x}-1=\frac{-2x}{1+x}$

But that's exactly what you get when you substitute (I've checked.) Check your arithmetic.

whoa geez, thanks. nva knew that the second term reduces, ty.

I take you're ironic :) Sorry, didn't mean to offend you, just to help. Anyway, pay attention to the arithmetic when you substitute - your solution is OK.

no really, i mean it, i always miss these things

[quote="Farenhajt"]Well, it seems to be quite elementary:

[hide]${{1-{{1-x}\over{1+x}}}\over{1+{{1-x}\over{1+x}}}}={{{2x}\over{1+x}}\over{2\over{1+x}}}=x$[/hide][/quote]

...  :mad: Now you're starting to offend me :P .

So, anyways, i should thank all of you guys for helping me, but i'm not finished yet :P

What would be different techniques in solving functional equations??

If the problem asks to prove no such function exists, plug in specific values (like $4x = 0$ or $y = 1$ or $x = y$) to try to arrive at a contradiction.

[quote="Virgil Nicula"]$1^{\circ}.\ \ f(x)+f\left(\frac{1}{1-x}\right)=x,\ x\not\in \{0,1\}.$[/quote]

We have that $f(x)+f\left(\frac{1}{1-x}\right)=x$ [b][1][/b]

Let $x=\frac1{1-x}\Rightarrow f\left(\frac1{1-x}\right)+f\left(\frac{x-1}{x}\right)=\frac1{1-x}$ [b][2][/b]

Let $x=\frac{x-1}{x}\Rightarrow f\left(\frac{x-1}{x}\right)+f(x)=\frac{x-1}{x}$ [b][3][/b]

So, plugging [b][1][/b] and [b][3][/b] in [b][2][/b], gives:

$x-f(x)+\frac{x-1}{x}-f(x)=\frac1{1-x}$

$2f(x)=x+\frac{x-1}{x}-\frac1{1-x}=\frac{x^3-x+1}{x(x-1)}\Rightarrow \boxed{f(x)=\frac{x^3-x+1}{2x(x-1)}}$


$a_1,a_2,\cdots a_n,b_1,b_2\cdots ,b_n\in [1,2]$ and $\sum_{i=1}^n a_i^2=\sum_{i=1}^n b_i^2$. Prove that
$\sum_{i=1}^n \frac{a_i^3}{b_i}\leq\frac{17}{10}\sum_{i=1}^n a_i^2$.
Determine the necessary and sufficient condition for which equality holds.

Note that we must have $(2a_i-b_i)(2b_i-a_i)\ge0\Rightarrow a_i^2+b_i^2\le\frac{5}{2}a_ib_i$. Thus,\[ \sum(\frac{a_i^3}{b_i}+a_ib_i)=\sum\frac{a_i(a_i^2+b_i^2)}{b_i}\le\frac{5}{2}\sum a_i^2=\frac{17}{10}\sum a_i^2+\frac{2}{5}\sum(a_i^2+b_i^2)\le\frac{17}{10}\sum a_i^2+\sum a_ib_i \]

a similar one was given in a tst from singapore wasn't it?

[quote="yptsoi"]Your solution is so nice, me!
Here is my solution...

Let $A=\displaystyle\sum \frac{a_{i}^3}{b_{i}}$, $B=\displaystyle\sum \frac{b_{i}^3}{a_{i}}$ and $C=\frac{17}{10}(\displaystyle\sum a_{i}^2)=\frac{17}{10}(\displaystyle\sum b_{i}^2)$.

Obviously if $nA+B \le (n+1)C$ for all natural n, then the inequality was proved. In fact it is because: if $A \le C$, solved; If $A = C+s (s>0)$, there must exists a sufficiently large N such that $NA+B = NC + Ns + B > (N+1)C$, contradiction!

To prove $nA+B \le (n+1)C$, it is sufficient to prove that
\[ n\frac{a_{i}^3}{b_{i}}+\frac{b_{i}^3}{a_{i}} \le \frac{17}{10}(\frac{21n-4}{17}a_{i}^2+\frac{21-4n}{17}b_{i}^2)  \]
, which is equivalent to
\[ (x-2)(x-\frac{1}{2})[10nx^2+(4n+4)x+10] \le 0 (x=\frac{a_{i}}{b_{i}})  \][/quote]

My solution is somehow similar to that of me, and I think your solution is really weird... what a strange approach..

But i think my approach is much more natural than yours... As yours can only be thought of by GENIUS!

mecrazywong, I love...... your solution! very nice! thanks for share  :D

nature solution. It depends on people. someone could think that his solution is the best. just solve it, we have done!

Actually my original solution is even more weird-I proved that one of $a_i,b_i$ must attain the boundary!

The following is the shortest, nicest and strangest solution. (Of course it doesn't come from me!)

We have the two inequalities
$\sum_{i=1}^n (2a_i-b_i)(2b_i-a_i)\geq 0$ and $\sum_{i=1}^n (2a_i-b_i)(\frac{a_i^2}{b_i}-2a_i)\leq 0$
$\Rightarrow \sum_{i=1}^n a_ib_i\geq \frac45\sum_{i=1}^n a_i^2$ and $\Rightarrow \sum_{i=1}^n \frac{a_i^3}{b_i}\leq\frac52\sum_{i=1}^n a_i^2-\sum_{i=1}^n a_ib_i$
The conclusion follows!


At Price's Market, four oranges cost one dollar. How much, in dollars, will 10 oranges cost at Price's Market?

I typed in $2.5$, however, it did not accept it.  Does anyone know why?

Probably because 2.5 is not an acceptable monetary unit.
If there are cents involved, you need to express your answer with two decimal places.


Evaluate $ 29 \times 31 \plus{} 19 \times 21$.

Use differences of squares.

$ (x\minus{}1)(x\plus{}1) \equal{} x^2 \minus{} 1$

So we have $ (30\minus{}1)(30\plus{}1) \plus{} (20\minus{}1)(20\plus{}1)$, or $ 30^2 \plus{} 20^2 \minus{} 2$.  This is $ 900 \plus{} 400 \minus{} 2$, or $ \boxed{1298}$.

We use difference of squares
$ 29\cdot 31 \equal{} 30^2 \minus{} 1 \equal{} 899$
$ 19\cdot 21 \equal{} 20^2 \minus{} 1 \equal{} 399$
899+399=1298

Edit: Beaten