Putnam calculus derivative trigonometry function geometry MIT

So, how was it? Easier or harder than last year? Easier even than 1998? Harder even than 1999? Somewhere between those extremes? And was that a lot of calculus (A6, B2) and differential equations (A2, B5)?

This topic is for general discussion, score prediction, etc. Specific comments on specific problems should go on the 12 other topics that are being posted along with this.

I found the exam was easier than last's year(Some people might have different opinion of course).
I did feel that  this year's putnam exam had a lot of calculus.
I think the scores this year will be slightly higher than last year's.Honorable mention will probably be around 60 though again :)

While proctoring, I thought of (but didn't write up) solutions for A1, A2, A3, A6, B1, B2, B4, and B5 - but I wasn't that happy with what I had for B1 and B5.

General sentiment among some of my less-experienced students: "Everyone should get A1." Not that everyone did, of course.

A3 felt cheap. Too cheap. (But only if you saw it.)

I was definately more confident this year than last.  I still didn't do well however... :blush: .

A1: I CANT BELIEVE I MISS THIS ONE...smh. I over-thought the problem; I basically turned it onto a algebra problem...*sigh* when all you had to do was look at 6 squares centered at a point P!

A2: I turned in a partial (very partial) solution for this one. Did anyone else approach this using logarithmic derivatives?

A3: Couldn't think of anyway to attack this problem during the test.  However, on the train ride home I did think about using the Taylor series for cosine... :oops:

A4: Read problem, didn't feel like investigating it...

A5: Read problem, couldn't remember (or didn't want to) much abstract algebra to proceed on this problem.

A6: Didn't even looked at this problem. (actually, I still haven't really "look" at it)

Overall, I spent most of my time on part A trying to solve A2...

[quote="sundiata82"]I was definately more confident this year than last.  I still didn't do well however... :blush: .

A2: I turned in a partial (very partial) solution for this one. Did anyone else approach this using logarithmic derivatives?

[/quote]

Yeah, I also got stuck using the log derivatives.

I got $ g\equal{}\frac {f^{11}}{h^7}$, and then got $ \frac {f^9}{h^6} \equal{} 3x \plus{} 1$.

But when I substitue the results into the first equation, I got a diff equation that I had no idea how to solve...

A3 was even easier than A1... It took no time

Got A1,A2,A4,B1,B2 .. the wording in A3 did a good job in being misleading (I realized the limit was zero, but didn't think the sequence would be identically zero), and tried too hard to solve an ugly recursion I came up with on B3. I also accidentally forgot to prove the upper bound on B2, I wonder how the graders will treat this?

Many people I talked to got 3 questions on part A, usually {1,2,4} or {1,3,5} or the like.

Part B:

B1: Yay!, I solved this one...

B2: I took me a while to understand what they were asking for on this...

B3: See B2

B4: This problem made me think of multivariable calculus; not something I like...so I focused my energy else where...

B5: *SIGH*...Spent the majority of the second half on this one.  I'm curious to see the solution.  The proctor basically had to pry the A5 booklet from my sweaty palms at the end of the exam... :rotfl: .  I think I'm gonna have nightmares about this one for a while...lol

B6: Read it once; felt my efforts could be invested else where (obviously B5)....

Well, this is my second and last time taking the Putnam.  I won't be eligible next year since I'm graduating.  Even though I wont be able to officially take it anymore, I will be keeping a look out for future Putnam Competitions.....

Nobody else talks about the difficulty?

I think it as easy as that of last year. But the easiest problems are easier (B1, A3)

I had A1-A4 first part and thought I had B1-B5 in the second part, but I have this stupid mistake in the end of B3 that might or might not cost $ \geq 8$ points. I am hoping for an 80+..

For the difficulty I thought A1<B1<A3<A2<B2<A4<B3<B4<B5<A5<A6<B6

Uh, that was really the 2009 Putnam? The question writers must be running out of good ideas...

In terms of effect on the scoring level, I think that A1 was easier than A3. A3 was quick if you saw it, but the combination of matrices and trig functions may have intimidated people out of considering it, not to mention that the question asked (about a limit) was an act of misdirection. And not everyone knows about rank. But A1: yes, any solution will have to be multi-step, and it probably will matter how well you wrote it up. Still, it used no technical tools, had things that many inexperienced problem solvers would be willing to try - and if you tried very hard, you probably worked through it to the end.

Edit: my estimate of the difficulty level by problem looks a lot like Albanian Eagle's estimate.

In terms of difficulty, I would rate A6 along with A1 and B1 because it required barely any insight at all. I came up with a counterexample in less than three minutes (guess and check). But I assume few test takers attempted it because it was the last (and presumably hardest) problem on the A part.

This was my second time taking the Putnam and I am by far not a math major haha!  Well, I worked on A1, A6, and B3 and felt much more confident in what I submitted from last year.  I cannot really say that this year was easier than last year because for me the questions are always difficult, sorry i'm a chemistry major :roll: .  Did anyone else notice how last year there were a few questions involving circles and this year there were a few involving squares?  That was a nice change.

First time taking the putnam.

I think I only did B1 completely right. I probably could've done the same for A1 if the Putnam question maker specified that the side length of the square should be positive, or bigger than 0.
I turned A1 into a very trivial problem by assuming that the length can be 0.

I also did B2, but what I got is (0, 1] not (1/3, 1], but I'd be content if I can get 1 point on this.  :blush:

While Putnam grading is always a bit of a mystery, I'd be quite confident that an answer of $ (0,1]$ on B2 would get a score of zero. Anything which goes below $ \frac13$ would seem to miss the point of the problem.

[quote="WelcomeGhosts"]sorry i'm a chemistry major :roll: .[/quote]
One of the people I put on my college's team is a Chem/Chem E double major.

Pretty much easier question for question than 2008. Way too much calculus...where are you combinatorics, number theory, and (combinatorial) geometry?

[quote="Arachnotron"]where are you combinatorics, number theory...[/quote]A4, A5, B3, and B6 aren't enough for you?

I got through the whole thing very fast, but I'm pretty sure I'm better at this now than I was when I was an official competitor.

I was disappointed.  The problems didn't seem to require ingenuity so much as the ability to push through an argument quickly.  And I was stressed out about other things during Part B and did worse on that section even though the MIT people I've talked to said they had an easier time during Part B.

Just a note; there are no official solutions. Those solutions are just ones posted by (I believe) Kiran Kedlaya and are unofficial.

I'll agree with t0r. While in my case this is partially due to some poor arguments (you do not want to know what I put down for A3), I think I spent over 75% of the time writing stuff up and 25% actually figuring out problems. This is a far cry from USAMO days when it was more like 10-90. I'm not a fan.

What do you think the cutoffs will be for being a putnam fellow?
My prediction is $ \geq 104$.

That seems about right, maybe a bit lower, like 102  :)

WEEE!!!! THE BEST I EVER DID ON A PUTNAM!!! I LOVE ANALYSIS!!!!

A1: Actually I think this is the last one I solved... Pretty cute though.

A2: this is not that bad, but I never learned how to solve differential equations rigorously... (I always solve it the "physicist" way, where you say something like oh dy/dx is a fraction so you multiply both sides by dx...).  So my proof sort of skip the details of how I solve those...Hope that's good enough...

A3: Immediately think of induction, immediately brute force the case of n=3 and find that it's all 0... So cheap.

A4: After playing around a little I found out that only fractions of the form $ a^2/b\plus{}c$ can be obtained, but don't know how to prove that.  Ended up just constructing the set of all rational number except those of the form $ 2/3 \plus{} c$ and prove that it satisfies the constraints.

A5, A6: I was so happy that I solved 4 problems on A I don't even bother :P

B1: See a lame induction on the maximum prime factor.  Wrote it up cuz I don't want to spend too much time on this.

B2: FAVORITE PROBLEM ON THE TEST!!! Riemann Sum in disguise is soooo cool... Although I sort of forgot how to prove what a converging sequence converges too (I know it converges to 1/3), so I kinda BSed with the "by a well known theorem" technique...

B3: Hmm I thought I solved this one but my solution is different than the one posted... Don't know then...

Well so that's 6 problems at the least.  Not bad!!!

[quote="pkerichang"]
A4: After playing around a little I found out that only fractions of the form $ a^2/b \plus{} c$ can be obtained, but don't know how to prove that.  Ended up just constructing the set of all rational number except those of the form $ 2/3 \plus{} c$ and prove that it satisfies the constraints.[/quote]

plug in x = 1/4 in to rule (c) yields -16/3, adding 6 to that yields 2/3.

Btw,I think putnam authors need to put some weight on inequalities too,like for example A2/2003(even if that wasn't that hard).
There is so much creativity under the subject of inequalities.Even though most of the methods like Cauchy-Schwarz,Holder etc are well known they are hard to be applied but putnam authors "neglect"(in my opinion) those kind of problems.

Well,what I mean is that I would be really happy if problems appearing on putnam looked like the following proposed by Fedor Petrov
long time ago,but still its creativity surprises me:

http://www.mathlinks.ro/viewtopic.php?search_id=1184560432&t=3296

I don't know how easy or hard that problem is but i think putnam needs those kind of problems (A1-A2,B1-B2 maybe?).
Anyways,some people might disagree like the post above :)

Well, it was the first time for me, and with the advantage of olympiad training during high school I could get correctly A1, A2, B1, B2, B3 and write something (significant I hope) in A4, A5 and B5. Well A3 didn't seem trivial to me, probably because I hadn't seen nothinng like that before, instead B3 sound quite familiar during the exam. Can somebody give a good guess about score-ranks?

[quote="jaws1216"][quote="kunny"]Solution is Up!

http://www.unl.edu/amc/a-activities/a7-problems/putnamindex.shtml[/quote]

A "Kent Merryfield" plug in the solutions. :D[/quote]

As you may have guessed, the AoPS forums are one of my sources, besides email from some usual suspects like Noam Elkies, talking to MIT students after the exam, and even working on the problems myself.

I just posted a new version of the solutions with some additions and corrections. I'll probably post at least one or two more versions later this week, so keep an eye out.

Well, I'm a high schooler, so I took it unofficially.

I solved A,B 1-3 fully, I believe. (Got A3 with 15 min. left  :D ) I thought I had A4 and B5, but turned out there were flaws in my proofs. (A4: forgot that $ a$ could be $ \pm$ a quadratic residue $ b$ for $ \frac ab$ in the set, so had $ \frac23\not\in S$, B5: forgot the case where $ f$ was bounded between two numbers and did not converge.) However, I could easily fix each flaw, but oh well... I also knew that A6 was false but couldn't find a sufficient counterexample.

I was slightly annoyed by how the problems were easy to solve but very,very tedious to write up...

[quote="fibointepi"]

I was slightly annoyed by how the problems were easy to solve but very,very tedious to write up...[/quote]

Right, that is one the reasons that makes putnam competition more difficult than it seems.

Time is a key factor.

[quote="Elemennop"]That's because Muirhead destroys any inequality.[/quote]

Why don't you try using Muirhead to solve this legendary inequality proposed by Vasc a long time ago on this forum:

Given $ a,b,c$ are positive real numbers. Prove that:
$ (a^2\plus{}b^2\plus{}c^2)^2\geq 3(a^3b\plus{}b^3c\plus{}c^3a)$


[quote="ppt"]Can someone explain me the difference between $ \sum_{cyc}$ and $ \sum_{sym}$?
Thanks![/quote]

I don't know how best to explain but cyclic sum means write the entities on a circle and go from first to last along the circumference either clockwise or anticlockwise.

Symmetric sum means if one entity bears some relation to another entity, then it does bears the same relation with every other entity.

Example
$ \sum_{cyc}a^2bc\equal{}a^2bc\plus{}ab^2c\plus{}abc^2$

$ \sum_{sym}a^2bc\equal{}a^2bc\plus{}a^2cb\plus{}ab^2c\plus{}cb^2a\plus{}abc^2\plus{}bac^2\equal{}2(a^2bc\plus{}ab^2c\plus{}abc^2)$

You explained clearly. I've understood. Thank you, Hell_ever!


function
logarithms
Putnam
LaTeX
limit
integration
real analysis

Does there exist a continuous function f defined for x>=1 such that f(1)+f(2)+...=infinity, but for any a>1 we have  \sum (n>=1) a^n*f(a^n)< infinity? I believe the answer is yes, but I cannot find a function.

Hi Harazi,

please be more specific about the source. I would like to make you add the problem number and maybe even the volume/year where the problem or the solution can be found...  :)

Oh, believe me, I would put the explicit source if I knew. Unfortunately, I found this problem in one of my materials and the only source written there is AMM.

Here's something which makes me think that such a function doesn't exist (it's no proof of course, actually, it's far from that, but it gave me some doubts).

I tried to find an infinite set of intevals s.t. for any $a>1$, only finitely many $a^n$ are inside the intervals. If I could find some intervals like that then maybe I could find them such that infinitely many integers lie in those intervals, and I could make the function $0$ or as small as I want everywhere outside the intervals, so we would be done. Also, If I can find intervals like these, by taking $\ln$ of the set $[1,\infty)$, we get the set $[0,\infty)$ and an infinite set of intervals s.t. for any $a>0$ only finitely many $na$ are inside the intervals.

However, I found a problem on the Internet, Putnam 1964/B3, stating that if $f:R\rightarrow R$ is a continuous function s.t. $lim_{n\rightarrow \infty}f(na)=0$ for all $a>0$ then $lim_{x\rightarrow \infty}f(x)=0$. You can easily see why this contradicts the fact that I can find an infinite set of intervals included in $[0,\infty)$ s.t., for any $a>0$,only finitely many $na$ lie inside the intervals, so my assumption was wrong.

Maybe I'm also wrong about this, but my ideas concerning this problem are now totally messed up, so sorry if I'm posting some pretty stupid remarks right now.. :D

I feel that I must spam this topic too :( anyway, a quick latex advice grobb, use $\lim_{n\to\infty}$ which is \lim_{n\to\infty} or better yet $\displaystyle \lim_{n\to\infty}$ which is \displaystyle \lim_{n\to\infty} :)

also works for \int $\int^1_0$ or determinants \det $\det (A-B)$, or any other known function :P

The function harazi asked about does exist. Is anyone still interested in this problem? Also do you prefer a complete solution or a series of hints?

Anything that could lead to a complete solution. Thank you very much.

Then I'll just split the problem into a series of simpler questions and ask them one after another hoping that somebody will come up with an answer. Here goes the first part. It is fairly clear that the function, if it exists, should look like a sequence of narrow spikes near integers being identically $0$ everywhere else. So let $h_m=f(m)$ be the hight of the spike at $m\in\mathbb N$ and let $d_m$ be the half-width of the spike at $m$. Our first task is to choose the heights $h_m$. The first condition to satisfy is $\sum_m h_m=\sum_m f(m)=+\infty$. The second condition deals with integer $a>1$: we must have $\sum_n a^n h_{a^n}<+\infty$ for $a=2,3,\dots$. The last condition comes from grobber's observation that no matter how small $d_m$ are, we shall always have an $a>1$ such that $|a^n-m|<d_m$ for infinitely many pairs $n$ and $m$. This implies that we must have $m h_m\to 0$ as $m\to+\infty$ at the very least. So here is the first question: how to choose $h_m$   satisfying these three conditions?

For those three purposes, we could take $f(n)=\frac 1{n\ln n}$ whenever $n$ is not a perfect power, and $0$ for perfect powers.

OK, let us keep this choice. Now let us look at arbitrary $a$. Let $\mathbb M$ be the set of positive integers that are not powers of other integers. With our choice of $h_m$, the sum $\sum_n a^n f(a^n)$ is about the same as the sum $\sum_{n\in N(a)}\frac 1{n\ln a}$ where $N(a)$ is the set of $n$ for which there exists $m\in\mathbb M$ such that $|a^n-m|<d_m$. The set $N(a)$ cannot be made finite, but we do not need that much. So here comes the second question: what is enough to ask from $N(a)$ to ensure convergence?

harazi asked me to post a complete solution. So, here goes. The answer to the last hint question is "It suffices to demand that $N(a)$ contains at most one element in any interval $[k,2k]$ (or, if you prefer, $[k^2,(k+1)^2]$) for all sufficiently large $k$". This ensures that the elements of $N(a)$ grow fast enough to guarantee the convergence of the series $\sum_{n\in N(a)}\frac1n$. Now comes the main lemma:
[b]Lemma:[/b]
Let $1<A<B<+\infty$. Let $k$ be any positive integer. Then there exists $\delta>0$ such that for any number $a\in[A,B]$ not more than one of the numbers $a^n$ with $k\leq n\leq 2k$ can be within distance less than $\delta$ from $\mathbb M$ (where $\mathbb M$ is the set of integers that are not perfect powers as defined before).

I want somebody else (say, harazi or grobber) to try to prove the main lemma by himself (if nobody figures it out, I'll post a proof in a few days). And now I'll just use it to finish the proof. Choose two sequences $A_k\to 1+$ and $B_k\to+\infty$ such that $A_k^k\to+\infty$. Now choose the corresponding $\delta_k>0$ given by the lemma. For $m\in\mathbb M$ let $d_m$ be the minimum of $1/3$ and all $\delta_k$ with $A_k^k<m+1$ (this is a finite set, so the minimum exists). For any $a>1$, we have $a\in [A_k,B_k]$ for all sufficiently large $k$ and, therefore, there may be only one $n\in[k,2k]$ for which there exists $m\in\mathbb M$ such that $|a^n-m|<d_m$. The end.

I think I can see how that works.

There are only finitely many elements of $\mathbb M$ which, when raised to the power $\frac 1n$ for some $k\le n\le 2k$, fall in $[A,B]$. Call this finite set $\mathbb M_0$. Now perform the following operation: given $\delta>0$, take the intervals of length $2\delta$ centered at the points of $\mathbb M_0$, and raise them to the powers $\frac 1k,\frac 1{k+1},\ldots,\frac 1{2k}$. We get finitely many intervals. What we want is that no two such intervals intersect inside $[A,B]$. Since there are finitely many intervals, we can make $\delta$ small enough for our purpose.

Is it correct?

Almost. It will be completely correct if we show that no two elements of $\mathbb M_0$ coincide when raised to the powers $\frac1k,\dots,\frac 1{2k}$ (otherwise the corresponding intervals overlap no matter how small $\delta$ is).

Right. I missed that, somehow :? Sorry about that.

Assume we have two elements of $\mathbb M$, written as $a=\prod_{i=1}^tp_i^{a_i}$ and $b=\prod_{i=1}^tp_i^{b_i}$ s.t. $a^m=b^n$. This means that for all $i$'s, we have $ma_i=nb_i$. In our situation, $m,n\in[k,2k]$. If they're $k$ and $2k$, then we easily get a contradiction (one of $a,b$ would be a square). If not, then none of them divides the other one, so we get, for instance, $1<\frac m{(m,n)}|b_i,\ \forall i$, meaning that $b$ is a $\frac m{(m,n)}$'th power, which is false.


Which of the following is equal to the product
\[ \frac {8}{4}\cdot\frac {12}{8}\cdot\frac {16}{12}\cdots\frac {4n \plus{} 4}{4n}\cdots\frac {2008}{2004}?
\]$ \textbf{(A)}\ 251 \qquad \textbf{(B)}\ 502 \qquad \textbf{(C)}\ 1004 \qquad \textbf{(D)}\ 2008 \qquad \textbf{(E)}\ 4016$

[hide]
Note that everything cancels out except for $ \frac{2008}{4}\equal{}502$.  Answer is $ B$.
[/hide]

All the numbers cancel out except for the first denominator and the last numerator, so the answer is $ \frac{2008}{4}\equal{}502 \Rightarrow \boxed{B}$

502 as said above everything cancels out except for the 2008 and the 1/4

Reduce each fraction by $ 4$ and you get that the fraction is \[ \frac{2\cdot 3\cdot...\cdot 502}{1\cdot 2\cdot...\cdot 501}\equal{}\frac{502!}{501!}\equal{}502\] so the answer is $ \boxed{B}$.

[quote=mm999aops]wait I don't get it why are you posting simple AMC12 questions you should go to aops wiki. It is much more organized and you can post your own solutions there.
And by the way it is obviously 502.
I like @above red fire truck's simple solution[/quote]

This is a contest collections thread. Please actually figure out what you are talking about before saying to move this to the wiki.

[quote=OTMS_MATHCLUB][quote=RedFireTruck]$\frac{2008}{4}=\boxed{502}$[/quote]

You know you are supposed to explain your solution instead of just writing a division statement, right?[/quote]

telescoping


Or showing that \[ \sum\frac{x_i^3}{x_i^2+x_ix_{i+1}+x_{i+1}^2}=\sum\frac{x_{i+1}^3}{x_i^2+x_ix_{i+1}+x_{i+1}^2} \] And after you just have to show that:
$\sum\frac{x_i^3+x_{i+1}^3}{x_i^2+x_ix_{i+1}+x_{i+1}^2}\geq \frac{2(\sum x_i)}{3}$
And it can be done easily by easy ineq:
$\frac{x_i^3+x_{i+1}^3}{x_i^2+x_ix_{i+1}+x_{i+1}^2}\geq \frac{x_i+x_{i+1}}{3}$


Evaluate the sum
$ \sum_{n\equal{}0}^{\infty}\frac{1}{F_{2^n}}$
where $ F_m$ is the $ m$ th term of fibonacci sequence.


Hey,

I am not sure if I am in the right place to post this. First time as a student in the US. I need everybody's help!\
I am trying to complete a project for my statistics class and I need some opinions about:

IF YOU HAD ONE HOUR OF FREE TIME TO VISIT AN STORE, WHAT KIND OF STORE WOULD IT BE (like clothing, electronics, music, outdoors, department...)

I need 15 nice guys and 15 nice girls that help me soon. IT is kind of urgent, the deadline is eating me.

THANKS  A  LOT!!! :(

Best Buy or Circuit City, so electronic stores

[quote="themonster"]Book store for me.  I hope you realize this survey is going to have a tech/intelectualtendency.  AoPS is definitely NOT a representative sample of the population as a whole.[/quote]

hey.. just because some of us are techies and stuff doesnt mean we cant ..... oh wtvr.. Electronic store lol.  - but a store like Modells or a restaurant would come in a close second. Music store maybe too.

er... um. gosh, this is really hard. i guess the book store.

Book store for me too
I don't think it's a good idea to do a survey on a board for mostly math people, unless your intention is to describe the habits of people who go to this forum specifically.
Maybe you should print something out, give it to people in your school, and ask them to return it to a certain locker or to your teacher.

Electronics store, I'm a technology geek.

one hour...

then it would be a book store for me as well.

Probably electronics. I live in front of my computer.

got to be a sporting goods store, like dicks

[quote="Spoon"]just an hour? I could easily waste away my life in a book store...but I don't anymore because I don't like wasting time -_-. Put me down for book store.[/quote]
We got several large book buildings in Beijing. If I go to one of them, I shall at least spend one hour finding the books I want, at least......

ack....i'd say book or music.......put me down for music.

well. a book store... assuming it has a computer with internet access  where you can download music :P . but if it doesnt, it still works for me.


Does anyone know how to transfer microsoft word documents into pdf form?
I found a word file version of the Super Mario Theme sheet music. Can anyone transfer into pdf it for me?

Note: This is [u][i][b]NOT[/b][/i][/u] the real version written by the nintendo company. This is a free version that someone who is experienced in music composed based on the mario tune and posted on a website.

This version is for piano.

On the author part, it says it was composed by Koji Kondo and transcribed by Aaron Ackerson. That is because Koji Kondo composed it based on the mario tune and Aaron Ackerson did some editing and polishing.
[url=http://individual.utoronto.ca/auyeun/vgsheetmusic.html]This is a site with alot of video game sheet music[/url]

Use Open Office. It's free and has an Export as PDF option.

Enjoy.

If you dont want to use open office due to it's size and all that, you could even try cutePDF....

http://www.cutepdf.com/Products/CutePDF/writer.asp

If it's just text that you need , a good idea might be to write the whole thing in TeXnic Center and then just export the whole file as  a pdf....If you need to convert some other file (like a word document or a powerpoint presentation) to PDF, you can use software like CutePDF to "Write" PDF files.

There are programs which will write pdfs and work just like a printer.

I think it's called Win2PDF or something.

You can try [url=http://openoffice.org/]Openoffice[/url], the new [url=http://www.microsoft.com/office/preview/beta/overview.mspx]Office 2007 beta[/url], or PDFCreator [google it].  (all free, but Office 2007 expires next February or so)


If $ a,b,c>0$, the

\[ \sqrt{\frac{a\plus{}b}{c}}\plus{}\sqrt{\frac{b\plus{}c}{a}}\plus{}\sqrt{\frac{c\plus{}a}{b}}\geq2\sqrt[4]{\frac{6(a\plus{}b\plus{}c)^{3}}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}}\]

[quote="lipako"]If $ a,b,c > 0$, the
\[ \sqrt {\frac {a \plus{} b}{c}} \plus{} \sqrt {\frac {b \plus{} c}{a}} \plus{} \sqrt {\frac {c \plus{} a}{b}}\geq2\sqrt [4]{\frac {6(a \plus{} b \plus{} c)^{3}}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}}\]
[/quote]

By Am-Gm we have:
$ 8abc \le (a\plus{}b)(b\plus{}c)(c\plus{}a) \le \frac{8}{27}(a\plus{}b\plus{}c)^3$
So:
$ LHS \ge 3\sqrt{2} \ge RHS$

[hide][quote="lipako"]If $ a,b,c>0$, the

\[ \sqrt{\frac{a+b}{c}}+\sqrt{\frac{b+c}{a}}+\sqrt{\frac{c+a}{b}}\geq2\sqrt[4]{\frac{6(a+b+c)^{3}}{(a+b)(b+c)(c+a)}}\][/quote][quote="sandumarin"][quote="caubetoanhoc94"][quote="lipako"]If $ a,b,c > 0$, the
\[ \sqrt {\frac {a + b}{c}} + \sqrt {\frac {b + c}{a}} + \sqrt {\frac {c + a}{b}}\geq2\sqrt [4]{\frac {6(a + b + c)^{3}}{(a + b)(b + c)(c + a)}}\]
[/quote]

By Am-Gm we have:
$ 8abc \le (a+b)(b+c)(c+a) \le \frac{8}{27}(a+b+c)^3$
So:
$ LHS \ge 3\sqrt{2} \ge RHS$[/quote]
____________
I think it's a mistake!
$ LHS\ge 3\sqrt{2} $ and $ RHS\ge 2\sqrt[4]{\frac{6(a+b+c)^3}{\frac{8(a+b+c)^3}{27}}}=3\sqrt{2 }$[/quote][/hide]
Then what about [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3504908#p3504908]this[/url] one?  :lol:

This is stronger:
$ a,b,c>0\Rightarrow\sqrt{\frac{b+c}{a}}+\sqrt{\frac{c+a}{b}}+\sqrt{\frac{a+b}{c}}\ge\sqrt{\frac{8(a+b+c)^3}{(a+b)(b+c)(c+a)+4abc}} $
_____
Marin Sandu

$ a,b,c>0\Rightarrow\sqrt{\frac{b+c}{a}}+\sqrt{\frac{c+a}{b}}+\sqrt{\frac{a+b}{c}}\ge $
$ \ge (a+b+c)\sqrt{\frac{6}{ab+bc+ca}}\ge 2\sqrt[4]{\frac{6(a+b+c)^3}{(a+b)(b+c)(c+a)}} $
______
Marin Sandu

$ a,b,c>0\Rightarrow\sqrt{\frac{b+c}{a}}+\sqrt{\frac{c+a}{b}}+\sqrt{\frac{a+b}{c}} $
$ \ge 2(a+b+c)\sqrt{\frac{3}{a^2+b^2+c^2+ab+ba+ca}}$
________
Marin Sandu

[quote="arqady"][quote="sandumarin"]This is stronger:
$ a,b,c>0\Rightarrow\sqrt{\frac{b+c}{a}}+\sqrt{\frac{c+a}{b}}+\sqrt{\frac{a+b}{c}}\ge\sqrt{\frac{8(a+b+c)^3}{(a+b)(b+c)(c+a)+4abc}} $
[/quote]
It's just Holder:
$\left(\sum_{cyc}\sqrt{\frac{b+c}{a}}\right)^2\sum_{cyc}c(a+b)^2\geq8(a+b+c)^3$ and we are done![/quote]
________
Thank You! It's just my solution!

The following inequality is also true.
Let $a$, $b$ and $c$ be positive numbers. Prove that:
\[\sqrt{\frac{b+c}{a}}+\sqrt{\frac{c+a}{b}}+\sqrt{\frac{a+b}{c}}\ge\sqrt{\frac{9(a+b+c)^3}{(a+b)(b+c)(c+a)+5.5abc}}\]
It's wrong. I mistaken.  :(

[quote="arqady"]The following inequality is also true.
Let $a$, $b$ and $c$ be positive numbers. Prove that:
\[\sqrt{\frac{b+c}{a}}+\sqrt{\frac{c+a}{b}}+\sqrt{\frac{a+b}{c}}\ge\sqrt{\frac{9(a+b+c)^3}{(a+b)(b+c)(c+a)+5.5abc}}\][/quote]
Here it is the first step :

Or $(a,b,c)=(1,1,4)$. I proved it by Holder!  :D  :(  Thank you szl6208.


geometry
inequalities
IMO Shortlist
circumcircle

In an acute triangle $ABC$, let $D$, $E$, $F$ be the feet of the perpendiculars from the points $A$, $B$, $C$ to the lines $BC$, $CA$, $AB$, respectively, and let $P$, $Q$, $R$ be the feet of the perpendiculars from the points $A$, $B$, $C$ to the lines $EF$, $FD$, $DE$, respectively.

Prove that $p\left(ABC\right)p\left(PQR\right) \ge \left(p\left(DEF\right)\right)^{2}$, where $p\left(T\right)$ denotes the perimeter of triangle $T$ .

[i]Proposed by Hojoo Lee, Korea[/i]

Notice that $p(ABC) = 8R \prod cos{\frac{A}{2}}$, $p(DEF) = 8R \prod sinA$, $p(PQR) = 4R \prod sinA$.

[quote="treegoner"]Notice that ........ $\boxed{\ p(PQR) = 4R \prod sinA\ }$.[/quote][size=134][color=red][b]Why ?[/b][/color] I found $\boxed{\ p(PQR)=R\sum \sin 2A\sqrt{1-\sin 2B\sin 2C}\ }\ \ !$.[/size]

Sorry! I misread the problem  :blush: . I will spend some time for this problem later.

[quote="Amir.S"]In an acute triangle $ABC$, let $D, E, F, P, Q, R$ be the feet of perpendiculars from $A, B, C, A, B, C$ to $BC, CA, AB, EF, FD, DE,$ respectively. Prove that $p(ABC)p(PQR) \ge p(DEF)^{2}$, where $p(T )$ denotes the perimeter of triangle $T$ .[/quote]

$BC=a$, $QR= \cos A \sqrt{a^{2}-4bc \cos B \cos C \sin^{2}A}$, $EF=a \cos A$, so $BC \cdot QR \geq EF^{2}\iff a \cos A \sqrt{a^{2}-4bc \cos B \cos C \sin^{2}A}\geq a^{2}\cos^{2}A$
$\iff \sqrt{a^{2}-4bc \cos B \cos C \sin^{2}A}\geq a \cos A \iff a^{2}\sin^{2}A \geq 4bc \cos B \cos C \sin^{2}A$
$\iff a^{2}\geq 4bc \cos B \cos C$, which is true as $a=b \cos C+c \cos B$.

So $p(ABC)p(PQR)=(BC+CA+AB)(QR+RP+PQ)$
$\geq (\sqrt{BC \cdot QR}+\sqrt{CA \cdot RP}+\sqrt{AB \cdot PQ})^{2}\geq (EF+FD+DE)^{2}=p(DEF)^{2}$.

This is G7 in ISL 2005. It was proposed by me :D The key idea is to use the inequality Sung-yoon stated. Check out the proof in the new edition of my weblication 'Topics in Inequalities - Theorems and Techniques' which is now availabe at  my new homepage at  http://ultrametric.googlepages.com/tin2006new.pdf

In fact, more is true. Look at the solutions in the official IMO short list book.

[hide="Solution"]
Let $ \angle BAC \equal{} A$, $ \angle ABC \equal{} B$, $ \angle ACB \equal{} C$, $ BC \equal{} a$, $ AC \equal{} b$, and $ AB \equal{} c$.  Furthermore, let $ O$ be the circumcenter of $ \triangle ABC$.  Since $ \angle OAC \equal{} 90 \minus{} \angle ABC \equal{} 90 \minus{} \angle AEF \equal{} \angle PAE$, we see that $ APO$ is a line.  Similarly, $ BQO$ and $ CRO$ are lines.  Now, let the midpoints of $ AC$, $ BC$, and $ AB$ be $ Y$, $ X$, and $ Z$ respectively.  Notice that since $ OZ\perp BC$, we have that $ OD \ge OZ$.  Since $ OR\perp DE$ and $ OQ\perp DF$, we have that $ OQDR$ is cyclic with diameter $ OD$, so\[ QR \equal{} OD\sin \angle QOR \ge OZ\sin 2 A \equal{} 2\sin A\cos A\cdot \frac {BZ\cos A}{\sin A} \equal{} a\cos^2 A \]Doing the same with the others, we get that\[ (QR \plus{} PQ \plus{} PR)(BC \plus{} AB \plus{} AC) \ge (\sum a\cos^2 A)(\sum a)\ge (\sum a\cos A)^2\]by Cauchy.  Notice that $ \triangle AFE \sim \triangle ACB$ with ratio $ \frac {AE}{AB} \equal{} \cos A$.  Hence, $ \frac {FE}{BC} \equal{} \cos A\implies FE \equal{} a\cos A$ and we may conclude that $ p(ABC)p(PQR)\ge [p(DEF)]^2$. [/hide]

[hide="Slightly stronger"]In fact, a much stronger inequality is true. According to Fagnano's problem, the perimeter of $\triangle DEF$ is less than or equal to the perimeter of the medial triangle, which has perimeter $\frac{p(ABC)}{2}$. It is therefore sufficient to show that $p(DEF) \leq 2 p(PQR)$.

Because $\triangle DEF$ is the orthic triangle of $\triangle ABC$, $C$ is the $F$-excenter of $\triangle DEF$, so $P$ is the point where the $F$-excircle is tangent to $\triangle FED$. Similarly, $Q$ and $R$ are the tangency points of the $E$-excircle and $D$-excircle to $FD$ and $EF$, respectively. Hence, we may let $PE = QF = x$, $RF = PD = y$, and $QD = RE = z$. We wish to show that $x+y+z \leq PQ + QR + RP$.

By the law of cosines,
\begin{align*}
\cos D
&= \frac{(x+y)^2 + (z+x)^2 - (y+z)^2}{2(x+y)(x+z)} \\
&= \frac{2x^2 + 2xy + 2xz- 2yz}{2(x+y)(x+z)} \\
&= 1 - \frac{2yz}{(x+y)(x+z)},
\end{align*}
so $PQ = \sqrt{y^2 + z^2 - 2 \left(1 - \frac{2yz}{(x+y)(x+z)}\right)} = \sqrt{(y-z)^2 + \frac{4y^2 z^2}{(x+y)(x+z)}}$. $PQ + QR + RP \geq x+y+z$ therefore becomes
\[ \sum_{cyc} \sqrt{(y-z)^2 + \frac{4y^2 z^2}{(x+y)(x+z)}} \geq x+y+z, \]
or
\[ \sum_{cyc} \sqrt{(y-z)^2(x+y)(y+z)(z+x) + 4y^2 z^2 (y+z)} \geq \sqrt{(x+y)(y+z)(z+x)}(x+y+z).\]
Squaring both sides, our inequality becomes
\[\sum_{cyc} (y-z)^2(x+y)(y+z)(z+x) + \sum_{cyc} 4y^2 z^2 (y+z) + 2A \geq (x+y)(y+z)(z+x)(x+y+z)^2,\]
where
\begin{align*}
A
&= \sum_{cyc} \sqrt{(y-z)^2 (x+y)(y+z)(z+x) + 4y^2 z^2 (y+z)} \sqrt{(x-y)^2 (x+y)(y+z)(z+x) + 4x^2 y^2 (x+y)} \\
&\geq \sum_{cyc} \sqrt{4y^2 z^2 (y+z)} \sqrt{4x^2 y^2 (x+y)} \\
&= 4xyz \sum_{cyc} y \sqrt{(y+x)(y+z)}.
\end{align*}
Observing that $\sqrt{(a+b)(a+c)} \geq a + \sqrt{bc} \iff a(\sqrt{b} - \sqrt{c})^2 \geq 0$, we find that
\[ 4xyz \sum_{cyc} y \sqrt{(y+x)(y+z)} \geq \sum_{cyc} 4x^3 yz + \sum_{cyc} 4x^{\frac{3}{2}} y^{\frac{3}{2}} z^2, \]
so it is enough to show that
\[ \sum_{cyc} (y-z)^2(x+y)(y+z)(z+x) + \sum_{cyc} 4y^2 z^2 (y+z) + 8 \sum_{cyc} x^3 yz + 8 \sum_{cyc} x^{\frac{3}{2}} y^{\frac{3}{2}} z^2 \geq (x+y)(y+z)(z+x)(x+y+z)^2. \]

We also have
\begin{align*}
(x+y)(y+z)(z+x) \left( \left( \sum_{cyc} (y-z)^2\right)  -(x+y+z)^2 \right)
&= (x+y)(y+z)(z+x)(x^2 + y^2 + z^2 - 4xy - 4yz - 4xz) \\
&= (\sum_{sym} x^2 y + 2 xyz)(x^2 + y^2 + z^2 - 4(xy + yz + zx)) \\
&= \sum_{sym} x^4 y + \sum_{sym} x^3 y^2 + \sum_{sym} x^2 y^2 z + \sum_{sym}x^3 yz - 4 \sum_{sym} x^2 y^2 z \\
&\, \, \, \, -4\sum_{sym}x^3 y^2 - 4 \sum_{sym} x^2 y^2 z - 4 \sum_{sym} x^3 yz \\
&= \sum_{sym} x^4 y - 3 \sum_{sym} x^3 y^2 - 7 \sum_{sym} x^2 y^2 z - 3 \sum_{sym} x^3 yz
\end{align*}

Henceforth, we shall denote by $[a,b,c]$ the sum $\sum_{sym} x^a y^b z^c$. Then we may thus rewrite the desired inequality as
\[ [4,1,0] + 4[3,2,0] + 4[3,1,1] + 4[2,\frac{3}{2}, \frac{3}{2}] \geq 3[3,2,0] + 7[2,2,1] + 3[3,1,1], \]
which is clearly equivalent to
\[ [8,2,0] + [6,4,0] + [6,2,2] + 4[4,3,3] \geq 7[4,4,2]. \]
By Schur's inequality with $r=2$, we have $[4,0,0] + [2,1,1] \geq 2[3,1,0]$, so $[4,3,3] \geq 2[5,3,2] - [6,2,2]$. Hence, it is enough to show that
\[ [8,2,0] + [6,4,0] + [6,2,2] + 8[5,3,2] \geq 4[6,2,2] + 7[4,4,2]. \]
By Muirhead's inequality, $[5,3,2] \geq [4,4,2]$ and $[6,4,0] \geq [6,2,2]$, so it is sufficient to show
\[ [8,2,0] + [5,3,2] \geq 2[6,2,2]. \]
Now, we have
\[ \sum_{cyc} (x^2 - y^2)^2 (x^4 + y^4) z^2 \geq 0,\]
which expands to
\[ [8,2,0] \geq 2[6,2,2] - [4,4,2]. \]
Thus, it is sufficient to show
\[ 2[6,2,2] - [4,4,2] + [5,3,2] \geq 2[6,2,2], \]
or
\[ [5,3,2] \geq [4,4,2], \]
which follows from Muirhead.
[/hide]
(dnkywin informed me that this stronger result is true.)

We note that $\angle{PAF}=\angle{QBF}=90-C$ so the lines $AP,BQ,CR$ concur at the circumcenter $O$ of $\triangle{ABC}$.Thus we note that $POQF$ is a cyclic quadrilateral with $OF$ as its diameter,so $PQ=OF\sin2C \ge d(O,AB)\sin2C =R\cosh\sin2C$.We get $PR$ and $QR$ analogously.We also note that $p(ABC)=2R\sum{sinA}$ and $p(DEF)=R\sum{sin2A}$(Note that the pedal triangle has circumradius $\frac{R}{2}$ so applying sine rule will make the fact clear).Thus the problem is equivalent to showing that

$2 \sum{sinA} \sum{sin2CcosC} \ge (\sum{sin2A})^2$.

$\Leftrightarrow (\sum{(\sqrt{2sinA})^2})(\sum{(\sqrt{sin2AcosA})^2}) \ge (\sum{sin2A})^2$

which is obvious by Cauchy.It can easily be seen that equality holds if and only if $\triangle{ABC}$ is equilateral.

Surprisingly simple for a G7.

[hide]
We adopt the standard conventions $BC=a, CA= b, AB=c, \angle BAC = A, \angle ABC = B, \angle BCA = C.$ Let $R$ be the circumradius of $\triangle ABC$ and let $O$ be its circumcenter.

[b]Claim: [/b] $AP, BQ, CR$ are concurrent at $O.$

[b]Proof: [/b] Note that $BCEF$ is cyclic since $CF \perp AB$ and $BE \perp AC.$ Hence, $\angle PBC = 90^{\circ} - \angle BDF = 90^{\circ} - A = \angle OBC,$ so $A,P,O$ are collinear. Similarly $B,Q,O$ and $C,R,O$ are also collinear. The result follows. $\blacksquare$

Since $OP \perp DP$ and $OQ \perp DQ,$ $OPDQ$ is cyclic with diameter $OD.$ By sine rule, $PQ = OD \sin POQ = OD \sin 2A.$ Let $M$ be the midpoint of $BC.$ Since $OM \perp BC,$ $OD \geq OM.$ From right angled $\triangle OMB,$ $OM= R \sin (90^{\circ} - A) = R \cos A.$ Hence, $PQ \geq R \sin 2A \cos A.$ We get similar inequalities for $QR$ and $RP.$

By sine rule on $\triangle DEF,$ $\dfrac{DF}{\sin DEF} = R \implies \dfrac{DF}{\sin (\pi - 2A)} = R \implies DF = R \sin 2A,$ where we have used the fact that the radius of the nine point circle is half the radius of the circumcircle (an alternative way to compute $DF$ would be to use sine rule on $\triangle BDF,$ but that is slightly more tedious). We obtain analogous expressions for $EF$ and $DE.$

The desired inequality transforms to
\[(2 \sin A + 2 \sin B + 2 \sin C)(\sin 2A \cos A + \sin 2B \cos B + \sin 2C \cos C)   \geq  (\sin 2A + \sin 2B + \sin 2C)^2 \\
\iff \\ (\sin A + \sin B + \sin C)(\sin A \cos ^2 A + \sin B \cos ^2 B + \sin C \cos ^2 C)  \geq  (\sin A \cos A + \sin B \cos B + \sin C \cos C)^2,\]
which is just Cauchy Schwartz on the sequences $(\sqrt{\sin A}, \sqrt{\sin B}, \sqrt{\sin C})$ and $(\cos A \sqrt{\sin A}, \cos B \sqrt{\sin B}, \cos C \sqrt{\sin C}).$

For equality to hold, the midpoints must coincide with the feet of perpendiculars, so $\triangle ABC$ must be equilateral. Plugging $A=B=C  = \dfrac{\pi}{3}$ in this equation, we see that equality indeed does hold. Thus, equality holds iff $\triangle ABC$ is equilateral. $\blacksquare$

[/hide]

It's about time.

[b] Lemma 1: [/b] In a triangle $ABC$, we have $\sin A + \sin B + \sin C \ge \sin 2A + \sin 2B+ \sin 2C$
Proof: Sum to product yields $\frac{\sin 2A + \sin 2B}{2} = \sin(A+B) \cos (A-B)$. Now just sum up cyclically.

[b] Lemma 2: [/b] In a triangle $ABC$ with excircles touching sides $BC, CA, AB$ at $D, E, F$, respectively, the perimeter of triangle $ABC$ is at most twice that of triangle $DEF$ (USA TSTST 2011, P7).
Proof: Consider the projection of $EF$ onto side $BC$. Suppose $E$ is projected to point $E'$ and $F$ is projected to point $F'$ on side $BC$. Basic triangle facts yield that $EF \ge E'F'=a-(s-a)(\cos B + \cos C)$. So we simply want \[ \sum_{cyc} a_(s-a)(\cos B + \cos C) \ge \frac{a+b+c}{2} \]. This rearranges to \[ a+b+c \ge 2a \cos A + 2b \cos B + 2c \cos C \]By LoS, it remains to show \[ \sin A + \sin B + \sin C \ge \sin 2A + \sin 2B + \sin 2C \]Observe that sum to product yields $\frac{\sin 2A + \sin 2B}{2} = \sin(A+B) \cos (A-B) \le \sin C$, from which summing yields the desired.

First, I claim $p(ABC) \ge 2 p(DEF)$. By extended LoS, we have $EF=2R \cos A \sin A$. Thus we want to show that \[ 2R \sin A + 2R \sin B + 2R \sin C \ge 2 (2R \sin A \cos A + 2R \sin B \cos B + 2R \sin C \cos C) \]. Cancelling $2R$ and using $2\sin A \cos A = \sin 2A$, the inequality reduces to [b]Lemma 1 [/b].

Next, I claim $p(PQR) \ge \frac{1}{2} p(DEF)$. Let's shift our reference frame to triangle $DEF$. Observe that $P, Q, R$ are just the touch points of the $D, E, F$ excenters (which are $A, B, C$ respectively) in triangle $DEF$. Then this inequality is true by [b] Lemma 2 [/b].

Now multiplying our two claims, we have $p(ABC) p(PQR) \ge (p(DEF))^2$, as requested.

Quite straight-forward for a G7 :)

Note that $P, Q, R$ are the projections of the circumcenter $O$ (of $\triangle ABC$) onto the lines $EF, FD, DE$ respectively. Secondly, $$QR=OD\cdot \sin \angle EDF=OD\cdot \sin 2A \ge 2R\sin A\cos^2A=BC\cos^2A.$$ Finally, we observe that $\frac{EF}{BC}=\cos A$ so $$BC\cdot QR \ge EF^2$$ and the given inequality follows from the CS inequality. Equality holds for equilateral triangles only.

Project $Q,R$ onto $BC$ at $Q',R'$. Note that $Q'D=\cos^2 A\cdot BD$, and $R'D=\cos^2 A\cdot CD$, so $QR\ge Q'R'=BC\cos^2 A$. Also, by similar triangles, $EF=BC\cdot \frac{AE}{BE}=BC\cos A$. So,
$$QR\cdot BC\ge Q'R'\cdot BC=BC^2\cos^2 A =EF^2$$
Now, by Cauchy Schwarz, $$(PQ+QR+RP)(AB+BC+CA)\ge (P'Q'+Q'R'+R'P')(AB+BC+CA)\ge (DE+EF+FD)^2$$ as desired.

[quote=AMN300][hide=*]It's about time.

[b] Lemma 1: [/b] In a triangle $ABC$, we have $\sin A + \sin B + \sin C \ge \sin 2A + \sin 2B+ \sin 2C$
Proof: Sum to product yields $\frac{\sin 2A + \sin 2B}{2} = \sin(A+B) \cos (A-B)$. Now just sum up cyclically.

[b] Lemma 2: [/b] In a triangle $ABC$ with excircles touching sides $BC, CA, AB$ at $D, E, F$, respectively, the perimeter of triangle $ABC$ is at most twice that of triangle $DEF$ (USA TSTST 2011, P7).
Proof: Consider the projection of $EF$ onto side $BC$. Suppose $E$ is projected to point $E'$ and $F$ is projected to point $F'$ on side $BC$. Basic triangle facts yield that $EF \ge E'F'=a-(s-a)(\cos B + \cos C)$. So we simply want \[ \sum_{cyc} a_(s-a)(\cos B + \cos C) \ge \frac{a+b+c}{2} \]. This rearranges to [/hide]\[ a+b+c \ge 2a \cos A + 2b \cos B + 2c \cos C \][hide=**]By LoS, it remains to show \[ \sin A + \sin B + \sin C \ge \sin 2A + \sin 2B + \sin 2C \]Observe that sum to product yields $\frac{\sin 2A + \sin 2B}{2} = \sin(A+B) \cos (A-B) \le \sin C$, from which summing yields the desired.

First, I claim $p(ABC) \ge 2 p(DEF)$. By extended LoS, we have $EF=2R \cos A \sin A$. Thus we want to show that \[ 2R \sin A + 2R \sin B + 2R \sin C \ge 2 (2R \sin A \cos A + 2R \sin B \cos B + 2R \sin C \cos C) \]. Cancelling $2R$ and using $2\sin A \cos A = \sin 2A$, the inequality reduces to [b]Lemma 1 [/b].

Next, I claim $p(PQR) \ge \frac{1}{2} p(DEF)$. Let's shift our reference frame to triangle $DEF$. Observe that $P, Q, R$ are just the touch points of the $D, E, F$ excenters (which are $A, B, C$ respectively) in triangle $DEF$. Then this inequality is true by [b] Lemma 2 [/b].

Now multiplying our two claims, we have $p(ABC) p(PQR) \ge (p(DEF))^2$, as requested.[/hide][/quote]
$$a \sin\frac{A}{2}+b \sin\frac{B}{2}+c \sin\frac{C}{2}\geq \frac{a+b+c}{2}\geq a \cos A +  b \cos B +  c \cos C$$
$$\sqrt{1+sinBsinC}\cdot sinA+\sqrt{1+sinCsinA}\cdot sinB+\sqrt{1-sinAsinB}\cdot sinC> sinA+ sinB+ sinC$$

The key idea is the same, but I saw it by spiral similarity  :D

[b][color=#f00]Main claim.[/color][/b] $BC\cdot QR\ge EF^2$, and the analogous results hold.

[i]Proof.[/i] Let $M$ be the midpoint of $BC$.

[asy]
defaultpen(fontsize(10));
size(220);

pair A=dir(70);
pair C=dir(-30);
pair B=dir(210);
pair D=foot(A,B,C);
pair E=foot(B,A,C);
pair F=foot(C,B,A);
pair P=foot(A,F,E);
pair Q=foot(B,D,F);
pair R=foot(C,D,E);
pair M=midpoint(B--C);

dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$D$", D, dir(D));
dot("$E$", E, dir(E));
dot("$F$", F, dir(F));
dot("$P$", P, dir(P));
dot("$Q$", Q, dir(Q));
dot("$R$", R, dir(E));
dot("$M$", M, dir(M));

draw(A--B--C--cycle);
draw(D--F);
draw(D--E);
draw(B--Q);
draw(M--Q, dashed+blue);
draw(M--R, dashed+blue);
draw(M--F, dashed+blue);
draw(M--E, dashed+blue);
draw(E--F, blue);
draw(R--Q, blue);
draw(C--R);
draw(circumcircle(D,E,F), yellow);
[/asy]

As $M$ is the midpoint of the arc $\widehat{EMF}$ on the nine-point circle and $Q$, $R$ are the points in which the $E$ and $F-$excircles touch $FD$ and $DE$, respectively, we know that $M$ is the center of the spiral similarity that sends $QR$ to $EF$. Thus, $$\dfrac{QR}{EF}=\dfrac{QM}{MF}=\dfrac{BM\cdot \frac{\sin{\angle MBQ}}{\sin{\angle BQM}}}{\frac{BC}{2}}=\dfrac{\cos{A}}{\sin{\angle BQM}}\ge\cos{A}=\dfrac{EF}{BC}$$Hence, $BC\cdot QR\ge EF^2$, as desired.

Finally, by Cauchy-Schwarz Inequality and the above claim, $$p(ABC)\cdot p(PQR)=(BC+CA+AB)(QR+RP+PQ)\ge (\sqrt{BC\cdot QR}+\sqrt{CA\cdot RP}+\sqrt{AB\cdot PQ})^2\ge(EF+FD+DE)^2=p(DEF)^2$$and we are done. $\square$


how hard are word problems on the amc 10 rather than on Mathcounts?

AMC 10 is designed for high schoolers while MC is designed for middle schoolers


Perhaps we should see about getting an AoPS group at AwesomeMath.  Certainly, there are enough of us to make it worthwhile.  Probably the first step should be to decide on a place to meet early on.  Could some one who was there last year make a suggestion?

last year we all had to stay in this room for like 5 hours or somethin. maybe meet at the pool outside?  it might be different though

I presume our primary goal is to: "Chillax" with people we sort of know.

That's cool, but our days are kind of booked, except for saturday...  :?:

[quote="Fanatic"]Well, for those of us that were in mathcounts and got AoPS shirts, we could wear those...[/quote]

umm.. i didn't get one of those..
could whoever wants to organize wear a shirt like that, or some kind of AoPS ID and we could just join him/her?
that maybe would work better
what do you guys think

[quote="Aryth"]YEAH!  :D
haha, that was an awesome trade.[/quote]
If my state hadn't forgotten pins, I would've easily traded 57 pins for 57 shirts.

I never saw anyone trading pins for shirts there.  :wallbash_red:  :wallbash_red:  :wallbash_red:

gosh i actually bought the aops shirt from this website

Man, I always wanted a t-shirt for some reason and now that I have one I don't feel special anymore  :(

welll is there gonna be a group or not??! :|


The vertices of a regular $2005$-gon are colored red, white and blue. Whenever two vertices of different colors stand next to each other, we are allowed to recolor them into the third color.
(a) Prove that there exists a ﬁnite sequence of allowed recolorings after which all the vertices are of the same color.
(b) Is that color uniquely determined by the initial coloring?

(b) The color is uniquely determined by the initial coloring.
Instead of red, white and blue color the vertices $0,1,2$ (respectively) notice that whenever we recolor, the sum of the labels over all vertices remains invariant modulo $3$. since $2005\equiv1(mod 3)$ the initial sum modulo $3$ is the last label.

Let $A,B,C$ be the colors. Observe that we can make $BAAB \mapsto BACC \mapsto BBBC$, $CAAB \mapsto CACC \mapsto CBBC \mapsto AAAA$, $BABA \mapsto CCCC, BAAA \mapsto CCAA \mapsto CBBA \mapsto CBCC \mapsto CAAC \mapsto BBBB$, $ABC \mapsto AAA$. The same is valid for cyclic positions. Hence, for any $4$ consecutive vertices, it's possible to turn the first three of them of the same color. Thus, do this operation, until we have many consecutive groups of $3$ consecutive vertices of the same color, and a single vertex $v$ with an unkwown color.

Now, observe that we have the operation $AAAB \mapsto AACC \mapsto ABBC \mapsto CCBC \mapsto CAAC \mapsto BBBB$ (the same holds for cyclic positions). Hence, doing this operation many times clockwisely, starting at $v$, it's possible to make all the vertices with the same color as $v$, which completes item $(a)$.

For item $(b)$, oberve that if at the beggining we have $x$ vertices of color $A$, $y$ vertices of color $B$ and $z$ vertices of color $C$, $x,y,z$ are not all equal $\pmod{3}$, since $2005 \equiv 1 \pmod{3}$. Assume WLOG $y \equiv z \pmod{3}$. If we make a movement switching $BC \mapsto AA$, the amount of vertices of colors $B$ and $C$ keeps the same $\pmod{3}$. If we make a movement switching $AB \mapsto CC$ , the amount of vertices of colors $B$ and $C$ keeps the same $\pmod{3}$, since $-1 \equiv 2 \pmod{3}$. The same is valid when we make $AC \mapsto BB$. Hence, the amount of vertices of colors $B$ and $C$ will always keep the same $\pmod{3}$. Hence, if at some point, the colors are all equal, it must be of color $A$, so we are done.

$\blacksquare$


$P$ is a point inside a parallelogram $ABCD$. Construct a line through $P$ paralel to $AD$ and meets $AB$, $CD$ at $E$, $F$. Constuct another line through $P$ that is paralel to $AD$ and meets $AD$, $BC$ at $G$, $H$.

Prove $AH$, $CE$, $DP$ are concurrent.

Are you sure about this? Several points are not used, and I don't think what I understand is true.


function
integration
vector
trigonometry
symmetry
real analysis
real analysis unsolved

the following use the inner product space $ C[ \minus{} \pi,\pi]$ of all continuous, real-valued functions on $ [ \minus{} \pi,\pi]$, together with the inner product

                                     $ \langle f,g \rangle \equal{} \frac {1}{\pi} \int_{ \minus{} \pi}^{\pi} f(x)g(x) dx$.

1. Show that the set $ \{\frac {1}{\sqrt {2}} , \sin(x), \sin(2x), \ldots , cos(x), cos(2x), \ldots\}$is an othogonal set. Show also that all vectors have length 1.

2. Find the Fourier expansion of $ f(x) \equal{} x$

3. Use Parseval's Identity : $ \langle f,g \rangle \equal{} \langle \hat{f}, \hat{g} \rangle$  to show that

                 $ \sum_{n \equal{} 1}^{\infty} \frac {1}{n^2} \equal{} \frac {\pi^2}{6}$

the first two were pretty easy. Except I don't what to do with the third one. I ve seen what Parseval's identity is, but I dont see how it applies to the our definition of $ \langle f,g \rangle$. does anyone have any advice/hints?
(the reason I posted the first two questions even though i already figured them out is becasue I was wondering if the first two pertained to the third in any way)

Thanks!

Hint: just use $ f \equal{} g,$ so that $ \langle f,f\rangle \equal{} \frac1{\pi}\int_{ \minus{} \pi}^{\pi}[f(x)]^2\,dx$

Thanks Kent for the hint. I pondered it for a while, but I still don't see where I need to go. I even looked up other sources of what Parseval's identity was, and I can see why we should have $ \langle f,f \rangle$, but I still don't see how to get the result.

All right, so what is the inner product on the other side?

Suppose we can write $ f(x)\equal{}\frac{a_0}{\sqrt{2}}\plus{}\sum_{n\equal{}1}^{\infty}(a_n\cos nx\plus{}b_n\sin nx)$, then

$ \langle f,f\rangle\equal{}\frac1{\pi}\int_{\minus{}\pi}^{\pi}f^2\,dx\equal{}
a_0^2\plus{}\sum_{n\equal{}1}^{\infty}(a_n^2\plus{}b_n^2)$

In order to get the concrete computation asked in this problem, you'll need to compute $ a_n$ and $ b_n$ for this particular function. That's what problem 2 above asks for. (Hint: by odd/even symmetry, $ a_n\equal{}0$ for all $ n.$ So you only need to work to compute $ b_n.$)

thanks kent

I do have a question out of curiosity though

under problem number 2) I left out the fact that

$ \hat{f} : \equal{} \langle f(x),\frac{1}{\sqrt{2}} \rangle\cdot\frac{1}{\sqrt{2}} \plus{} \sum_{n\equal{}1}^{\infty}\langle f(x), \sin(nx) \rangle \sin(nx) \plus{} \sum_{n\equal{}1}^{\infty}\langle f(x),\cos(nx)\rangle\cos(nx)$.

in which my teacher left out the note : find the coefficients for $ \hat{f}$.

thus, my question is how did you know to equate

$ \langle f,f\rangle\equal{}\frac1{\pi}\int_{\minus{}\pi}^{\pi}f^2\,dx\equal{} a_0^2\plus{}\sum_{n\equal{}1}^{\infty}(a_n^2\plus{}b_n^2)$

without knowing what $ \hat{f}$ was? to be more specific did you equate the two expressions as a consequence of the fourier transform?

also, why was it not necessary for me to compute $ \langle \hat{f},\hat{f} \rangle$ ?

lastly, what's the difference between $ : \equal{}$ and $ \equal{}$ ?

sorry for all the questions, I think I m going through some kind of sugar rush.

[quote="mstrip"]what's the difference between $ : \equal{}$ and $ \equal{}$ ?[/quote]

$ : \equal{}$ is a definition.  That is, $ f(x) : \equal{} x^2$ means "define the expression $ f(x)$ to be $ x$ squared."  (I believe $ : \equal{}$ is from MAPLE/Mathematica; to be more proper, use $ \equiv$.)  $ \equal{}$ is just an equality.  For example, $ x^2 \minus{} 5x \plus{} 6 \equal{} 0$ means "$ x$ is a value such that $ x^2 \minus{} 5x \plus{} 6$ is numerically equal to $ 0$."

[quote](I believe $ : \equal{}$ is from MAPLE/Mathematica; to be more proper, use $ \equiv$.)[/quote]
I've never seen $ \equiv$ used for definitions. I would understand $ f(x)\equiv x^2$ as an indication that equality holds for all values of $ x$, but not necessarily by definition (i.e. it may be a property of $ f$ to be proved). The notation $ : \equal{}$ goes back at least to [url=http://en.wikipedia.org/wiki/Assignment_%28computer_science%29]ALGOL[/url].

Hmm.  I know I read a paper that used $ \equiv$ for definitions while doing my research at RSI, but I don't remember which paper it was.  I suppose it's less common than I thought.

There's really no standard here- all of our symbols for defining something are borrowed from computer languages. $ : \equal{}$ has the advantage of being unambiguous, so it's the one I'd use.

As for the question pf why Kent did what he did? This is a standard exercise, and he's seen/used it so many times he automatically filled in the blanks.


How many inches are in the longest segment of integral length that can be drawn on a 6'' by 10'' rectangular sheet of paper?

i'm not really sure about this one but i think it is 6^2+10^2 because it is the diagonal
so 6^2=36 +10^2=100 =136 and so it is sqrt136 i think :?:
Since it is the largest [u]integer line,[/u] then it is 11

If the side lengths are 6 and 10, than the diagonal of the paper is sqrt136. This is less that sqrt144, which is 12, so the longest possible integral length is 11.

[quote="mrsaturns"]i'm not really sure about this one but i think it is 6^2+10^2 because it is the diagonal
so 6^2=36 +10^2=100 =136 and so it is sqrt136 i think :?:[/quote]

A good example of why you need to read the question more than once. Remmber that it says integral  :) .

No, this is fine. I'm just giving you a tip.


AMC
USA(J)MO
USAMO
geometry
inequalities
number theory
graph theory

What should I study in order to get a decent score on the USAMO?

Study for fun, not for score. Score should be a secondary goal. I do tons of math but not all of it is preparation for contest X.

That said, do old olympiad problems. Read Zeitz and/or Engel. At least, that's what I'm doing.

hmm my practice: just relax and pray for a 7 (probably the best score ill get...)

i have tons of homework and multiple AP tests to prepare for so unfortunently, i cant do too much USAMO/proof problems.

I pretty much lucked out this year, state holiday monday = week of break next week.

Any studying of olympiad math I do is purely for score. Why? Because studying for fun means not doing olympiad math.

That said, there are four main categories of olympiad problems:
1. Algebra
2. Geometry
3. Combinatorics
4. Number Theory

Here, functional equations and inequalities fall under algebra and graph theory under combinatorics.

What you should do is try to be able to consistently solve #1s and #4s. If you can do this, you will pretty easily get a 14, which will place you well into red MOP (if a freshman) and a good way towards blue MOP (if not). Then, find your strengths and learn to solve #2s and #5s in those areas. Once you do that, you have a choice. You can either attempt to solve #3s and #6s in your strong areas or solve #2s and #5s in your weak areas. If you can do either of these consistently, you have a decent chance of making black MOP or even the team.

Personally, I feel I can solve problems 1 and 4 in any subject, and 2 and 5 in anything except geometry. So, I'm going to attempt to solve a #3/6 algebra problem, and on failing that a #2/5 geo problem.

[quote="Altheman"][quote="Hamster1800"]Any studying of olympiad math I do is purely for score. Why? Because studying for fun means not doing olympiad math.[/quote]

Wow. I do Olympiad math for fun. I'm surprised that you have such a hardcore view.[/quote]

I agree, most of my preparation comes from working on problems for fun.  I rarely sit down and work on math for a long amount of time, but I usually have an Olympiad problem or two in my head to think about and try to solve.  This is probably why I suck at geo, because you can't do geo problems in your head  :P .

Still, if you're trying to make a push to get ready for this year's USAMO, Hamster's advice is really good.

"Decent" is entirely relative.  You've already done better than "decent" by most people's standards just by qualifying, so cut yourself some slack.  :)

That having been said, there are [url=http://wangsblog.com/jeffrey/?page_id=155]good resources[/url] available (courtesy of paladin8).

[quote="Hamster1800"]What you should do is try to be able to consistently solve #1s and #4s. If you can do this, you will pretty easily get a 14, which will place you well into red MOP (if a freshman) and a good way towards blue MOP (if not). Then, find your strengths and learn to solve #2s and #5s in those areas. Once you do that, you have a choice. You can either attempt to solve #3s and #6s in your strong areas or solve #2s and #5s in your weak areas. If you can do either of these consistently, you have a decent chance of making black MOP or even the team.

Personally, I feel I can solve problems 1 and 4 in any subject, and 2 and 5 in anything except geometry. So, I'm going to attempt to solve a #3/6 algebra problem, and on failing that a #2/5 geo problem.[/quote]

That sounds just like my strategy, except I rule at geometry and suck at number theory.

[quote="Altheman"][quote="Hamster1800"]Any studying of olympiad math I do is purely for score. Why? Because studying for fun means not doing olympiad math.[/quote]

Wow. I do Olympiad math for fun. I'm surprised that you have such a hardcore view.[/quote]
My point is that I do math for fun, but since preparation for USAMO is a secondary goal, I end up doing USAMO math. I wouldn't study if I didn't find it fun as I'm a relatively lazy person. As it is I end up playing chess instead of studying most of the time.


MATHCOUNTS
AMC
AMC 10
AMC 10 A
AIME
ARML
trigonometry

Hi, I'm a new mod in the Ohio Forum.

I'm k00lperson! Well, I'm an 8th grader who lives in Ohio. Taking part in Mathcounts, AMC 10..life's stressful, lawl. Anyway, if you would like to introduce yourself, go ahead! Maybe we'll bump into each other at some competition.

-k00l

W00T, sup dawg! (or should I say, cat?)

So yeah, we should totally make this Ohio forum more active.

So I'm veezbo and I'm in 10th grade. And I did mathcounts for 3 years, while never really understanding math, and failed. 8th grade was the only year I made it to state, and I got 16th there (btw, I'm from Sycamore, who's math team seems to be drowning)

I did OCTM and OHMIO last year. Although I was the top freshman at our school, I failed.

I also did the AMC10A and AMC12B. Didn't qualify for AIME, 'nuff said.

And that's basically my math career! I really only discovered AoPS at the end of last year, and I think I've finally gotten the hang of doing some competition math (but still need to get better)

My goals are in my sig. I really hope I can reach them...

yes. I agree that this forum [b]should[/b] be more active. (about the cat, lol, my brother says that all the time). Anyway, yes  everyone should have goals. Don't you think? :P

lol, your brother is awesome

hmm, yeah, without goals, I would probably just be a mindless zombie the rest of my life.

so to anyone reading this, do your part in preventing the zombie apocalypse. Please have goals.  :D Thank you!

haha. Well, everyone probably [b]do[/b] have goals, but we must be careful not to bring them to something we cannot achieve in our life. But it might be possible for other people, and I'm not saying we are uncapable :P.
Anyway, yes I know my brother is awesome. Oh wait do you even know who my brother is? :P

Yeah, there is a certain human extent to everything, so I guess oftentimes, its just best to say "I'm going to do the best I can, and try my hardest to avoid making regrettable decisions"

Then again, if you have a controlled ego, it might be beneficial to set goals that you believe you can never reach so that you try even harder to extend your "best" possible attempt.

Lol, I don't even know who you are, much less your brother.  :P  Does he have an AoPS account?

ok, I feel like leaving this discussion :)
Anyway, yes my brother does have an aops account

Good idea. I should probably stop procrastinating (haven't started my winter break homework)

Anyways, I think I have a good idea who your brother is  :D

Yes, the ohio forum is active, because there is another mod :).
Well, why don't we post some math then :).

[quote="tjhance"]wow, the ohio forum is suddenly active again :O

we need more math here[/quote]

This guy is back
Awesome
give me math probz plox

Check the new topic!

And K00lperson's brother is awesome.
I think everyone knows it.  :D

Oh, hi! I'm Royalflower.
I'm in 9th grade, and I like flowers (duh) and dancing (duh if you have read my blog).
Um.... I really have no goals that would be of interest to you people but I can post them if you want.

wow activity! :|

Well, I'm eqjj168 and I am from Dublin, OH.
I am in 8th grade and I have done MathCounts for the last 2 years and I'm doing it this year too. Last year I made it to States.
I also took the AMC 8 this year, doing worse than I expected, but got 2nd in my school.

Yeah I'm gonna do a new intro since the other Ohio one is like two years old... >_>

I'm from Cincinnati and... fail at most math I try to do. Finally made AIME this year (barely) as a junior aaaaand... got a 1. Yeah.  :rotfl:

Also, if you do ARML, you may know who I am...  :ninja:

Hello, I'm dragon724 (I would prefer you call me drakonzeta, though). Anyway, anyone know why SSA is not a theorem?  I've proved it using the law of Sines, and It is just as useful as SSS, SAS, ASA, and AAS.  Quick demonstration of my proof.  Feel free to criticize, complement, or overall review it.
1. Triangles ABC and DEF with <D congruent to <A, BC congruent to EF and AC congruent to DF.
2. mEF=mBC mAC=mDF.
3. Sin<A=Sin<D
4. Sin<A/mBC=Sin<B/mAC.  Sin<D/mEF=Sin<E/mDF.
5. Sin<E/mDF=Sin<A/mBC=Sin<B/mAC (Just compiled Transitive and Substitution
 into one step. In an official proof, that wouldn't happen.
6. Sin<E/mDF=Sin<B/mDF
7. Sin<E=Sin<B
8. <E is congruent to <B
9. Triangle ABC is congruent to triangle DEF

The problem is that your statement 7 does not imply statement 8. If $ \sin\angle E \equal{} \sin\angle B$, then you either have $ \angle E\equal{}\angle B$ OR $ \angle E \equal{} 180 \minus{} \angle B$. So, in fact, there may be two possibilities of a triangle ABC if you know BC, AC, and $ \angle A$.

Oh, OK.  Didn't think about that.  Oh, well I was looking for constructive critizism.  I'm new to AoPS (The time that the account has been on here is not accurate!  I didn't use it until now!) and I'm just learning.  Thanks for bringing that to my attention.  *Thinks about ways to prove it if m<E=180-m<B*

SSA only works for right angled triangles, and that follows from adding the limitation to your proof.

I cam to that conclusion relatively quickly. Oh well.  Anyway...

SSA also works here:

If $ AC \equal{} DF$, $ BC \equal{} EF$, $ \angle A \equal{} \angle D$, and $ AC \le BC$, then $ \triangle ABC\cong\triangle DEF$.

Also here:

If $ AC \equal{} DF$, $ BC \equal{} EF$, $ \angle A \equal{} \angle D$, and $ \angle A \ge 90$, then $ \triangle ABC\cong\triangle DEF$.

That requires additional limitations to the proof, but, it's Ok.  I was looking for some constructive critism.  I've got a class in a little bit. Bye.


find all increasing functions(not strictly!)  $f: N\to N$ with$f(n)+f(n+1)=f(2n)$ and $f(n)*f(n+1)=f([n^{2}/2])$ for every natural number n.

n=1 give f(1)+f(2)=f(2), therefore f(1)=0, but $0\not \in N$.

i am very sorry.of course i wanted to say $[n^{2}/2]$

It is easy to chek, that $f(n)=[\frac{n}{2}]$ is solution.

it is indeed very easy to check that $f(n)=0$ and $f(n)=[n/2]$ are solutions (and they are the only solutins) but i think it isn't so easy to prove  :(

I don't know why $f(2)=1$ gives $f(n)=[\frac{n}{2}]$? :(

to rust:$f(n)=0$ is easy to find and to prove but the hard part is to prove $f(n)=[n/2]$.....


Find all $ m$ so that equation $ x^4\plus{}2x^2\plus{}2mx\plus{}m^2\plus{}2m\plus{}1\equal{}0$ have at leat a root!

Ehhhh I'm not sure if this is right... (I'll point out the loophole in the proof)

Regard the equation as a function of $ a$. Then we have a quadratic.

The determinant:

$ 4(x \minus{} 1)^2 \minus{} 4(x^2 \plus{} 1)^2 \geq 0$
$ x(1 \minus{} x)(x^2 \plus{} x \plus{} 2) \geq 0$

The third element in the product is never zero, so the inequality holds when $ 0 \leq x \leq 1$

LOOPHOLE: perhaps there is a critical point within this range.

Proof of Loophole:

$ \frac{df}{dm} \equal{} 2m \plus{} 2x \plus{} 2$
Critical point when $ m\plus{}x\plus{}1\equal{}0$. As x varies from $ 0$ to $ 1$, $ m$ takes on values from $ \minus{}1$ to $ \minus{}2$, which is congruent to the range we derive below.

Or else:

When $ x \equal{} 0$, $ m \equal{} \minus{} 1$
When $ x \equal{} 1$,$ m \equal{} \minus{} 2$

Thus $ \minus{} 2\leq m \leq \minus{} 1$

[quote="iniquitus"]When $ x \equal{} 0$, $ m \equal{} \minus{} 1$
When $ x \equal{} 1$,$ m \equal{} \minus{} 2$

Thus $ \minus{} 2\leq m \leq \minus{} 1$[/quote]
Why?

[quote="Altheman"]Solve for $ m$ instead of $ x$...[/quote]
Could you please post your solution?


HMMT MIT college Harvard Princeton trigonometry Duke

GUYS I exist. Let's go to pumac and hmmt. Particularly the latter. Who is going to deal with them?

I'll work on figuring something out for HMMT.  Some basics:

It's February 7 at MIT (not that Harvard is that far away from MIT).  A flight is going to be at least 250 USD/person, and train fare would be about 200 USD, so I presume that driving is our best bet.  (I presume that gas from here to there is less than 1500-ish USD.)  The drive is about 12 hours straight, so if we leave on Friday afternoon at 4-ish we can expect to arrive Saturday morning at 6-ish.  Based on previous years, we need to be there by 8:30 or so.  The competition ends at 5:00 or so, so we could be back by 8 Sunday morning.  We'd need two drivers, and we could almost certainly take only one team.  Assuming 10 MPG, the cost would be the 10 USD/student registration fee plus about 40 USD for gas, meaning a combined cost of about 50 USD if we register by Dec. 1.

I don't think you realize how far away Boston is and how bad the traffic is in New England. I would be far more interested in flying. If we get stuck driving, you need to take into account worst-case traffic delays (even without meals, this kills the extra 2 hours and more), gas, toll roads (google maps lists 4), meals, bleh. If you don't like flying or driving, there's always this: http://maps.google.com/maps?f=d&saddr=Durham,+NC&daddr=Boston,+MA&hl=en&geocode=&mra=ls&dirflg=w&sll=41.536695,-72.53191&sspn=1.813213,3.515625&ie=UTF8&ll=40.413496,-74.223633&spn=14.740473,28.125&z=5, which costs considerably less.

Very nice.......we'll meet you there, OK?

Seriously, if we're considering flying, let's go by train instead.  It's >50 USD cheaper, plus we end up in the Back Bay and that's a nifty little place.  I know my way around Boston well enough to get us to MIT.

If we go Amtrak, it's a total cost of 208 USD/person.  Plus we can leave at 11:06 AM on Friday, arrive at 7:52 AM Saturday, compete, leave at 9:50 PM Saturday, and return at 4:42 PM Sunday.  We hang around in DC for about five hours in the middle of the trip.  It's about a 35-minute walk from the stop to MIT, so we should be able to make it assuming the train is on time.  That I'll have to look in to.  If we don't want to cut it that close, we could leave Raleigh at 8:54 AM and arrive at 12:13 AM Saturday, but we'd have to hang out around MIT.  I can get you all into some buildings where it's a little easier to sleep, and where the security guards won't bother you, so that shouldn't be too much of a problem.  Also, you can get a hot breakfast in the building I have in mind, so that's a definite plus.

Ummm. If you're determined enough to deal with this, then figure out who's going and poll them on what they want. Personally, I am always partial to flying and spending the night (and not in a soup kitchen), and depending on how college visitations are working, I may end up doing that regardless of our plans.

I would like to go probably, but first I'm gonna have to find the money :maybe:

I'm definitely interested in going to both.

For HMMT, I don't think trains / driving are very realistic.  Riding on a train all night won't leave you even close to rested, and it is a long time.  For the extra $ \$$ 50, it's probably worth it to fly.

I might be able to go, especially if I use the opportunity to tour Harvard / MIT at the same time.

I'd be interested in going. As far as transportation goes for Boston, I think flying then taking the train to get around is the best way to go. @Teki, Don't you think driving is really annoying around Boston?

Only having been to MA, I wouldn't know but yea I don't imagine it being much fun.

If we fly in, then we can take the T (is that what you meant, Tenoreoz?) around Boston.  I can give you all a semi-informal tour of MIT if you want, and I suppose we could walk around Harvard afterwards.  From the train station, the T would barely help in terms of time.  The only reason I don't advocate flying in is that in terms of schedules, there are no flights from here to Boston that arrive early Saturday morning.  As for sleeping possibilities, I think that b-flat underestimates how hard it can be to sleep on a plane.

Also, I've driven to Boston, so I know how the driving can be.  That said, it was during the summer.  I'm not sure how thousands of college students would affect that.

@jb05: I like the visitation idea, so long as it happens after HMMT.  I've been to both schools before, but only during the summer.  I'd like to see what the campuses are like when there are a lot more students around.

Dude. If you want to go to HMMT, you'd better be ready to spend some money. By which I mean plane tickets AND lodging. We fly in Friday, spend the night, and fly back Saturday after the contest. Please be reasonable.

Regarding visitations, last year I spent Thursday-Sunday looking around Cambridge. Unfortunately, it wasn't formal at all, so I'm still going to need interviews and such. It's definitely an idea I would advocate if you are trying to decide where to go to school.

[quote="jb05"][quote="Tenoreoz"]@Teki, Don't you think driving is really annoying around Boston?[/quote]

In somewhat more generality, don't you think driving is really annoying around New England?[/quote]

In even more generality, don't you think that driving itself is annoying?

Not necessarily.  It depends both on the journey and the destination.

I'll spend money on HMMT if PuMAC is cheap, which it should be, given how it was last year.

HMMT attempt failed last year.
BUT I WANNA GO!!!!!
and flying is definitely more realistic than driving or train, as the latter two will give you ~2 hours of sleep that night when you need ~7

[quote="math=fun"]I want to go to HMMT too!
We should form teams, go as a group together, and we need get at least two adults/teachers to go with us.  Maybe we can meet at somewhere or be on AOPS at the same time and talk about going as a group.
If this team thing work out it would be great, because I want to go with somebody I know!![/quote]

It appears that I was missed in the discovery of psychic powers of somehow knowing who you are with 1 post...

I think that jb's plan works best, considering how everyone seems to want to take a look around the place. So we fly in early Friday I'm guessing?

@Teki, yea the T was what I meant. Is Harvard really not that bad of walking distance from MIT? Last time for me, taking the T to get to Harvard wasn't exactly that short of a ride if I remember correctly.

[quote="PI-Dimension"]It appears that I was missed in the discovery of psychic powers of somehow knowing who you are with 1 post...[/quote]

Oh man. Parsing this sentence is loads of fun.

@Tenoreoz: Harvard and MIT only have a stop between them on the metro. It's a really short ride. The walk is definitely a walk, but it's reasonable. I seem to recall very roughly half an hour to an hour. I am too lazy to pull up google maps.

Are there any coaches going? (Maybe I could get Noland, or we could get Goebel or Rash again)

I mean, Princeton at least is driving distance, but HMMT is a bit far. Ugh...

I could talk to Rash if we decide to ask him.

On a related note, do you guys also want to do iTest again like last year? It seems that they've disallowed teams from multiple states this year...not that it would affect us in anyway.

hmm yeah I can sign us up for that. who's interested? jb05, b-flat, teki-teki, tenoreoz, and I?

Maybe, maybe not. Depends on how much time I have to waste. Do someone talk to coaches, though.

Given my light schedule this year (compared to last, though last year's wasn't exactly heavy...), I should be able to do iTest.

I'll be able to make some time for iTest so I should be able to do it too.

As for PuMAC/HMMT, I'll talk to Rash about it this week. So we want to go to both this year correct?

So I kind of need to know whether this is happening in case I need to find another team.

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1255784#1255784[/url]

EDIT: Damn. This post is no longer relevant cause I was misinformed

However, guys. Get your affairs in order.

From MIT to Harvard is probably a 20 minute walk, maybe 30 if you walk slow.  It's about the same in the other direction as well.

@John: if you still want to do the college visit thing, here's a plan that my parents seem to like: we could fly up Thursday afternoon, spend Thursday night in a local hotel, spend Friday at Harvard and Friday night in a Harvard dorm, and then head to MIT on Saturday for the tournament.  Saturday night could be spent in an MIT dorm and you could tour MIT on Sunday.  (If we do this, I might head out after HMMT on Saturday as I've seen a lot of MIT already.)

I will deal with my travel plans myself.

My question is whether anybody is making an effort to get a team together.

I talked to Rash a while back and he said he would talk to Goebel about it (help with travel expenses, a coach, etc...). We just to need to finalize team members and we'll be set for registration.

Guys come to MIT it is awesome!  Don't go to Harvard  :wink:.  Feel free to drop in on me and Ben if you ever come to Boston.

i'm interested in joining. when are the teams going to be finalized?

@Tenoreoz

In order of importance:

1. Has Goebel or anyone else actually definitively agreed to coach?
2. Do we have transportation, and what is it?
3. Do we have any financial aid?

If yes, I would like to be in contact with possible team members and rescue people (e.g., dnkywin and me) from the realm of utter confusion.

@jb

I haven't talked to him about this in a while, but here's what he said to me before. I'll talk to him soon to get an update.

1. He said he would try to find someone to coach similar to last year. He seemed certain about getting one for PuMAC but I'm not sure about HMMT.
2. Didn't discuss this much, but it seems like flights are the way to go to not get too behind on schoolwork and such.
3. He mentioned us receiving some financial aid from NCCTM, but I don't know how much.

Ok that works out a lot better. The drive is no worse than the penn state drive, so flying vs. driving is mostly a question of finances and the willingnes of the coach to drive for a long time. I guess I will talk to Mr. Goebel.

dude guys I exist.

And I could get Noland (seeing not_trig chickened out of being with the underdogs... I had to explain like for 10 minutes what balls and urns was in our first meeting???)

Guys, if we still wants to go to HMMT, we should think about begin to register.  Has anyone gotten more imformation from the coaches or who's going with us?

Bleargh I suspect HMMT is in the slightly distant future; we do have a PUMaC team decided though, probably

in other news Duke is next month. darn? I should not fail this year :(

Let's make this clear. We DO NOT have a pumac team until people start doing things. There's this strange consensus going around that something is happening which is not.

Duke is Nov. 15.  I plan to get to the tiebreaker round and then fail, as in the past two years.

I will go to PuMAC if we drive, and I have no preference between not staying in a hotel/spending the night.

So that makes John and me and who else?  Patrick?  Pom?  Damien?  Bryce?  Kevin?  We'd need what, eight people?

We also need at least 2 coaches, I believe.  I assume Rash will come if we organize it: Damien/Pom/Patrick, you need to ask him.  Goebel may go; the same with Noland.  I can ask Noland; somebody needs to contact Goebel.  Patrick/John, can you do this?

whoa whoa dude did you see the email? it has all the team members etc

Please just respond to the email. I talked to Mr. Rash and was planning to ask Goebel as soon as a nontrivial set of people responded.

what should I do if I want to go to the princeton contest with you all?


In triangle ABC, 3sinA+4cosB=6 and 4sinA+3cosA=1. Then what is angle C in degrees?

[quote="The QuattoMaster 6000"][hide]
Say sinA=x.  Since $x^{2}=1-(cos^{2})A$, we have that (from the second equation), $4x+3\sqrt{1-x^{2}}=1$.  Solve for x and get that $x=\frac{4+6\sqrt{6}}{25}$ (discard the minus in the quadratic since 0<x<180).  Plug in this value to the first equation and get that $cosB=\frac{138-18\sqrt{6}}{100}$.  Find Angle C by taking 180-(cos^-1)B-(sin^-1)A. [/hide][/quote]

That's pretty messy, is there a prettier way? or is the answer just ugly?


Prove that the sum of 2 odd number's squares  can not be a perfect  square.

[hide]Since the square of an odd integer is congruent to $ 1\bmod 8$, the sum of two odd squares is congruent to $ 2\bmod 8$. But $ 2$ is not a quadratic residue $ \mod 8$.[/hide]

[quote="Elio (n)"]I have a question :

Why  (mod 8) ? Why not (mod 4) ?[/quote]
I think the probelm can be proved by using mod 4.

[hide="This may be true"]Since $ x^2\equiv 1\pmod4$ if $ x$ is odd.

sum of square of any two odd numbers $ x^2\plus{}y^2\equiv 2\pmod4$ if $ x, y$ are odd.Say ,$ 4p\plus{}2, p\equal{}0,1,2,3....$

If $ 4p\plus{}2$ is written as a perfect square $ m^2$, m cannot  be an odd and should be even ,say $ 2k$

i.e. $ 4p\plus{}2\equal{}(2k)^2\equal{}4k^2 \implies 2p\plus{}1\equal{}2k^2$ , it is also impossible . Since any odd $ \not \equal{}$an even .[/hide]

[hide="solution"]$ (2m \plus{} 1)^2 \plus{} (2n \plus{} 1)^2 \equal{} 4m^2 \plus{} 4n^2 \plus{} 4m \plus{} 4n \plus{} 1 \plus{} 1$. A square must be in the form $ (2n)^2$ or $ (2n \plus{} 1)^2$, and they are in the form $ 4n^2$ and $ 4(n^2 \plus{} n) \plus{} 1$, or it's a multiple of four or one more than a multiple of four. $ (2m \plus{} 1)^2 \plus{} (2n \plus{} 1)^2$ is neither. Thus it is never a square.[/hide]

[quote="Elio (n)"]I have a question :

Why  (mod 8) ? Why not (mod 4) ?[/quote]

mod 4 also works since mod 4 is a subset of mod 8. Actually since every odd square is 1 mod 8, they are also 1 mod 4. And since 2 is not a quadratic residue mod 8 it isn't a quadratic residue mod 4 either.

[quote="nayel"]And since 2 is not a quadratic residue mod 8 it isn't a quadratic residue mod 4 either.[/quote]  No, this implication does not hold.  For example, 5 is not a quadratic residue mod 8, but 5 is a quadratic residue mod 4.


Find the number of nine digit numbers that are integral multiples of 11 if no digit is equal to zero and no two digits are alike.

[hide]The number that satisfy these conditions has (120)(24) = 2880 permutations including itself. I don't know what to do next.[/hide]

For a multiple of 11, you want the sums of the digits in the odd and even positions to differ by a multiple of 11 (meaning 0, 11, 22, ...). For example, the 2nd, 4th, 6th and 8th digits could be taken from the set {2, 3, 4, 8}, and the remaining digits would come from {1, 5, 6, 7, 9}. Once you have a choice of sets, you can calculate permutations, like you obviously figured out, but there is more than one choice of set pairs.
Hint:
[hide]Since the choices are not too many, you can list them.
The sum of all nine digits is 45. Let S, and s be the largest and smallest of the sums of the digits in the odd and even positions. You can set systems of equations to figure out what values S and s can take.[/hide]


group theory
abstract algebra
linear algebra
matrix
calculus
integration
superior algebra

Here is a great result that was comunicated by a friend: prove that any subgroup of $GL_n(Q)$ is conjugated with a subgroup of $GL_n(Z)$.

Which subgroup of $GL_n(\mathbb{Z})$ is $GL_n(\mathbb{Q})$ conjugated to?

Please reply!
Do you mean finite, do you mean proper?

Sorry, I wasn't able to enter mathlinks this week. I think it is finite, but I will try to check and see.

i'm sorry but GLn(Z) is that the group of integer matrices with integer inverse

(because invertible integer matrices don't necessarily have integer inverses)

i thought that group would be the group of integer matrices with determinant 1 or -1

now if any subgroup of GL(Q) is conjugated with one of GL(Z) wouldn't that imply that every rational invertible matrix is congruent with a matrix with determinant 1 or -1

while conjugation preserves determinant??

i'm sorry i'm just trying to understand the problem [/i]

Then try to solve it for finite subgroups of $GL_n(\mathbb{Q})$ since if $A$ is a matrix in a finite subroup of $GL_n(\mathbb{Q})$ then $A$ has finte order which quickly leads to $\det A=\pm 1$.

Or try to solve the original problem for $SL_n(\mathbb{Q})$ and $SL_n(\mathbb{Z})$.

I'm pretty sure the groups under consideration should be finite. This is the version I'll try to prove anyway.

Let $G\le Gl_{n}(\mathbb Q)$ be a finite group, and let $e_{1},\ldots,e_{n}$ be the standard basis of $\mathbb R^{n}$. Now let $\Lambda$ be the subgroup of $\mathbb R^{n}$ generated by the finite system of elements of the form $ge_{i},\ i=\overline{1,n},\ g\in G$. $\Lambda$ is a subgroup of $\frac{1}{m}\mathbb Z^{n}$ for some positive integer $m$ (we can take $m$ to be the product of all the denominators of the entries in the elements $g$ of $G$, for example), so it's a discrete subgroup of $\mathbb R^{n}$; this implies that it's free abelian. On the other hand, since it contains $\mathbb Z^{n}$, its rank must be $n$.

Now let $(f_{i})_{i=1}^{n}$ be a basis for $\Lambda$. It is also a basis for $\mathbb R^{n}$ and each $g\in G$ sends each $f_{i}$ into $\Lambda$, which consists of linear combinations of the $f_{j}$'s with integer coefficients. This means that the matrices corresponding to the elements of $G$ with respect to the basis $(f_{i})_{i}$ have integer entries, and we're done.

grobber,
I think you are right
[quote="grobber"]I'm pretty sure the groups under consideration should be finite... [/quote]
Do you have an example of non finite subgroup of $\mbox{GL}_{n}(\mathbb{Q})$ which is not conjugated with a subgroup of $\mbox{GL}_{n}(\mathbb{Z})$?

Any group which contains a matrix with non-integral trace (because conjugation preserves trace) or non-integral determinant. In particular, $Gl_{n}(\mathbb Q)$ itself.

Quite obvious, wasn't it? :)

Do you mean to say that given any subgroup H of GLn(Q) there is a subgroup H' of GLn(Z) and an inner automorphism of GLn(Q) taking H onto H' ?


inequalities
invariant
geometry proposed
geometry

We well know the nice inequality
Given triangle for all $M$ then
$\frac{MA}{a}+\frac{MB}{b}+\frac{MC}{c}\ge\sqrt{3}$
More we can prove it with weight
For all $x,y,z>0$ and for all $M$ prove that
$(y+z)\frac{MA}{a}+(z+x)\frac{MB}{b}+(x+y)\frac{MC}{c}\ge 2\sqrt{xy+yz+zx}$
Is it nice ?  :D

[quote="gemath"]
Given triangle for all $M$ then
$\frac{MA}{a}+\frac{MB}{b}+\frac{MC}{c}\ge\sqrt{3}$
[/quote]
Where is it? I need it, before slove your problem  :wink:

It's very nice one gemath,but I think it still work with an old method  :wink:

It is my own, are you clear N.T.TUAN ?  :D  :D

No no I do not think that the inequality $\sum \frac{MA}{a}\geq \sqrt{3}$ is yours. It's too old and known one. I also post it in the forum. It's also Vietnam TST 2003, but I think this must be appear sooner than 1998.
By the way let me re-call a question I posted long time ago, but I don't remember that if anyone solve it or not:
Let $M,A,B,C$ are on a circle. Prove that $\sum \frac{MA}{a}\geq2$ and $\sum \frac{MA^{2}}{a^{2}}\geq2$

[quote="N.T.TUAN"][quote="gemath"]It is my own, are you clear N.T.TUAN ? [/quote]
We can slove problem follow
Let $x,y,z$ are positive reals and triangle $ABC$, Find $M$ such that $xMA+yMB+zMC$ is minimum.[/quote]

It does not help us much in solving the second inequality.

[quote="pluricomplex"]

It does not help us much in solving the second inequality.[/quote]
But If use it , We can prove second inequality. I think  :wink:

[hide=" to pluricomplex"]Dạo này thế nào hả Hưng? Trang này hay phết nhể? :P [/hide]

Hi some Vietnamese!  :D
The first, I said it is my own it means the inequality
$\sum (y+z)\frac{MA}{a}\ge 2\sqrt{xy+yz+zx}$ is my own ! Are you clear two mans? And the inequality $\sum \frac{MA}{a}\ge \sqrt{3}$ I think some children can do it (and they can know it isn't my own!  :D )

The second, my problem isn't base on find min $\sum xMA$ because it is simple to see in the problem finding min the triangle is invariant, but my problem is for arbitrary triangle!

And the last, the best you should post all solution if you thought it is too easy :D:D
[quote="pluricomplex"]I think it still work with an old method  :wink:[/quote]
[quote="N.T.TUAN"]
But If use it , We can prove second inequality. I think  :wink:
[/quote]

and [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=103823]here[/url] is the problems that pluricomplex said!

[quote="gemath"]Hi some Vietnamese![/quote]
Yes, Hi some London  :D

[quote]
The second, my problem isn't base on find min $\sum xMA$ because it is simple to see in the problem finding min the triangle is invariant, but my problem is for arbitrary triangle!
[/quote]

You are not read my posts!

[quote="gemath"]Maybe I didn't read clearly! But now stop spams! Please!  :D[/quote]
OK!

To Mods: I am sorry!  :(

Given $ABC$ and $A'B'C'$ then

$\sum a' \frac{MA}{a}\geq 2\sqrt{r'(4R'+r')}$

Replace $a' = y+z, ...$  :P

That's right treegoner it is my ideal but actually there is another way to prove it.
But now pluricomplex, I think some old methods of you are so useful that A week elapse and there isn't any solution!  :D  :D
[quote="pluricomplex"]I think it still work with an old method  :wink:[/quote]

[color=purple]i have a way to the solution gemath i haven't tried it yet but i post it and i hope you can use it for the 2nd problem
  Let K is the point in the plane such that$(y+z)\vec{KA}+(z+x)\vec{KB}+(x+y)\vec{KC}=\vec{0}$.And you prove similar than the 1st problem
 P/s:gemath you are from London do you like Arsenal "The gunner" ? :D  [/color]

[quote="evarist"]you can use it for the 2nd problem
  Let K is the point in the plane such that$(y+z)\vec{KA}+(z+x)\vec{KB}+(x+y)\vec{KC}=\vec{0}$.
 [/quote]
For non wordy if you think you can do and "your old method" is better than  pluricomplex then post your solution please, or else you shouldn't spam. :D
[quote="evarist"]And you prove similar than the 1st problem
  [/quote]
And do you know the way that I do the first ? :D
It is more simple than I can think it is for Vietnam TST 2003
$(\sum\frac{MA}{a})^{2}\ge 3\sum\frac{MBMC}{bc}\ge 3$
by the way I like M.Utd  :D

[color=purple]We have $a.m_{a}\leq \frac{a^{2}+b^{2}+c^{2}}{2\sqrt{3}}thus a GA\leq \frac{a^{2}+b^{2}+c^{2}}{3\sqrt{3}}$
 It's easy to check that $\frac{MA}{a}=\frac{MA.GA}{aGA}\geq \frac{\vec{MA}\vec{GA}}{aGA}\geq \frac{3\sqrt{3}GA^{2}}{a^{2}+b^{2}+c^{2}}+3\sqrt{3}\frac{\vec{MG}\vec{GA}}{a^{2}+b^{2}+c^{2}}$
  With two similar problem we have the proff for the 1st problem.So i [u][b] hope[/b][/u] the 2nd problem can solve if you use the lemma
  $\sum (x+y)\vec{KA}=\vec{0}$
  I hope and i can't solve ? What about you gemath?[/color]

[quote="gemath"]I known that solution! But obvious we can't have any relation between $K$ you defined and the inequalities you used for $G$. Therefore I guess we can't do similarly, but if you thouht so why don't you try it. And you shouldn't say that is a lemma, it is only definition for a point :D
By the way treegoner have solution for this problem, I will ask him to post it here ![/quote]
 [color=purple] Ok i'll waiting for you.Thanks very much if you post it early! [/color]


How to simplify this:

[img]http://tutorial.math.lamar.edu/AllBrowsers/3401/Linear_files/eq0040M.gif[/img]

And solve the equation?

hello, i can not see any linear differential equation.
Sonnhard.


Hey!

Does anybody have any problems simmilar to A3 from Vietnam 1994        ( http://www.kalva.demon.co.uk/vietnam/viet94.html ) .. or knows any place I could find some like that! I would really need some... 10x... :D

cheers! :D

by the way... can someone post a solution for that problem? :D

cheers

Define the sequence x0, x1, x2, ... by x0 = a, where 0 < a < 1, xn+1 = 4/p2 (cos-1xn + p/2) sin-1xn. Show that the sequence converges and find its limit.

The limit of x(n) is 1. (Showing that x(n) is a bounded monotonic increasing sequence and 1 works as fixed point).

And more precisely the limit of (1-x(n))/(1-x(n+1))  is Pi^2/8

Starting with x(0)=1/2, the limit of (1-x(n))*(Pi^2/8)^n=1.13310671311454040122106763765010355585124640491698637889408425136093959115463...is mysterious!

B.C.

I suddenly became interested in this problem! :D :D

how do you actually prove that The limit of x(n) is 1. (Showing that x(n) is a bounded monotonic increasing sequence and 1 works as fixed point). ?

bye bye!!


geometry
rectangle
search
3D geometry
prism

Consider a rectangle l units long, and w units wide and the lw unit squares found in it. A diagonal is drawn from one corner to the opposite. The diagonal cuts through certain unit squares. Find in terms of l and w the number of unit squares that are cut through.

There is a formula. If the length is $ l$ units long, and the width is $ w$ units long, then the number of squares the diagonal cuts through is \[ l\plus{}w\minus{}GCD(l, w)\]

You can experiment with smaller $ l$ and $ w$, and see if you get the same answer the formula gives.

ah. I don't know how to prove it :(

there are several threads about this type of problem.

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1773065499&t=117544]here[/url], [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=572521091&t=178840]here[/url], [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1622578094&t=269198]here[/url], and [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1176704#1176704]here[/url]

For a cube, the number of cubes the diagonal passes through is the side length of the cube. I'm still working on trying to find a general formula for a prism :fool: Fail visualizing


modular arithmetic
number theory proposed
number theory

Let $n$ be a positive integer. Prove that the equation
\[x+y+\frac{1}{x}+\frac{1}{y}=3n\]
does not have solutions in positive rational numbers.

Just to add, that this comes from Baltic way 2002 contest.
SEE [url]http://www.math.su.se/bw2005/history.shtml[/url]

Let $x=\frac{a}b,y=\frac{c}d$ where $(a,b)$ and $(c,d)$ are pairs of coprime natural numbers. The equation becomes $\frac{a}b+\frac{c}d+\frac{b}a+\frac{d}c=3n$, or $(a^2+b^2)cd+(c^2+d^2)ab=3nabcd$. So $ab|cd$ and $cd|ab$, which implies $ab=cd$. Exactly 0 or 2 of $a,b,c,d$ are multiples of 3, so $a^2+b^2+c^2+d^2\equiv1\text{ or }2\pmod3$. But we have that $a^2+b^2+c^2+d^2=3nab$, which is impossible since $3\not|a^2+b^2+c^2+d^2$.


An odd integer $ n \ge 3$ is said to be nice if and only if there is at least one permutation $ a_{1}, \cdots, a_{n}$ of $ 1, \cdots, n$ such that the $ n$ sums $ a_{1} \minus{} a_{2} \plus{} a_{3} \minus{} \cdots \minus{} a_{n \minus{} 1} \plus{} a_{n}$, $ a_{2} \minus{} a_{3} \plus{} a_{3} \minus{} \cdots \minus{} a_{n} \plus{} a_{1}$, $ a_{3} \minus{} a_{4} \plus{} a_{5} \minus{} \cdots \minus{} a_{1} \plus{} a_{2}$, $ \cdots$, $ a_{n} \minus{} a_{1} \plus{} a_{2} \minus{} \cdots \minus{} a_{n \minus{} 2} \plus{} a_{n \minus{} 1}$ are all positive. Determine the set of all `nice' integers.

I think there is a small typo and $ a_{2} \minus{} a_{3} \plus{} a_{3} \minus{} \cdots \minus{} a_{n} \plus{} a_{1}$ is really meant to be $ a_{2} \minus{} a_{3} \plus{} a_{4} \minus{} \cdots \minus{} a_{n} \plus{} a_{1}$.

Suppose that $ n$ is nice and let $ A \equal{} (a_1, \ldots, a_n)$ be a witness to the niceness of $ n$.  Any cyclic permutation of $ A$ is a witness, as well, so we may assume that $ a_1 \equal{} 1$.  Then we have that $ a_1 \minus{} a_2 \plus{} a_3 \minus{} \ldots \minus{} a_{n \minus{} 1} \plus{} a_n \equal{} 1 \plus{} (a_3 \plus{} a_5 \plus{} \ldots) \minus{} (a_2 \plus{} a_4 \plus{} \ldots)$ and $ a_2 \minus{} a_3 \plus{} a_4 \minus{} \ldots \minus{} a_n \plus{} a_1 \equal{} 1 \plus{} (a_2 \plus{} a_4 \plus{} \ldots) \minus{} (a_3 \minus{} a_5 \minus{} \ldots)$ are both positive, so in fact $ a_2 \plus{} a_4 \plus{} \ldots \equal{} a_3 \plus{} a_5 \plus{} \ldots$.  This implies that $ 1 \plus{} 2 \plus{} \ldots \plus{} a_n \equal{} 1 \plus{} a_2 \plus{} a_3 \plus{} a_4 \plus{} a_5 \plus{} \ldots \plus{} a_n \equal{} 1 \plus{} 2(a_2 \plus{} a_4 \plus{} \ldots)$ is odd and so $ n \equiv 1 \pmod 4$.

To finish the problem, it suffices to construct a permutation of the desired sort for these $ n$, but unfortunately I haven't managed to do this yet  :oops:.

[b]Construction[/b]:$ (a_1,...a_n) \equal{} (1,3,5,...,4k \plus{} 1,4k,4k \minus{} 2,...,2)$
Easily to check this permutation verify all of conditions :)


If $ \alpha_1,\alpha_2,\alpha_3$ are the measures of angles in the triangle $ A_1A_2A_3$ prove the inequality:
\[ \frac{20}{8}<\sum_{i\equal{}1}^3 [ \sin^4 \frac{\alpha_i}{4} \plus{} \cos^4 \frac{\alpha_i}{4}]<\equal{}\frac{21}{8}\].

What's the source of the problem?

This problem has been posed in Japanese university entrance exam.

[color=blue]1984 Toyama University entrance exam/Pharmaceutical[/color]

Suppose that positive real numbers $ \alpha ,\ \beta ,\ \gamma$ satisfy $ \alpha \plus{} \beta \plus{} \gamma \equal{} \pi$.

(1) Prove that $ \cos \alpha \plus{} \cos \beta \plus{} \cos \gamma \equal{} 1 \plus{} 4\sin \frac {\alpha }{2}\sin \frac {\beta }{2}\sin \frac {\gamma}{2}$.

(2) $ \frac {5}{2} < \sin ^ 4 \frac {\alpha}{4} \plus{} \cos ^ 4 \frac {\alpha}{4} \plus{} \sin ^ 4 \frac {\beta}{4} \plus{} \cos ^ 4 \frac {\beta}{4} \plus{} \sin ^ 4 \frac {\gamma}{4} \plus{} \cos ^ 4 \frac {\gamma}{4}\leq \frac {21}{8}$.

Notice that you may use $ 0 < \sin \frac {\alpha}{2}\sin \frac {\beta}{2}\sin \frac {\gamma}{2}\leq \frac {1}{8}$.

The source for this problem is Gazeta Matematica.Another nice proof for this problem you could see here:
[hide]
\[ F(\alpha_1,\alpha_2) \equal{} \prod_{i \equal{} 1}^3 \cos \alpha_i \equal{} \cos \alpha_1 \plus{} \cos \alpha_2 \plus{} \cos (\pi \minus{} \alpha_1 \minus{} \alpha_2) \equal{} \cos\alpha_1 \plus{} \cos\alpha_2 \minus{} \cos(\alpha_1 \minus{} \alpha_2)
\]
We have the sistem:
\[ \frac {\partial F}{\partial \alpha_1} \equal{} \minus{} \sin \alpha_1 \plus{} \sin(\alpha_1 \plus{} \alpha_2) \equal{} 0
\]

\[ \frac {\partial F}{\partial \alpha_2} \equal{} \minus{} \sin \alpha_2 \plus{} \sin(\alpha_1 \plus{} \alpha_2) \equal{} 0
\]
from where $ \sin \alpha_1 \equal{} \sin \alpha_2$ therefore $ \alpha_1 \equal{} \alpha_2$ or $ \alpha_2 \equal{} \pi \minus{} \alpha_1$. The second solution implying $ \alpha_3 \equal{} 0$ must be excluded. It results imediately that $ \sin^2 \alpha_1 \minus{} \sin \alpha_1 \equal{} 0$,$ \sin \alpha_1 ( \cos \alpha_1 \minus{} 1) \equal{} 0$ from where $ \alpha_1 \equal{} \frac {\pi}{3},\alpha_2 \equal{} \frac {\pi}{3},\alpha_3 \equal{} \frac {\pi}{3}$.The point $ (\frac{\pi}{3},\frac{\pi}{3})$ will be maximum by a similar algorithm with the one from $ \prod_{i \equal{} 1}^{3} \sin \alpha_1$.Puting $ \alpha_1 \equal{} \frac {\pi}{3} \plus{} \epsilon$,$ \alpha_2 \equal{} \frac {\pi}{3} \plus{} \epsilon$ we will estimate
\[ F(\frac {\pi}{3} \plus{} \epsilon,\frac {\pi}{3} \plus{} \epsilon) \equal{} 2\cos(\frac {\pi}{3} \plus{} \epsilon) \minus{} \cos(\frac {2\pi}{3} \plus{} \epsilon) \equal{} \cos \epsilon \minus{} \sqrt {3}\sin \epsilon \minus{} \frac {1}{2}(1 \plus{} 2\sin^2 \epsilon \minus{} 2\sqrt {3}\sin \epsilon \cos\epsilon) \plus{} 1 \equal{} \cos \epsilon \plus{} \frac {1}{2} \plus{} \frac {1}{2}(2\sqrt {3}\sin \epsilon \cos \epsilon \minus{} 2\sqrt {3}\sin \epsilon \minus{} 2\sin^2 \epsilon) < \frac {3}{2} \minus{} \frac {1}{2}\sin^2\epsilon
\]
It results that $ F(\frac {\pi}{3} \plus{} \epsilon,\frac {\pi}{3} \plus{} \epsilon) < F(\frac {\pi}{3},\frac {\pi}{3}) \equal{} \frac {3}{2}$ and therefore $ (\frac {\pi}{3},\frac {\pi}{3})$ is the maximum point,the unique one in the definition domain.[/hide]

[quote="kunny"]
This problem has been posed in Japanese university entrance exam.

[color=blue]1984 Toyama University entrance exam/Pharmaceutical[/color]

Suppose that positive real numbers $ \alpha ,\ \beta ,\ \gamma$ satisfy $ \alpha \plus{} \beta \plus{} \gamma \equal{} \pi$.

(1) Prove that $ \cos \alpha \plus{} \cos \beta \plus{} \cos \gamma \equal{} 1 \plus{} 4\sin \frac {\alpha }{2}\sin \frac {\beta }{2}\sin \frac {\gamma}{2}$.

(2) $ \frac {5}{2} < \sin ^ 4 \frac {\alpha}{4} \plus{} \cos ^ 4 \frac {\alpha}{4} \plus{} \sin ^ 4 \frac {\beta}{4} \plus{} \cos ^ 4 \frac {\beta}{4} \plus{} \sin ^ 4 \frac {\gamma}{4} \plus{} \cos ^ 4 \frac {\gamma}{4}\leq \frac {21}{8}$.

Notice that you may use $ 0 < \sin \frac {\alpha}{2}\sin \frac {\beta}{2}\sin \frac {\gamma}{2}\leq \frac {1}{8}$.[/quote]

Here is my solution in attached files.


$ a_1, a_2 , \cdots , \forall i,  j  \in Z_\plus{} (i  \neq  j ),0<a_i \leq m, \left| a_i \minus{}a_j \right| \geq \frac 1{i\plus{}j}, \Longrightarrow m \geq 1.$

[quote="jcc0107"]$ a_1, a_2 , \cdots , \forall i, j \in Z_ \plus{} (i \neq j ),0 < a_i \leq m, \left| a_i \minus{} a_j \right| \geq \frac 1{i \plus{} j}, \Longrightarrow m \geq 1.$[/quote]
IMO SL 2002 A2


Sa se rezolve in numere naturale ecuatia $\ [n^{\sqrt n}]=2n$, unde $\ [a]$ reprezinta partea intreaga a numarului real $\ a$.

Solutia este atasata!
Sorry de Latex! :D

da...exact asa ma gandeam si eu sa comparam cu n^2...Misto solutie:)

Din nou ceva cu parte intreaga...o problema a domnului profesor Adrian Zanoschi:
   Demonstrati ca pentru orice $\ n\in \mathbb{Z}$ exista numerele intregi $\ a,b$ astfel incat $\ n=[a\sqrt 2]+[b\sqrt 3]$.
$\text{Shortlist JBMO 2000}$ :D