1)Prove the ineq:

sum((log_k (k+1)^2)>=n^2/(n-1) for any natural n>=5. k=1,n;

2)Need a hint for the following problem:
Show that there is no x in [0,oo) so that:

(8^3)^x/(3^2)^x+(27^2)^x/(2^3^x)=12.

bye bye!

Your Hint Georg is here! :D :D
try to prove that
(8^3)^x/(3^2)^x+(27^2)^x/(2^3^x)>12 for any x>=0.

cheers! :D :D

HI! I see that for the 1st one you don't ask for a hint so I guess you want a solution! :)

Solution:

apply Cauchy for the ineq:

sum((log_k (k+1)^2)>=(sum(log_k (k+1)))^2/(n-1). and you
get that sum(log_k (k+1))>=n which is obious by induction! :D

cheers! :D :D


inequalities
search
inequalities theorems

Search harder.
http://en.wikipedia.org/wiki/Category:Inequalities
I think Karamata is explained quite well by darij somewhere on the forum. Use teh search function.

[quote="José"]I saw loads of times, in proofs, users using AM-GM, Cauchy (Cauchy-Schwarz?), Jensen, Karamata and Schur inequalities (e.g.:"applying AM-GM yields..."). Now, can anyone explain me what do all this ineq's state? I searched, but without success. Thanks.[/quote]

AM-GM: http://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means

Schur: http://en.wikipedia.org/wiki/Schur%27s_inequality

What I'm looking about Cauchy is not that ineq, is the ineq with real numbers ($a_{1},a_{2},...,a_{n},b_{1},b_{2},...,b_{n}$). Any link will be appreciated.

About Karamata: I'll search again.

Any help?

The version of Cauchy-Schwarz-Bunyakowski, as well as the versions of Minkowski, Holder, and Young's inequalities, on Wikipedia are exactly the inequalities you're looking for. Just use the standard Euclidean norm:
$f = (a, b, c)$
$\|f\| = \sqrt{a^{2}+b^{2}+c^{2}}$
etc.

Karamata: [url]http://www.mathlinks.ro/Forum/viewtopic.php?highlight=karamata&t=14975[/url]. Jensen is also described there.

Interesting topic: http://www.mathlinks.ro/Forum/viewtopic.php?t=65385

Sorry if I post old topics, but I'm trying to start solving inequalities. It seems really hard, but I'm trying to go, like the song, "little by little".


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

$ \sum(\sqrt{a}\plus{}\sqrt[3]{b^2}\plus{}\sqrt[6]{c^5})\leq2(a\plus{}b\plus{}c)\plus{}3$

[quote="filomen"]If $ a,b,c > 0$, then

$ \sum(\sqrt {a} \plus{} \sqrt [3]{b^2} \plus{} \sqrt [6]{c^5})\leq2(a \plus{} b \plus{} c) \plus{} 3$[/quote]

just need to prove

$ \sqrt {a} \plus{} \sqrt [3]{a^2} \plus{} \sqrt [6]{a^5} \le 2a\plus{}1,$

let $ a\equal{}x^6$,

$ \iff (x\minus{}1)^2 (2 x^4\plus{}3 x^3\plus{}3 x^2\plus{}2 x\plus{}1) \ge 0$


HMMT calculus integration logarithms probability geometry trigonometry

What did you all think about this year's HMMT?

I found the questions to be much more approachable on the two rounds I took (Algebra and Calculus) than last year, but I thought that took some of the fun out of the rounds as well. Our team died on the team round.

The guts round, as expected, was really fun - last year it was fun for me because it was my first time, but this year I sorta knew what I was doing. I was also happy to see my team's guts round score break 120 (I don't think it's happened in a while... yeah, I know, some teams double that...). I thought that this year's guts round questions were a bit more straightforward as questions - last year, they had the limericks and the interdependent solutions and stuff.

On a more positive note, I thought the rooms for guts were nicer this time =D Overall, an enjoyable experience (and unfortunately my last).

I thought Algebra and Combinatorics were about the same difficulty as the ones from past years that I looked at. Combinatorics was probably easier actually.

I should probably get used to the "guts round was awesome!" being rubbed in my face. Damn flight cancellations... I hope Quagga did well.

Do any of you onsite people know results or when they're coming? I understand the remote issue probably caused substantial delays.

Guts was awesome!! Oh wait, we didn't do it. (trip got canceled by county)

Well, I took combinatorics and calculus. I made a dumb mistake on #4 combo and #6 calculus (on the calculus, it was something like $ \frac {n}{n \minus{} 1} \to e$. On combo, it was like I had squares that weren't valid and so I subtracted them from the squares that were counted, but they were different squares!) Yeah, #10 combo was pretty easy, but I have no idea how to do the sudoku one still. Also have no idea how to do #10 on calculus.

Well that's a lie. I can do #10 on calc if I could do $ \int y^x\ln x\, dx$, but that's a really hard integral to take.

Yeah, so what I've gotten from the people on site is that I got first overall and in calc, Jacob Steinhardt got second overal, third in combinatorics, first algebra. Daniel Li got 1st geo. TJ got 4th team & sweepstakes (even with a 0 on guts)

Hey, does anyone have the answers for the tests?  I'm mostly interested in combinatorics because I already know algebra is hopeless:

I got:1.  60
2. 1004.5
3. 100800
4. 10201
5. 128
6. ?1/9
7. 323
8. ?61
9. ?.5
10. 1540

The question marks mean random guess; I think I checked the rest with my team and they seem right.

And thanks to all the people that made the remote testing possible, it was really a great test!

Um, like remote people won everything. Huhs.

That was pretty obnoxious. Guts was pretty hardcore.

I mean, I'm not usually one to brag much, but that was probably the most proud I've ever been of a team performance. We didn't miss a single question until about the 10th set. Between us and Exeter, it got really close. We apparently got the closest answer for one of the last "hey guys, guess a number" question to make us win by like 40ish points.

I think it went 269-224-220 between Quagga A, PEA, and Alex Zhai.

That was so frustrating though, for sweepstakes. 40 points. like. whoa. thats so close, in the grand scheme of things. Thats like... 3 individual questions, or a team round question, or something on the guts. I tip my hat to the PEA team. It was an impressive performance. TJ A. Oh my god. They would have OWNED everyone overall in sweepstakes. they got like 1300ish points without the guts round.

I still don't like the idea of the remote locations. It was unfortunate. I'm about to snowball fight it up though. So I can't really complain.

btw, Palmer, call me.

8 is $ 2025 \equal{} 45^2$. Notice that you can consider the $ 1$ and $ 2$ separately, so if you're guessing you should guess a square.

My answer to #9 is $ \frac {1}{2} \plus{} \frac {1}{2\cdot 5^{2008}}$, but other people disagree. Oh, yeah, I'm wrong because I repeatedly misread the question as even sum rather than both even.

And, uh, last year TJ A and Exeter were separated by 4 points, with WOOTlings only being ~30 points ahead of both.

Oh, that's cool (for #8); I didn't look at #9, but I saw "probability" so I guessed something less than 1 :) and positive.

Well for the sudoku one, like, the bottom row of the 3x3 with the 1 and the 2 doesn't contain the 3 with probability 4/7, and then like, the probability of a 3 being in that square instead of the other 5 in that row (that it could be) with probability 1/6 so is it 2/21?  (I know somebody that got that).

The competition was a lot of fun. I will type up a longer post later, but it was great. The guts round was so fun. I have to say, it was quite intense, like two runners striving to beat each other in a race, we were neck and neck with Exeter until they missed a few problems. I solved so many geometry problems haha... I wish that I had looked at the second to last problem set instead of brute-forcing that one with the angle. We could have easily gotten an extra 36 points on that set in say 5 mins. I thought that the round was more reasonable this year. Like it was actually possible to solve the later problems.

Can we get results if anyone remembers? Who won combo? Matt Superdock?

Well, HMMT was certainly fun.  After winning Sweepstakes last year (my first year), it was a bit of a letdown that neither I nor the team was competitive for placing, but the problems were interesting.

For the sudoku problem, the way I did it was to say that in the middle top square (the one containing the ?), there are 6 places for the 1 and 6 for the 2.  So if P is the probability that 3 (or 4, or 5 . . . ) is on the ?, then $ 1\equal{}2\frac{1}{6}\plus{}7P$, giving $ P\equal{}\frac{2}{21}$.

I am sort of mad about the arctangent problem on algebra.  I saw how to do it but I switched a negative sign and got $ \tan{20\minus{}45}$ instead of $ \tan{45\minus{}20}$ . . . Other than that, it was one of my favorite problems.

The team round was pretty tough for my team.  I did the lattice walks part, and we did okay on the second part (forget what it was) , but the incircles were a disaster.  We didn't get to any of the last 4 problems, which killed our score.

Guts . . . was interesting.  It was exciting, of course, but it was frustrating to keep falling farther and farther behind.  Incidentally, how do you do the problem with the nested square roots equal to $ 1\plus{}\sqrt{x}$?  I kept squaring it, and stuff canceled, but I got $ \frac{1}{2^{2007}}$ as the answer and it was apparently wrong.

Somehow, despite getting blown away on the important rounds, my team for relays managed to get 2nd place . . .

The awards ceremony was truly one of the strangest things I've ever seen.  I was in the secondary lecture hall, and watching the awards being announced on a screen was rather odd, as was the fact that many times people would have to cross between lecture halls to get awards.  (Not to mention the fact that so many awards went to TJ which wasn't even there . . .)

So overall it was good, I got to see a lot of people from AoPS and Mathcamp.  I want to thank the tournament directors for hosting it despite the weather and for putting on such a great competition!

EDIT: The winner of combo was Brian Basham of PEA.

Anyone know when the problems and solutions will be posted

I thought I did really well until I found out that TJ was taking it remotely and I would be competing against Jacob Steinhardt. I thought I had zero chance of winning because beating him in combo is comparable to beating god in a "who's the biggest baller" contest.
well, toot toot goes the Basham horn :)
On the other hand, I got a grand total of 16 points on algebra.

p.s. I miss you Jacob!

I think the Algebra and Geometry this year are slightly harder than in other years (as you can see, the top Algebra scorer got less than 40)
By the way, does anyone know the result of the Team Round? Our team did badly individually, but we were quite good as a team.

Lol, well I couldn't have been beaten by a cooler guy. Too bad you had to have an Exeter kid beat TJ's string of 1st places in individual, though =P. Hopefully see you at MOP this year.

[quote="Pakman2012"]I'm probably missing a few teams, but you know that happens.[/quote]

lexington ftw

who epitomized Simpson's Paradox by finishing 6th in guts, 6th in individual, and out of the top 10 in the team round, and finished 5th overall

So I'm not sure what the purpose of the Exeter rant was, since I myself already pointed out that Exeter beat us in the individual round. Exeter and TJ have a friendly rivalry. Let's try to keep it friendly by not making rants against each other on AoPS.

So, gighiuhui may or me not be more commonly known as "David Bay Rush"

I apologize.  And, try reading that post again except put it into his voice and it suddenly becomes humorous :)

That is actually quite true. When I think of David Rush saying that it doesn't seem quite so inflammatory any more.

Also, I keep getting random calls from Exeter. Is that you guys, or is Zuming trying to reach me for some reason?

EDIT: Also David you need to stop getting 14s on the Mandelbrot. It is making it impossible for anyone to catch up with you. That said, I probably just jinxed myself for the last Mandelbrot and will proceed to get a 1.

No idea about the calls, probably Zuming, but I'm not sure why he would be calling you.

[quote="greentreeroad"]What date will next year's HMMT be?[/quote]
I think they said Feb. 7th for next year, maybe not though.

If Feb.7th is Saturday, then it probably is HMMT date. :D

[quote="TCAtect"]Let's hope it won't snow so much this time...[/quote]
The snow was rather enjoyable for those of us who came up from Florida :)

And rather typical, yet still unwanted from those in New England...

Well, if it is going to snow, we can go to MIT one day earlier next year.
NO school :rotfl:

i suppose this would be the right place to start the following discussion

my school lake braddock doesnt do any high school compeititions >.> so i tried to get them started and gave my math teachers some links but they didnt do anything  :mad: and they didnt even sign up for amc 10/12 so their trying to send me to TJ to take the test
also since the teacher never set up a team for HMMT i tried to start one myself but its too late  :( so im gonna have to solo the team and guts round as an 8th grader  :o (epic phail) but i suppose it would be a nice experience to see a competition where the problems are above mathcounts/amc level
just hope i can make top ten in the general test assuming all the smart people will be taking subject tests instead  :P

Wow good luck with that...  :o

Well you're an eighth grader, so its not surprising that your school doesn't do high school competitions. There's not many competitions left, so I'd just wait it out. Next year will probably be better?  :)


Always have to make them hard(er) eh JBL :-x

jk... it's always better to extend a problem as much as possible and plus the new twist(if you want to call it that) makes you think harder(I'm just saying this off of first instinct I haven't actually worked the problem yet)

Now there must be an easier way, but AM-HM works fairly smoothly, and maybe you can do something like integrate 1/x for a lower bound.  But that's [i]not[/i] getting started.

[quote="Magnara"]Anyone know the formula for the sum of the recipricols from 1 to n?  I know it exists....and it involves gamma....[/quote]

I'm 99% sure it doesn't exist, except that you can estimate it.

lim n-> infinity of ln(n) - harmonic sum(n) = euler's constant (gamma)

How do you calculate lim n-> infinity of ln(n)?

You don't calculate them separately. You can't say

[tex]\displaystyle \lim_{n\to\infty} \Bigg(\sum_{k=1}^n \frac{1}{k}-\log(n)\Bigg)=\lim_{n\to\infty} \sum_{k=1}^n \frac{1}{k}-\lim_{n\to\infty} \log(n)[/tex]

because the second expression is meaningless. But it makes sense that they should converge together:

[tex]\displaystyle \gamma=\lim_{n\to\infty} \Bigg(\sum_{k=1}^n \frac{1}{k}-\int_1^n \frac{1}{x}\; dx\Bigg)[/tex],

which intuitively should converge since the difference between the sum and the integral becomes negligible as n gets large. (You can prove it by taking a slightly different integral and sum.)

However, that doesn't tell you anything about this problem.  The functions f(x) = 0 and f(x) = sin(x)/ln(x) both have the same limit at infinity, but that doesn't tell you how they relate to each other anywhere not "near infinity."  This problem deals with somewhere not near infinity, so the limit isn't particularly helpful at all.

First I claim that
[tex]\displaystyle{\sum_{k=2^{i-1}+1}^{2^i} \frac{1}{k} \geq \frac{1}{2} \qquad 0<i}[/tex]
[tex]\displaystyle{\sum_{k=2^{i-1}+1}^{2^i} \frac{1}{k} = \frac{1}{2^{i-1}+1} + \frac{1}{2^{i-1}+2} + \ldots + \frac{1}{2^i}}[/tex]
Where
[tex]\displaystyle{\frac{1}{2^{i-1}+1} > \frac{1}{2^{i-1}+2} > \ldots > \frac{1}{2^i}}[/tex]
There are [tex]\displaystyle{2^i-(2^{i-1}+1)+1=\frac{2^i}{2}}[/tex] terms. Sum the smallest term, [tex]\displaystyle{\frac{1}{2^i}}[/tex], [tex]\displaystyle{\frac{2^i}{2}}[/tex] times, yields [tex]\displaystyle{\frac{1}{2}}[/tex]. QED

Now, clearly, summing this summation [tex]n[/tex] times...
[tex]\displaystyle{\sum_{i=1}^{n} \sum_{k=2^{i-1}+1}^{2^i} \frac{1}{k} = \frac{1}{2} + \frac{1}{3} +\ldots + \frac{1}{2^n} \geq \frac{n}{2}}[/tex]

We wish to show:

[tex]\displaystyle{\frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{500} > 4}[/tex]

Letting [tex]n=8[/tex],

[tex]\displaystyle{\sum_{i=1}^{8} \sum_{k=2^{i-1}+1}^{2^i} \frac{1}{k} = \frac{1}{2} + \frac{1}{3} +\ldots + \frac{1}{256} \geq 4}[/tex]

I really am just learning how to do proofs, and I'm not sure if this is even a proof... any advice?

Good.  Your notation is a bit messy, but your proof is good.

here is a good one...

prove

[tex]\displaystyle{\Big(1+\frac{1}{2}+\frac{1}{3}+\ldots +\frac{1}{2004}\Big) > 8}[/tex]

Where did you get that from?
Did you make a harmonic sequence program with the calc? If so, the proof might be very [brutal]


Do there exist real numbers $b$ and $c$ such that each of the equations $x^2 +bx + c = 0$ and $2x^2 + (b + 1)x + (c + 1) = 0$ has two integer roots?

yes like $(b,c)=(-4,0)$
 (eq 1 has solutions $0,4$; eq 2 has solutions $1,\frac1{2}$)

And obviously $\frac{1}{2}$ is an integer... :D

ow. let's say it is trying very hard to be an integer. :p

I don't think it has solutions then, but is there a nice way to show it?

In order for the second equation to have integer roots, $\frac{b+1}{2}$ must be an integer.  Also, $\frac{c+1}{2}$ must be an integer.  Meaning both $b$ and $c$ have to be odd.  But such a condition cannot be met.  In the first equation, for there to be integer roots, $b$ has to be an integer.  If it is an odd integer, it must be the sum of even and odd integer roots.  The product of the roots, $c$, would then be even.  So it is impossible for both equations to have integer roots.

You could also notice that $b$ and $c$ must be odd and $b^2 - 4c$ must be a perfect square, however $b^2 - 4c = 5\ (\textrm{mod}\ 8)$, contradiction.


http://www.mathlinks.ro/viewtopic.php?t=203087
Let a,b,c>0  Prove that
$ a^3b^3c^3(a\plus{}b\plus{}c)^3\geq(ab\plus{}bc\plus{}ca)^3[(\minus{}a\plus{}b\plus{}c)(a\minus{}b\plus{}c)(a\plus{}b\minus{}c)]^2$

[quote="tran300491"]http://www.mathlinks.ro/viewtopic.php?t=203087
Let a,b,c>0  Prove that
$ a^3b^3c^3(a \plus{} b \plus{} c)^3\geq(ab \plus{} bc \plus{} ca)^3[( \minus{} a \plus{} b \plus{} c)(a \minus{} b \plus{} c)(a \plus{} b \minus{} c)]^2$[/quote]
Try $ a\rightarrow0^\plus{}.$ :wink:

try a+b+c=p
    ab+bc+ca=q
     abc=r
I think it can solve this problem


show :D

$ \lim_{x\to 4^{\plus{}}}\;\sum_{k\equal{}0}^\infty\;\left\{\dfrac{1}{2k\plus{}1}\left(\frac{x\minus{}5}{x\minus{}3}\right)^k\right\}\;\equal{}\;\boxed{\frac{\pi}{4}}$

[quote="misan"]show :D

$ \lim_{x\to 4^{ \plus{} }}\;\sum_{k \equal{} 0}^\infty\;\left\{\dfrac{1}{2k \plus{} 1}\left(\frac {x \minus{} 5}{x \minus{} 3}\right)^k\right\}\; \equal{} \;\boxed{\frac {\pi}{4}}$[/quote]
obviusly by theorem of T'epica

why can't you just plus in x=4? it looks like it should be continous.

I explain my idea:
I was not correct when I spoke about T'eplica (this is about regular of serie ) misan's rpoblem is true by method Abel-Puason:
this is about sserie $ \sum a_{n}t^{n}$ when $ t\minus{}>1$ ( and there is also some conditions ), then $ \sum a_{n}t^{n}\minus{}>\sum a_{n}$
in our case we have
$ \sum \frac{(\minus{}1)^{k}}{1\plus{}2k}t^{k}\minus{}> \sum \frac{(\minus{}1)^{k}}{1\plus{}2k}\equal{}\frac{\pi}{4}$


trigonometry
ratio
probability
real analysis
real analysis unsolved

let x1,x2,...,xn n real numbers, can we find a real µ so that for every i :sin(µx_i) >=0 ?

I believe that [i]yassinus[/i] was asking about the existence of a nontrivial (nonzero) value for $ \mu$. My opinion is that the conjecture isn't true.

Let two elements in $ (x_i)$ be $ a$ and $ \minus{}a$. Then, notice that $ sin(\minus{}\mu a)\equal{}\minus{}sin(\mu a)$ for each nonzero $ \mu$, so they never have the same sign.

Use \sin for $ \sin$ (likewise \cos, \tan, etc.).

Actually, that's not enough for a counter-example: we could still use $ \mu \equal{} \frac {2\pi}{a}$, among others.  We can make it into a counter-example by taking two such pairs $ a, b, \minus{} a, \minus{} b$ such that the ratio $ \frac {a}{b}$ is irrational.

What if we take our sequence from $ [0, 2\pi)$?  And, if there are still counter-examples, what is the probability that a random sequence from $ [0, 2\pi)$ has this property?

thank you JBL , good remarq ,I'm stupid because I was searching to prove it ans I didn't thought that the statement could be not true... if the xi are in  [0,2pi[ , any µ suffisantly small works ( µ=1/2 works too ) !! I think that I dont understand the question !!?

Oh, good point, I should have noticed that  :oops:.  (That also eliminates my other thought, restricting the $ x_i$ to $ [0, \infty)$.)  So I guess we could ask for $ x_i$ in one of these ranges and $ \mu > 1$, but then we have wandered fairly far from the wording of the original question.  Hopefully now it's not trivial to prove or disprove?

now , it's seems to be dificult  :wink:  but we should give a distrubution to the xi if we want to ask about probability ?

Maybe this will work? By dirichlet's theorem $ \forall N\ge 1, \exists q\in \math{N}$, and integers $ p_1,\ldots,p_n$ such that $ \big| x_i-\frac{p_j}{q}\big|<\frac1{Nq}$.
In particular, $ qx_1,\ldots, qx_n$ differs from an integer by less then  $ \frac1N$. Let $ \mu=q\frac\pi{2}$. Choosing $ N$ sufficiently big, we have $ \sin{\mu x_i}=\sin(\frac{q\pi}2x_i)\ge 0$

I dont remember the theorm but are yoy sure that q is the same for all xi ?

but if some p_i is =3(mod 4) that doesn't work !!! am I wrong ?

You are right. The argument must be modified.


Hm, you're right. I think it's because $0!=1!=1$, so you actually in majorazing should count all 1s as 0s. Now everything should be fine.

What about (4,2) and (3,3)? \[\frac{4!2!}{3!3!}= \frac{4}{3}.\] It seems to me the problem isn't that simple.

It seems that there must be another conditions on $a_{i},b_{i}$ or $n$ but unfortunately  I don't know how to repair it for now  :(


geometry
inequalities
analytic geometry
invariant
ratio
trigonometry
parallelogram

Let $Q$ be a point in the interior of a triangle $ABC$. Let $M$, $N$, $P$ be points on the sides $BC$, $CA$, $AB$, respectively, such that $MN$ parallel to $AQ$, $NP$ is parallel to $BQ$, and $PM$ is parallel to $CQ$. Prove that $\text{Area}\left(MNP\right)\leq\frac13\cdot\text{Area}\left(ABC\right)$.

One case of equality is when $Q$ is the centroid and $M,N,P$ trisect their respective sides.

We rewrite the inequality as $\frac{|MNP|}{|ABC|}\le\frac13$, where $|XYZ|$ denotes area.

For the proof, use barycentric coordinates: $M=(0,m,1-m),\, N=(1-n,0,n),\, P=(p,0,1-p),\, Q=(a,b,c)$.
We have $\frac{|MNQ|}{|ACQ|}=\frac{MN}{AQ}\cdot\frac{AN}{AC}=\frac{m}{b}\cdot n$ and $\frac{|ACQ|}{|ABC|}=b$, so $\frac{|MNQ|}{|ABC|}=mn$ and $\frac{|MNQ|+|NPQ|+|PMQ|}{|ABC|}=\frac{|MNP|}{|ABC|}=mn+np+mp$ (it is easy to show that $Q$ is inside $MNP$).

We also have that $|MNP|=|ABC|-|ANP|-|BPM|-|CMN|$ by dissection. $\frac{|ANP|}{|ABC|}=\frac{AN}{AC}\cdot\frac{AP}{AB}=n\cdot(1-p).$ The other two are cyclically similar, so $\frac{|MNP|}{|ABC|}=1-n(1-p)-p(1-m)-m(1-p)=1+mn+np+pm-m-n-p.$
Equating the two expressions, $m+n+p=1.$
Now we have a standard algebraic inequality; since $(m+n+p)^{2}=1$ and $m^{2}+n^{2}+p^{2}\ge\frac13$, $mn+np+pm\le\frac13.$

This is a purely affine problem; coordinates work well for this type of question.

[color=blue][b]Problem.[/b] Let Q be a point inside a triangle ABC, and let M, N, P be points on the sides BC, CA, AB, respectively such that MN || AQ, NP || BQ and PM || CQ. Prove that $\frac{\left|MNP\right|}{\left|ABC\right|}\leq\frac13$.[/color]

Though Jmerry gave a very nice solution of this problem, let me still show another one:

At first we note that the problem is invariant under any (real) affine transformation (since the problem uses only the - affinely invariant - notions of incidence, arrangement, parallelism and ratio of areas). Also, by http://www.mathlinks.ro/Forum/viewtopic.php?t=6134 post #2 Lemma 3, since the point Q lies inside triangle ABC, there exists an affine transformation mapping the triangle ABC to a new triangle and this point Q to the orthocenter of this new triangle. Hence, in order to solve the problem, it will be enough to solve it for the case when the point Q is the orthocenter of triangle ABC.

In this case, the lines AQ, BQ, CQ are the altitudes of triangle ABC, and thus perpendicular to its sides BC, CA, AB, respectively. The lines MN, NP, PM, being parallel to these lines AQ, BQ, CQ, must therefore also be perpendicular to the sides BC, CA, AB, respectively. Using directed angles modulo 180°, we thus have < (MN; BC) = 90°, < (NP; CA) = 90° and < (PM; AB) = 90°. Hence,

   < MNP = < (MN; NP) = < (MN; BC) + < (BC; CA) - < (NP; CA)
               = 90° + < (BC; CA) - 90° = < (BC; CA) = < BCA.

Similarly, < NPM = < CAB and < PMN = < ABC. Thus, the triangles MNP and BCA are similar. Let k be the ratio of similitude, i. e. the ratio of any side of the former triangle to the respective side of the latter:

$k=\frac{MN}{BC}=\frac{NP}{CA}=\frac{PM}{AB}$.

Then, of course, the ratio of the areas of these triangles MNP and BCA equals $k^{2}$; in other words, we have $\frac{\left|MNP\right|}{\left|BCA\right|}=k^{2}$. This rewrites as $\frac{\left|MNP\right|}{\left|ABC\right|}=k^{2}$. But the problem requires us to show that $\frac{\left|MNP\right|}{\left|ABC\right|}\leq\frac13$. Thus, it will be enough to prove that $k^{2}\leq\frac13$.

Now, using non-directed angles again, we note that, since $NP\perp CA$, the triangle ANP is right-angled, and thus $AP=\frac{NP}{\sin\measuredangle NAP}=\frac{NP}{\sin A}$. Since $\frac{NP}{CA}=k$, we have $NP=k\cdot CA$, so that $AP=\frac{k\cdot CA}{\sin A}=\frac{kb}{\sin A}$. On the other hand, since $PM\perp AB$, the triangle BPM is right-angled, and therefore $PB=PM\cdot\cot\measuredangle PBM=PM\cdot\cot B$. Since $\frac{PM}{AB}=k$, we have $PM=k\cdot AB$, and thus $PB=k\cdot AB\cdot\cot B=kc\cdot\cot B$. Hence,

$c=AB=AP+PB=\frac{kb}{\sin A}+kc\cdot\cot B$.

Dividing this equation by kc, we get

$\frac{1}{k}=\frac{b}{c\sin A}+\cot B$.

But after the law of sines, $\frac{b}{c}=\frac{\sin B}{\sin C}$, and thus,

$\frac{b}{c\sin A}=\frac{b}{c}\cdot\frac{1}{\sin A}=\frac{\sin B}{\sin C}\cdot\frac{1}{\sin A}=\frac{\sin B}{\sin C\sin A}$
$=\frac{\sin\left(C+A\right)}{\sin C\sin A}$        (since sin B = sin (C + A), as we have B = 180° - (C + A) by the sum of angles in triangle ABC)
$=\frac{\sin C\cos A+\cos C\sin A}{\sin C\sin A}=\frac{\cos A}{\sin A}+\frac{\cos C}{\sin C}=\cot A+\cot C$,

so that

$\frac{1}{k}=\frac{b}{c\sin A}+\cot B=\left(\cot A+\cot C\right)+\cot B=\cot A+\cot B+\cot C$.

By a well-known inequality ( http://www.mathlinks.ro/Forum/viewtopic.php?t=3607 problem 4), we have $\cot A+\cot B+\cot C\geq\sqrt3$. Thus, $\frac{1}{k}\geq\sqrt3$, so that $k\leq\frac{1}{\sqrt3}$. Consequently, since k > 0, we can conclude that $k^{2}\leq\left(\frac{1}{\sqrt3}\right)^{2}=\frac13$, and we are done.

  Darij

[quote="jmerry"]We have $\frac{|MNQ|}{|ACQ|}=\frac{MN}{AQ}\cdot\frac{AN}{AC}=\frac mb\cdot n$[/quote]

Can you tell me how you derive these?

There's an intermediate triangle $ANQ$. $\frac{|ANQ|}{|ACQ|}=\frac{AN}{AC}$ should be obvious. The other half is slightly less obvious, but $MNQ$ and $ANQ$ have two pairs of parallel sides: $MN$ and $AQ$, $NQ$ and $NQ$. The area ratio is then $\frac{MN\cdot NQ}{AQ\cdot NQ}=\frac{MN}{AQ}$. In both cases, you could see it as "same height, different base"; the ratio of the areas is then the ratio of the bases.
One key rule in affine calculations: the ratio of two parallel segments is meaningful, but no length is on its own, and a ratio of non-parallel segments is nonsense.

It's all standard affine manipulations. I could derive a full coordinate area formula if I wanted to, but that was unnecessary here. Actually, I could go back and show that $m=b,n=a,p=c$, so $ANMQ$ is a parallelogram (and this works for any point $Q$ inside the triangle), but none of that was necessary.


A triangle  ABC with bisector AD.It  are known $ AB\plus{}AD\equal{}CD$,$ AC\plus{}AD\equal{}BC$ ,Find the  angle triangle ABC. :)

[b]nice problem!

[/b]I think the answer are : A=120      B=40     C=20
Is it true?
I can't solve this problem.
If u have a solution , please tell us your solution.

Thanks for your nice problem pirkuliev. :lol:


calculus
integration
trigonometry
function
real analysis
real analysis unsolved

Find

$\int\frac{1}{x}\tan\left(x+\frac{1}{x}\right)$

Can any one help ? I use integration by parts and substitution , but all doesnt work .  :?

uhm... actually, wolfram integrator fails in computing it, so i guess there's not a primitive in terms of elementary functions, though i don't have any proof of it...
sure you didn't mean a defined integral? (i mean.. finding the integral between 0 and 1, or whatever else?)

no , it is undefinite integrate .... I guess it cant be done then ....

Undefinite = Indefinite?

Does it mean anything to anybody that:
$\int^{b}_{a} \frac{1}{x} \tan(x+\frac{1}{x})\,dx = -\int^{1/b}_{1/a}\frac{1}{x}\tan(x+1/x)\,dx$
Maybe you can't solve it using integration techniques in a straightforward way ,but if you are able to develop enough information about the function then you could algebraically solve for it.
I'm just throwing ideas out there, since I'm only in multivar.

If F(y) is the indefinite integral, then $F(b)+F(1/b) = F(a)+F(1/a)$ for all a,b.
Does that help much?


(Anyone who has seen this before, lay back and let others do it; this is an exercise to see how people will approach this.)

Given a Set $ S$ of size $ n$ where all elements are integers, write an algorithm that returns $ \mbox{true}$ if a non-empty subset of $ S$ has the property that the sum of its elements is exactly 0. Else, return $ \mbox{false}$.

For example, given Set $ S_1 \equal{} \{3, \minus{} 5, 6, 4, 2, \minus{} 7, \minus{} 11\}$, the algorithm should return $ \mbox{true}$ since the subset $ \{ \minus{} 5, 6, 4, 2, \minus{} 7\}$ sums to zero.

You are not required to find the subset, only to determine if such a subset exists.

1 point for a $ O(N^N)$ solution.
2 points for a $ O(N2^N)$ solution.
3 points for a $ O(N\sqrt {2^N})$ solution
4 points and a million dollars for a $ O(N^k)$ solution where k is a constant independent of N.

(Runtimes are worst-case runtimes.)

The $ N \cdot 2^N$ solution is the brute-force one; what would it possibly take to think up a $ N^N$ algorithm?

(Anyway, in case the reference escapes (edit: anyone else reading this thread), the 4-point remark is due to the fact that this problem is NP-complete.)

You again! XD

[hide="Spoiler"]
[quote="t0rajir0u"]The $ N \cdot 2^N$ solution is the brute-force one; what would it possibly take to think up a $ N^N$ algorithm?[/quote]

Not much, I might say. Some person I was tutoring managed to come up with a $ N^N$ runtime algorithm. Worse yet, it also took up an amount of memory proportional to $ N^N$.

It involved sorting the set, adding positive-negative pairs, HASHING said result to an ArrayPair object (containing an array of the elements remaining and another array representing the elements used up), then creating n-ary trees converted to binary trees that represented all the branching possibilities from there on, etc...

Surprising why he didn't see the brute force method immediately with all that knowledge about data structures.[/hide]

$ O(N \sqrt{2^N})$ can be achieved with meet-in-the-middle. Split $ S$ into $ S_1$ and $ S_2$ each with (about) $ N/2$ elements. Brute force all subset sums of $ S_1$ in $ O(N\sqrt{2^N})$ time and store them in a hashset. Then brute force all subset sums of $ S_2$ in $ O(N\sqrt{2^N})$ time, checking for the negative of each in the hashset in $ O(1)$ time, returning true if a match is found. Obviously, if any of the subset sums is zero, we also return true.

Actually it's not that hard to come up with an O(N^N) solution :D. Just check all the possible sequences of numbers from [1, N] of length at most N, discarding the ones that have repeating elements.


limit
logarithms
calculus
integration
real analysis
real analysis unsolved

$Find \ \displaystyle \lim_{n\to+\infty} u_n \ \   where$ :
\[
u_n= \left( \displaystyle \prod_{k=1}^n\sqrt {1+\frac {k} {n}} \right)^{\sin(2\pi \sqrt {n^2+1})}
\]

Just take the logarithm, use the fact that $\sin(2\pi\sqrt{n^2+1})$ is equivalent to $\frac \pi n$, and then it's simply a Riemann sum converging to an integral on $[0,1]$ (I don't want to do the computations :)).

OK Grobber
\[
 \displaystyle \lim_{n\to+\infty} u_n=\left (\frac {4} {e}\right)^{\frac {\pi} {2}}
\]


geometry
circumcircle
Euler
geometric transformation
reflection
geometry proposed

Consider a non-iscoceles acute triangle $ ABC$ such that $ AB^2 \plus{} AC^2 \equal{} 2BC^2$. Let $ H$ and $ O$ be the orthocentre and the circumcentre of triangle $ ABC$, respectively. Let $ M$ be the midpoint of $ BC$ and let $ D$ be the intersection of $ MH$ with the circumcircle of $ ABC$, such that $ H$ lies between $ M$ and $ D$. Prove $ AD$, $ BC$, and the Euler Line of triangle $ ABC$, are concurrent.

This was from Math Reflections 2008 Issue 1, the deadline for sending solutions is over so I think I can post it now.

Call G the centroid. we have that
$ GO^2 \plus{} GM^2 \equal{} \frac {1}{9} \left [ 9R^2 \minus{} a^2 \minus{} b^2 \minus{} c^2 \frac {1}{4}\left (2(b^2 \plus{} c^2) \minus{} a^2 \right ) \right] \equal{}$
$ \equal{} \frac {1}{9} \left [9R^2 \minus{} 3a^2 \minus{} \frac {3a^2}{4} \right ] \equal{} R^2 \minus{} \frac {1}{4}a^2 \equal{} R^2 \minus{} \left ( \frac {a}{2} \right )^2 \equal{} OM^2$
So for pitagora $ OG \perp AM$. Now call E the second intersection of HM with the circumcircle; clearly AE is a diameter so $ MD \perp DA$
Call $ P: DA \cap BC$, the H is the orthocenter of $ \triangle AMP$ and so the euler line is an altitude and it pass through P.


What are the rules for inscribing polygons above quadrilaterals in circles? The standard opposite angles summing to 180 obviously doesnt work. Ive looked online, but I cant seem to find anything like this. They usually only list quads.

Pick any 4 vertices and they need to be concyclic.

LOL.... DOH! I cant believe I didnt think of that. Thanx. *laughs at meself*


French team for IMO 2004 :

Pierre Bertin
Raphael Beuzart-Plessis (raf on the forum)
Carlo Defranchis
Vincent Jug
Igor Kortchemski (me :-))
Antony Lee (Antony on the forum)

so the team's out :P congrats to all ... btw which one of you didn't solve problem 6 from the tst? :P:D


I want to type

The arc ABC

in LaTeX, and in math form. How do I do that? \arc{ABC} doesn't do it...

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690]LaTeX FAQ 10. How can I make an arc over AB?[/url]


MATHCOUNTS
geometry
FTW
probability
area of a triangle
number theory

I always seem to get to problem 25 on sprints, and then run out of time. Any suggestions on how to improve my speed?

Ah, so people have already started practicing for the 2010 Mathcounts? Im a bit behind. :lol:

And to improve your speed, you should read up on the types of problems that give you trouble. AoPS volume 1 is a great book. Also, dont bother rechecking the first 10 or so.

[quote="Thunder365"] Also, dont bother rechecking the first 10 or so.[/quote]

I was perfect in the first 14 at Nationals in 7th grade; I missed one from the first 10 this year, and two others in the next four. That, along with not reading the word "different" on target #3 kicked me out of countdown (three points off while going to the bathroom during the sprint  :roll: ), along with some other stupid stuff (not dividing by two when finding the area of a triangle, starting at 150 instead of 139, etc) but the point is, if you can, check every problem. If it's in the first 10, chances are it didn't take very long anyway.

For speed, it depends; what types of problems do you spend most time on? If it's difficult problems, try practicing on harder problems, like some AMC 10/AMC 12 (the ones that use Mathcounts concepts)/AIMEs. If it's a specific type of problem, such as geometry, practice lots of geometry questions. If you just go slow on each problem, FTW might be a decent way to get a bit faster, although it doesn't work that well once you get pretty fast. Also, do lots and lots of problems; you'll probably find many similar ones.

The most important thing is to have great fundamentals; if you do, you should be able to solve these problems pretty quickly.

[quote="ged3.14"]Any suggestions on how to improve my speed?[/quote]

1. Do math

2. When you get sick and tired of #1, do #3

3. Do more math

4. When you get sick and tired of #3, do #1

Srsly

Focus on which type of problems slow you down the most. Probability Geometry,whatever. Then read about the subject and work on specific problems that improve your skills. Also some problems are repeated a lot but using different numbers. Anaylze different tests and find out which problems seem to come up a lot so you know wat to do as soon as you see the problem

[hide="My suggestion"]
2/10 AoPS forum
1/10 FTW/Alcumus
3/10 Problems
3/10 Reading textbooks/learning things (looking over mistakes, reading about things you are weak in, ...)
1/10 Just think about a topic (prove something, explore a method of doing a problem, ...)

That is just a guideline I am going to use for myself, and it is going to vary day to day.

First, you should do some problems and get familiar with the type of test you are practicing for (MATHCOUNTS=problem solving, AMC=concepts, ...). Do some of the easy problems first (school MATHCOUNTS level), so you feel confident (good thing). Then take it one step higher (State MATHCOUNTS level). Look at your mistakes very carefully, and find more info on that topic. Next, do some more of those types of problems. Then start doing some medium level problems again (Chpater MATHCOUNTS level). Keep repeating this cycle until you find that most of your mistakes are silly mistakes (misreading, computational error). Then move on to a little harder problems (State MATHCOUNTS level) and do the same thing. Then start learning tips and tricks to stop silly mistakes and do problems in a better way. Then do some hard tests (National MATHCOUNTS level). Look at your errors and do the same thing you did with easier problems. After you feel you have mastered the topics and done your best to stop silly mistakes, then start improving speed, as this will help with double-checking at the end to prevent some more silly mistakes. Once you are through almost all your tests and have mastered everything above, just do tests and memorize common problems that come up a lot. Once your done this, you will just have to practice, and I think it from here on it is just luck.[/hide]

Oh and like practice in bad conditions, that is:

No scratch paper
Bad tables (like the kind your pencil pokes through paper when you write)
Frigidness
Annoying siblings

Then, nats seems like no pressure and optimal conditions.

Or you could practice with a fever, infected ears and a throat ache.

You could try to just do a lot of problems to improve how fast you can compute, or practice right when you wake up or when you're really tired. Also, if you've been to nationals before, you could try to simulate the conditions and practice with that.
If you really can't improve your speed no matter what, if you just concentrated on the first 25 and could get all those right, that would probably be pretty good. (I don't know your standards)

[quote="xpmath"][quote="Mewto55555"]

Then, nats seems like no pressure and optimal conditions.[/quote]

Be prepared for everything!

Also, for targets, since you have more time, try to underline important words that are important for the problem. For example, not reading the word "different" in target #3 at nats this year might be bad. For sprints, that probably isn't such a good idea. Honestly, just try to find what you've been doing wrong and fix it.[/quote]

Practice drinking water so you have to go to the bathroom during it. I had this problem in 08, which kept me out of the top 57, and an acquaintance of mine had it this year which kept him out of cd   :oops:

I would practice and work in terrible conditions.
Lots of commotion
bad area of work
ect.
and get practice tests hope it helps :D

does anyone know where to get National level practice problems (esp counting/probability & geometry) :?:

by the way, how do you get your profile rated?

You can trade for them in the MATHCOUNTS Resources HERE! sticky at the top this forum. When other people rate your posts, you get a rating. It is the average of your posts' rating.

This site:
http://www.agmath.com/57427.html

Has a file of "10 essentials" where past national mathcounts problems are organized into subjects. It also has a unit on number theory, counting, and probability. (counting and probability are almost finished.)

Where is your rating shown?

[quote="BOGTRO"]Your rating is right under your username, where it says Posts, Location, etc.

Anyway, If you can do the first 25 on sprints perfectly, you shouldn't have any problems. But if you are trying to get more:

You receive 40 minutes. Skip the problems that seem like they may take you a while and come back to them if you have time. For example, if you are very bad at geometry, and #3 is a geometry problem, don't worry about it. If you are astounding at algebra and horrible at geometry, dont' feel like you HAVE to get the first 5. Get #30, an algebra problem instead of #3, a geometry problem. They are all worth 1 point.

That being said, try to do the first 10 quickly and always save at least 10 minutes for checking. If you cant do a problem, you can't do it, regardless of it's number. Practice with subjects you aren't good in.

Basically, practice, practice, and practice. And when you practiced a lot, practice more. And when you did all of the mathcounts problems, do old AMC 10's and stuff like that.

--BOGTRO[/quote]
Yes, reviewing AMC 10s would be good advice, but if you can, try to solve the problems WITHOUT looking at the answer choices.  This will make it more like real MathCounts(since, unfortunately, MC is not multiple choice).  Also, reviewing recent MC Chapter and State Competitions(which you can get for free), you should be prepared, assuming you review.  Of course, getting to MC Nats takes a lot of luck(something I must have lacked last year)


Let $ f\in\mathbb{Z}[x]$ be a nonconstant polynomial. Prove that the sequence $ f(3^{n})\pmod n$ is not bounded.

Obviosly we can consider only polinoms $ P(x)\equal{}a_{l}x^{l}\plus{}...\plus{}a_{1}x\plus{}a_{0}$ with $ a_{0}\equal{}0$. Let $ P(1)\not \equal{}0$.
 Consider $ n\equal{}2^{k}3^{m}$. We have $ 3^{n}\equal{}1(mod 2^{k})\to P(3^{n})\equal{}P(1)\not \equal{}0\mod 2^{k}$ and $ P(3^{n})\equal{}0\mod 3^{m}$, therefore
$ P(3^{n})\equal{}3^{m}B\mod n$, were $ 3^{m}B\equal{}P(1)\mod 2^{k}$. Therefore we can $ k\to\infty$ and choose $ m\equal{}m(k)$, suth that
$ 3^{m}B\to\infty,\ B\equal{}P(1)3^{\minus{}m}\mod 2^{k}$.


If $ A \subset \mathbb{R}^{n}$ is compact and convex, then A is a deformation retract.

The nearest-point map of $ \mathbb R^n$ onto $ A$ is a retraction. The straight-line homotopy shows that it's a deformation retraction.


I have been working through the AoPS "Introduction to Number Theory" book and i have came across something i dont understand.

(If you have it its on page 103)

The question is

Find the product of all the positive divisors of $ 450$ that are multiples of $ 3$.

I dont really see how their solution works =)

Thanks everyone for helping me out   :)   :lol:

$ 450\equal{}2 \cdot 3^{2} \cdot 5^{2}$. First we can find the product of all of its divisors which is $ 450^{\frac{(2\plus{}1)(2\plus{}1)(1\plus{}1)}{2}}\equal{}450^{9}$. Now we can find the product of all factors not divisible by $ 3$ and divide. So we do the same process with $ 50$, giving that it is $ 50^{3}$. Therefore the answer is $ 3^{18}\cdot 5^{12} \cdot 2^{6}$.

Another way to do it is to think of it the following way. The factor must be of the form $ 3^{a} \cdot 5^{b} \cdot 2^{c}$ where $ 1\le a \le 2$, $ 0 \le b \le 2$ and $ 0 \le c \le 1$. Multiplying all of the possible values together also gives the above expression.


$ a_1\equal{}a_2\equal{}a_3\equal{}1$ and $ a_{n\plus{}1}\equal{}\frac{1\plus{}a_na_{n\minus{}1}}{a_{n\minus{}2}}$

show that for every $ n\equal{}1,2,3....$ $ a_n$ is integer

:)

[quote="moritzplatz"]maybe i don't understand the problem....
but:
$ a_1 \equal{} 1 \newline a_2 \equal{} 1 \newline a_3 \equal{} 1 \newline a_4 \equal{} \frac {1 \plus{} (1 \cdot 1)}{ \minus{} 1} \equal{} \minus{} 2 \newline a_5 \equal{} \frac {1 \plus{} ( \minus{} 2 \cdot 1)} { \minus{} 4} \equal{} \frac {1 \minus{} 2}{ \minus{} 4} \equal{} \frac {1}{4} \newline$

wich is not integer....

but i'm certanly wrong  :blush:[/quote]
$ \boxed{a_4 \equal{} \frac {1 \plus{} (1 \cdot 1)}{ \minus{} 1} \equal{} \minus{} 2}$

$ \minus{}1$ in the denominator , how comes?

[quote="vinskman"][quote="moritzplatz"]maybe i don't understand the problem....
but:
$ a_1 \equal{} 1 \newline a_2 \equal{} 1 \newline a_3 \equal{} 1 \newline a_4 \equal{} \frac {1 \plus{} (1 \cdot 1)}{ \minus{} 1} \equal{} \minus{} 2 \newline a_5 \equal{} \frac {1 \plus{} ( \minus{} 2 \cdot 1)} { \minus{} 4} \equal{} \frac {1 \minus{} 2}{ \minus{} 4} \equal{} \frac {1}{4} \newline$

wich is not integer....

but i'm certanly wrong  :blush:[/quote]
$ \boxed{a_4 \equal{} \frac {1 \plus{} (1 \cdot 1)}{ \minus{} 1} \equal{} \minus{} 2}$

$ \minus{} 1$ in the denominator , how comes?[/quote]
 :o  :o
you are right!

i readed $ a_n \minus{} 2$ but it is $ a_{n \minus{}2}$

sorry :oops:  :oops:

I do think that this problem is a kind of generalization to the problem posted here. [url]http://www.mathlinks.ro/viewtopic.php?t=241278[/url] .I supose that the proof is similar with those given there

$ a_{n\plus{}1}a_{n\minus{}2} \equal{} 1 \plus{} a_{n}a_{n\minus{}1}$
$ a_{n}a_{n\minus{}3} \equal{} 1 \plus{} a_{n\minus{}1}a_{n\minus{}2}$

Subtracting the equations,

$ a_{n\plus{}1}a_{n\minus{}2} \plus{} a_{n\minus{}1}a_{n\minus{}2} \equal{} a_{n}a_{n\minus{}3} \plus{} a_{n}a_{n\minus{}1}$ $ \equal{}>$

If n is even:

$ \frac{a_n}{a_{n\plus{}1}\plus{}a_{n\minus{}1}} \equal{} \frac{a_{n\minus{}2}}{a_{n\minus{}1}\plus{}a_{n\minus{}3}} \equal{} \frac{a_2}{a_3\plus{}a_1} \equal{} \frac{1}{2}$.

If n is odd:

$ \frac{a_n}{a_{n\plus{}1}\plus{}a_{n\minus{}1}} \equal{} \frac{a_{n\minus{}2}}{a_{n\minus{}1}\plus{}a_{n\minus{}3}} \equal{} \frac{a_3}{a_4\plus{}a_2} \equal{} \frac{1}{3}$.

Therefore, for even n,  $ a_{n\plus{}1} \equal{} 2a_{n}\minus{}a_{n\minus{}1}$,
for odd n, $ a_{n\plus{}1} \equal{} 3a_{n}\minus{}a_{n\minus{}1}$
and in either case, $ a_{n\plus{}1}$ will be an integer and we are done.

[quote="cipos2"]i think it is this, but i may be wrong

oh and sorry but i don't know how to use latex

if we solve for a(n) we get

a(n+1) * a(n-2) -1 / a (n-1 )  = a (n)

and solving for a(n-1) we get

a(n+1) * a(n-2) -1 / a (n )  = a (n-1)

so a (n ) is divisible by a(n-1) and vice versa

so if one is an integer the other one is

and since we are given the first three terms as integers we can expect it to continue.[/quote]

$ \boxed {\text{I demo your statement in Latex. If you search the contents, you may click "quote" and get to see .}}$ :roll:

[u]The "quote" icon is the right-hand lower corner of this board.[/u]

if we solve for $ a_n$ we get

$ a_{n \plus{} 1} a_{n \minus{} 2} \minus{} \frac {1} {a _{n \minus{} 1}} \equal{} a_{n}$

and solving for $ a_{n \minus{} 1}$ we get

$ a_{n \plus{} 1} a_{n \minus{} 2} \minus{} \frac {1} {a _{n \minus{} 1}} \equal{} a_{n}$

so $ a_n$ is divisible by $ a_{n \minus{} 1}$ and vice versa

so if one is an integer the other one is

and since we are given the first three terms as integers we can expect it to continue.[/u]


Let $ n$ be a nonzero positive integer and $ a>1$ a real number such that $ a^k \not\in \mathbb{N}$ for any $ k\in \mathbb{N}^*$ and $ m\equal{}[a^n]$. Prove that:

$ \displaystyle\sum_{k\equal{}1}^{n}[a^k]\plus{}\displaystyle\sum_{k\equal{}1}^{m}[log_a k]\equal{}m\cdot n$


Hmm, you lot of ppl are so smart...
Just, wondering? Who thinks the new SAT's are gonna be harder? Since they are changing them in 2005, which is when I'm supposed to take them (im freshman in high school)... i kinda wanted to see what everyone else thought...
the math is still easy.. it's only up to algebra II
verbal.. no more analogies... ahh pooey, those are easy...
essay.. i hate essays
no more quantitative comparisons.. oh well.. whatever
I found the old versions really easy.. got 1350 in 7th grade, and 1470 in 8th.. not sure about the new ones tho.. and what's with full score of 2400?? i mean, honestly...

gah the essays look really hard as in the prompts they give you and where you can draw your information from. like if you want to seem smart you have to know history and politics and literature and sigh...i hope they grade easy.

yeah, I agree.. sooo, i guess the best way to prepare is to start reading up on literature and stuff huh?

mystic, have you seen any example essays? If so, can you link me? Personally, I'm really dreading the new version, mainly because of the essay. Also, about them grading the essays, I heard somewhere that about 1 million SATs were taken each year, which is an insane number of essays to grade...how on earth do they think they're going to grade all those? I just think the essay is a really stupid idea (and my class is the first one to have to take it).  :-x  :shock:

The Writing portion of the new SAT I will be essentially like the current SAT II Writing test, except that it will be only 50 minutes instead of 60.  The essay prompts may be a bit longer, because students were finding it difficult to write on a topic out of context.

Here is a [url=http://www.collegeboard.com/student/testing/sat/lc_two/writ/essaysamp/essaysamp.html]link to some sample essays[/url] on the College Board web site, and how they scored.

[i]Moderator's note: I edited Ravi B's reply only to put the link into hypertext coding around the obvious link text, to avoid horizontal scrolling. Thanks for sharing the link.[/i]

thanks for the info.. i haven't taken the sat ii writing.. don't want to either.. =)[/quote]

No, I haven't taken it for real.  I am a tutor, but not for writing (yet).  I find the grammar questions on the Writing test relatively easy.  I get a little self-conscious about writing essays, especially in a time-limited setting.

[quote="MagnusRex"]Ah, thanks. Ravi, have you or has anyone else here taken the SAT II writing test? What did you get?[/quote]
I took it, and got a 750. Here's my complete list of scores:

ACT: 32
SAT I Math: 780
SAT I Verbal: 710
SAT II Writing: 750
SAT II Physics: 750
SAT II Math IIC: 800

Anyway, I'm class of '04, and rather glad that I don't have to take the new, more annoying, SAT, specifically dumbed down for the kids that can't do math or think logically. (Reduced importance on math, still no importance placed on science, got rid of the analogies)

Wow, nice scores.

[quote]
new, more annoying, SAT, specifically dumbed down for the kids that can't do math or think logically
[/quote]

Lol, I agree.

Grr... new SAT... and my class is the first to be affected by it. I don't see why we need an essay on the SAT - there's already the SAT II Writing and the college essays...ought to be enough proof of our writing skills for the colleges.

I took the SAT as a 7th grader and i got a 1360 (760 math, 600 verbal), and i took the ACT this year, and got a composite 32 (33 math, 28 reading, 29 writing, 36 science)  and i'm not sure if this is true, but are colleges going to start favoring the ACT, as opposed to the SAT?  (hmm...before me, the last response was a looong time ago, oh well....)

[quote="justinian"]I'm not sure if this is true, but are colleges going to start favoring the ACT, as opposed to the SAT?[/quote]

I think there is so much tradition behind which colleges favor which test that it will be a very slow process to see much change in the favor given to each test. After all, most colleges and universities in the United States draw from all regions of the country, and have applicants each year who submit differing tests. Supposedly my alma mater favors the ACT, for example, but I submitted only SAT scores to get into it.

If you have particular worries, go look at websites or e-mail admissions officers.

[quote="billzhao"]Is it true that some universities (in particular the Ivy League ones) will not accept the old SAT scores from the class of '06?[/quote]

There is so much else to applying to such schools besides SAT scores that that wouldn't be the only issue. A lot of members of class of '06 who will be applying to Ivy League schools begin the first of two years of full-time college enrollment under the label of "high school" in the coming school year. For example, the [url=http://www.mheso.state.mn.us/mPg.cfm?pageID=797]Post-Secondary Enrollment Options[/url] program here in Minnesota allows high-school age students to take college (university) courses to fulfill secondary education requirements, on the state's dime.

For what it's worth, I would think being on the Canada IMO team will look pretty decent on a college application, so the SAT test is really a small-potato worry for some of the applicants who post regularly on this forum.

[url]http://www.collegeboard.com/newsat/colleges/require.html[/url]

This is a list from Collegeboard of colleges that will want to use the new writing section. On the other hand, I do not completely understand what the site's message implies. I wonder if you can apply with an old SAT score and a SAT II writing score? Of course, the SAT II writing will disappear soon.

[quote="tetrahedr0n"][url]http://www.collegeboard.com/newsat/colleges/require.html[/url]

This is a list from Collegeboard of colleges that will want to use the new writing section.[/quote]

Thanks for sharing the very useful link. There are already several of the top math schools on that list, and more be added week by week.

[quote]have you or has anyone else here taken the SAT II writing test? What did you get?[/quote]

I got a 700 and I got a 720 verbal on the SAT.  But Mystic -- don't worry -- they grade REALLY easy on the essay section.  On the writing SAT II they have "subscores," and I got an 800 for the essay and a 610 for the grammar.  That shows how easy they grade.

If you're worried about the essay, check here...
http://www.princetonreview.com/college/testprep/testprep.asp?TPRPAGE=557

and here for answers
http://www.princetonreview.com/college/testprep/testprep.asp?TPRPAGE=558&TYPE=SAT-LEARN

(i think those are correct links, if not, click around the site - it's worth it)

anyway, the princeton review makes it seem like the essay is a joke - the college board only cares about structure, organization, grammar - they don't care much about ideas (which, for me, would be the hard part - lucky me.)

[quote="People at the Princeton Review"]Will I need to have outside knowledge on the topic to do well on the essay?

No. Essay readers will award a score based on a combination of structure, grammar, and organization of thought. The readers know that you have only 25 minutes to write, so they will grade your essay for what it really is - a first draft written on demand under timed conditions. Creativity won't count, so focus on the basics. [/quote]

also

[quote="People at the Princeton Review"]
Will colleges only accept the new SAT for the class of 2006?

If you are in the Class of 2006, 2007 or beyond, ONLY take the New SAT. Colleges and universities across the country will accept the new test. Plus, despite what you may have heard, youll probably find the New SAT easier than the old one. In fact, weve raised our guarantee from 100 points to at least 200 points for the New SATthats how confident we are that you can beat the new test. [/quote]

I'm just annoyed that I already took the "old" test already *sigh* what a waste of a beautiful Saturday.

[quote="JS1527"][quote]
I got a 700 and I got a 720 verbal on the SAT.  But Mystic -- don't worry -- they grade REALLY easy on the essay section.  On the writing SAT II they have "subscores," and I got an 800 for the essay and a 610 for the grammar.  That shows how easy they grade.[/quote][/quote]

awesome, congrats :) more comforting thoughts.

I guess this is somewhat related, but they're changing the ACT too, right? adding an "optional essay"? what is that going to do? does that sound strange to anyone beside me?

[quote="MagnusRex"]Ah, thanks. Ravi, have you or has anyone else here taken the SAT II writing test? What did you get?[/quote]

Class of 2004:

(one shot for all)

SAT 1 Math: 800
SAT 1 Verbal: 730
SAT 2 Writing: 800
SAT 2 Math 2c: 800
SAT 2 Physics: 780

[quote="billzhao"]Is it true that some universities (in particular the Ivy League ones) will not accept the old SAT scores from the class of '06?[/quote]

Yufei, unless you kill someone between now and then, you can get into the college of your choice.

I got into Princeton primarily on the strength of my math related achievements, and I wish that I were half as good as you.

[quote="billzhao"]Is it true that some universities (in particular the Ivy League ones) will not accept the old SAT scores from the class of '06?[/quote]
For the class of 2006, the following colleges have announced they will not accept the old SAT: Boston College, Brandeis, Dartmouth, NYU, Rice, U. California, U. Illinois, U. Michigan, U. North Carolina, Vanderbilt, and Washington U.

The following colleges have not yet decided: Emory, Haverford, Trinity College, Tufts, U. Penn, and U. Virginia.

Most of the other leading colleges have announced they will accept either the old SAT or the new SAT.  Some will allow splits; for example, they may consider your best math score from an old SAT and your best verbal score from a new SAT.  For details, as Joel said, you should check the web sites of your favorite colleges.

I took the SAT I this Saturday and I had a verbal section that was not experimental that used the short passages (all were 1 parragraph in length).  I usually have trouble with reading comprehension, but those questions seemed a lot easier than the questions with the long passages.  I hope that's a plus for you guys who have to take the new SAT.


I am going to try to create a game. I will explain the rules when the game starts. I need 10 people. You [i]must[/i] sign up in PM or else you won't be playing. If you have played Whoonu before, then you will have some small idea of what's in store.

Please post rules when signups begin, not later!

Nobody wants a surprise?

Oh well.

First Phase(10-4 people): I post a topic and people send in answers. I arrange the answers from most favorite to least favorite. The sender of the least favorite is eliminated.

Second Phase(4 people): We have a match with the final four people, who are split into two groups of two. The match is a classic round of Whoonu: I give them each five cards, and they send me two cards of which they think that I will like the most. In each pair of people I will post them in order, and they will get points based on that: favorite gets four points, second gets 3, etc. Ties will be broken by the first phase results. If the tie is still not broken then they will go into Rock Paper Scissors contest.

Third Phase(top two): They will guess a number that I have picked from 1 to 50. The closer one gets 1/2 point.
Then they will each guess a word with points a=1, b=2, c=3, etcetera. The closer total of the word to my total of my chosen word gets 1 point.
Now they will go through a Rock Paper Scissors *insert item here* round. They know what wins and loses with Rock, Paper, and Scissors, but they do not know what the mystery item is and how it wins. But it is guaranteed that it is possible to win and lose with it. Best 2/3, winner gets 1/2 point.
If nobody has reached 1.5 points before this phase is over, they will have to guess my favorite color, in hexadecimal. The closer one when written in base 10 wins the game.

At least you know what's coming[size=0]Beat the Mod...[/size]

/in :(  :(  :( there are probably few people playing.

[quote="1=2"]I am going to try to create a game. I will explain the rules when the game starts. I need 10 people. [size=200]You [i]must[/i] sign up in PM[/size] or else you won't be playing. If you have played Whoonu before, then you will have some small idea of what's in store.[/quote]

Read more carefully next time!

Guys nobody's signing up. Are they [i]chicken[/i]?


calculus
integration
function
symmetry
calculus computations

hi,

i would like that somebody helps me to compute this integral :

${f(x) = \frac{\sqrt{Ln(9-x)}}{\sqrt{Ln(9-x)}+\sqrt{Ln(x+1)}}}$

compute : $\int_{2}^{4}f(x)dx$

i have another small question :my teacher told us that nobody has found a function S that that $S'(x) = \frac{sin(x)}{x}$,is that right ?

thanks in advance

Are you sure that's not $\int_{2}^{6}f(x)\,dx$? There's a symmetry trick involved...

To your other question: It's not "nobody has found"; it's that the antiderivative is not an [url=http://en.wikipedia.org/wiki/Elementary_function]elementary function[/url].

sorry,but it says that is  2 -> 4 not 2 -> 6 .


There are a number of charts that I've posted in this forum in years past that I should update with this year's data. I haven't done the work yet, but I will. In the meantime, I think jmerry has an item or two that he could post.

jmerry, in terms of the hardest individual questions, it looks like 8 people solved #10 this year.  In 2006, 3 people solved #8.  In 1997 and 2007, 6 people solved #8.  This year's problem is tied with 1997 I-2 (!) with 8 solvers.  (I'm getting the 1997 data from the green book.)

Here are some other numbers.

Best secondary teams (not the top team in their region):
1. Thomas Jefferson B1 156
2. SoCal A2 152
3. Bay Area A2 148
4. Georgia A2 146
5. Lehigh Ice 144
6. New York City N 132
7. San Diego Sun 130
8. New York City Y 129

Standings before the relay rounds:
1. Lehigh 190
2. NYC 188
3. Montgomery 182
4. SoCal 180
5. SFBA 179
6. North Carolina 178
7. San Diego 172
8. Exeter 171
9. AAST 170
10. Chicago 163
11. Michigan 163
SFBA, Exeter, and Indiana all leapfrogged quite a few teams on the relay to get to their final positions.

Division A will be growing again next year with the addition of the primary teams from Phillips Academy, Central Jersey, Utah, and Maine, and secondary teams from Thomas Jefferson, Lehigh Valley, and San Diego.

Finally, I think congratulations are due to San Diego, who went from not existing in 2006, to winning B division national awards in 2007, to winning national awards in [i]both[/i] divisions in 2009!

[quote="Sly Si"]...San Diego, who went from not existing in 2006, ...[/quote]
There's a slight pre-history there. If you go well back into ARML history in the 1990's you'll find an occasional "Torrey Pines" team - but those teams died out. Then, there were four San Diego residents on the Southern California teams in 2004, two on the A team and two on the B team, and there were two San Diego residents on the 2005 SoCal teams, one on the A team and one on the B team. Mostly, these were 11th or 12th graders at the time, although one of the 2004 participants was an 8th grader at the time. (He went on to contribute to the North Carolina ARML team for many years.) One 2006 9th grader attended SoCal ARML meetings through most of the 2005-2006 school year; when it became clear that a San Diego team would be created, he joined that team and has been one of its leaders throughout its entire existence.

But yes, that is a remarkably swift rise from no organization to becoming one of the handful of strongest organization in the country.

That's good to know about the 1997 problems. I'm using points per team as the measure, so this year's #10 still ranks ahead of them.

I'm not entirely sure about #7 (Maine) moving up; it was six last year.
There will be at least one team moving down: Chicago A3 was in their last year and failed to requalify. Also, if it's only the top 6 in Div. B moving up, Wisconsin Red drops out of Div. A. If it's seven, their 128 this year is barely enough to requalify.
The Alabama team didn't come this year, so we lost an established A team that way. They might come back next year. Texas is in disarray, and one or both of their A teams might not make it next time.
Assuming nothing else changes, that gives 46-51 A teams next year, up from 44 this year.

Acutally, next year I think I'm going to register all 6 NYC teams in the A division, so the A division will grow by an additional 3 teams.

You do realize that nobody [b]ever[/b] does that? OK, Ontario entered as an A team back in 1989, and AAST inexplicably listed their freshman team (97th overall) as an A team in 2008. In the last 15 years, there are more examples of the division B winner not competing in division A the next year than there are of division A teams which didn't get there by finishing high enough in recent years.

(Those examples: Ontario B1 in 2001 and Peninsula/South Bay in 2000)

I've been thinking about it and have reached the conclusion that because New York City fields an A division team, all its teams should be in the A division.

Hmmm... what if Jimmy's attitude were expressed as a universal rule? I smell a "what if".

Meet your new Division B champions:

Andover 150 (But see note 1.)
Central Jersey 136
Utah 129
Maine 126
Nassau 124
Western Penn 117
Grissom 115 (But see note 2.)
Howard County 113
Suffolk 112
Nebraska 111

Note 1: Andover was formed by secession from within the territory of a Division A organization (Eastern Massachusetts).

Note 2: Grissom is a partial replacement for a Division A organization that failed to field a team this year (Alabama).

Relsut Team Viet Nam??Can anyone help me?
I can not find them.

Did Viet Nam enter this year? I don't see them listed in either International division. I checked the ARML internal scoring site, not just the published results.

Vietnam began participating in year 2008. I am extremely surprised to find himself not to join within 2 final exam and haven't result.

[quote="Kent Merryfield"]Here's an exercise in the long-term ranking of teams. I have solid data back to 2003, which is 7 years. In the following table, the numbers are overall ranks (paying no attention to whether the team was in Division A or Division B), with ties averaged. The column "7-year rank" is then the sum of the ranks. All teams are A1 teams unless indicated otherwise. In some cases, I have granted continuity across different names, such as giving Ontario West the history of Massey Ontario and this year's Grissom team the history of Alabama. (I'm not sure what's going on with my history of Chicago and NYC teams other than their A1 teams - internal names have changed and I may be drawing the wrong line.)

(stuff)

Edit: I recombined the Chicago scores following some suggestions made by Coach Fritz. The new version clearly shows Chicago A2 (and its various previous names) as the best #2 team of an organization over these 7 years, despite its lower than usual finish in 2009.

Edit again: I corrected the standings for Exeter and for every other team affected by the Exeter/Andover flip. And for the fact that I somehow left out Montgomery, Iowa, and Grissom/Alabama. (Sorry!)[/quote]

I would be interested to see this with a harmonic mean or similar.  This would penalize teams with a single bad year less (i.e. a theoretical 1/1/1/1/20 would beat a theoretical 2/2/2/2/2).

Cool, thanks!  The (rank+10) is better, I didn't think of that.

Here's a scheme using the same 7 years of data as in my post above. For a team of rank $ r,$ give $ 2^{(1 \minus{} r)/10}$ points. Add points. Zero for teams that didn't exist. The exponential decay of points gives less credit for finishes outside the top 20 or so than jmerry's scheme. Teams like San Diego and Florida are included, with zeros for the years when they had no team.

1. Exeter 5.611
2. NYC 5.340
3. Mngmry 5.198
4. Chicago 5.067
5. Lehigh 5.026
6. SFBA 5.000
7. N.Car 4.723
8. TJ 4.701
9. Georgia 3.988
10. Texas 3.881
11. AAST 3.494
12. E. MA 3.325
13. SoCal 3.277
14. UpstateNY 2.691
15. Minn 2.470
16. Wash 2.230
17. Mich 2.122
18. Indiana 1.949
19. Ont. West 1.911 (Includes Massey Ontario)
20. Missouri 1.471
21. Alabama 1.353 (Includes 2009 Grissom)
22. Chi. A2 1.318 (Top A2 team)
23. Wisc. 1.301
24. Florida 1.259
25. San Diego 1.179
26. Iowa 1.111
27. CT 1.092
28. Colorado 1.016
29. NYC N 0.974 (A2 team)
30. Georgia B 0.955 (A2 team)
31. SFBA B 0.882 (A2 team)
32. Ohio 0.859
33. TX Silver 0.856 (A2 team)
34. TJ B 0.842 (A2 team - now)
35. S. Car. 0.819
36. Taiwan 0.765 (In only 4 of the 7 years)
37. NYC Y 0.687 (Top A3 team)
38. SoCal A2 0.661 (A2 team)
39. Chspk 0.632 (Includes Maryland Homeschool)
40. Chi. C 0.623 (A3 team)

The 10 in the $ 2^{(1\minus{}r)/10}$ is arbitrary - you could try making that 6, or 12, or whatever.

Well, some of the problems are, except that there's a tighter time restriction on ARML rounds. However most problems are like easy to medium AIME level.

Also, AIME is a purely individual contest, while ARML is mainly team-oriented.  Only one of the rounds is taken individually; the other three (Team, Power, and Relays) are taken by the entire team.


topology
advanced fields
advanced fields unsolved

let $ X \equal{} \{0\}\cup \left\{1,\frac {1}{2}\ ,\frac {1}{3}\,....\right\}$  and $ Y$ be the countable discrete space .prove that $ X$ and $ Y$ don't have the same homotopy type.

There is no room for homotopy in either space, so "homotopy equivalent" is the same as "homeomorphic" for them. And they are not homeomorphic: there is no injective continuous map from $ X$ into $ Y$.

Are we viewing $ X$ as a subspace of $ \mathbb{R}$?

I assume $ X$ has topology induced from $ \mathbb R$.
Are there other interpretations?

Oh, sorry I wasn't thinking about the fact that $ \{0\}$ was adjoined and hence I was thinking that every singleton was open...


[quote="lingomaniac88"]On a piano, the frequencies (in Hertz) of adjacent keys form a geometric progression.  The keys are named as follows, by their pitch:

...C, C#, D, Eb, E, F, F#, G, Ab, A, Bb, B, C, C#, D, Eb, E, F, F#, G, Ab, A, Bb, B, C, C#, D, Eb...

There exists an A that has a frequency of 440 Hz.  The next highest A has a frequency of 880 Hz.

Humans can hear sounds with frequencies between 20 Hz and 20,000 Hz.  What is the lowest note that a human can hear?[/quote]
[hide]
should be E
the lowest A humans can hear is A27.5
Since there are 12 half steps in an octave, each half step higher multiplies the frequency by $2^{\frac{1}{12}}$. Each half step lower multiplies the frequency by $2^{\frac{-1}{12}}$. An octive higher would be double the frequency, while an octive lower would be half the frequency.
$27.5*2^{\frac{-5}{12}}=20.60172230705437060849011006541$
5 half steps below A is E
that is around the lowest frequency heard.(20Hz)
the nearest note below E is Eb, which is too low (19.4Hz)
[/hide]


What is the largest of three consecutive odd integers whose sum is 57?

Dividing $ 57$ we get $ \boxed{19}$.

Edit: Ah, failed...... Forget it said largest, so the answer should be $ 19\plus{}2$ or $ \boxed{21}$.

Let's make the first integer x, the second x +2, and the third x+4. Adding all these together gives us 3x+6=57. Thus, the first integer is 17 and because we're looking for the largest of these integers, 17+4=21.


logarithms
quadratics
algebra
polynomial
quadratic formula

[hide="solution."]That can be re-written as

$\log_{3}x^{2}=\log_{3}\left(10x+8\right)-1=\log{3}\frac{10x+8}{3}$

So

$x^{2}=\frac{10x+8}{3}$

$3x^{2}=10x+8$

We use the quadratic formula, or factor, and get x=-2/3 or 4, but the first solution is extranneous(I know, I can't spell), so the only solution is x=4[/hide]

[hide="@ Glenn:"]You can't have a logarithm of a negative.[/hide]

so I was right, just wrong int he fact that I included the negative root >_<

actually it never says x is real, meaning a complex number solution probably exists

[quote="Bill The Maniac"]Fundamental Theorem of Algebra says two complex roots of a quadratic equation. They are 4, -2/3. No other roots appear.[/quote]
The fundamental theorem of algebra says that every polynomial has at least one complex root I thought.
But -2/3 doesn't work anyways, I was wrong (when I attempted it)
So 1=2 was right.


A convex polygon $\mathcal{P}$ in the plane is dissected into smaller convex polygons by drawing all of its diagonals. The lengths of all sides and all diagonals of the polygon $\mathcal{P}$ are rational numbers. Prove that the lengths of all sides of all polygons in the dissection are also rational numbers.

When replying to the problem, I ask that you make posts for solutions and submit comments, jokes, smilies, etc. separately. Please use LaTeX for posting solutions. Thanks.

[hide]Lemma 1: The problem's conclusion holds in the case where the polygon is a quadrilateral.

Proof of Lemma 1: Suppose a quadrilateral $ABCD$ (whose diagonals intersect at $O$) has rational sides and diagonals.  We wish to prove that $OA$, $OB$, $OC$, and $OD$ are all rational.

Let $\angle AOB=\theta$, so $\angle COD=\theta$ and $\angle BOC=\angle AOD=180^\circ-\theta$.  Note that $\cos(\theta)=-\cos(180^\circ-\theta)$.  Using this knowledge and applying the Law of Cosines to triangles $AOB$, $BOC$, $COD$, and $DOA$, we get

\begin{eqnarray} (AO)^2+(BO)^2-2(AO)(BO)\cos(\theta)&=&(AB)^2\\ (CO)^2+(DO)^2-2(CO)(DO)\cos(\theta)&=&(CD)^2\\ (BO)^2+(CO)^2+2(BO)(CO)\cos(\theta)&=&(BC)^2\\ (AO)^2+(DO)^2+2(AO)(DO)\cos(\theta)&=&(AD)^2 \end{eqnarray}

Taking $(3)+(4)-(1)-(2)$ and factoring, we get

\[ 2((AO)+(CO))((BO)+(DO))\cos(\theta)=2(AC)(BD)\cos(\theta)=(AD)^2+(BC)^2-(AB)^2-(CD)^2  \]

Since $AC$, $BD$, $AD$, $BC$, $AB$, and $CD$ are all rational, it follows that $\cos(\theta)$ is also rational, since $AC$ and $BD$ must be non-zero, so $\cos(\theta)$ irrational would imply that $2(AC)(BD)\cos(\theta)$ is irrational while $(AD)^2+(BC)^2-(AB)^2-(CD)^2$ is rational, an impossibility.

Now, take $(1)-(3)+(4)-(2)$.  We get

\[ (AO)^2-(CO)^2+(AO)^2-(CO)^2-2(BO)((AO)+(CO))\cos(\theta)-2(DO)((AO)+(CO))\cos(\theta)=(AB)^2+(AD)^2-(BC)^2-(CD)^2  \]

Factoring this and condensing, we obtain

\[ ((AO)-(CO))(2(AC)-2(BD)\cos(\theta)=(AB)^2+(AD)^2-(BC)^2-(CD)^2  \]

$\cos(\theta)$, $AC$, $BD$, and the RHS of the above are all rational, so $(AO)-(CO)$ must also be rational.  Thus, we have $(AO)+(CO)$ rational and $(AO)-(CO)$ rational; it follows that both $AO$ and $CO$ are themselves rational.  Likewise, $BO$ and $DO$ are rational, and we're finally done with this case.

End Lemma 1

Consider an $n$-gon with rational sides and diagonals.  Specifically, consider one of its diagonals (call it $XY$), and its intersections with various other diagonals.  Travelling from $X$ to $Y$, let us label these intersections $A_1$, $A_2$, and so on, up to $A_n$.  Let the diagonal that intersects with $XY$ to form point $A_1$ be $P_1Q_1$, and consider quadrilateral $XP_1YQ_1$.  Since our $n$-gon has all rational sides and diagonals, it follows that all the sides and diagonals of this quadrilateral are rational, since they are all either sides or diagonals of the $n$-gon.  By Lemma 1, we find that $XA_1$ is rational.  Likewise, we can find that $XA_2$ is rational, $XA_3$ is rational, and so on.  Thus, for any $p$, $q$, with $p>q$, we must have $A_pA_q$ rational, since $(A_pA_q)=(XA_p)-(XA_q)$.  Therefore, all the small segments cut off on this diagonal are rational.  Likewise, all the small segments cut off on any other diagonal are also rational.  Therefore, the sides of all the small polygons in the dissection of the $n$-gon are all rational.[/hide]

Sorry if this is considered reviving an old, dead thread, but this was a Contest Problem without a solution...

We can also prove the lemma as follows :

By Law of Cosines, $\cos BAC = \frac{AB^{2}+AC^{2}-BC^{2}}{2 AB \cdot AC}$, which is rational.  Similarly, $\cos CAD$ and $\cos BAD = \cos BAC \cos CAD-\sin BAC \sin CAD$ are rational.  It follows that $\sin BAC \sin CAD$ is rational, as well as $\sin^{2}CAD = 1-\cos^{2}CAD$, so $\frac{\sin BAC \sin CAD}{\sin^{2}CAD}= \frac{\sin BAC}{\sin CAD}$ is rational.  Since $\angle BAC = \angle BAO$ and $\angle CAD = \angle OAD$, $\frac{AB \sin BAO}{AD \sin OAD}= \frac{\tfrac{1}{2}AB \cdot AO \cdot \sin BAO}{\tfrac{1}{2}AO \cdot AD \cdot \sin OAD}= \frac{[BAO]}{[AOD]}= \frac{BO}{OD}$ is rational.  Since $BO+OD= BD$ is also rational, $BO$ and $OD$ must also be rational.  The rest follows by symmetry. $\blacksquare$

Suppose $AB$ is a side of a polygon in the dissection, lying on diagonal $XY$, with $X$, $A$, $B$, $Y$ in that order. Then $AB = XY - XA - YB$. In this way, we see that it actually just suffices to prove the result for a quadrilateral.

To do this, we apply barycentric coordinates. Consider quadrilateral $ABDC$, with $A=(1,0,0)$, $B=(0,1,0)$, $C=(0,0,1)$. Let $D = (x,y,z)$, with $x+y+z=1$. By hypothesis, each of the numbers
\begin{align*}
	-a^2yz + b^2(1-x)z + c^2(1-x)y &= AD^2 \\
	a^2(1-y)z + b^2zx + c^2(1-y)x &= BD^2 \\
	-a^2(1-z)y - b^2(1-z)x + c^2xy &= CD^2
\end{align*}
is rational. Let $W = a^2yz + b^2zx + c^2xy$. Then,
\begin{align*}
	b^2z + c^2y &=  AD^2 + W \\
	a^2z + c^2x &= BD^2 + W \\
	a^2y + b^2x &= CD^2 + W.
\end{align*}
This implies that $AD^2 + BD^2 + 2W - c^2 = 2S_C z$ and cyclically (as usual $2S_C = a^2+b^2-c^2)$.  If any of $S_A$, $S_B$, $S_C$ are zero, then we deduce $W$ is rational. Otherwise, we have that \[ 1 = x+y+z = \sum_{\text{cyc}} \frac{AD^2 + BD^2 + 2W - c^2}{2S_C} \] which implies that $W$ is rational, because it appears with coefficient $\frac{1}{S_A} + \frac{1}{S_B} + \frac{1}{S_C} \neq 0$ (since $S_{BC} + S_{CA} + S_{AB}$ is actually the area of $ABC$).

Hence from the rationality of $W$, we deduce that $x$ is rational as long as $S_A \neq 0$, and similarly for the others.  So at most one of $x$, $y$, $z$ is irrational, but since $x+y+z=1$ this implies they are all rational.

Finally, if $P = \overline{AD} \cap \overline{BC}$ then $AP = \frac{1}{y+z} AD$, so $AP$ is rational too, completing the proof.

Clearly, it suffices to show that for points $A,B,C,D$ with all side lengths and diagonals between them rational and $AC,BD$ intersecting at $P$, $AP$ is rational. But since $\frac{AP}{BP}=\frac{AD}{BC}$, $BP$ is a rational multiple of $AP$. Similarly $CP$ is a rational multiple of $BP$, so $CP+AP=AC$ is a rational multiple of $AP$. But $AC$ is rational, so $AP$ is as well.

EDIT: Oops, totally misread, thought it was cyclic. The same lemma should work, I'll see if I can bash out the details later (should be much worse now :/)

[quote=pi37]$\frac{AP}{BP}=\frac{AD}{BC}$[/quote] Unless I'm misunderstanding, this isn't true in general? For example, take $ABCD$ to be a parallelogram which is not a rectangle.

One can also use cartesian coordinate to prove the same lemma.

It's suffice to prove the result for a quadrilateral.

Let $P=AC\cap BD$, since $AP+CP$ is rational, we only need to prove $\dfrac {AP}{CP}=\dfrac {[BAD]}{[BCD]}$ is rational. Let $\sqrt {a}=[BAD]$, where $a$ is a rational number, define $\sqrt {b}, \sqrt {c}, \sqrt {d}$ similarly. Since $\sqrt {a}+\sqrt {c}=\sqrt {b}+\sqrt {d}$, so if $\dfrac {\sqrt {a}}{\sqrt {c}}$ is not rational, then we must have $\{ a, c\} =\{ b, d\} $, which means $ABCD$ must be a trapezoid. The result is obviously true for the case $ABCD$ is a trapezoid.$\Box $

[quote=Leooooo]$\sqrt {a}+\sqrt {c}=\sqrt {b}+\sqrt {d}$, so if $\dfrac {\sqrt {a}}{\sqrt {c}}$ is not rational, then we must have $\{ a, c\} =\{ b, d\} $[/quote]
[hide=My explanation for this part]From $\sqrt{a}+\sqrt c= \sqrt b+ \sqrt d$ then $\left( \sqrt{a/c}+1 \right)^2=1/c \left(\sqrt b+ \sqrt d \right)^2$ or $2\sqrt{bd}/c= 2\sqrt{a/c} +\frac{a+c-b-d}{c}$. If $a+c \ne b+d$ then if we square this we obtain $\sqrt{a/c}$ is rational. If $a+c=b+d$, come back to $(\sqrt a+ \sqrt c)^2= \left( \sqrt b+ \sqrt d \right)^2$ and get $ac=bd$. Thus, pairs $(a,c)$ and $(b,d)$ are solutions to same quadratic equation so $\{ a,c \}= \{b,d \}$.[/hide]

Clearly it suffices to show the result for a quadrilateral $A_1A_2A_3A_4$. Let $X=A_1A_3\cap A_2A_4$, $a_i=A_iA_{i+1}$, and $x_i=XA_i$. Let $\alpha_i=\angle XA_iA_{i+1}$, and $\beta_i=\angle XA_{i+1}A_i$. By the law of cosines, we have $\cos\alpha_i,\cos\beta_i\in\mathbb{Q}$. We have the equations
\[x_i\cos\alpha_i+x_{i+1}\cos\beta_i=a_i,\]
where the indices are taken mod $4$.

We see that $x_1+x_3$ and $x_2+x_4$ are rational, so letting $x_1=p$ and $x_2=q$, it suffices to show that $p$ and $q$ are rational. Indeed, the four equations transform to
\begin{align*}
p\cos\alpha_1 + q\cos\beta_1 &= r_1 \\
-p\cos\beta_2+q\cos\alpha_2 &= r_2.
\end{align*}
If the determinant of this system is nonzero, then we're done, so WLOG we may assume that we have
\[\cos\alpha_1\cos\alpha_2 + \cos\beta_1\cos\beta_2=0.\]
By symmetry then, if this was nonzero if we replaced $1$ with $i$ and $2$ with $i+1$, then we'd also be done, so we may assume that
\[\cos\alpha_i\cos\alpha_{i+1} + \cos\beta_i\cos\beta_{i+1}=0\]
for all $i$. Note that $\alpha_i+\beta_i+\alpha_{i+1}+\beta_{i+1}=\pi$, so using product to sum, we get
\[\cos(\alpha_i-\alpha_{i+1}) + \cos(\beta_i-\beta_{i+1}) = 0.\]
For a given $i$, we then either have
\[\alpha_i-\alpha_{i+1} + \beta_i-\beta_{i+1} = (2k+1)\pi\]
or
\[[\alpha_i-\alpha_{i+1} -( \beta_i-\beta_{i+1} )= (2k+1)\pi. \]
Letting $\theta=\angle A_iXA_{i+1}$, the first case implies
\[\pi-\theta - (\theta) = (2k+1)\pi,\]
so $\theta= -k\pi$, which is absurd as $0<\theta<\pi$. Thus, we always have the second case, so
\[(\alpha_i-\beta_i)-(\alpha_{i+1}-\beta_{i+1})=(2k+1)\pi\]
for some integer $k$. Thus,
\[(\alpha_i-\beta_i)-(\alpha_{i+2}-\beta_{i+2}) = 2k\pi\]
for some integer $k$. However, $-\pi<\alpha_i-\beta_i<\pi$, so we in fact have
\[\alpha_i-\beta_i = \alpha_{i+2}-\beta_{i+2}.\]
But we clearly also have $\alpha_i+\beta_i=\alpha_{i+2}+\beta_{i+2}$, so in fact $\alpha_i=\alpha_{i+2}$ and $\beta_i=\beta_{i+2}$. This means that $ABCD$ is a parallelogram.

Thus, as long as $ABCD$ is not a parallelogram, we're done. Let's now solve the problem in the case that it is a parallelogram. In this case, $X$ is just the midpoint of $AC$ and $BD$, so $XA=AC/2$ and so on, which are clearly rational.

We will only prove the result for a quadrilateral; it's easy to see that this suffices.

Let the quadrilateral be $ABCD$ and $AC\cap BD=P$. Let $\theta$ be the acute angle between $AC$ and $BD$. We will prove the following claim.
-----
[color=red][b]Claim:[/b][/color] $\cos\theta\in\mathbb Q$.

[i]Proof:[/i] If $AC\perp BD$, the result is obvious. Otherwise, let $H_B, H_D$ be the projections of $B,D$ onto $AC$. Note by Law of Cosines that $\cos\angle BAC\in\mathbb Q$; thus,
$$AH_B = AB\cos\angle BAC \in\mathbb{Q}.$$
Similarly, $AH_D\in\mathbb Q$, so $H_BH_D\in\mathbb{Q}$. Observe that $BD = H_BH_D\cos\theta$, done. $\blacksquare$
-----
For any positive reals $a,b$, we say that $a\sim b$ iff $a/b\in\mathbb{Q}$. If $\alpha=\angle PAB$, and $\beta=\angle PBA$, we get that
$$\mathbb Q\ni \cos\theta = |\cos(\alpha + \beta)| = |\underbrace{\cos\alpha\cos\beta}_{\in\mathbb Q} - \sin\alpha\sin\beta|$$
or $\sin\alpha\sin\beta\in\mathbb{Q}$. However, $\sin^2\alpha = 1-\cos^2\alpha\in\mathbb{Q}$. Thus,
$$\sin\alpha\sin\beta \sim \sin^2\alpha \implies \sin\alpha\sim\sin\beta.$$
Moreover, the law of sine of $\triangle ABD$ gives
$$\sin\angle PAB \sim \sin\angle PBA \sim \sin \angle DAB$$
Hence, $[ABP]\sim [ABD]$ or $BP\sim BD$. Hence, $BP\in\mathbb{Q}$ as desired.

Without loss of generality, consider any diagonal $u$. If we prove that any other diagonal $v$ that intersects with $u$ at a point which is a rational distance away from the endpoints if $u$, then all segments on $u$ will be rational, which we can extend to all diagonals. Let $A,C$ be endpoints of $u$ and $B,D$ be endpoints of $v$ and we see that we have reduced the problem to the same problem on not a polygon but a quadrilateral $ABCD$.
\\$~$\\
Let $AB=w,BC=x,CD=y,DA=z,AC=p,BD=q.$ These numbers are all rational numbers. Now, let $AC$ and $BD$ intersect at $E$, and let $AE=p_1,CE=p_2,BE=q_1,DE=q_2.$ By Stewart's theorem, we have the relations
\begin{align*}
p_1pp_2+q_1^2p &= w^2 p_2 + x^2p_1 \\
p_1pp_2+q_2^2p &= z^2 p_2 + y^2p_1 \\
q_1qq_2+p_1^2q &= w^2 q_2 + z^2q_1 \\
q_1qq_2+q_1^2q &= x^2 q_2 + y^2p_1
\end{align*}
If we subtract the first two and the last two, we get
\begin{align*}
pq(q_1-q_2) &= p_2(w^2-z^2)+p_1(x^2-y^2) \\
pq(p_1-p_2) &= q_2(w^2-x^2)+q_1(z^2-y^2)
\end{align*}
Now, if we let $c_1=pq,c_2=w^2-z^2,c_3=x^2-y^2,c_4=w^2-x^2,c_5=z^2-y^2$ then we can use the identity $c_2p_2+c_3p_1=\left(\tfrac{c_2+c_3}{2}\right)(p_1+p_2)+\left(\tfrac{c_2-c_3}{2}\right)(p_1-p_2).$ Let $r_1=p_1-p_2,r_2=q_1-q_2$ and let $s=\tfrac{c_2-c_3}{2}=\tfrac{c_4-c_5}{2}$ then
\begin{align*}
c_1r_1 &= \text{rational} + sr_2 \\
c_1r_2 &= \text{rational} + sr_1
\end{align*}
which has a unique solution if and only if $c_1\neq s.$ If it doesn't have a unique solution, it either has no solution, in which case we are fine since the quadrilateral doesn't exist, or it has infinite solutions, in which case we are still fine because all solutions are rational. Therefore, we are done.


I looked at 1=2's profile and it says he has only 2 posts, but I remember he used to have a crazy number of posts in the 2000's area. Is this a glitch on the server or only on my computer?


${ 3(a_{n+1}-a_n})^2 = 2(a_{n+1}+a_n)$
$ a_0 = 0$

What are the tricks for solving recurrence in general?

Coin $ 3(a_{n\plus{}1}\minus{}a_n)^2\equal{}2(a_{n\plus{}1}\plus{}a_n)$ and subtract both sides this from the given reccurence.

Sorry, but what do you mean by "the meaning turn out"... English was not my first language...

let $ b_n \equal{} a_{n\plus{}1}\minus{}a_n$.  Then we have

$ 3b_n^2 \equal{} 2a_0 \plus{} 2\sum_{i\equal{}0}^n b_i$
$ 3b_n^2 \minus{} 2b_n\plus{}2/3 \equal{} 2/3 \plus{} 2a_0 \plus{} 2\sum_{i\equal{}0}^{n\minus{}1} b_i$
$ (b_n\minus{}1/3)^2 \equal{} 1/3 (2/3 \plus{} 2a_0 \plus{} 2\sum_{i\equal{}0}^{n\minus{}1} b_i)$
$ b_n \equal{} 1/3 \pm \sqrt{1/3 (2/3 \plus{} 2a_0 \plus{} 2\sum_{i\equal{}0}^{n\minus{}1} b_i)}$

doesn't seem to help... so this shows how you could calculate $ a_n$ but not solve...

another try:

$ 3a_{n\plus{}1}^2 \minus{} 6a_{n\plus{}1}a_n \plus{} 3a_n^2 \equal{} 2a_{n\plus{}1} \plus{} 2a_n$

stuck... you can try playing around with the subscripts, such as using $ n\plus{}1$ and $ n\plus{}2$ instead of $ n$ and $ n\plus{}1$ and playing with the two equations somehow.  The issue here is that to make any major progress towards an explicit formula, we need a workable recursive formula.  and I don't see one yet.

Let's go back to that strange title you had -- $ 3(s \minus{} r)^2 \equal{} 2(s \plus{} r)$. Solving for $ s$, we get:

$ s \equal{} \frac {3r \plus{} 1 \pm \sqrt {12r \plus{} 1}}{3}$

The important part of that is the $ \pm$. In particular, for any $ a_n$, there are two possible $ a_{n \plus{} 1}$ that follow. Thus, we do not have a unique sequence (in fact, we have an infinite number)!

Having said that, let $ a$ be of finite degree, and let $ D$ be the degree of $ s,r$. We know the degree of $ s \minus{} r$ is $ D \minus{} 1$ and the degree of $ s \plus{} r$ is $ D$. Thus, $ 2(D \minus{} 1) \equal{} D \implies D \equal{} 2$, ie $ a$ can be approximated by a quadratic for most $ n$. Hunting for quadratics that fit, we find that $ a_n \equal{} \frac {n^2 \plus{} n}{3}$ and $ a_n \equal{} \frac {n^2 \minus{} n}{3}$ are two particular sequences that work.

$ 3(a_{n \plus{} 2} \minus{} a_{n \plus{} 1})^2 \equal{} 2(a_{n \plus{} 2} \plus{} a_{n\plus{} 1})$
$ 3(a_{n \plus{} 1} \minus{} a_n)^2 \equal{} 2(a_{n \plus{} 1} \plus{} a_n)$

Subtracting both sides gives $ a_{n \plus{} 2} \equal{} a_n$ or $ a_{n \plus{} 2} \minus{} 6a_{n \plus{} 1} \plus{} 5a_n \equal{} 0$.

kunny, I don't see where you get your answers there... please recheck it and post your work.

anyway, lets assume $ a_n$ is a quadratic.  Then $ a_n \equal{} an^2 \plus{} bn \plus{} c$.  Plugging this in, we have

$ 3(a(n \plus{} 1)^2 \plus{} b(n \plus{} 1) \plus{} c \minus{} an^2 \minus{} bn \minus{} c)^2 \equal{} 2(a(n \plus{} 1)^2 \plus{} b(n \plus{} 1) \plus{} c \plus{} an^2 \plus{} bn \plus{} c)$
$ 3(2an \plus{} a \plus{} b)^2 \equal{} 2(2an^2 \plus{} 2(a \plus{} b)n \plus{} a \plus{} b \plus{} 2c)$
$ 3(4a^2n^2 \plus{} 4a(b \plus{} a)n \plus{} (b \plus{} a)^2) \equal{} 4an^2 \plus{} 4(a \plus{} b)n \plus{} 4c \plus{} 2b \plus{} 2a$
$ 12a^2 n^2 \plus{} 12a(b \plus{} a)n \plus{} 3(b \plus{} a)^2 \equal{} 4an^2 \plus{} 4(a \plus{} b)n \plus{} 4c \plus{} 2b \plus{} 2a$

equating coefficients
$ 12a^2 \equal{} 4a \implies 3a^2 \equal{} a$ so $ a \equal{} 0, 1/3$.

assuming $ a \equal{} 1/3$
$ 4(b \plus{} a) \equal{} 4(a \plus{} b)$ no new infro
$ 3(b \plus{} 1/3)^2 \equal{} 4c \plus{} 2b \plus{} 2/3$
$ c \equal{} 1/4(3(b \plus{} 1/3)^2 \minus{} 2b \minus{} 2/3) \equal{} 3/4 b^2 \minus{}1/4$

so we have solutions of the form $ a_n \equal{} 1/3 n^2 \plus{} bn \plus{} 3/4b^2\minus{}1/4$ for some number $ b$.

[quote="mathwizarddude"]What are the tricks for solving recurrence in general?[/quote]
Most of these tricks rely on a clever use of [url=http://www.math.upenn.edu/~wilf/gfology2.pdf]generating functions[/url]:  good examples are the derivations of the generating functions of the Catalan numbers and the Bell numbers.  Unfortunately, the nature of these tricks makes them behave very badly with respect to squaring, so it's unlikely that they're of much use here.  (Squaring makes recurrences very hard - usually [url=http://mathworld.wolfram.com/QuadraticMap.html]impossible[/url] - to solve in closed form.)

My biggest issue is the lack of a unique answer.  What's the source?

[quote="kunny"]Sorry for my poor English, I mean that I used coin as make or produce artficially.[/quote]  The verb "coin" is used primarily for linguistic references, e.g., "to coin a word" or "to coin a phrase."  It's unusual to see it outside of that context (except, perhaps, in its original meaning, viz., "to produce a coin").

[quote="t0rajir0u"]My biggest issue is the lack of a unique answer.  What's the source?[/quote]
Hey, when did you become a moderator?  :wink:
I came up with this recurrence while solving an AMC problem, but it seems there are much better solutions.

[quote="kunny"]$ 3(a_{n \plus{} 2} \minus{} a_{n \plus{} 1})^2 \equal{} 2(a_{n \plus{} 2} \plus{} a_{ n\plus{} 1})$
$ 3(a_{n \plus{} 1} \minus{} a_n)^2 \equal{} 2(a_{n \plus{} 1} \plus{} a_{n})$

Subtracting both sides gives $ a_{n \plus{} 2} \equal{} a_n$ or $ a_{n \plus{} 2} \minus{} 6a_{n \plus{} 1} \plus{} 5a_n \equal{} 0$.[/quote]

This should be that

$ 3(a_{n \plus{} 2} \minus{} a_{n \plus{} 1})^2 \equal{} 2(a_{n \plus{} 2} \plus{} a_{n\plus{} 1})$
$ 3(a_{n \plus{} 1} \minus{} a_n)^2 \equal{} 2(a_{n \plus{} 1} \plus{} a_n)$

Subtracting both sides gives $ a_{n \plus{} 2} \equal{} a_n$ or $ a_{n \plus{} 2} \minus{} 2a_{n \plus{} 1} \plus{} a_n \equal{} \frac {2}{3}$.

$ 1^\circ$ $ a_{n \plus{} 2} \equal{} a_n \Longleftrightarrow a_n \equal{} a_1 \equal{} 0,\ \frac {2}{3} (\text{n: odd}),\ a_n \equal{} a_0 \equal{} 0\ (\text{n: even})$

$ 2^\circ$ $ a_{n \plus{} 2} \minus{} 2a_{n \plus{} 1} \plus{} a_n \equal{} \frac {2}{3}$, Let $ a_{n \plus{} 1} \minus{} a_n \equal{} b_n$ we have $ b_{n \plus{} 1} \minus{} b_n \equal{} b_1 \minus{} b_0$.

Here the reccurence gives $ a_1 \equal{} 0,\ \frac {2}{3}$, thus we get $ a_2 \equal{} 0,\ \frac {2}{3}$ for $ a_1 \equal{} 0$ or $ a_2 \equal{} 0,\ 2$ for $ a_1 \equal{} \frac {2}{3}$.

That follows several case work.

[quote="facis"]anyway, lets assume $ a_n$ is a quadratic.  [/quote]

That's meaningless. :P

kunny:

I'm still not following your subtraction.

and the quadratic is going off of TZF's work.  The solution doesn't have to be a quadratic, but it if it is a polynomial the solution is either a quadratic in $ n$ or else $ a_n\equal{}0$ for all $ n$.

ok, i see that $ 3\{(a_{n \plus{} 2} \minus{} a_{n \plus{} 1})^2 \minus{} (a_{n \plus{} 1} \minus{} a_n)^2\} \equal{} 2(a_{n \plus{} 2} \minus{} a_{n})$ (though I expanded it... which proved not very useful)

now you have $ \frac{(a_{n\plus{}2}\minus{}a_n)(a_{n\plus{}2}\minus{}2a_{n\plus{}1} \plus{}a_n)}{a_{n\plus{}2}\minus{}a_n} \equal{} 2/3$

oh.... ok.  I didn't see where you were getting the two cases before.

but isn't it possible that in for some $ n$ the sequence satisfies $ a_{n\plus{}2} \equal{} a_n$ and for other $ n$ it satisfies $ a_{n\plus{}2} \minus{} 2a_{n\plus{}1}\plus{}a_n \equal{} 2/3$?  because your casework assumes that only one of these recursions holds for all $ n$, which I don't think is a valid assumption.

on a side note, I forgot about the condition $ a_0\equal{}0$ in my previous posts... so the quadratics I found need to be further restricted.