Given the equation
\[ x^4 \plus{} px^3 \plus{} qx^2 \plus{} rx \plus{} s \equal{} 0\]
has four real, positive roots, prove that
(a) $ pr \minus{} 16s \geq 0$
(b) $ q^2 \minus{} 36s \geq 0$
with equality in each case holding if and only if the four roots are equal.

Let's call the four roots $ y_1,y_2,y_3,y_4$

(a) is equivalent to $ y_1y_2y_3y_4(y_1\plus{}y_2\plus{}y_3\plus{}y_4)(\frac{1}{y_1}\plus{}\frac{1}{y_2}\plus{}\frac{1}{y_3}\plus{}\frac{1}{y_4})\geq 16y_1y_2y_3y_4$ by Viete's formulas. We see that this clearly follows from Am-Hm  :)

(b) is equivalent to $ (y_1y_2\plus{}y_1y_3\plus{}y_1y_4\plus{}y_2y_3\plus{}y_2y_4\plus{}y_3y_4)^2\geq 36y_1y_2y_3y_4$ which follows from Am-Gm

Too easy for INMO. Just need to know relations between roots and coefficients and AM-GM-HM inequality.

Let the roots be $\alpha_1, \alpha_2, \alpha_3, \alpha_4$  where  $\alpha_i \in \mathbb{R}^+$
The four equation that we get after applying Vieta's relation are:\\
$\sum \alpha_1=p$\\
$\sum \alpha_1 \alpha_2=q$\\
$\sum \alpha_1 \alpha_2 \alpha_3=r$\\
$\alpha_1 \alpha_2 \alpha_3 \alpha_4=s$\\
Now apply means inequality to the sum of roots which gives us:
\[\dfrac{\sum \alpha_1}{4} \geq \sqrt[4]{\alpha_1 \alpha_2 \alpha_3 \alpha_4} \geq \dfrac{4}{\frac{1}{\alpha_1}+\frac{1}{\alpha_2}+\frac{1}{\alpha_3}+\frac{1}{\alpha_4}} \]
Now, converting expressions with coefficients from Vieta's relation in the inequality gives:
\[\dfrac{p}{4} \geq \sqrt[4]{s} \geq \dfrac{4s}{r}\]
(a) $AM \geq HM$ gives us  $pr \geq 16s$ which henceforth yields $pr-16s \geq 0.$\\
(b) Now by  applying $AM-GM$ inequality to sum of roots taken two at a time, we get:
\[\dfrac{\sum \alpha_1 \alpha_2}{6} \geq \sqrt[6]{\prod \alpha_1 \alpha_2}\]
Which transforms to
\[\dfrac{q}{6} \geq \sqrt[6]{s^3}\]
which simplifies to
\[\dfrac{q}{6} \geq s^{\frac{1}{2}}\]
Now, by squaring both sides
\[q^2 \geq 36s \Rightarrow q^2-36s \geq 0\]


$ f: R\to R$
if $ f(x \plus{} y) \equal{} f(x) \plus{} f(y) \minus{} 2f(xy)$
find $ f(2004)$

Let $ y \equal{} 1$ then we have $ f(x\plus{}1) \equal{} f(x) \plus{} f(1) \minus{} 2f(x) \equal{} f(1) \minus{} f(x)$. We can easily check that $ f(1) \equal{} f(3)$ and that $ f(2) \equal{} f(4) \equal{} 0$. Asume that $ f(2k\minus{}1) \equal{} f(1)$ and $ f(2k) \equal{} 0$ for every $ k \leq n$. Then $ f(2n\plus{}1) \equal{} f(1) \minus{} f(2n) \equal{} f(1)$. By the other hand $ f(2n\plus{}2) \equal{} f(1) \minus{} f(2n\plus{}1) \equal{} 0$ and inductively we get that $ f(2k) \equal{} 0$ for every $ k \in N$. It is enough to say that $ f(2004) \equal{} 0$.


AMC USA(J)MO USAMO AIME Putnam inequalities ratio

[quote="rem"]btw, the 2nd day of USAMO is on same day as Euclid. That is not good. :([/quote]
Is this true? :(  Same thing happened last year, I think.

[quote="Elyot"][quote="rem"]We have been discussing CMO and APMO, so we should discuss USAMO, I suppose.
So, I'll start. I don't have much to talk about so I will start with a relatively stupid question: Is it possible to run out of time on USAMO for a person who is pretty good at math?
btw, the 2nd day of USAMO is on same day as Euclid. That is not good. :([/quote]

No, it's not that great, but every other Canadian writing it will face the same handicap that you face, so it's fair.  It's probably more of a help than a hindrance.  Euclid gets you in the mood to pwn USAMO.[/quote]Not for those further west. . . . I'll be writing the USAMO first. Which probably isn't good, because I'll be lucky to solve one problem.

[quote="rem"]How many problems do you usually need to solve to get 1st?[/quote]

[url=http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2005-ua/05usamo-results.html]USAMO 2005 Mark Distribution[/url]

[url=http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2004-ua/04usamo-results.html]USAMO 2004 Mark Distribution[/url]

[url=http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2003-ua/03usamoscoring.html]USAMO 2003 Mark Distribution[/url]

long story short, perfect or close to perfect.

[quote="rem"]How many problems do you usually need to solve to get 1st?[/quote]

heheh.. go rem go

i predict yufei, rem will make top 3

[quote="rem"]Are the problems solved using "basic theorems" or do we usually need to know crazy identities(ie are crazy formulas necessary to solve the problems?)?
By "crazy" i mean formulas that are not known to many people and that deal with a case with many restrictions.[/quote]

I solved 5 problems on USAMO last year knowing probably less than what you know.  The one I didn't get was a crazy number theory one that you could solve in mod 19 or something crazy.  You really don't need a lot of theorems and stuff.  The problems demand lots of analysis, but no crazy formulae, and not too much chugging.  Expect some of the solutions to be longer though... it's not like APMO where most problems can be solved in under a page or so.

just out of curiosity, Elyot, I didn't see ur name on qualifying USAMO officially last year, yet u wrote it, how does that work?

[quote="Elyot"] The one I didn't get was a crazy number theory one that you could solve in mod 19 or something crazy.[/quote]

Funny, that was the one problem on last year's USAMO that I thought was easy, a give away seven points. To each his own, I guess.

[quote="Mathgod"]He probably wrote it unofficially.  ;)[/quote]

Yes.  My school didn't offer AMC or AIME or anything like that.  I just wrote USAMO unofficially for IMO-qualification purposes.  I would have placed top 10 for sure, but sadly I wans't official.

And yea, I really really sucked at number theory.  I missed IMO #4 last year as well (damn IMO at having 2 number theory problems and a combinatorics problem that you needed number theory to solve!!!!  WTF?!?!).  I'm better at number theory now, but it's not really on Putnam in the same way.

[quote="Elemennop"]

Funny, that was the one problem on last year's USAMO that I thought was easy, a give away seven points. To each his own, I guess.[/quote]
Wow, congrats on getting that problem.  Only a handful in North America solved that problem.

I believe I have read that you need to be US citizen or permanent resident to be official on the USAMO. Is that really true because I see pretty much everybody over here writing it like Yufei, Rem,... . Elyot, since your school didn't have the AMC-12 test, how did you proceed unofficialy? I mean where did you write them if not at your school? :?  My school used to have the AMC-12, but then the teacher decided against it claiming it lowers ppl's performance on Euclid.

I scored 8, I'm pretty satisfied with that.

[quote="Mathgod"]Ya good job Yufei! I just applied for Awesomemath, so if I make it in I would see you :P. I only did 8 questions for the quiz.... do you think that's enough? :P[/quote]

I would think so. (Though, truthfully, I don't know anything about how they select the people. It shouldn't be too competitive).

[quote="rem"]Yeah good job everyone.

I am so angry at myself. I'll have to wait for another year, at least, that's so long. I will be sooooooo bored and will do nothing throughout that year and will achieve nothing. I am fed up with life. those countless hrs of preparation, practising forever, it was all pointless. If I did this bad on USAMO, I definitely cant write up solutions, so I probably scored below 10 on APMO and the same on CMO. My life is ruined.I can't get over this.
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
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AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
I won't have any fun this year..................And I thought I was good at Math.silly me
:o :stretcher: :suicide: :death:[/quote]

Rem, don't lose hope yet. The year I first made IMO, I got 15/35 in APMO, 13/42 in USAMO and 2.5 questions in CMO. Your scores in all three contest are obviously higher than the scores that got me in the IMO team when I was grade 10. Thus, I think you still have a very good chance.

And even if you don't make it, you still have 2 more years. So cheer up!

[quote="rem"]EDIT: How do we find out how we did compared to other Canadians?[/quote]

I got 36. (made many stupid but fatal typos in my #3, as I had to rush the write up in 20 min, didn't get n=7 in #4, and made some very stupid but trivially fixable mistake in #5 that unfortunately destroyed the proof.) But overall, I'm very happy about my score.

[quote="pengshi"][quote="rem"]EDIT: How do we find out how we did compared to other Canadians?[/quote]
typos[/quote]

So you're typing up your contests now, eh peng?  Typographical errors?  Hahahahah.

[quote="pengshi"]I got 36. (made many stupid but fatal typos in my #3, as I had to rush the write up in 20 min, didn't get n=7 in #4, and made some very stupid but trivially fixable mistake in #5 that unfortunately destroyed the proof.) But overall, I'm very happy about my score.[/quote]it seems you made so many mistakes that you could have easily avoided (maybe on a better day, you wouldn't have made any of those mistakes?)

so my question is, how are u still happy, 'very happy', about your score?  were u expecting to get worse than 36 and did better than you thought?  or are you just a very together person who feels no need to get upset, and sees the experience as just...... fun?

i just started doing these math contests uhhh whenever AMC 12 happened.  so far i've done three, AMC 12, AIME, and Euclid.  anything really stupid/silly that went wrong in any of them, i was pretty pissed off.  i don't kno how you can be so happy.

[quote="rem"][quote="weileaf"]Not to put pressure on anyone, I think Canada will be in top 10 in IMO this year.[/quote]
The reason why Canadians this year are so good is beacuse 5 of the ppl on the IMO team last year have not graduuated and they are ownage. That's why the situtation is that those 5 are making IMO this year and everyone else is competing for the 6th spot.
Anyways, nice job everybody how ppl owned USAMO. weileaf is right. If Canadian ppl did that good compared to Americans then the Canadian IMO team will do nearly as well as USA IMO team which is top 10 for sure. Congratulations everyone! :)

btw Since there have been at least one gr 10 on IMO team for the last 3 or more years, do the organizers like to choose young ppl on the team? :?[/quote]

They might, since most of the team is going extinct this year.  I think if there's a tie, they'll take the younger student.

[quote="rem"]36 on USAMO is amazing. If somebody gets that they should be happy no matter how stupid their mistakes were. ;)[/quote]ah, i see.  impressive.

[quote="weileaf"]wow, I normally don't post on this forum, but I'm REALLY impressed with how well the Canadians has done in USAMO this year. Is it just me who sees that Canadians are stronger in Math than 3, 4 years ago? Since USAMO is the closest indicator we have for how well we can do in Olympics, I'm expecting great things from this year's IMO team. Not to put pressure on anyone, I think Canada will be in top 10 in IMO this year.[/quote]
last year pengshi and yufei were the 3rd and 4th on usamo werent they?

[quote="nirvanatear"]last year pengshi and yufei were the 3rd and 4th on usamo werent they?[/quote]

Last year, Peng and I tied at 3rd.

[quote="The Master"]
so my question is, how are u still happy, 'very happy', about your score?  were u expecting to get worse than 36 and did better than you thought?  or are you just a very together person who feels no need to get upset, and sees the experience as just...... fun?

i just started doing these math contests uhhh whenever AMC 12 happened.  so far i've done three, AMC 12, AIME, and Euclid.  anything really stupid/silly that went wrong in any of them, i was pretty pissed off.  i don't kno how you can be so happy.[/quote]

Well, I haven't been doing math lately (busy with college applications and robotics), so I was expecting to bomb USAMO like how I bombed CMO. (As well, I was sick with fever the night before the second day so I'm happy that it didn't affect much.) Also, for such contests, I usually make some fatal mistakes, and I'm glad that this time, the effect of this is minimized. (At the last IMO, because of mistakes, I got 2 "2"'s for questions that I actually pretty much solved.) This is why I'm very happy about my score.


These were some old individual ARML problems that I found but I dont have answers for them.  I was just wondering if someone can check or do the problem to see if my answers are correct.  Thanks.

1. The successive sides of a quadrilateral are 2,6,9, and x.  If the diagonals of the quadrilateral are perpendicular compute x

2. If the 3-digit positive integer n=ABC=AB+AC+BA+BC+CA+CB,
where A,B,C are digits, compute the largest possible value for n

3. Consider the sequence 1,2,2,3,3,3,4,4,4,4... where the integer n appears n times starting with 1.  Compute the 1992nd term of this sequence.

I got [hide]
for number 1, you find that 6^2-2^2=9^2-x^2, so x^2=49, [b]x=7[/b]
for number 2, you will get 100A+10B+c=22(A+B+C), so 78A-21B-12C=0, you find that even when B and C are their maximum (9), A's maximum must only be 3.  So you try when A=3, and you get 21B+12C=243.  You try B=9 (to maximize n) and you can get that C=6 so I got [b]396[/b]
for number 3, I found that the (n^2+n)/2 term is the last term for when it equals n(eg. the 1st term is 1, the 3rd term is 2, the 6th term is 3...but the 2nd term is 2, 4th germ is 3, 7th germ is 4..).  So when (n^2+n)/2=1992, n is about 62 something.  So when I do (62^2+62)/2, thats is 1953, so I know that the 1954th germ must be 63.. and since (63^2+63)/2 is greater than 1992, then the 1992nd term must be [b]63[/b][/hide]

I dont know what I did is right (because I dont have an answer key) so can someone go over them again please.  Thanks.

[hide]1) Draw an xy plane. Put point A on the x axis (positive), point B on the y axis (negative), point C on the x axis (negative), and point C on the y axis (positive). Let the origin be O. Let OA = a, OB = b, OC = c, OD = d. Let AB = 2, BC = 6, CD =9, and DA is the length we are looking for. We can write:

a^2 + b^2 = 2^2
b^2 + c^2 = 6^2
c^2 + d^2 = 9^2
a^2 + d^2 = ?

c^2 = 81-d^2 so b^2 + 81-d^2 = 36 and b^2 = d^2-45;a^2 + d^2-45 = 4 so a^2 + d^2 = 49.

sqrt(49) = 7 and that is the answer. :)

2) We can re-arrange this as 100A + 10B + C = 22A + 22B + 22C. Combine like terms: 78A = 12B + 21C. Divide by 3: 26A = 4B + 7C. We want A to be as large as possible, but it must equal 4B + 7C. 4 is too large (because even if B and C were 9, that would only make 99) so that leaves us with A = 1, 2, or 3. Let's try 3: 78 = 4B + 7C; try B = 9 which leaves C = 6. 396 is your answer.

3) 1 + 2 + 3... + n >= 1992
n(n+1)/2 >= 1992
n(n+1) >= 3984
sqrt(3984) = 63.11
so n = 63

I think all your answers are correct, although I think you meant 21B + 12C = 234, not 243.[/hide]

Good practice. :)

[quote="GlenthemanN"][hide]
Call the unknown side x, and the four perpendiculars a, b, c, and d.
Then we have

$a^{2}+b^{2}= 2^{2}$
$b^{2}+c^{2}= 6^{2}\Rightarrow b^{2}= 36-c^{2}$
$c^{2}+d^{2}= 9^{2}\Rightarrow c^{2}= 81-d^{2}$
$d^{2}+a^{2}= x^{2}\Rightarrow d^{2}= x^{2}-a^{2}$

Plugging into the first equation

$a^{2}-36-c^{2}= 2^{2}\Rightarrow a^{2}-45+d^{2}= 2^{2}\Rightarrow a^{2}-45+x^{2}-a^{2}= 2^{2}\Rightarrow x^{2}-45 = 2^{2}\Rightarrow x^{2}= 49$
$\boxed{x = 7}$
[/hide]

egh a simple mistake made me think it was 11 and not 7.. i hate how wrong solutions sometimes seem so right  :oops:[/quote]

Adding the 1st and 3rd equations then taking away the 2nd is a pretty quick way to go about it :)


suppose $ABCD$ is a paralellogram. prove that:

$AB^{2}+DC^{2}+AD^{2}+BC^{2}=AC^{2}+BD^{2}$

(i decided to post this problem here, not in intermediate topics :wink: )

[hide="Hint"]Label the vertices such that the perpendicular from $A$ on $CD$ meets $CD$ outside the parallelogram, and label this point $F.$ Again, drop a perpendicular from $D$ on $AB$ such that it meets $AB$ internally and label this point $E$. After that it is just a matter of using the Pythagorean theorem.

I think this problem should be in the Intermediate section.
[/hide]

thanks my dear for solution.

if u look at my post, i said that i decide to post this easy problem here.

ure solution is near to one of the solutions for this problem by projection.

add $2\cdot ab \cdot \cos (\alpha)+2\cdot ab\cdot \cos (\pi-\alpha)=0$ and use Carnot :)

To resolve this question we will use ‘just $ALKACHI's$ theorem  :
We have $AC^{2}=AB^{2}+BC^{2}-2AB.BC.\cos\widehat e{A}$, $[1]$
                $BD^{2}=AD^{2}+AB^{2}-2AD.AB.\cos \widehat e{B}$, $[2]$.
Then by assuming $[1]$ and $[2] we find$AC^2+BD^2= AB^2+BC^2+ AD^2+(AB^2)
                                        -2AB((BC)\cos\widehate{B}+AD\cos\widehate{A})
    Since  $BC$=$AD$  and  $AB$=$DC$    we will find  $AC^{2}+BD^{2}= AB^{2}+BC^{2}+AD^{2}+(CD^{2})-2AB.BC(\cos\widehat e{B}+\cos\widehat e{A})$
      We have then  $2(A+B)=2\pi$      then    $A+B=\pi$   then  $\cosh=\cos(\pi-A)=-\cosh$
  We find that         $(\cos\widehat e{B}+\cos\widehat e{A})=0$.

 :rotfl:

using stewart's theorem
then it becomes obvious

Try using law of cosines for an easy proof.  :)

and finally using paralell projection and pytagorian theorem will do it. :wink:


Για $ x,y,z$ πραγματικους ετσι ωστε $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 2$ τοτε
$ xyz \plus{} 2\geq x \plus{} y \plus{} z$

Πρώτο ποστ στο mathlinks για μένα (το έμαθα από το silouan στην κύπρο).. Χτες έκανα εγγραφή και απαντάω στο πρώτο ποστ που έγινε από τότε.

Έχουμε $ x\plus{}y\plus{}z\minus{}xyz\equal{}x(1\minus{}yz)\plus{}(y\plus{}z) \leq \sqrt{[x^2 \plus{} (y\plus{}z)^2] \cdot [(1\minus{}yz)^2\plus{}1]}$ από B.C.S.

Τώρα απλά πρέπει να δείξουμε ότι $ 4\geq [x^2 \plus{} (y\plus{}z)^2] \cdot [(1\minus{}yz)^2\plus{}1]$.

Κάνοντας τους επιμερισμούς στο δεύτερο μέρος παίρνουμε:

$ (xyz)^2\minus{}2x^2yz\plus{}2x^2\plus{}y^4z^2\minus{}2y^3z\plus{}2y^2\plus{}y^2z^4\minus{}2yz^3\plus{}2z^2\plus{}2(yz)^3\minus{}4(yz)^2\plus{}4yz$

$ \equal{}2(x^2\plus{}y^2\plus{}z^2)\plus{}[(xyz)^2\plus{}y^2z^4\plus{}y^4z^2]\minus{}2yz\cdot(x^2\plus{}y^2\plus{}z^2)\plus{}2(yz)^3\minus{}4(yz)^2$

$ \equal{}4\minus{}2(yz)^2\plus{}2(yz)^3$

Άρα πρέπει να δείξουμε ότι: $ yz\leq1$
Όμως $ 2yz \leq (y^2\plus{}z^2) \leq x^2\plus{}y^2\plus{}z^2 \equal{} 2$

Απεδείχθη  :)

niko de mas pige to matsaki xthes...gmse ta....

[quote="lemonidas"][quote="NickNafplio"]Χαιρετω τον ΗΜΜΥτη της ομαδας :P

Edit: Λαθος λογο βιασηνης...

ΥΓ: Μεγαλη πικρα ο αγωνας χτες  :([/quote]

Είπα κι εγώ.. Κάτι μου πήγαινε πολύ λάθος..  :P
Πάνω που ήμουν έτοιμος να ρωτήσω τον σιλουανό "τι στο καλό κάνει ο Σεμφίτης" το διόρθωσες...

ΥΓ: Άστα να πάνε.. Είχα και τον άλλο να στέλνει "αντε γεια" στο msn....[/quote]

Βασικα ηταν μια καλη ιδεα της στιγμης η οποια θα μπορουσε να βγαλει λυση αφου ηταν οκ με την περιπτωση ισοτητας, αλλα χαλαγε στο προσημο της τελευταιας παραστασης και εγω λογο βιασυνης μπερδευτηκα και δεν το προσεξα :wallbash: . Ευτυχως το εσβησα πρωτου με παρουν με τις ντοματες :P

ΥΓ: Θα ειμαι και σημερα μεσα στο Μπασκετ :P

Σεμφίτη σε πείραξε η ήττα που φαίνεται ( αλλά το spam βλέπω το έχεις στο αίμα σου  :P ) Να μια διαφορετική λύση :

Αν κάποιος από τους $ x,y,z$ είναι αρνητικός τότε η ανισότητα είναι προφανής (γτ ?)
Επομένως υποθέτουμε λόγω συμμετρίας ότι $ 0<x\leq y\leq z$ . Αν $ z\leq 1$ τότε $ 2\plus{}xyz\minus{}x\minus{}y\minus{}z\equal{}(1\minus{}z)(1\minus{}xy)\plus{}(1\minus{}x)(1\minus{}y)\geq 0$

Αν $ z>1$ τότε $ z\plus{}x\plus{}y\leq\sqrt{2(z^2\plus{}(x\plus{}y)^2)}\equal{}2\sqrt{1\plus{}xy}\leq 2\plus{}xy\leq 2\plus{}xyz$

QED (quod erat demonstrandum ) (αυτό το γράφω για το Σεμφίτη που μου έγραφε quite easily done :rotfl: )

[quote="wallace"]εκει που νόμιζα ότι το φόρουμ είχε καλα παιδιά (κόκκινα), να σου και οι βάζελοι...xaxa...[/quote]
me too :P  :P
Για να δουμε τι θα κανει σημερα και η Barca...

Υ.Γ.ωραια λυση Σιλουανε...

[quote="Protonios"]Παρακαλώ οι Βάζελοι να μην ποστάρουν ούτε ασκήσεις ούτε λύσεις. Δεν ανακτούν έτσι τη χαμένη αξιοπρέπειά τους και επιπλέον χαλάει η ομοιογένεια του φόρουμ. :roll:

Btw ίδια ακριβώς άσκηση πολωνία 1991... να μην τις εκμεταλλευτούμε τις shortlist... :rotfl:[/quote]

kosta gavros eisai? koita an den thes na beis stin omada kanonizetai....boro na kano ena tilefonimataki ston tyrli kai na tou thymiso oti eisai gavros!!!kai ksereis poso tous subathei o turlis tous olubiakous eee?

ante pali o fellouris tha se kselasposei!!

just kidding of course...

niko mi se foverizei o moderator me ta spam...thimitheitai to thryliko rallis-despotakis!!!!LOLOLOLOL.....


function
algebra
functional equation
real analysis
real analysis unsolved

prove that there is not function F : R-->R such as, for all realities x, y >0:
[url=http://img143.imageshack.us/my.php?image=fgcw8.png][img]http://img143.imageshack.us/img143/1528/fgcw8.th.png[/img][/url]

[hide]With $y=0$ you get $f(x)>f(x)^{2}$, hence $0<f(x)<1$ for all $x$.
But $f(x+y)\to+\infty$ as $y\to+\infty$,[/hide] a contradiction.

Edit: the functional equation need not hold for $y=0$...  :huh: (Thanks Ottem.)

yes this is what i 'm seeing
if$y=0$ we have$f(x)=f(x)$it's nothing

OK, I was totally wrong the first time. [hide="Second try:"] $f$ must be unbounded from above. Let $a>0$ be such that $f(a)>1$. Define an increasing sequence $\{x_{n}\}$ as follows: $x_{0}=a$, $x_{n+1}=x_{n}+\frac{1}{f(x_{n})^{2}}$. This definition is cooked up so that $f(x_{n+1})>f(x_{n})+1$. In particular, $f(x_{n})>n+1$ for all $n\ge 0$. Therefore, \[\sum_{n=0}^{\infty}(x_{n+1}-x_{n})\le \sum_{n=0}^{\infty}\frac{1}{(n+1)^{2}}<\infty,\] which means that the sequence $\{x_{n}\}$ has a finite limit, say, $b$. Since $f$ is increasing, $f(b)>f(x_{n})>n+1$ for all $n$, a contradiction.[/hide]


inequalities
function
inequalities unsolved

Prove that for any possitive integer $k> 1$ there exist possitive integer $x, y< k^{2}$ such that
$0<{\sqrt{x}}-{\sqrt{y}}< \frac{k-1}{k(k^{2}-k-1)}$
It may be an number problem but i think it's always a ineq problem.

I think for sufficiently large $k$, the inequality don't hold. [If x and y are integers]

$\sqrt{x}-\sqrt{y}>0 \Rightarrow x>y$

Let $x=y+z$ with $z$ is a natural number, then $\sqrt{x}-\sqrt{y}\geq \sqrt{y+1}-\sqrt{y}$.

Let $f(y)=\sqrt{y+1}-\sqrt{y}=\frac{1}{\sqrt{y+1}+\sqrt{y}}$, then f(y) is  a decreasing function for $y$ is natural number. Since $y \leq x-1 <k^{2}-1$,  $f(y) \geq f(k^{2}-2) = \sqrt{k^{2}-1}-\sqrt{k^{2}-2}$

Let $k=100$, $\sqrt{x}-\sqrt{y}\geq f(y) \geq \sqrt{9999}-\sqrt{9998}> \frac{100-1}{100(100^{2}-100-1)}= \frac{k-1}{k(k^{2}-k-1)}$

Contradiction.

So sorry guys, I have got some mistake.
I now repost my problem.
Prove that for any possitive integer  there exist possitive integer  such that
$0<(\sqrt{x}-[\sqrt{x}])-(\sqrt{y}-[\sqrt{y}]) < \frac{k-1}{k(k^{2}-k-1)}$
Here $[x]$ denote by the integerpart or real number x.
(I can't type the fractionpart).


Let be given [tex]p\ge 3[/tex] is prime number.
Sequence [tex](a_n),n=0,1,2,...[/tex] defined by
[tex]a_n=n,n=0,1,...,p-1[/tex]
[tex]a_n=a_{n-1}+a_{n-p},n=p,p+1,...[/tex]
Find remainde when divide [tex]a_{p^3}[/tex] by [tex]p[/tex]

I'm pretty sure I've seen this before around here. And maybe Pierre solved it. :? I'm not sure.

I have already posted this one, and it was solved by Pierre, and also Grobber did something[url]http://www.mathlinks.ro/Forum/viewtopic.php?highlight=number+theory+sequence&t=23201[/url]

Yes, this time I remember it too.  :)

Pierre.


Multiplying Probabilities Part 3

We investigate the probability that two events both occur when one event depends on the other.  We thereby learn the importance of "independent" in multiplying the probabilities of "independent" events.

Is this an ad or a lesson? :rotfl:  :rotfl:  :rotfl:

Sad: we are abandoned on cube world  :(  .... and square world.

Of course, it's a lesson. Also, I thought the video would be really helpful for begginers I thought the video was excellent!

What if you don't buy the book? Then the probability that you understand probability is 2/3. What about that number?  :?:  :huuh:


[quote="Shark_"]As you know, the Grissom Math Tournament is going to be held in less than 15 days. I did HORRIBLE on the Vestavia Hills Tournament and was hoping anyone here had some practice papers. Other than the Old Grissom ones.

Thanks,
Shark_[/quote]

first of all, you would obviously do horribly on an english test, since you should have said "i did horribly, not horrible."

just saying.  what place did u get a vestavia? i got 5th.

second of all, there are links to the 03 and 05 tests [url]http://www.geocities.com/grissom_math/[/url]

[quote="now a ranger"]

I bet some Vestavia kids would say they did horrible if they got 3rd.  :lol:[/quote]

yea who would ever think of setting personal goals as opposed to childish and arbitrary comparisons on ridiculously narrow scopes.

sure you did  :rotfl:  you got 1st with a 91, and 2nd was an 83

How is that horrible? Horrible= making careless mistakes and getting a low score

horrible = making the choice to even go to grissom...

well you kind of had to go because of the ARML qualifier didn't you?

but then again, you could probably edge people out skipping out on one...

haha the trophies were SMALL  :(

I liked the geometry test though : it was so easy to make mistakes...

Random,but are there any online websites for Vestavia Hills Math Tournament


[quote="shadysaysurspammed"]If $a,b,c$ are primes such that $a^b+1=c$,then find $a,b,c$(with proper reasoning)[/quote]
[hide]
If they are all prime then $a$ can't be odd beacuse then $c$ is even. If c is even then c=2 but thats not possible. So then a=2 and $2^b+1=c$. So then we have a=2, b=2, c=5. [/hide]

[hide="Solution"]We can immediately see that $a^b$ can't be odd because otherwise $c$ would be an even number greater than $2$ and would thus be composite. The only even prime is $2$; therefore $a = 2$. So we now have $2^b + 1 = c$. If b is odd, then let $b = 2k + 1$, and $2^b + 1 \equiv 2^{2k + 1} + 1 \equiv 2 \times 4^{2k} + 1 \equiv 2 \times 1^{2k} + 1 \equiv 0 \ \mbox{(mod 3)}$, which makes c greater than 3 and thus composite, so $b$ must be even. Thus $b$ must be $2$, the only even prime, and so $c = 5$.[/hide]

[quote="shadysaysurspammed"]If $a,b,c$ are primes such that $a^b+1=c$,then find $a,b,c$(with proper reasoning)[/quote]I'm pretty sure I posted this same problem a few months ago...

[EDIT]Yes I did, here is the link [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=84915[/url]

[hide="A variant"]If $b$ is odd, then $a^b+1$ is factorizable into $(a+1)(a^{b-1}-a^{b-2}+\dots+1)$, hence can't be prime. Therefore $b=2$. Consequently, $a^2+1$ is always even for an odd $a$, therefore $a$ is also even and equal to $2$, giving $c=5$.[/hide]


Could you all please post some telescoping problems please.

$ \frac{1}{1*2*3}\plus{}\frac{1}{2*3*4}\plus{}\frac{1}{3*4*5}......\frac{1}{n(n\plus{}1)(n\plus{}2)}$
Could someone please post a solution. :blush:

$ \sum_{k \equal{} 1}^{n} \frac {1}{k(k \plus{} 1)(k \plus{} 2)} \equal{} \frac {n(n \plus{} 3)}{4(n \plus{} 1)(n \plus{} 2)}$ can be shown with partial fraction decomposition, I'm just not sure how.

[hide="As an example:"]

$ \frac {1}{1*2} \plus{} \frac {1}{2*3} \plus{} ... \plus{} \frac {1}{n(n \plus{} 1)} \equal{} ???$

The idea is the same for this kind of telescopies:

$ \frac {1}{k(k \plus{} 1)} \equal{} \frac {1}{k} \minus{} \frac {1}{k \plus{} 1}$

Then:

$ \sum^{n}_{k \equal{} 1}{\frac {1}{k(k \plus{} 1)}} \equal{} \sum^{n}_{k \equal{} 1}{\frac {1}{k} \minus{} \frac {1}{k \plus{} 1}} \equal{} 1 \minus{} \frac {1}{n \plus{} 1}$
[/hide]

[hide="For your problem we could do something similar:"]

$ \frac {1}{k(k \plus{} 1)(k \plus{} 2)} \equal{} \frac {1}{2}(\frac {1}{k(k \plus{} 1)} \minus{} \frac {1}{(k \plus{} 1)(k \plus{} 2)})$

Then:

$ \sum^{n}_{k \equal{} 1}{\frac {1}{k(k \plus{} 1)(k \plus{} 2)}} \equal{} \sum^{n}_{k \equal{} 1}{\frac {1}{2}(\frac {1}{k(k \plus{} 1)} \minus{} \frac {1}{(k \plus{} 1)(k \plus{} 2)})} \equal{}$

$ \frac {1}{2}\sum^{n}_{k \equal{} 1}{\frac {1}{k(k \plus{} 1)} \minus{} \frac {1}{(k \plus{} 1)(k \plus{} 2)}} \equal{} \frac {1}{2}(\frac {1}{1*2} \minus{} \frac {1}{(n \plus{} 1)(n \plus{} 2)}) \equal{}$

$ \frac {n(n \plus{} 3)}{4(n \plus{} 1)(n \plus{} 2)}$   as our fellows above have already said.
[/hide]

[hide="more challenges"]
A] $ \frac {1}{1*2*3*4} \plus{} \frac {1}{2*3*4*5} \plus{} ... \plus{} \frac {1}{n(n \plus{} 1)(n \plus{} 2)(n \plus{} 3)} \equal{} ??$

B] $ \frac {1}{1*2*3*...(p \plus{} 1)} \plus{} \frac {1}{2*3*...(p \plus{} 1)(p \plus{} 2)} \plus{} ... \plus{} \frac {1}{n(n \plus{} 1)...(n \plus{} p)} \equal{} ??$

C] let $ x\equal{}1\plus{}\frac{1}{\sqrt{2}}\plus{}\frac{1}{\sqrt{3}}\plus{}...\plus{}\frac{1}{\sqrt{1,000,000}}$. Find $ \lfloor{x}\rfloor$

[/hide]

 :)

I remembered seeing a topic on telescoping sums, but couldn't find it. Refer to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=76327]this[/url] for a generalization by Virgil Nicula.


1. How many different anagrams (including nonsensical words) can be made from SUCCESS given that the first and last letters must both be S?

I think that because SUCCESS already has last and first letters S, we can work with the 5 remaining letters UCCES for 5!/2! ways.

2. How many different anagrams (including nonsensical words) can be made from FACETIOUSLY given that all six vowels must remain in alphabetical order (but not necessarily contiguous with each other).

I think that there are 11! anagrams to the word FACETIOUSLY, but there is only 1 out of the possible 6! arrangements of the vowels such that they are in order (y is considered a vowel). So 11!/6!?

Can anyone either check or correct my thinking?

Thanks.

1. Yeah, it is 5!/2, because 5! ways to arrange, then because there are 2 C's, you counted everything twice, so divide by 2.

2. Your way works.  Nice method!
The way I would do it is take the 5 consonants, and see that there are 5! ways to arrange them.  Then, with the vowels in fixed order, for each permutation of the consonants, how many ways are there to put them into the 7 slots?  That's doable.  But I like your method more  :)


[hide]

So the equation will be something like $  y = -4x + b  $ where b is any number that will be added to make the equation come true. We know $ y = 5  $ and $  x = -1  $ So we immediatly know that $  y = -4x  $ does not work. In this case, $ -4x = 4 $ so we have to add 1 to get 5

$  y = -4x + 1  $[/hide]

[quote="math92"]Wrine ann equation of a line in standard formm that has a slope of -4 and contains the opint M(-1, 5)[/quote]

[hide]In slope intercept form, the equation of the line would be $y=-4x+?$
Using the point we have to evaluate what ? is:
$5=-4(-1)+?$
$5=4+?$
$?=1$

So the equation in slope intercept form would be $y=-4x+1$
rearranged, it would be $4x+y=1$[/hide]


Show that if an odd prime number $ p$ can be expressed in the form $ x^5\minus{}y^5$ for some integers $ x,y,$ then:

$ \sqrt{\frac{4p\plus{}1}{5}}\equal{}\frac{v^2\plus{}1}{2}$ for some odd integer $ v$.

[quote="moldovan"]Show that if an odd prime number $ p$ can be expressed in the form $ x^5 \minus{} y^5$ for some integers $ x,y,$ then:

$ \sqrt {\frac {4p \plus{} 1}{5}} \equal{} \frac {v^2 \plus{} 1}{2}$ for some odd integer $ v$.[/quote]

$ p>0 \Rightarrow x>y$

$ p\equal{}x^5\minus{}y^5\equal{}(x\minus{}y)(x^4\plus{}x^3y\plus{}x^2y^2\plus{}xy^3\plus{}y^4)$ is a prime.

$ x^4\plus{}x^3y\plus{}x^2y^2\plus{}xy^3\plus{}y^4\equal{}(x\plus{}y)(x^3\plus{}y^3)\plus{}x^2y^2\equal{}1 \Rightarrow x\equal{}1,y\equal{}\minus{}1 \Rightarrow p\equal{}2$, contradiction.

$ x^4\plus{}x^3y\plus{}x^2y^2\plus{}xy^3\plus{}y^4\equal{}(x\plus{}y)(x^3\plus{}y^3)\plus{}x^2y^2\equal{}\minus{}1 \Rightarrow$ no solution.

So we must have $ x\minus{}y\equal{}1$,
$ p\equal{}(y\plus{}1)^5\minus{}y^5\equal{}5y^4\plus{}10y^3\plus{}10y^2\plus{}5y\plus{}1$

$ \sqrt {\frac {4p \plus{} 1}{5}} \equal{} \sqrt{4y^4\plus{}8y^3\plus{}8y^2\plus{}4y\plus{}1}\equal{}2y^2\plus{}2y\plus{}1\equal{}\frac{(2y\plus{}1)^2\plus{}1}{2}$, done.


A hybrid orbital does not distribute electron density symmetrically around the nucleus of an atom.  Therefore, a lone pair in a hybrid orbital contributes to the overall polarity of a molecule.  On the basis of these statements and the fact that NH3 is a very polar molecule and NF3 is a nearly nonpolar molecule, justify the notion that the lone pair of electrons in each of these molecules is held in an sp3 hybrid orbital.

N[b]F[/b]3?
isnt it obvious why its nearly non-polar?

Is it because F is a lot more electronegative than H


let $ a_1,a_2,\cdots,a_n;b_1,b_2,\cdots,b_n\in\mathbb{R}^\plus{}$ such that $ \sum_{i\equal{}1}^{n}{a_i^2} \equal{}\sum_{i\equal{}1}^{n}{b_i^2}\equal{}1$

find $ M\equal{}\max \big\{\min_{1\leq i\leq n}{\{\frac{a_i}{b_i}\}}\big\}$

It is either obvious or I missed something...
If for all $ 1\leq i\leq n$,we have $ a_i>b_i$,then $ \sum_{i\equal{}1}^{n}a_i^2>\sum_{i\equal{}1}^{n} b_i^2$,contradiction.
Hence there exist at least one index $ i$,such that $ a_i\leq b_i$,hence $ M\leq 1$.
If $ a_1\equal{}a_2\equal{}\dots\equal{}b_1\equal{}b_2\equal{}\dots\equal{}b_n\equal{}\frac{1}{\sqrt{n}}$ then $ M\equal{}1$.

$ M \equal{} 1.$ It is easy to show ;) Oh.. I'm late  :D

$ M \equal{} \max \big\{\min_{1\leq i\leq n}{\{\frac {a_i}{b_i}\}}\big\} \equal{} \frac {a_k}{b_k}\leq\frac {a_1}{b_1}\leq ... \;\textbf{all except k}$.

$ a_i\geq M \cdot b_i$

$ 1\geq M^2$ :)


Counting the Number of Arrangements of Letters in a Word

We learn how to count the number of possible arrangements of letters in words with repeated letters by counting the arrangements of SUSIE, SELLS, and SEASHELLS.

that must have taken a long time to right seashells  :omighty:  :ddr:  :winner_first:  :starwars:

When we were working on SEASHELLS, the answer was 9!/2*2*3!.
Can both two's be counted as 2!'s?

Yea, because that's what they're meant to be.

Since $2!=2$, it doesn't really change your answer at all.


Let $ x, y, z \geq 0$, $ x\plus{}y\plus{}z\equal{}1$, $ a \leq \frac{32}{9}$. Prove this:

$ \sqrt{1\minus{}axy}\plus{}\sqrt{1\minus{}ayz}\plus{}\sqrt{1\minus{}azx} \geq \sqrt{9\minus{}a}$

It is not easy.

[quote="arqady"]See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=156002[/quote]

There are some ideas to prove my inequality:
[url]http://www.mathlinks.ro/viewtopic.php?p=1362595#1362595[/url]


Show that for all integers n > 1, the expression n^4 + 4^n is not prime.

[hide]If $ n$ is even then by obviousity, $ 2|n^{4}\plus{}4^{n}$.

If $ n$ is odd, then $ n\equal{}2k\plus{}1$. We then have $ n^{4}\plus{}4^{n}\equal{}n^{4}\plus{}4\cdot(2^{k})^{4}\equal{}(n^{2}\plus{}2(2^{k})^{2}\plus{}2\cdot n\cdot 2^{k})(n^{2}\plus{}2(2^{k})^{2}\minus{}2\cdot n\cdot 2^{k})$.[/hide]


inequalities
Euler
number theory proposed
number theory

For any $n\ge 1$
\[\sum_{1\le k\le n}\frac{\varphi(k)}{k}\ge \frac{n+1}{2}\]

Proof:
http://www3.telus.net/ldh/math/phi_ineq.pdf

:showoff:

(update) oops, I had already posted this thing:
http://www.mathlinks.ro/Forum/viewtopic.php?t=88433

Wonderful proof !. I will need to read it more carefully, tough
P.S : My approach was to consider 'truncated' Euler products ...


induction
combinatorics proposed
combinatorics

Each square of a chessboard contains $ 0$. It is allowed to select any two squares connected by knight move and to increase numbers in them by $ 1$. Is it possible to obtain the chessboard filled with integers $ 1, 2, . . . , 64$?

Since knight's tours exist, the answer to this question is "yes" if the answer to the following question is "yes":

64 points are placed on a line, each beginning with value 0.  In a single move, it is allowed to increase the value of two adjacent points by 1.  Is it possible to have at some point the numbers $ 1, 2, \ldots, 64$?

We answer in the affirmative, by induction: with 4 points, we increase the first pair once, the second pair once and the third pair thrice.  With $ 4k \plus{} 4$ points, we increase the first pair once and every other pair twice.  The the first point is numbered 1, the second 3, the $ 4k \plus{} 4$th 2 and every other point 4.  Then carry out this process on the 3rd through $ 4k \plus{} 2$nd points (by induction) and we're done.

Editted to fix a small typo.

Or may be we can prove with a bit different solution:
-) With $ 4$ points: we increase the first pair once, the second pair once, the third pair three: $ 0000\rightarrow 1100\rightarrow 1210\rightarrow 1243$.
-) With $ 4k\plus{}4$ points: we do similarly with all group $ 4$ points, after that with second group $ 4$ points we do by this way: increase the first pair $ 4$ times, and the third pair $ 4$ times $ 1243\rightarrow 5687$.
Similarly for all other group!


Let $ X \equal{} \{1;2;3;...;15\}$. How many subset $ A\subset X$ have $ 5$ elements so that $ A$ have at least $ 2$ consecutive numbers.
Example: $ A \equal{} \{1;2;4;5;7\}$ have $ 2$ pair consecutive numbers!
$ B \equal{} \{1;3;5;7;9\}$ isn't satisfy!

It's very unhappy because nobody care my problem! It's not so hard then I decide post it in this forum not in olympiad forum!  :(
My solution:
A set $ A \equal{} \{a;b;c;d;e\}$ with $ a < b < c < d < e$ don't satisfy condition is five number so that $ 1\leq a < b \minus{} 1 < c \minus{} 2 < d \minus{} 3 < e \minus{} 4\leq 11$, then have $ C_{11}^5$ sets!
Then we have $ C_{15}^5 \minus{} C_{11}^5$ sets satisfy my problem!


As I suggested on the other post -- try expanding out each of the three terms (1 + x)^a, (1 + x)^b, and (1 + x)^(a + b).  What do you know about two polynomials which are equal?

First, I know how to expand each term:

(1 + x)^a = C(a, 0)*x^0 + C(a, 1)*x^1+...+C(a, a)x^a

the problem is that I must get some value for x, like x = 0, 1, or 2...
(I think that bigger than 2 won't solve the problem) and then prove that I reach to equal equations or two ways to count the same number of objects...so now I know that you can see what I mean...

I know using Vandermonde's equation it's really easy...

Thanks...

Why do you have to plug in a value for x?  Just multiply the polynomials together, and since you have equal polynomials you must have equal coefficients.  Checking the values for any coefficient on one side or the other gives you the formula you wanted.

Hey!!!, that's the same way to get x = 1...then, the variable will disapear and thus compare only the coefficients. And, to compare polinomial equations I use to compare those coefficients...Thanks...


geometry
function
modular arithmetic
perimeter
Putnam
number theory proposed
number theory

We proceed by contradiction.

Assume there exist a set of four points,  $ A(x_1, y_1), B(x_2,y_2), C(x_3,y_3), D(x_4,y_4)$

Consider the function $ S(x,y) \equal{} x \plus{} y \pmod{2}$ for all lattice points.

[b]Lemma 1[/b]If $ \sqrt {x} \in \mathbb{Z}$ and is odd then $ x$ is odd

Consider the distance between $ A$ and the remaining points.

Clearly $ (x_1 \minus{} x_i)^2 \plus{} (y_i \minus{} y_1)^2$ is odd then $ x_1 \minus{} x_i$ and $ y_i \minus{} y_1$ are of different parities

ie $ x_1 \minus{} x_i \equiv y_i \minus{} y_1 \plus{} 1 \pmod{2}$

Therefore $ x_1 \plus{} y_1 \equiv x_i \plus{} y_i \plus{} 1 \pmod{2}$

So $ S(x_1, y_1) \not \equal{} S(x_i, y_i)$ for all $ i$

But by a similar logic, this holds for $ B, C, D$ which is a contradiction.

[b]Mistake[/b] I suspect I've made a mistake, since the same proof suggests that this also holds for triangles?

[quote="thaithuan_GC"][quote="stvs_f"]Show there do not exist four points in the Euclidean plane such that the pairwise
distances between the points are all odd integers.[/quote]
In a triangle, if the length of three sides are integers then the perimeter of the triangle is an even integer.[/quote]
What if the triangle has sides $ 3,4,6$ :wink:

Just a formality, it is PEN R2 [Putnam 1993/B5] and as such has been posted in the PEN subforum

http://www.mathlinks.ro/viewtopic.php?t=150960

[quote="SimonM"]
Clearly $ (x_1 \minus{} x_i)^2 \plus{} (y_i \minus{} y_1)^2$ is odd then $ x_1 \minus{} x_i$ and $ y_i \minus{} y_1$ are of different parities[/quote]

$ x_1 \minus{} x_i$ and $ y_i \minus{} y_1$ are not necessarily integers, so it doesn't make sense to talk about their parity.


1. Chindea Filip - 110 pct
2. Bumbacea Radu - 83 pct
3. Filip Laurian - 81 pct
4. Tiba Marius - 79 pct
5. Ciolan Alex - 74 pct
6. Muntean Alex - 68 pct

Daca aveti toate lista cu rezultatele pe probleme sa o postati  :P

Voiam sa urez si eu felicitari celor 6 si bafta la JBMO, in special lui PhilAndrew, dar si lui Laurian care a dovedit o revenire excelenta. (PS: sa nu ii uitam pe compatriotii mei din Vianu)

[quote="andyciup"]Felicitari celor 6 si bafta la balk :)

Stie cineva rez. pe probleme pt seniori?
PS. 3 mathlinkeri si 3 colegi de liceu la IMO :)[/quote]Care sunt rezultatele la seniori pana la urma?

Eu stiu ca sunt calificati dzeta, maky, ciprian (astia-s mathlinkeri), si livia, madalina si dumi (astia-s colegii)
 :)

Ciprian? Doamne, ce fericit trebuie sa fie... Eu unul imi doream din tot sufletul sa se califice la OIM. Multe, multe felicitari lui dar si celorlalti.

Anyway, rezultate oficiale are cineva?

Multumesc pentru felicitari si mult succes echipelor de OIM/Yakutzia :) . Uitati si fisierul cu rezultatele la juniori (la seniori nu erau înca afisate).

hai ca o sa pun eu ultimele noutati la seniori :)
sper sa iasa bine tabelul asta ca e cam greu de scris
$\begin{tabular}{|l|ccc|ccc|ccc|c|}\hline & \makebox[0.4in]{G1}& \makebox[0.4in]{G2}& \makebox[0.4in]{G3}& \makebox[0.4in]{P1}& \makebox[0.4in]{P2}& \makebox[0.4in]{P3}& \makebox[0.4in]{P4}& \makebox[0.4in]{P5}& \makebox[0.4in]{P6}& \makebox[0.6in]{Total}\\ \hline Zahariuc Adrian & 7 & 0 & 7 & 7 & 7 & 5 & 7 & 7 & 7 & 153 \\ Ilie Livia & 5.5 & 7 & 7 & 6 & 3 & 0 & 0 & 7 & 0 & 106.5 \\ Turea Lucian & 7 & 7 & 7 & 7 & 0 & 0 & 1 & 7 & 4 & 105 \\ Dumitrescu Mihai & 7 & 5 & 5 & 7 & 7 & 0 & 0 & 5.5 & 1 & 97 \\ Persu Madalina & 7 & 5 & 7 & 7 & 0 & 1 & 0 & 7 & 0 & 97 \\ Ionescu Ciprian & 7 & 7 & 7 & 7 & 0 & 1 & 0 & 7 & 1 & 96 \\ \hline Prodescu Corneliu & 7 & 1 & 7 & 0 & 7 & 3 & 0 & 7 & 1 & 93 \\ Dobriban Edgar & 7 & 7 & 7 & 7 & 1 & 0 & 0 & 0 & 4 & 92 \\ Padureanu Victor & 7 & 7 & 5 & 7 & 1 & 0 & 0 & 4.5 & 0 & 86 \\ \hline Munteanu Sinziana & 7 & 0 & 7 & 2 & 1 & 0 & 0 & 7 & 0 & 85 \\ Rosu Eugenia & 7 & 0 & 7 & 7 & 1 & 0 & 0 & 0 & 1 & 78 \\ Deneanu Andrei & 1 & 0 & 6 & 2 & 0 & 0 & 0 & 7 & 1 & 71 \\ \hline \end{tabular}$

sper ca n`am gresit la adunarea punctelor  :blush: .
primii 6 merg la OIM, urmatorii 3 la yacutia.

edit : nu imi merge downloadul la juniori. e de la mine sau.... ?

Multumesc pentru felicitari si pentru rezultatele .xls. Cine e philandrew?

Fenomen Zahariuc, la 46.5p in fata locului doi :| Doamne. Si Padureanu e tare rau la varsta lui.

Cand se posteaza subiectele de la testele astea 3? Barajul de geometrie? Pls!  :)

Vreau sa va urez felicitari si bafta la OIM la toti, good job,  si extra-felicitari pentru Adrian.
Maky cred ca ai rabdare foarte mare din moment ce ai facut tot tabelul ala.  :omighty:

[quote="Anitta"]Fetele sunt super tari! Oare a mai avut Romania doua fete in echipa vreodata?   :lol:  Oricum, cinste lor ca s-au calificat si au aratat misoginilor de ce sunt in stare!   :showoff:
Mult succes ambelor echipe, in Vietnam si Bulgaria! Cheers! :D[/quote]Au fost 2 in total. Cu una (Ana Caraiani) am fost si eu de 2 ori in echipa :)

Care e adresa la JBMO 2007 din Bulgaria?

Si deci echipa noastra a luat locul I la JBMO 2007? Multumesc.

Se pare ca iarasi Bulgaria a castigat si jBMO-ul. Felicitari echipei noastre, in special lui PhilAndrew si Bumbacea Radu pentru scorurile perfecte.

M-am uitat acuma pe subiecte, poate stie cineva ce s-a intamplat cu problema 4 ... $ 7p+3p-4$ ... e vre-un typo aici?

Da, in cele din urma, este $ 7p+3^{p}$ acolo, eu cand le-am postat, le-am luat de pe site-ul oficial: [url]http://www.jbmo.eu/[/url] , deci greseala nu imi apartine.  :D

Bulgaria are doua loturi ca tara gazda. Poate totusi noi am castigat.

Într-adevar, rezultate excelente pentru echipa României! Totuși, ochiometric vorbind, cred că bulgarii ne-au luat cu câteva puncte.


How are the Wiki WoWs decided? And should there be a "AoPSWiki WoW Proposal Thread?" It would be useful, since AoPSWiki desperately needs some topics and people who browse through certain sections of the forums a lot can better know what good wiki articles there needs to be on. (bleh, I need to learn grammar. Am I making any sense?)

Also, since the question has been asked some many times, I feel that there needs to be a topic about $ 0.\overline{9}$. Although how that would be titled I have no idea. Alternatively, people could start using the search function, which is practically another AoPSWiki.

[quote="Nerd_of_the_Ages"]there needs to be a topic about $ 0.\overline{9}$.[/quote]

Start it yourself! [[0.999...]] looks like a good place.

I think Temperal has been doing all of the recent WoWs. While there is no specific request thread, there is [[Special:Wantedpages]], though it has been weighed down by contest problems ..

I've started the article at that page ([[0.999...]]).

It is possible to redirect articles (like [[0.99...]]) using the syntax
[code]
#redirect [[0.999...]]
[/code]
on the page to be redirected.

Of course, it's impossible to redirect all of those possible names, but it's really only necessary to take care of the more reasonable possibilities. I can't imagine any person going to the wiki and trying to search for 0.999999999999999999; even .9 rep is a stretch.

I think a search for 0.99 would stumble upon the article named 0.999 ...

Is the AoPSWiki WoW ticker on the forum page automated? It doesn't seem to be (I believe I asked Mr. Rusczyk to change it twice). It would be nice to have it update automatically.


ratio
search
number theory proposed
number theory

Hello everyone; I had a question which my friend proposed to me. He said he knew an infinite product representation of the wonderful phi.

Is this possible, and if it is, what exacly is the infinite product?

(unfortunately he did not tell me what the product was, but when he does ill post it, to let hear what you guys think about it)

Thanks!

Well you can come up with lots of different infinite products to express any number you like, but how 'interesting' or nontrivial they are is a rather subjective question of aesthetic appeal.  For example you could express $ \phi$ as
\[{ \phi=\prod_{n=1}^\infty{\rho_n}(1+\rho_n})\]
where $ \rho_n=\frac{F_n}{F_{n+1}}$ and $ F_n$ is the $ n$th Fibonacci number.  But since its convergence to $ \phi$ is rather easy to demonstrate, I suppose most would say it's not very interesting.

neat; im pretty sure that wasn't the right representation that my friend was talking about; he mentioned that his infinite product contained pretty ordinary; run of the mill integers. Where can i go to see more infinite products that converge to phi?

The infinite representations of $ \varphi$ with which I am familiar (neither of which are products) are

$ \varphi \equal{} \sqrt{1 \plus{} \sqrt{1 \plus{} \sqrt{1 \plus{} ...}}}$

and

$ \varphi \equal{} 1 \plus{} \frac{1}{1 \plus{} \frac{1}{1 \plus{} \frac{1}{1 \plus{} ...}}}$.

A quick search has not turned up an infinite product representation.  Could you ask your friend what he has in mind?


Support
function
calculus
derivative
integration
real analysis
real analysis theorems

Hello all,

I'm reading this university printout of Real analysis. I'm familiar with compact groups and the properties of continuous functions defined on them.
Ok, say we have $A \subset \mathbb{R}^n$, $f: A \to \mathbb{R}$ and define the support of $f$ to be spt $f = A \cap \overline{\{x \in A: f(x) \ne 0\}}$.
Let $U \subset \mathbb{R}^n$ be open, $k \in \mathbb{N}$ and $f: U \to \mathbb{R}$. Now $f \in C_0^k(U) \Longleftrightarrow f \in C^k(U)$ and spt $f$ is compact. Where, of course, $C^k(U)$  is the space of functions $U \to \mathbb{R}$ that are $k$ times continuously differentiable.

I ask: what is the deep idea behind the support? The printout now continues to use $C_0^k$ functions without further introducing the idea behind the definition of the support. It is obvious that if, for example, $g \in C_0(\mathbb{R}^n)$, then $\exists \max g$. But how do we automaticly, in an obvious way, now know that $g$ is, for example, uniformly continuous on $\mathbb{R}^n$? I know that $g|\textrm{spt}\, g$ is uniformly continuous on spt $g$. Also how do we know that if $h \in C_0^k(\mathbb{R}^n)$ then all its partial derivatives are also uniformly continuous on $\mathbb{R}^n$?

Thanks!

[quote]I know that $g|\textrm{spt}\, g$ is uniformly continuous on spt $g.$[/quote]
Just remember that for $x$ not in the support of $g,\,g(x)=0.$ The function which is identically zero is certainly also uniformly continuous, and that makes $g$ uniformly continuous on all of $\mathbb{R}^n.$

If you're wondering about estimating $|g(x)-g(y)|$ for $x\in\text{spt}(g)$ and $y\not\in\text{spt}(g),$ then remember that there is point $z\in\text{spt}(g)$ (in fact, $x$ is on the boundary of the support) such that $z$ is closer to $x$ than $y$ is and $g(z)=g(y)=0.$ Hence this is controlled by the estimate we have from $g$ being uniformly continuous on its support.

The same argument applies to each of the partial derivatives.

The practical idea: there are operators - integral operators, differential operators - whose very existence when applied to arbitrary functions may be in doubt, but which can at least be shown to exist for functions which are smooth enough and/or which vanish sufficiently rapidly at infinity. We start by defining our operator only for such nice functions, but we also seek to find norm identities for it in whatever normed function space ($L^p,$ Sobolev, etc.) is appropriate. If we succeed in proving such a norm estimate, we can extend our operator by continuity, becuase $C_0^{\infty}$ is dense in $L^p$ for $p<\infty$ or in most of the other normed spaces we're interested in.


Show that the following series converges.
$\sum_1^{\infty}[k(f(1/k)-f(-1/k))-2f'(0)]$ where $f$ is thrice differentiable with continuous third derivative.

Suppose that on $[-1,1],\,|f'''(x)|\le M.$ (This is guaranteed by the continuity of the third derivative.

By Taylor's Theorem,

$f(x)=f(0)+f'(0)x+\frac12f''(0)x^2+\frac16f'''(c)x^3.$

Using this estimate,

$k\left[f\left(\frac1k\right)-f\left(\frac1k\right)\right]-2f'(0)= \frac16(f'''(c_1)-f'''(c_2))\frac1{k^2}.$

Thus, $\left|k\left[f\left(\frac1k\right)-f\left(\frac1k\right)\right]-2f'(0)\right|\le \frac{M}{3k^2}.$

The series converges by comparison to $\sum_{k=1}^{\infty}\frac1{k^2}.$

The hypothesis $f\in C^3$ is a stronger than needed to prove the theorem. It would be sufficient to assume that $f'''(0)$ exists.


PAMO=pan african mathematical olympiad, morocco only participates in the IMO=african +european+american+asian+australian+...MO ;)  :)

why strange Arne?
the top 2 countries (mathematically speaking) in Africa are Morocco and south africa,probably with a big difference compared to the other countries what makes PAMO too easy for the official imo teams,that's why south africa makes an other selection for PAMO but in morocco mathematical contests are not given a big importance(hardly for selecting the imo team)...so they make do with imo :)

are you a member of omm?
will you participate in the next stage on 05april


Hello,

This is an optimization problem. The idea is to find a path of words from one word to other with the minimal weight possible. Now the rules are the next ones:

(1) Each word in the path must be an actual word. In order to decide what is a word and what is not, we will go to [url=http://dictionary.reference.com/]dictionary.reference.com[/url]. You can use plurals, participles, etc, you can not use proper names (as people names, places names, etc)
(2) You can change [b]one[/b] letter in the word by step.
(3) You can permute the letters in a word.
(4) You can not make words neither shorter nor longer (all the words in the path has the same number of letters).
(5) Rules (2) and (3) cannot be mixed, that is, you cannot permute and change letters in the same step

The idea is to find the shortest path between two words. Let's see what we can do from [b]dusk[/b] to [b]dawn[/b] :P (Yes, it can sound a little suggestive) :D
Well, we can go like
[list]D U S K
D U N K
D U N E
D I N E
D I M E
D A M E
D A M N  (Pardon my french :P )
D A W N[/list]
therefore there are seven steps. Any one has a shorter path?

On the other hand, does any one wants to go from [b]Problem[/b] to [b]Solving[/b]?

[quote="sheepwarrior"]dusk
dunk
dank
dawk
dawn
:D
dawk is a word.. i hope i followed the rules correctly[/quote]
Yes it does, all the words are in the indicated dictionary. That is a lot shorter (3 steps :P ) Someone going from [b]Problem[/b] to [b]Solving[/b]?

I'm pretty sure that it is impossible to go from [b]problem [/b]to [b]solving[/b], as I can't think of any words that connect with either [b]problem [/b]or [b]solving[/b].

[quote="JBL"]I can think of how to get off of solving: Salving, for example, to halving or slaving and onwards from there.  An exhaustive search leads me to believe that there is no way to actually get there from problem, however.  David?[/quote]
Well, I don't have one. The game itself is pretty well known, and I just thought that given one of the portals of the site is ArtofProblemSolving, the proposal was interesting...

Now, apparently if you take a standar dictionary and form the graph given by the rules already explained, it seems like you are right and Problem is an isolated point. May be we should ask some rule?


inequalities
integration
trigonometry
function
calculus
real analysis
real analysis unsolved

Prove that $ |\int_0^x\frac {\sin t}{t}dt| \leq \int_0^\pi \frac {\sin t}{t}dt \leq \pi$

I think the left inequality doesn't always hold, at $ x\equal{}2\pi$, for example.

For the right one, we know that $ \forall t\in [0,\pi],\sin t\leq t$ by drawing a figure.

Hence, $ \int_{0}^{\pi}\frac{\sin t}{t}dt\leq\int_{0}^{\pi}\frac{t}{t}dt\equal{}\pi$ :)

The function $ g(x)\equal{}\int_0^x\frac{\sin t}{t}\,dt$ is an odd function that is positive for $ x>0$ and negative for $ x<0.$ For the purposes of this problem, we may assume WLOG that $ x>0.$

$ g(x)$ is a smooth function (entire, in fact) everywhere. It has local maxima at $ x\equal{}(2k\plus{}1)\pi$ for all $ k\ge 0$ and local minima at $ x\equal{}2k\pi$ for $ k\ge 1.$

The first inequality is the statement that the local maximum at $ \pi$ is in fact the global maximum. Despite what Timestopper_STG says, it is true.

One way to approach this: let $ b_n\equal{}\int_{n\pi}^{(n\plus{}1)\pi}\frac{\sin x}{x}\,dx$ and show that $ \sum_{n\equal{}0}^{\infty}b_n\equal{}\sum_{n\equal{}0}^{\infty}(\minus{}1)^n|b_n|$ satisfies the hypotheses of the alternating series test.

The right-hand inequality is an estimate - a very crude and generous estimate - of the height of that global maximum. Here, Timestopper_STG has given the simplest argument for this particular estimate. Doing a little more work (I recommend using power series) shows that that maximum is approximately $ 1.18$ times $ \frac{\pi}2,$ where $ \frac{\pi}2$ is significant as the whole improper integral (that is, the limit of $ g(x)$ as $ x\to\infty.$)

Key words: Gibbs phenomenon, Fourier series, Fourier transform, sinc function.

May I ask some silly question ... what does WLOG stand for?  :blush:

It stands for "without loss of generality." This accompanies an assumption that reduces a problem to one particular case, but with the declaration that all other cases would work out in the same way.

I used it here when I assumed that $ x>0.$ But since $ g(x)$ is an odd function, we have $ |g(x)|\equal{}g(\minus{}x)$ for $ x<0,$ so what we learn for positive $ x$ will apply to all $ x.$

Hello

there is one maple illustration for this inequalitie


calculus
integration
trigonometry
calculus computations

[quote="fedja"]If you use the substitution $x=\cos y$, you'll get the integral $\int \cos^2 y\cdot y\cdot (-\sin y)\,dy$. Can you do that?[/quote]

Use this, you are to calculate with integral by parts.

Put  $x=\cos\theta$

$\Rightarrow~~$  $\int {x^2\cos^{-1}x\,dx =\int \theta \cos^2\theta d\cos\theta= -\int \theta (1-\sin^2\theta )\sin\theta\,d\theta}$

$=~-\int\theta\sin\theta d\theta+\int\theta\sin^3\theta d\theta ~=~\int\theta d\cos\theta-\int\theta (1-\cos^2\theta) d\cos\theta$

$=~{1\over 3}\int\theta d\cos^3\theta~~=~~{1\over 3}\theta\cos^3\theta-{1\over 3}\int \cos^3\theta d\theta ~=~{1\over 3} \theta\cos^3\theta-{1\over 3}\int\cos^3\theta d\theta d\theta$

${=~{1\over 3}\theta\cos^3\theta -{1\over 3}\sin\theta+ {{\sin^3\theta}\over 3}+C~~=~~{{-3+\sin^2\theta}\over 9}\sin\theta +{{\theta\cos^3\theta}\over 3}+C}$

$=~~-{{2+1-\sin^2\theta}\over 9}\sin\theta+{{\theta\cos^3\theta}\over 3}+C~~=~~ -{{2+\cos^2\theta}\over 9}\sin\theta+{{\theta\cos^3\theta}\over 3}+C$

$=~~-{{2+x^2}\over 9}\sqrt{1-x^2}+{x^3\over 2}\arccos (x)+C$

where  $x=\cos\theta$


View this: http://www.youtube.com/watch?v=yysnkY4WHyM

That is really cool. :) Can someone explain it to me?

He tells you that it occurs due to conservation of momentum and coupling via the base.  Also recall that these are waves and waves can influence each other.

It struck me as interesting that the metronomes where placed on a board with "wheels".  Last semester, we had to solve an interesting problem which asks that if the wheels are spinning at some constant angular velocity, at what frequency does the board oscillate?


function
induction
algebra unsolved
algebra

Find all function $ f: \mathbb R \to \mathbb R$ that satisfy \[ f(x + \cos(2007y))=f(x)+2007\cos(f(y)),\quad \forall x,y\in \mathbb R.\]

[quote="duythuc_lqd"]Find all function $ f: R \to R$ satisfy $ f(x \plus{} cos(2007y)) \equal{} f(x) \plus{} 2007cos(f(y)), \forall x,y\in R$[/quote]

Let $ P(x,y)$ be the assertion $ f(x \plus{} \cos(2007y)) \equal{} f(x) \plus{} 2007\cos(f(y))$

Subtracting $ P(0,y)$ from $ P(x,y)$, we get $ f(x \plus{} \cos(2007y)) \equal{} f(x) \plus{} f(\cos(2007y)) \minus{} f(0)$

Using then $ g(x) \equal{} f(x) \minus{} f(0)$, we get $ g(0) \equal{} 0$ and $ g(x \plus{} y) \equal{} g(x) \plus{} g(y)$ $ \forall x, \forall y\in[ \minus{} 1,1]$

From this, it is easy to show thru induction that $ g(py) \equal{} pg(y)$ $ \forall y\in[ \minus{} 1,1]$ $ \forall p\in\mathbb Z$
Then a new induction establishes $ g(x \plus{} py) \equal{} g(x) \plus{} pf(y) \equal{} g(x \plus{} py)$, $ \forall x$, $ \forall y\in[ \minus{} 1,1]$ and $ \forall p\in\mathbb Z$

And so $ g(x \plus{} y) \equal{} g(x) \plus{} g(y)$ $ \forall x,y$

But $ P(0,x)$ $ \implies$ $ f(\cos(2007x)) \equal{} f(0) \plus{} 2007\cos(f(x))$ and so $ g(x)\in[ \minus{} 2007, \plus{} 2007]$ $ \forall x\in[ \minus{} 1,1]$

So $ g(x)$ is a solution of Cauchy's equation bounded on $ [ \minus{} 1, \plus{} 1]$ and so $ g(x) \equal{} ax$ and $ f(x) \equal{} ax \plus{} b$

Plugging this back in the original equation, we get $ a\cos(2007y) \equal{} 2007\cos(ay \plus{} b)$ and so the solutions :

$ f(x) \equal{} 2007x \plus{} 2k\pi$
$ f(x) \equal{} \minus{}2007x \plus{} (2k\plus{}1)\pi$


[hide] Since every number has a "pair" in the set except for 6, every number removed removes a pair of numbers except for 6. Thus, removing 6 makes the probability higher. [/hide]

[quote="Quevvy"][quote="espark52"]two different numbers are randomly selected from the set S={1,2,3,4,5,6,7,8,9,10,11}. the probability that their sum is 12 would be greater if the number n had first been removed from set S.[/quote]
[hide]The probability of the sum being 12 is
$\frac{5}{\binom{11}{2}}=\frac{5}{55}=\frac{1}{11}$
If you removed the number 6, the probability is:
$\frac{5}{\binom{10}{2}}=\frac{5}{45}=\frac{1}{9}$[/hide][/quote]

That's valid, but it's kinda cheating. How would we have magically guessed that what you did was going to end up that way? humblemastermind's solution addresses that.

[quote][b]13375P34K43V312 wrote:[/b] That's valid, but it's kinda cheating. How would we have magically guessed that what you did was going to end up that way? humblemastermind's solution addresses that.
[/quote]

i agree with humblemastermind.


How can I estimate the magnitude of an star just looking at it? I know some reference stars, but it is very difficult to compare the flux of two stars. Is that star two times brighter than the other, or 2.5 times? Is there some trick to do this more efficiently?

hey do it by carmalien method
thats the method used by nasa i suppose
but dont forget to correct te aberration by the formula
M=m+5-5logr
r=radius
m=reflector angle
hope this might help u !
kudos!
bye!

Finally, a topic that is not too old to reply to.  :)
Funny, I did not notice this topic before. It must have been moved.


vector
linear algebra
linear algebra unsolved

My linear alebra class has just begun, and my having difficulties to solve this exercice:

 A. Let V be a vector space. Let x and y be elements of V (x and y are vectors) and a is an element of R, assume that ax=0. Show that either a=0 or x=0.

 B.Use this to show that if ax=bx and x is different from vector 0, then a=b.

 Any idea?

 Oh, now I am editing this post after 20 minutes... I found a solution that is long, you may have a shorter solution.

A. If $a\neq 0$ then $ax=0\Rightarrow \frac{1}{a}.ax=0\Rightarrow x=0$.

B. $ax=bx \implies (a-b)x=0$ use the A.

Yes, these are the ideas I have used. But what you have written is not enough.

The proof posted by "leibniz" is indeed sufficient.

We as asked to prove that $a=0$ or $x=0.$ There are two cases: either $a=0$ or $a\ne0.$ If $a=0,$ we are already done. If $a\ne0,$ then "leibniz" shows that $x=0.$

That $a=0,x=0$ is possible need not be explicitly shown; it has never been ruled out.

If you want to phrase it as truth tables, then:

"If P then Q" is identical to "(not P) or Q." We could rearrange this to say that "A or B" is indentical to "if not A, then B." And the latter is exactly the structure of the proof "leibniz" gave.

[quote="Cartesian-mind"]You have to add to A the contrapositive: If x different from 0, then a=0. You still need to tell that a and x can be both null which is easy to prove.[/quote]
I hope that the Kent merrryfield answer will be enough now :D


Given a circle with a chord $AB$ and a point $W$ on the chord. Suppose that $KL$ nad $MN$ be two chords through $W$. If chords $AB$ and $KN$ meet at $P$ and chords $AB$ and $ML$ meet at $Q$, then
\[
\frac{1}{PW} - \frac{1}{QW} = \frac{1}{AW} - \frac{1}{BW}
\]

This is a pretty known result. See, e. g., Grobber's posts in http://www.mathlinks.ro/Forum/viewtopic.php?t=3498 .

  darij

I haven't carried out the computations, but we should be able to derive it from the fact that the cross-ratio of $NA,NB,NK,NM$ is equal to the cross-ratio of $LA,LB,LK,LM$. We just translate this to $(A,B;P,W)=(A,B;W,Q)$.

[Edit: Darij was here before me :)]


Let $4x+3=a$, so that the difference between $a^y$ and $2^y$ is a perfect square. If we have $y=1$ and $a=3$, then $3-2=1$. $3=3+4x \Rightarrow x=0$. So we have $(0,1,1)$ as one solution. Are there several solutions?  :?

yes, there are several solns.

as posted above me, by taking mod 4 we exclude the case $y \geq 2$ odd, and we get some nice things for $y$ even

there seem to be infinitly many solutions  :?
[hide]$(x,y,z)=(n^2+n,1,2n+1)$
are there any others?
edit-with $n\in \mathbb{N}$ of course ;) [/hide]

[quote="Psycho"]anyone feels like proving that $3^y-2^y$ is never a perfect square, except for $y=1$ :P
if you can prove this, the only correct solution is mine :)
edit-and $(0,0,0)$ too of course ;)[/quote]
For $y \ge 2$, mod 4 forces y to be even.
mod 3 forces y to be odd. contradiction.
It remains to check $y \le 2$.

meep [o_o] never mind what i last said  :?  i don't know how i came to that...
although [hide]$(x,y,z)=(n^2+n,1,2n+1)$ or $(0,0,0)$ remains the only answer according to me...[/hide]

well you are right. Consider $y \ge 2$, then mod 4 and y is even. Then $4x+3 = m^2+n^2$ for some m,n which is impossible, no solution.
Then y=1 or 0. If y=0, x can be anything while z must be 0.
If y=1, $4x+1=z^2$, write $z=2k+1$, then $x=k^2+k$, so the only solutions are $(k^2+k, 1, k)$ and $(k,0,0)$

[quote="Psycho"][quote="siuhochung"]well you are right. Consider $y \ge 2$, then mod 4 and y is even. Then $4x+3 = m^2+n^2$ for some m,n which is impossible, no solution.
[/quote]
how come $4x+3$ has to be of the form $m^2+n^2$ then?  :?
sorry if it sounds stupid... i just don't seem to see it  :blush:[/quote]
it's pythagorean triple.

oh  :oops:
nice solution, man  :omighty:  :thumbup:

i made some minor mistakes actually.
fit should be $2^{y/2}=2mn$ and $(4x+3)^{y/2} = m^2+n^2$. If both m,n are even, then the latter one must not be fulfilled. So one of them must be 1. Then $(4x+3)^{y/2}=  4^{y/2} + 1$. If $ x \ge 1$, l.h.s. is too large. Else r.h.s. is too large, so there is no solution.
The final subcase of this case is that y=2 and m=n=1. But it is very easy to verify that there is no solution for this. So for the remaining it just follows my previous post.


[i][b]Bertrand's Theorem[/b] states that for every $x>1$, there exists a prime number between $x$ and $2x$.  Use this fact to show that every positive integer can be written as the sum of distinct primes.  (For this result, assume that one is a prime.)[/i]

Masoud Zargar

[hide="Solution"] Let $x = x_{0}$.  If it is prime, we are done.

Suppose it is odd.  Then $r_{0}= \lfloor \frac{x_{0}}{2}\rfloor+1$ has the property $2r_{0}-x_{0}= 1$, so there exists a prime $r_{0}< p_{0}< 2r_{0}$ such that $2p_{0}> x_{0}$.  Let $x_{1}= x_{0}-p_{0}$, which is less than half of $x_{0}$, and repeat the algorithm.

Suppose it is even.  Then there exists a prime $\frac{x_{0}}{2}< p_{0}< x_{0}$, and then let $x_{1}= x_{0}-p_{0}$ as before.  Repeat the algorithm.

In each case we have $2x_{k+1}< x_{k}$, so every new prime $p_{k}$ generated is guaranteed to be distinct.  The algorithm is guaranteed to terminate because each successive term is at most half the previous.

Once it terminates, $\sum p_{k}= x_{0}$.  QED. [/hide]

Now please use strong induction to solve the problem.

Ooh, that's fun too.

[hide="Solution 2"] The inductive hypothesis holds true for $1$.  Suppose it holds true for $1, 2, 3, 4, ... k$.

Then there exists a prime $p$ such that $\lfloor \frac{k+1}{2}+1 \rfloor < p < 2 \lfloor \frac{k+1}{2}\rfloor+2$, and since the inductive hypothesis is true for $(k+1)-p < p$, which cannot have $p$ in its unique prime representation, it holds true for $k+1$.  Hence our inductive hypothesis is true.  QED. [/hide]


Could someone walk me through a proof that if B and D are bases of some vector space V, then the change of basis $ P_{B \minus{} > D}$ is the inverse of the change of basis matrix $ Q_{D \minus{} > B}$?

I am having trouble getting started, since I am pretty new to block multiplcation of matrices and I keep getting lost in notation. By the way, I am assuming that what I am learning is pretty much standard but I could be worng, so if any of the terms need more definition let me know.

Pick a transformation $ f$ from some space $ W$ (with basis $ E$) to $ V$. Let $ M_1$ be its matrix with respect to $ E$ and $ B$, and $ M_2$ be its matrix with respect to $ E$ and $ D$. We have $ M_2\equal{}P\cdot M_1$ and $ M_1\equal{} Q\cdot M_2$, so $ M_2 \equal{} P\cdot Q\cdot M_2$ and $ M_1 \equal{} Q\cdot P\cdot M_1$.
Those $ M_1$ and $ M_2$ could be anything with the right first dimension, so $ PQ$ and $ QP$ are a (left) multiplicative identity. For multiplication on the right, just use transformations from $ V$ to $ W$ instead.

So now $ (PQ)M\equal{}(QP)M\equal{}M(PQ)\equal{}M(QP)\equal{}M$ whenever $ M$ is a square matrix of the right size. Straight from the definition, $ PQ$ and $ QP$ are the identity.


2003 National Sprint Round

A fair, standard six-faced die is tossed eight times.  The sequence of the eight results is recorded to form an eight-digit number.  What is the probability that the number is a multiple of eight?  Express your answer as a common fraction.

8 evenly divides everything that is a multiple of , so we just have to figure out the probability on the last 3 digits. We know that any number may be formed that is smaller than 666, so we count up how many there are(83) and get for the probability $\frac{83}{666}$

Billy

For my solution, I just looked at the last three digits, which would yield $6^{3} = 216$ combinations. I then split the problem up into two halves, as one half would have odd numbers in the hundreds digit and the other half would have an even number in the hundreds digit. If the hundreds digit was one, three, or five, the two digit answers that would yield a multiple of  would be 12, 36, 44, and 52. If the hundreds digit was 2, 4, or 6, the numbers that would work would be 16, 24, 32, 56, and 64. This sums up to $3(5+4) = 27$. Thus the probablility is $27/216$, which equals $3^{3}/3^{3}*2^{3}$ which when simplified is 1/8


algebra
polynomial
quadratics
inequalities
ratio
function
modular arithmetic

The following problem occured a long time ago in Amer. Math. Monthly:

Does there exist  an integer polynomial $f \in \mathbb{Z}[x,y]$ in two variables which is a bijection $: \mathbb{Z}^2 \to \mathbb{Z}$?

This is a very intriguing problem. I spent some time on it trying to find such a polynomial, but after one hour I think there aren't such polynomials. Do you happen to know the answer?

I also think that the answer is "no." Although I still haven't had enough time thinking on this problem to obtain a rigorious solution, here is what I would suggest:

1)  Let $f$ has degree $k$ as a polynomial in $x$ and degree $l$ as a polynomial in $y$. Then, show by a density argument that if $f$ is surjective, then
\[
\frac{1}{k} + \frac{1}{l} \geq 1.
\]

2) If $k=1$ or $l=1$, then $f$ is clearly not injective. In this manner, we are left only with the case $k=l=2$ of quadratic forms.

3) Now, (this may be somewhat harder), I claim that if a binary quadratic form $\mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}$ is surjective, then its discriminant $\Delta$ is a perfect square. The key to this is that if $N$ is expressible under this quadratic form, then $\Delta$ is a quadratic residue modulo $4|N|$.

4) Thus, if our binary quadratic form $f$ is surjective, then it must be the product if two linear forms. This clearly contradicts injectivity.

What do you think of this?

A very nice approach. I also found that inequality with l and k, but I wasn't sure on it and also I have troubles with proving that thing with the discriminant. I will keep trying. Anyway, a very beautiful problem and very nice ideas. Thanks a lot.

Please, explain how did you get that inequality with l and k. :maybe:

Unfortunately, while trying to write the explanation, I realized my argument is false. Sorry. Maybe Vess could help us again!

In the same time there exists such polynomial $f:\mathbb{N}\times\mathbb{N}\to\mathbb{N}$.

Let $S_y(n)$ be the set of integers which are the image of the polynomial (when we fix $y$) and which are in the interval $(-\frac n2,\frac n2)$. Then the cardinal of $S_y(n)$ is, when $n$ is large, approximately $n^{\frac 1k}$ (i.e. their ratio tends to $1$). When we change $y$ we get other sets of the same cardinality, and when $y$ varies we get approximately $n^{\frac 1l}$ sets. Then when $x,y$ go around $\mathbb Z$, we get approximately $n^{\frac ik+\frac 1l}$ integers which are the images of the polynomial function over $\mathbb Z$ and which are in the interval $(-\frac n2,\frac n2)$. This means that when $n\to \infty$, the ratio between the number of images of the polynomial which are in $(-\frac n2,\frac n2)$ and $n$ approaches $\frac 1k+\frac 1l$. This is the density of the set, which must, at the same time, be $1$, so we get $\frac 1k+\frac 1l=1$. We actually get equality because the sets $S_y(n)$ obtained when we vary $y$ are disjoint, so I've also used the injectivity. Something similar would have found us $\ge$ instead of $=$, had we used only the surjectivity.

I know it's not rigorous, but I think the approach should be something similar (for the inequality; I don't know about that thing with the discriminant).

In fact it is not rigorous at all, but it was also my idea and since I have no other I will accept it as it is. It is quite intuitive fact, but I blocked when trying to make it rigorous and this is also the reason for which I said it doesn't work.

No. I used the same approach and made the same mistake. This idea of proving that $1/k + 1/l \geq 1$ for surjective polynomials certainly applies in the case when $k$ and $l$ are both even, or when $k = l$ (are odd.) While it is totally obvious, for example, that if all but finitely many positive integers are expressible under the form
\[
x_1^{k_1} + \cdots + x_m^{k_m}, \quad x_1,\ldots,x_m \in \mathbb{N}
\]
then $\frac{1}{k_1} + \cdots + \frac{1}{k_m} \geq 1$, it is certainly not clear at all why this would hold if we assume that $x_1,\ldots,x_m$ are *arbitrary* (instead of positive) integers.

I DISCLAIM THAT  if $f$ is surjective, then $1/k + 1/l \geq 1$. Can anyone explain why the polynomial
\[
x^2 + y^3
\]
is not surjective? (we have $\frac{1}{2} + \frac{1}{3} = \frac{5}{6} < 1$.) If you write down carefully the above-mentioned idea applied to the polynomial $x^2+y^3$, you'll quickly realize that it is wrong. Why is it obvious that infinitely many of the diophintine equations
\[
y^3 = x^2 + 2, y^3 = x^2 + 3, y^3 = x^2 + 4, \ldots
\]
are not soluble? This certainly cannot be done by a density argument.

Concerning this thing with the discriminant, let $\Delta$ be the discriminant of a surjective quadratic form. It is well-known that if $N$ is expressible under the quadratic form, then $\Delta$ is a quadratic residue modulo $|N|$ (Need I explain this?). Thus, $\Delta$ is a quadratic residue modulo all primes $p$. We claim that this is only possible when $\Delta$ is a perfect square. Suppose otherwise, and let $\Delta = xq^2$ with $x \in \mathbb{N}$ is square-free.  Because $\Delta$ is not a perfect square, we have $x \neq 1$. Now, let $p$  be any prime and $\big(  \frac{ \cdot }{p} \big)$ denote the Legendre symbol. Then,
\[
\Big( \frac{x}{p} \Big) = \Big( \frac{\Delta}{p} \Big) = -1,
\]
for all primes $p$.

the last thing is wrong...why must $\Delta$ not be a quadratic residue modulo p? for example, 7 is not a square, but 7 is a quadratic residue mod p=3.

Indeed, the proof that if a number is a quadratic residue for all sufficiently large primes is much more tricky!

Sorry, I incidentally submitted the post before finishing it.
[To
Peter Scholze: the  claim is that, if an integer is a quadratic residue modulo *all* primes, then it is a perfect square.]
Also, this last line should read
\[
\Big( \frac{x}{p} \Big) = 1,
\]
for all primes $p$.
 Here is the remainder of the proof that $\Delta$ is a perfect square, i.e. $x = 1$. It cannot be that $x=-1$, because $\big( \frac{-1}{p} \big) = -1$ for $p \equiv 3 \pmod{4}$. So, $x$ can be written in the form
\[
x = \varepsilon p_1 p_2 \cdots p_s,
\]
where $\varepsilon = \pm 1$ and $s \geq 1$ and $p_1,\ldots,p_s$ are distinct primes.

By quadratic reciprocity, we have
\[
(1) \quad \Big( \frac{p_i}{p} \Big) \Big( \frac{p}{p_i} \Big) = (-1)^{\frac{(p-1)(p_i-1)}{4}},
\]
for $1 \leq i \leq s$ and  all odd primes $p$, provided that $p_i$ is odd.

 In addition,
\[
(2) \quad \Big( \frac{2}{p} \Big) = (-1)^{\frac{p^2-1}{8}}
\]
can both take values $1$ and $-1$.

In view of (1), (2) and Dirichlet's theorem on primes in arithmetic progression, we may choose the prime $p$ in such a way that
\[
\Big( \frac{\varepsilon}{p} \Big) = -1,
\]
and
\[
\Big( \frac{p_i}{p} \Big) = 1, \quad \forall 1 \leq i \leq s.
\]
[I may give a further explanation if needed, but the post would become too long.]
For this $p$,
\[
\Big( \frac{x}{p} \Big) = \Big( \frac{\varepsilon}{p} \Big) \Big( \frac{p_1}{p} \Big) \cdots \Big( \frac{p_s}{p} \Big) = -1,
\]
a contradiction. Thus, $x=1$, i.e. $\Delta$ is a perfect square. This completes the proof.

One more correction. It is not true that we may choose $p$ so that $\Big( \frac{\varepsilon}{p} \Big) = -1$. Instead, we may choose $p$ so that:
\[
(1) \qquad \Big( \frac{p_1}{p} \Big) = -1
\]
and
\[
(2) \qquad \Big( \frac{p_i}{p} \Big) = \Big( \frac{\varepsilon}{p} \Big) = 1,
\]
for $2 \leq i \leq s$. Indeed, if $p_1$ is odd, then we may choose $p \equiv 1 \pmod{8p_2\cdots p_s}$ and $p \equiv -1 \pmod{p_1}$. And, if $p_1=2$, we may choose $p \equiv 5 \pmod{8}$ and $p \equiv 1 \pmod{p_2 \cdots p_s}$. The rest of the proof is the same.

Dear Vess, is it for sure that this problem has a solution? I mean, I read some papers which indicated that most of the problems related to finding polynomial bijections from Z^n to Z are still open. Do you know if there was any solution sent for this problem? It is so hard... Moreover, do you happen to know were I could find articles related to sets of values taken by polynomials in many nedeterminates? Thanks.

@vess: that was exactly the way I was proving it for myself(at least the main ideas sounds like that I didn't read every formula in your proof) but I didn't write it down(too lazy :( ). do you know any proof without using theorems like quadratic reciprocity and dirichlet? those make the task rather trivial.

To Harazi:

   No.  :P  Although, as far as I know, a comment on this problem has appeared in a later issue of the AMM, I have not seen a solution, and am not sure whether this problem is open or not. Unfortunately, I do not remember the exact reference to AMM, but I'm sure there was some article on polynomial bijections in  a subsequent issue.
  I would appreciate if you could tell me which articles have you read and what results have you found on that subject.
  Right now I cannot remember papers on the range of multivariate polynomials over $\mathbb{Z}$ or more general rings. Even questions on the range of a polynomial in one variable often turn out to be highly non-trivial. Take, for example, this one:

    Does there exist a non-linear polynomial $f \in \mathbb{Z}[x]$ whose range     contains all numbers of the form $2^p$ with $p$-prime?

To the best of my knowledge, no solution has been found to this, although the existence of infinitely many Mersenne primes would obviously ensure the non-existence of such a polynomial. Let me mention that the problem of finding all polynomials whose range contains all (integral) powers of $2$ appeared in Kvant recently, and was used as Problem 6 in the 2003 Bulgarian MO.

To Peter Scholze:

  There is hardly a simpler proof, in my opinion. The problem is to show that, given any square-free integer $x \neq 1$, there exists a prime such that $x$ is not a quadratic residue modulo this prime. How to explicitly find such a prime $p$? We need to use some result on the existence of primes, and we used Dirichlet's theorem. The mere infinitude of primes would clearly not suffice, but is obviously implied by our result. As for quadratic reciprocity, this is the simplest way to deal with $\Big( \frac{x}{p} \Big)$ for all large primes $p$.
  Yes, I agree that these two results trivially imply what we needed. I know a very different proof, which is *by no means* simpler --- it uses a density result of Chebotarev. This latter method, however, easily implies a more general result on arbitrary power residues:

 Claim. If $r \in \mathbb{N}$ and $x \in \mathbb{Z}$ are such that $x$ is an $l$th power modulo all sufficiently large primes, then $x$ is an $l$th perfect power.

Harazi, I would appreciate if you could post some relevant results that you've read.

--Vesselin

Well, I just read some pieces of information with Google. As far as I remember, Polya and Fueter found all quadratic polynomials which map N^2 into N and there are exactly those well-known. But their proof was based on very deep  and non-elementary results in transcendental numbers. But M.A. Vsemirnov seems to have found an elementary solution for this theorem. Unfortunately, I cannot find anywhere his article On n-tupling polynomials with small degree where much more information is given. If you can contact him , maybe you will find more information. His e-mail is vsemir@pdmi.ras.ru He seems to be an expert on such problems . I'm sorry, but I can't help you more.

@vess: you know the solution to problem 6 of the IMO 2003: prove that there exists a prime q for every prime p such that $n^p - p$ is not divisible by q for every integer n? there is an absolutely elementary solution to that(look, e.g. at http://www.kalva.demon.co.uk/imo/isoln/isoln036.html ). I thought there is maybe a solution to our problem in the same fashion, but after searching for it for one or two hours I gave up.

but since the first step of the proof to the original problem discussed here is wrong that doesn't matter at all in that topic.

[quote="Peter Scholze"]
but since the first step of the proof to the original problem discussed here is wrong that doesn't matter at all in that topic.[/quote]

At least we showed that $f$ cannot be a binary quadartic form.  :)

As for a more elementary proof of the fact that each integer which is not a square is a quadratic non-residue modulo arbitrarily large primes, I strongly believe that there is not a proof in the spirit of IMO 2003/6 (and I can explain why.) Note that this latter proof cannot be modified to show the existence of infinitely many primes $q$ for which $p$ is not a $p$th power $\mod{q}$. A density result of Chebotarev, on the other hand, immediately implies the infinitude.

Note that, by the way, the 2003/6 problem is immediately implied by the claim (which, as far as I can recall, is due to Ankeny and Rogers) made in my previous post.

Well, I must say that the claim you mentioned is wrong in that form; there are some counterexamples(but ONLY these):

If $8|l$, then $2^{\frac{l}{2}}x^{l}$ also has this property for every prime $p$. (Only for $p\equiv 1\mod 8$ we could get problems, but in that case 2 is a square, so still we have no problems).

See http://www.mathlinks.ro/Forum/viewtopic.php?t=85100 (last post on that page).

Peter

[quote="harazi"]Yes, this is why I said the problem is so intriguing. As far as I remember an example was the polynomial $ f(x,y)=x+\frac{(x+y)(x+y-1)}{2} $[/quote]
but this is not an integer polynomial...does somebody know if exists one?


calculus
derivative
calculus computations

I was wondering how you would find the second derivative of $x^3+y^3=9xy$.


calculus
limit
real analysis
real analysis solved

Does that mean that x_n/(1/n) -> 1, or, in other words, n*x_n -> 1?

I think that's what it means, and I assume that ]0,1[ means (0,1).

Let a_n = 1/x_n. it's clear that a_n is strictly increasing and it tends to infinity. We need lim n/a_n. We apply Cesaro-Stolz. To do this we need lim 1/(b_(n+1)-b_n) = lim x_n*x_(n+1)/(x_n-x_(n+1)) = lim x_n*x_(n+1)/x_n^2 = lim x_(n+1)/x_n = lim (1-x_n) = 1, so n*x_n = n/b_n -> 1.

There is a harder question

Find   lim n(n.x_n - 1)
        n->oo

Namdung

[quote="alekk"]x0 in ]0,1[ and x_(n+1)=x_n*(1-x_n)
prove that x_n~1/n[/quote]
this problem has been solved before on this forum. It seems that it was in 'solved problems' or 'proposed problems' section. And the solutions was first given by moubiool.

[quote="Namdung"]There is a harder question
Find   lim n(n.x_n - 1)
        n->oo
[/quote]
this one seems to diverge to -\infty.

well, this post should be moved to calculus section...

[quote="Namdung"]Liyi, you were right. Now, the more meaningfull question
Show that  n(1-nx_n)/ln(n) has finite and nonzero limit[/quote]
See [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=32602[/url] for a complete solution of a very similar problem.

Prove easily that: $\left(\forall\right) n\in N,\ 0<a_{n+1}<a_n<1;\ a_n\searrow 0,\ \frac{a_{n+1}}{a_n}\nearrow 1$.
$\frac{1}{a_{n+1}}-\frac{1}{a_n}=\frac{a_n}{a_{n+1}}\searrow 1\ \ \overrightarrow {\underrightarrow {\left| \ Cesaro-Stolz\ \right| }}\ \ \frac{1}{na_n}=\frac{\frac{1}{a_n}}{n}\rightarrow 1\Longrightarrow na_n\rightarrow 1.$

[u]Remark.[/u] $\lim_{n\to \infty}\ \sum\limits_{k=0}^n a^2_k=a_0.$


If you have an equation like:
$ a_n\cdot x^{n}\plus{}...\plus{}a_1\cdot x\plus{}a_0\equal{}0$, how do you solve these equations?
For integers, we see that $ x\mid a_0$, but the problem is for real numbers.
Can anyone help?

http://mathworld.wolfram.com/Polynomial.html

Try googling and by the way its a polynomial equation. You probably already know that for $ n\equal{}0,1,2,3,4$ there are methods for solving them and for higher ones there are no general method.

As BanishedTraitor pointed out, $ n \equal{} 0,1,2$ are easy.  $ n \equal{} 3$ exists but it's a bit too messy to flat our memorize.  $ n \equal{} 4$ is really convoluted.  Thanks to Galois and Abel, we know that there is no general expression for solutions when $ n\geq 5$.  However, the Fundamental Theorem of Algebra says that as long as $ f(z)$ is a nonconstant polynomial, these solutions necessarily exist in the complex plane.

[quote="MellowMelon"]There are some general methods for $ n \equal{} 5$ and above; they just use transcendental functions.[/quote]

... but they are still extremely convoluted and are not worth memorizing.

Thanks all!
I've seen many problems which substitute into a equation of this form, and usually those who solve that problems just write and don't solve these equations. That's the reason why I asked? BTW, of course I knew it's a polynom.


Solve:
$ 2abc\equal{}(a\plus{}2)(b\plus{}2)(c\plus{}2)$
for positive integers a,b,c.

we may assume WLOG that $ a\leq b\leq c$. we have
\[ \displaystyle 2 \equal{} \frac {a \plus{} 2}{a}\cdot \frac {b \plus{} 2}{b}\cdot \frac {c \plus{} 2}{c} \equal{} \left(1 \plus{} \frac2a\right)\left(1 \plus{} \frac2b\right)\left(1 \plus{} \frac2c\right)\leq \left(1 \plus{} \frac2a\right)^3. (\mbox{ * })
\]
but for $ a\geq 8$, $ (1 \plus{} \frac2a)^3 < 2$, which means we must have $ a\leq 7$. on the other hand, it's also clear from (*) that $ a\geq 3$.

1. $ a \equal{} 3$. then $ 5(b \plus{} 2)(c \plus{} 2) \equal{} 6bc\Leftrightarrow bc \minus{} 10b \minus{} 10c \equal{} 20\Leftrightarrow (b \minus{} 10)(c \minus{} 10) \equal{} 120$. this gives the solutions $ (b,c) \equal{} (11,130),(12,70),(13,50),(14,40),(15,34),(16,30),(18,25),(20,22)$.

2. $ a \equal{} 4$. then $ 3(b \plus{} 2)(c \plus{} 2) \equal{} 4bc\Leftrightarrow bc \minus{} 6b \minus{} 6c \equal{} 12\Leftrightarrow (b \minus{} 6)(c \minus{} 6) \equal{} 48$. this gives $ (b,c) \equal{} (7,54),(8,30),(9,22),(10,18),(12,14)$.

3. $ a \equal{} 5$. then $ 7(b \plus{} 2)(c \plus{} 2) \equal{} 10bc\Leftrightarrow 3bc \minus{} 14b \minus{} 14c \equal{} 28\Leftrightarrow b \equal{} 4 \plus{} \frac {2c \plus{} 84}{3c \minus{} 14}$. think about how to make this an integer. not hard, but a little trickier than the other cases.

4. $ a \equal{} 6$. then $ 2(b \plus{} 2)(c \plus{} 2) \equal{} 3bc\Leftrightarrow bc \minus{} 4b \minus{} 4c \equal{} 8\Leftrightarrow (b \minus{} 4)(c \minus{} 4) \equal{} 24$. so $ (b,c) \equal{} (5,28),(6,16),(7,12),(8,10)$.

5. $ a \equal{} 7$. then $ 9(b \plus{} 2)(c \plus{} 2) \equal{} 14bc\Leftrightarrow 5bc \minus{} 18b \minus{} 18c \equal{} 36\Leftrightarrow b \equal{} \frac {18c \plus{} 36}{5c \minus{} 18} \equal{} 3 \plus{} \frac {3c \plus{} 90}{5c \minus{} 18}$. proceed like in 3.


inequalities
trigonometry
inequalities proposed

Let $x,y>0$ such that $x^2+y^2=1$.Prove that $ xy+max (x,y) \leq \frac{3\sqrt3}{4} $ ;)

[quote="cezar lupu"]Let $x,y>0$ such that $x^2+y^2=1$.Prove that $ xy+max (a,b) \leq \frac{3\sqrt3}{4} $ ;)[/quote]

what is a,b ??? you mean max(x,y) ?

yes,I'm tired.Anyway thanks for the observation.I'll edit my post. ;)

well, here is the solution:

Let $x=cos \theta$ and $y=sin \theta$, where $\theta \in (0,\frac{\pi}{2})$. Clearly we have $x,y>0$ and $x^2+y^2=1$.
Assume that $x\geq y => max(x,y)=x$.
Let now $f$:$(0,\frac{\pi}{2}) \to {\mathbb R}$, $f(\theta)=cos\theta sin\theta + cos \theta = \frac{1}{2} sin(2\theta) + cos\theta$
We have $f'(\theta)=cos(2\theta)-sin\theta$. Solving $f'(\theta)=0$ we get the good sollution $\theta = \frac{\pi}{6}$.
On $(0,\frac{\pi}{6})$, $f$ is increasing, and on $(\frac{\pi}{6}, 0)$, $f$ is decreasing. So we get $f(\theta)\leq f(\frac{\pi}{6})=\frac{3\sqrt{3}}{4}$.

In the same way we make if $max(x,y)=y$

q.e.d.

I haven't checked it all,but at first glance it is ok!Do you have a solution at 9-grade level? ;)

9 gradish soln:

let $x=\sin \alpha$ and $y=\cos \alpha$; WLOG $x \geq y$.

$\sin \alpha \cos \alpha + \sin \alpha \leq \dfrac{3\sqrt{3}}{4}$

$\Leftrightarrow \sin \alpha ( 1 + \cos \alpha) \leq \dfrac{3\sqrt{3}}{4}$

$\Leftrightarrow \sin \alpha 2 \cos^2 \dfrac{\alpha}{2} \leq \dfrac{3\sqrt{3}}{4}$

$\Leftrightarrow 2 \sin \dfrac{\alpha}{2} \cos^3 \dfrac{\alpha}{2} \leq \dfrac{3\sqrt{3}}{8}$

$\Leftrightarrow \sin \dfrac{\alpha}{2} \cos^3 \dfrac{\alpha}{2} \leq \dfrac{3\sqrt{3}}{16}$

let $a=\sin \dfrac{\alpha}{2}$ and $b=$...

by am-gm $\sqrt[4] {(a^2) \dfrac13 b^2 \dfrac13 b^2 \dfrac13 b^2 } \leq \dfrac{a^2+b^2}{4}=\dfrac14$.

so we have $\sqrt {(a^2) \dfrac13 b^2 \dfrac13 b^2 \dfrac13 b^2 } \leq \dfrac{a^2+b^2}{4}=\dfrac{1}{16} ~ \Rightarrow ~ ab^3 \leq \dfrac{3 \sqrt{3}}{16}$ .             $\blacksquare$

is this ok?

yes,perfect_radio your  solution it's ok but a little bit long.anyway,you could have used AM-GM inequality from here:

  $\displaystyle\leftrightarrow\sin\alpha(1+\cos\alpha)\leq\dfrac{3\sqrt{3}}{4}$ and you will use the inequality

   $ \sin\alpha +\cos\alpha\leq\sqrt2$ and i think it's over from here. ;)

$Let\ 0<x\le y\ such\ that\ x^2+y^2=1.\ Must\ prove\ y(1+x)\le \frac{3\sqrt{3}}{4}.$
$\underline{Method\ 1^*.}\ \ max[y(1+x)]\ \ iff\ \ max[y(1+x)]^2\ \ iff$ $\ \ max[(1-x^2)(1+x)^2]\ \ iff\ \ max[(1-x)(1+x)^3]\ \ iff\ \ max[(1-x)(\frac{1+x}{3})^3].$
$\ But (1-x)+3.\frac{1+x}{3}=2\ (constant).\ Thus,\ we\ touch\ up\ the\ maximum$ $\ when\ 1-x=\frac{1+x}{3}=\frac 24$
$\ \ i.e\ \ x=\frac 12,\ y=\frac{\sqrt{3}}{2}\ and\ the\ maximum\ is\ max[y(1+x)]=\frac{3\sqrt{3}}{4}.$
$\underline{Method\ 2^*.}\ \ Let\ ABC\ a\ triangle\ for\  which\ AB\perp AC,AB=x,AC=y,BC=1.$
$\ I\ note\ S\ the\ symmetrical\ of\ A\ with\ respect\ to\ BC.\ We\ apply\ the\ wellknown$ $\ inequality\ 2s\le 3\sqrt{3}R$ $\ in\ the\ triangle\ ASC\ and\ we\ obtain\ (R=1/2):$
$2(h_a+y)\le 3\sqrt{3}.\frac 12,\ i.e.\ y(1+x)\le \frac{3\sqrt{3}}{4}.$

The inequality $\underline{\overline{|2s\le 3\sqrt{3}R|}}$ is wellknown.

Very nice,Levi. The inequality $2s\leq\3sqrt3 R $ is due to Mitrinovic.

This  problem  was published  at  issue  12/2004  of  M&Y  magazine . Im  sure  Vietnamese  members  of this  forum  will  post the  nice  solution  which  can  be found  there .


geometry
3D geometry
sphere
advanced fields
advanced fields unsolved

Assume that $X$ is a torus with one meridional circle shrunk as a point. How to compute $H_{1}(X)$?

This is essentially a sphere with two points identified. Use a CW decomposition. owk

Or use Van Kampen's theorem and that $H_{1}$ is the abelianisation of $\pi_{1}$ ;)

Using Mayer-Vietoris instead would be closer to the spirit of the question :D owk


I am sure I am missing something obvious here, but for some reason...

I am looking at the series [tex]\sum_{n=0}^{\infty}{\frac{2^n}{n^3+1}}[/tex].  I need to find out whether it converges or diverges.

I know that it diverges, but it's weird.  For large values of n, I know that 2^n always "dominates" n^3...so it diverges obviously.  I just don't know how to show it with one of those tests.  I thought about showing that [tex]\displaystyle\lim_{n\to\infty}{\frac{2^n}{n^3+1}}\neq0[/tex] so I tried using L'Hopital's rule:

[tex]\displaystyle\lim_{n\to\infty}{\frac{2^n}{n^3+1}}=\frac{\frac{2^n}{\ln{2}}}{3x^2}=\frac{(2^n)'}{(\ln{2}*3*x^2)'}=\frac{(2^n)'}{((\ln{2})^2*2*3*x)'}=\frac{2^n}{(\ln{2})^3*2*3}[/tex]

This equals [tex]\infty[/tex]

The problem is, I KNOW I am overdoing it...and should just be using a simpler test, but I can't seem to figure out which one...

just notice that $\displaystyle \frac{2^n}{n^3+1}>1$ for $n$ large enough.

Furthermore, if the series $\sum a_n$ converges, $a_n$ must converge to 0.
Because \[\lim_{n\to\infty} a_n=\lim_{n\to\infty} (\sum_{k=0}^n a_k - \sum_{k=0}^{n-1} a_k) = 0.\]
This is the fastest test...  :D

well, i don't think there's a simpler test...

but you don't need to take log to compute the limit.. just write $2^n$ as $(1+1)^n$ then expand it by binomial theorem...  ;)

[tex]\frac{2^n}{n^3+1}<\frac{2^n}{n^3}[/tex]
[tex]\forall n\geq{10}, \frac{2^n}{n^3}>1 \Rightarrow \lim_{x->\infty}\frac{2^n}{n^3}>1[/tex]
If the limit is greater than 1, then the series diverges, by the aptly named divergence test.


geometry
3D geometry
tetrahedron
sphere
analytic geometry
parameterization
algebra

Let be given a tetrahedron $ ABCD $ . Take a point $ M $ in the interior of triangle $ ABC $ and the points $ A', B' , C' $ lying on $ DA , DB , DC $ respectively so that $ MA' , MB' , MC' $ are parallel respectively to the planes $ (DBC) , (DCA) , (DAB) $ . Prove that the circumsphere of tetrahedron $ A'B'C'D $ passes through a fixed point distinct from $ D $ when $ M $ moves in the interior of triangle $ ABC $

Here's a sketch of the solution:

It is not difficult to see that
\[
\frac{DA^{\prime}}{DA}+\frac{DB^{\prime}}{DB}+\frac{DC^{\prime}}{DC}=1.
\]
In fact, since $M$ belongs to the interior of triangle $ABC,$ there exist $a,b,c>0,$ with $a+b+c=1$ such that
\[
a\overrightarrow{DA}+b\overrightarrow{DB}+c\overrightarrow{DC}=\overrightarrow
{DM}.
\]
On the other hand, we obviously have
\[
\overrightarrow{DA^{\prime}}+\overrightarrow{DB^{\prime}}+\overrightarrow
{DC^{\prime}}=\overrightarrow{DM}
\]
(see attached figure), thus $\overrightarrow{DA^{\prime}}=a\overrightarrow
{DA},$ etc.

The center of the sphere passing through points $A^{\prime},B^{\prime
},C^{\prime},D$ is at the intersection of the three planes $\pi_{A},\pi
_{B},\pi_{C},$ which are perpendicular to the line segments $DA^{\prime
},DB^{\prime},DC^{\prime}$ and pass through their midpoints.

We use coodinates. Suppose $D$ is the origin $(0,0,0)$ and the coordinates of a point $P$ are $\left(  x_{P},y_{P},z_{P}\right)  .$ The coordinates of $A^{\prime}$ are $\left(  ax_{A},ay_{A},az_{A}\right)  ,$ etc. The equation of the plane $\pi_{A}$ is
\[
\left(  x-ax_{A}\right)  ^{2}+\left(  y-ay_{A}\right)  ^{2}+\left(
z-az_{A}\right)  ^{2}=x^{2}+y^{2}+z^{2},
\]
which simplifies to
\[
2ax_{A}+2ay_{A}+2az_{A}=a^{2}\left(  x_{A}^{2}+y_{A}^{2}+z_{A}^{2}\right)  ,
\]
a linear equation, with parameter $a$.

We deduce that the solution of the system of equations of the planes $\pi_{A},\pi_{B},\pi_{C}$ -that is, the coordinates of the center of the sphere- will depend linearly on parameters $a,b,c.$ But since $a+b+c=1,$ the coordinates of the center depend linearly of two independent parameters. It follows that the center of the sphere lies on a (fixed) plane. If $D^{\prime}$ is the point symmetric to $D$ across this plane, then $D^{\prime}$ is a fixed point and the sphere obviously passes through $D^{\prime}.$


geometry
geometric transformation
rotation
trigonometry

A thin uniform rod passes under one fixed peg and over another.Show that if equilibrium is possible then the length of the rod must be greater than

        a(1 + tanB/m)  where a is the distance between the pegs ,B is the inclination of the rod to the horizontal  and m is the coeficient of friction at either peg

In order to avoid confusion with mass, I will label the coefficient of friction $\mu$. Let L be the length of the rod and $\rho$ its linear mass density, so that the mass of the rod is $m = \rho L$. Label the left and right rod ends L, R, its midpoint M, and A, B the above and below pegs. Clearly, the rod can be in an equilibrium only if the above peg A is on the same side of the midpoint M as the below peg B and MA > MB, otherwise, nothing would oppose the rotation of the rod around the below peg. Hence, a = MA - MB and $MB < MA < \frac L 2$. WLOG, assume that both pegs are between the left end L and the midpoint M of the rod. The gravitational force component $F_G$ in the direction of the rod is $F_G = mg \sin \theta = \rho L g \sin \theta$. This force has to be balanced by the friction forces $\mu F_A + \mu F_B$ at the pegs A, B, where $F_A, F_B$ are the peg forces on the rod (or rod forces on the pegs) perpendicular to the rod. These forces are obtained from the equality of moments. The torque of $F_A$ with respect to B is $T_A = a F_A$ and similarly, the torque of $F_B$ with respect to A is $T_B = a F_B$. For example, the torque $T_A$ is equal to the difference between the torque with respect to B of the gravitational force component perpendicular to the rod on the longer segment BR and the torque with respect to B of the gravitational force component perpendicular to the rod on the shorter segment BL:

$a F_A = T_A = \left(\frac L 2 + MB\right) \cdot \frac{\rho g \cos \theta}{2} \left(\frac L 2 + MB\right) - \left(\frac L 2 - MB\right) \cdot \frac{\rho g \cos \theta}{2} \left(\frac L 2 - MB\right) =$

$= \frac{\rho g \cos \theta}{2}\left[\left(\frac L 2 + MB\right)^2 - \left(\frac L 2 - MB\right)^2\right] = \rho g L \cdot MB\ \cos \theta$

and similarly, the torque $T_B$ is equal to

$a F_B = T_B = \rho g L \cdot MA\ \cos \theta$

The equilibrium condition is

$\rho L g \sin \theta = F_G \le \mu (F_A + F_B) = \frac{\mu \rho g L\ (MB + MA) \cos \theta}{a}$

$\frac{a \tan \theta}{\mu} \le MB + MA$

$a + \frac{a \tan \theta}{\mu} \le MB + MA + (MA - MB)$

$a\left(1 + \frac{\tan \theta}{\mu}\right) \le 2\ MA \le 2 \cdot \frac L 2 = L$


use\usepackage[sort, comma]{natbib}

style: plainnat

run latex, no error, and the output is fine.
run bibtex, it has 3 error.

i attached the errors it hinted.

I dont know how to do with it, pls help with it.

Not sure what you mean by running latex then running bibtex.
But anyway, Bibtex tells you there is a whitespace in line 2 of the aux file which you should investigate first.
If that doesn't help then, to offer a proper diagnosis, we need a [i]complete[/i] [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=157945]minimal example[/url] showing the errors before any diagnosis can be offered.


i have a question

Do i need US citizenship for taking USAMO ?

No:
[quote="USAMO Teachers Manual"]U. S. citizens and students legally residing in the United States and Canada (with qualifyng scores) are eligible to take the USAMO.[/quote]


linear algebra
matrix
trigonometry
Euler
linear algebra unsolved

is it true that every matrix $ R\in M_{3x3}(\mathbb{R})$ such that $ ^tRR\equal{}I_3$ and $ detR\equal{}1$ (where $ ^tR_{ij}\equal{}R_{ji}$) can be expressed as a product of
$ R_z\equal{}\begin{bmatrix} \cos\theta & \sin\theta & 0 \\ \minus{}\sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{bmatrix}$, $ R_y\equal{}\begin{bmatrix} \cos\phi &  0 & \sin\phi \\  0 & 1 & 0 \\  \minus{}\sin\phi &  0 & \cos\phi \end{bmatrix}$, $ R_x\equal{}\begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos\psi & \sin\psi \\ 0 & \minus{}\sin\psi & \cos\psi \end{bmatrix}$, with $ \phi, \theta, \psi \in [0,2\pi)$?
if it is, how can i prove it and generalize it to $ SO(n)$?

For $ n\equal{}3$ use Euler angles ; see:
http://mathworld.wolfram.com/EulerAngles.html
see the "xyz(pitch-roll-yaw)" convention.
I think that the result generalizes to $ n>3$, using Givens matrices.