Diophantine equation
number theory proposed
number theory

Find all positive integers $ n$ such that: $ n\plus{}1$ and $ 3n\plus{}1$ are simultaneously perfect squares.

$ n\minus{}1\equal{}a^2$ therefore $ n\equal{}a^2\minus{}1$
so $ b^2\equal{}3n\plus{}1\equal{}3a^2\minus{}2$ which leads us to pell-type equation $ b^2\minus{}3a^2\equal{}\minus{}2$
we seek all numbers of the form $ c\equal{}b\plus{}a\sqrt{3}$ which are solutions to the equation $ N(c)\equal{}\minus{}2$
the smalest such $ c$ is $ c_0\equal{}1\plus{}\sqrt{3}$. and all such $ c$ are than givven by $ N(c)\equal{}N(c_o)N(d)\equal{}\minus{}2$ where $ d$ are all solutions of pell equation $ N(d)\equal{}1$
to be short $ c_m\equal{}(1\plus{}\sqrt{3})(2\plus{}\sqrt{3})^m$ where $ m$ is a non-negative integer
which gives us all solutions for $ a\equal{}\frac{(1\plus{}\sqrt{3})(2\plus{}\sqrt{3})^m\minus{}(1\minus{}\sqrt{3})(2\minus{}\sqrt{3})^m}{2\sqrt{3}}$
and $ n\equal{}a^2\minus{}1$ for every non-negative integer $ m$


linear algebra
matrix
algebra
polynomial
linear algebra theorems

Hi, I have a question about a basic theorem in linear algebra.

Let $ F$ be a field, let $ M$ be a matrix over $ F$. Then if $ M$ has distinct eigenvalues, we can write the matrix $ A$ of the eigenvectors of $ M$, and the diagonal matrix $ B$ of its eigenvalues, and we have $ M \equal{} A B A^{\minus{}1}$, where $ A$ and $ B$ are possibly over some finite extension of $ F$. My question is, does this work even if $ M$ doesn't have distinct eigenvalues?

I think a related question is, if $ \lambda$ is a double root of the characteristic polynomial of $ M$, then does $ M \minus{} \lambda I$ have nullspace of dimension at least $ 2$?

I guess you are asking for condition of diagonalizability.

In general, we have algebraic multiplicity greater or equal to geometric multiplicity.
And when they are equal for every eigenvalue, the matrix is diagonalizable.

See http://planetmath.org/encyclopedia/GeometricMultiplicity.html

Could you please give an example of a matrix that is not diagonalizable?

Edit: The matrix \[ A \equal{} \left(
\begin{array}{ccc}
3 & \minus{}1 \\
1 & 1 \end{array}
\right)\] has eigenvalue $ 2$ with algebraic multiplicity $ 2$, but the nullity of $ A\minus{}2I$ is only $ 1$.


suppose you carry out the reaction between a tertiary halide and a base(big/small) under high temperature(assuming room temp. as low) and in DMSO.
then let $r_{p,su}$ =rate of substitution in case of prmary halide
            $r_{p,El}$=rate of elimnation in case of primary halides
            $r_{t,su}$=rate of substitution in case of tertiary halide
            $r_{t,El}$=rate of elimnation in case of tertiary halides
[b]for small bases[/b]
1.$r_{t,El}>>r_{p,El}>r_{p,su}>>r_{t,su}$
2.$r_{t,El}>>r_{p,El}>r_{t,su}>>r_{p,su}$
3.$r_{p,El}>>r_{t,El}>r_{p,su}>>r_{t,su}$
4.$r_{p,El}>>r_{t,El}>r_{t,su}>>r_{p,su}$

[b]for big bases[/b]

1..$r_{p,El}>>r_{t,El}>r_{t,su}>>r_{p,su}$
2..$r_{p,El}>r_{t,El}>r_{t,su}>>r_{p,su}$
3..$r_{p,El}>r_{t,El}>r_{p,su}>>r_{t,su}$
4..$r_{p,El}>>r_{t,El}>r_{p,su}>>r_{t,su}$
5.$r_{t,El}>>r_{p,El}>r_{p,su}>>r_{t,su}$
6.$r_{t,El}>r_{p,El}>r_{p,su}>>r_{t,su}$
7.$r_{t,El}>>r_{p,El}>r_{t,su}>>r_{p,su}$
8.$r_{t,El}>r_{p,El}>r_{t,su}>>r_{p,su}$

Choose the right options.. :D

For small bases the correct choice is 1. (the rate of substitution in a primary halide must be higher than in a tertiary one). For big bases, 3. is the correct one.


Putnam MIT college Harvard Princeton Yale Duke

Putnam Fellows:

Oleg Golberg, MIT
Mathew Ince, MIT
Daniel Kane, MIT
Ricky Liu, Harvard
Tiankai Liu, Harvard
Aaron Pixton, Princeton

Group N1:

Robert Barrington Leigh, Toronto
Thomas Belulovich, MIT
Richard Biggs, Carnegie Mellon
Steven Byrnes, Harvard
Alexander Fink, Calgary
Po-Ru Loh, Caltech
Alison Miller, Harvard [Elizabeth Lowell Putnam Prize]
Thanasin Nampaisarn, MIT
Eric Price, MIT
Kuat Yessenov, MIT

Group N2:

Ralph Furmaniak, Waterloo
Hyun Soo Kim, MIT
Zhiwei Lin, Chicago
Roger Mong, Toronto
Andrei Negut, Princeton
Dimiter Ostrev, Yale
Steven Sivek, MIT
Lingren Zhang, Duke

Teams:

1. Harvard
2. Princeton
3. Duke
4. MIT
5. Waterloo

Honorable mention:

Caltech
Carnegie Mellon
Stanford
Toronto
Yale

One note: of MIT's three Putnam Fellows and four N1's, only one (Daniel Kane) was on the MIT team.

I assume the Romanians who read this are going to ask about Ana Caraiani; she's on the Honorable Mention individual list (and was on the Princeton team.)

I'll have more analysis tomorrow when I get more time.

[quote="The Official Announcement"]A total of 3545 students from 500 colleges and universities in Canada and the United States participated in the Competition. There were teams from 395 institutions. More complete details will appear in the [i]American Mathematical Monthly.[/i][/quote]
In context of that "teams from 395 institutions" I'll proudly claim the team rank for California State University, Long Beach: 123. We also placed one individual (Chris Barrett) among the top 500 individuals.

[quote="cellik413"]Any idea where we can find the full results including all the honorable mentions? Thanks.[/quote]
I have them, if there's someone you're curious about. I didn't want to type in the full Honorable Mention list.

From the other recent thread on this topic:

[quote="DPatrick"]Eventually they will get placed here: http://www.unl.edu/amc/a-activities/a7-problems/putnam/[/quote]
[quote="kedlaya"]The AMC web site that David mentioned will be updated as soon as the mailing arrives at MIT and is passed on from Hartley Rogers to myself; I'm hoping that will happen sometime this week.[/quote]

[quote="Kent Merryfield"]JBL, you are on the Honorable Mention list.[/quote]

Okay, cool!  (We're on Spring Break, so I assume the math department won't tell me until Monday.)  If you have the time, it'd be nice to see some statistics/analysis -- I probably can wait until it goes up on K. Kedlaya's site, though.  :-)

[quote="JBL"][quote="José"]Any MathLinker????[/quote]

Many members of this forum (not me) are among those listed.  Nice work, Tom and Alison![/quote]

Thanks!

After several years of rapid increase, the number of persons taking the Putnam went down slightly this year:

Year: # of competitors
1998 2573
1999 2900
2000 2818
2001 2954
2002 3349
2003 3615
2004 3733
2005 3545

Of the 3545 competitors, 1657, or 46.7%, received a score of zero.

The median score was 1. The mean score was 7.9.

The Putnam Fellows scored 100, 98, 89, 86, 80, and 80.

Group N1 had scores ranging from 67 to 73.

Group N2 had scores ranging from 60 to 66.

The Honorable Mentions had scores ranging from 49 to 59.

Group I (the rest of the top 100) had scores ranging from 44 to 48.

Group II (the rest of the top 200) had scores ranging from 32.4 to 43.

(No, I don't completely understand that 32.4 - there's some kind of tiebreaking around rank 200, and it's always been of feature of Putnam ranks, but beyond that I can't explain it.)

The rest of the top 500 (actually 526) had scores ranging from 20 to 32.

Hi, if you can, how did the team of Mcgill did? Thanks.

[quote="DPatrick"]I am very happy to see Carnegie Mellon (my alma mater) in the top 10 -- I think that's a first.

Go Tartans![/quote]

They were 3rd in 1987.

The official announcement only lists the top 10 teams, so the placement of any other team is at best a guess. McGill might not have been too far away.

You have two Honorable Mentions: Gabriel Gauthier-Shalom and Mathieu Guay-Paquet; and two people in group II: you, and Ragnar Freij.

Two other people were in the top 500.

[quote="toad"]If you wouldn't mind I would appreciate if you could tell me my score.  Dustin Mayeda (University of Washington)[/quote]

He only gets individual scores for the students at his school.  Only your own math department will be able to tell you how you did.

I just picked up the list of winners this morning, and have finished the usual HTML transcription for the AMC web site:

http://www.unl.edu/amc/a-activities/a7-problems/putnam/-html/putnam2005results.html

I corrected a few obvious typos on the fly, but I probably introduced a few new ones; let me know (by email please) if you find any.

I haven't been given the stats sheet yet; I should have that up by early next week.

In the honorable mentions, I think "DYMTRO KARABASH" should be "DIMITRIY KARABASH," although it is possible that it's some third option.

I think theres a slight typo in the result list.

Alexander Fink is not from Chicago, I think hes from calgary

My prof was surprised when i asked him who Alexander Fink was.

[quote="toad"]If you wouldn't mind I would appreciate if you could tell me my score.  Dustin Mayeda (University of Washington)[/quote]
You are in the group of students ranked from 201-483.

I got the statistics sheet and have HTMLified it, so the updating of the AMC Putnam archive with the 2005 results is now complete.

Is the Fink comment about my results page? It seems to be correct there.

Karabash's name is spelled DMYTRO on the results announcement and also on Columbia's online directory, so I'm sticking with it.

Any other potential typos, please email me directly, since I don't check this forum regularly.

And befor e anyone asks: I don't have your individual score, and never will UNLESS you took the Putnam at MIT (and are not a Fellow!), in which case I'll probably have it by Monday.

Once again, the sheer peculiarity of the Putnam team ranking system has asserted itself. By most measures, the dominant school of the competition is MIT, and they only increased that dominance between 2004 and 2005. In some ways, Harvard didn't perform as well in 2005 as it had in 2004. And yet, Harvard is the first place team, while MIT slipped from first place all the way to 4th.

MIT had 3 Putnam fellows, 4 in group N1, 2 in N2, and 14 Honorable Mention.  The closest other school to that performance was Harvard, with two Putnam Fellows, 2 in N1, none in N2, and 4 Honorable Mention. MIT had 41 of the top ~200 students in 2004 and 42 of them in 2005; Harvard's share of those students slipped from 31 to 19.

By Putnam rules, a school must name its three-member team in advance - a month and a half in advance. Couple that with a rank-based scoring system that exaggerates the importance of the weakest member of the team, and you get the peculiar effects.

Of the top 10 teams, at least 8 of them did not name the optimal team, as you'll see below.

The following table lists for the top 10 teams the placement ranges of their official team, followed by what those ranges would have been had they chosen the optimal team. The team score is the sum of the ranks.

Key to the symbols (all but the * are symbols used by the Putnam itself):

P = Putnam Fellow, rank 1 - 5.5
N1 = next group, rank 7 - 16
N2 = next group, rank 17.5 - 23
H = Honorable Mention, rank 27 - 70
I = rest of the top 100, rank 81 - 97.5
II = rest of the top 200, rank 100.5 - 196
* = rest of the top 500, rank 201 - 483

[code]Team             Actual team     Best team
Harvard          P,  N1, H       P,  P,  N1
Princeton        P,  N2, H       P,  N2, H
Duke             N2, H,  II      N2, H,  I
MIT              P,  H,  I       P,  P,  P
Waterloo         N2, H,  II      N2, H,  H

Caltech          N1, H,  *       N1, H,  H
Carnegie Mellon  N1, H,  II      N1, H,  I
Stanford         H,  H,  II      H,  H,  H
Toronto          N1, I,  I       N1, N2, I
Yale             N2, H,  II      N2, H,  II[/code]
Although the five Honorable Mention teams were not explicitly ranked in the official announcement, it is possible to work out that Toronto necessarily ranked ahead of Caltech.

Had both Harvard and MIT selected their best teams, MIT could not have finished worse than second, but it would still be possible, albeit unlikely, that Harvard could have finished ahead of MIT. The extreme case of that: MIT's three Putnam Fellows could have ranked 4, 5.5, 5.5 (we'll never know that for sure) for a total rank of 15, while Harvard could possibly have had ranks 1, 2, and 7. As I said, unlikely, but possible.

As it is, choosing P, H, I, MIT's total rank was in the range 109-173. Duke (N2, H, II) was in the range 145-289, and yet, Duke finished ahead of MIT.

[quote="Kent Merryfield"][quote="FieryHydra"]Okay, okay, I wasn't in the top 5, or top 200, for that matter, but I scored 28 points and my rank is 324. So, I am very pleased with my result. What's more, our university (Southern Utah University) participated for the first time in the Putnam Contest and got a rank of 113, which is not bad at all, I think!

Cheers!
 :lol:[/quote]
Well, you finished ahead of us (Cal State Long Beach). And your individual score beat our top guy's score by two points. So, congratulations![/quote]

Prof Merryfield, thank you very much. I really appreciate it. In fact, a couple of months ago, I did think of transferring to the computer science department at Cal State, Longbeach, but for some reason the idea didn't materialise.

comment on alexander fink was directed to kent merryfield.

Yea, its correct on your page kedaya.

OK, fixed it. Hey, I'm a U. of Chicago alum (Ph.D. program).

Ah, then u must know Paul Sally ... We were wondering who he was when the results were posted, so I asked Paul Sally who 'Alexander Fink' was, and his reply was 'Alexander Fink? Who is Alexander Fink?' I tried to explain that he got about 12th, and was from Chicago. 'What? We need to continue this conversation later. You have to tell me who Alexander Fink is.'

Another story abt the math pirate :)

Hey guys, I'm just wondering, what happened with Jacob Tsimerman this year?  Did he graduate or something?

Tsimmerman was on the Toronto team and scored in group I, rank 81-97.5.

[quote="Xevarion"]How about Jason Ferguson? He always owned everyone from my school in high school competitions in Texas...

I think he's at Duke now?[/quote]

He was on Duke's team.

Which in turn implies from one of the earlier posts that he was between 100 and 200 overall.

Hi eveyone !
Hi Joel ! Congrats on honorable mention (no surprise though)
Speaking of Karabash, Dmitri is his name in Russian (which is his first language),
 Dmytro is his name in Ukrainian (which is where he's from and which is officialy stamped in the passport). The two are equivalent.

Hi Roman!  Thanks. :-)  I guess I've never seen him spell his name with the "o," but that certainly explains it.


Find a closed form for
\[ \binom{n}{1}+3\binom{n}{3}+5\binom{n}{5}+\cdots \]
This problem is from the combinatorics section in aops 2. I solved it combinatorically but when I tried to use generating functions I kinda got stuck. My work's below and any help is very much appreciated  :lol:

[hide="Using generating functions"]
Let's just consider the case when $ n$ is odd for now since the even case is pretty much the same.

Define a function $ f(x)=\binom{n}{1}+2\binom{n}{2}x+3\binom{n}{3}x^{2}+\cdots+n\binom{n}{n}x^{n-1}$. Let $ \omega$ be the second roots of unity, with $ \omega \neq 1$. Consider
\[ \begin{cases}f(1)=\binom{n}{1}+2\binom{n}{2}+\cdots+n\binom{n}{n}\\ f(\omega)=\binom{n}{1}+2\binom{n}{2}\omega+\cdots+n\binom{n}{n}\end{cases}\]
Now solving the problem boils down to finding $ \frac{f(1)+f(\omega)}{2}$. It is well known that $ f(1)=n\cdot2^{n-1}$. Now here's where I get stuck. I have no idea how to evaluate $ f(\omega)$.

[/hide]

$ k*\binom{n}{k}=n*\binom{n-1}{k-1}$. The problem has been reduced to one that we already know how to solve.


Δίνεται μια ακολουθία {an} πραγματικών αριθμών. Συμβολίζουμε με S το σύνολο των σημείων συσσώρευσης της {an}. Δείξτε ότι τα ακολουθα είναι ισοδύναμα.

1. υπάρχουν άπειρα n (φυσικοί) τέτοια ώστε an>s-ε (όπου s=supS και για κάθε ε>0)

2. υπάρχουν πεπερασμένα n (φυσικοί) τέτοια ώστε an>s+ε (πάλι s=supS και για κάθε ε>0)

Σκέψου τον ορισμό του σημείο συσσώρευσης και τί σημαίνει να είναι το $ s$ το supremum των σημείων συσσώρευσης...

Cheerio,

Durandal 1707

Νομίζω ότι μπορεί να λυθεί εύκολα με τις παρακάτω σκέψεις
Αν ισχύει πρόταση 1 και $ s\equal{}supS$ υπάρχει μια τουλάχιστον υπακολουθία της $ a_n$ που συγκλίνει στο $ s$
Αν ισχύει πρόταση 2 υπάρχουν άπειροι όροι της ακολουθίας που είναι μικρότεροι του $ s$ και επειδή το $ s$ είναι σημείο συσσωρεύσεως της ακολουθίας θα υπάρχει μια τουλάχιστον υπακολουθία της $ a_n$ που συγκλίνει στο $ s$

Ίσως έτσι που το έγραψα να δημιούργησε παρερμηνείες, τώρα που το ξαναδιάβασα κατάλαβα καλύτερα τι ρωτάει

Δίνεται μια ακολουθία {an} πραγματικών αριθμών. Συμβολίζουμε με S το σύνολο των σημείων συσσώρευσης της {an}. Δείξτε ότι τα ακολουθα (1) και (2) είναι ισοδύναμα.

1)  s=supS

2)  Για κάθε ε>ο ισχύουν τα παρακάτω
     α. υπάρχουν άπειρα n (φυσικοί) τέτοια ώστε an>s-ε (όπου s=supS και για κάθε ε>0)
      β. υπάρχουν πεπερασμένα n (φυσικοί) τέτοια ώστε an>s+ε (πάλι s=supS και για κάθε ε>0)

Νομίζω πως από το (1) στο (2) το απέδειξα θεωρώντας ότι εφόσον s σημείο συσώρευσης θα είναι το όριο κάποιας υπακολουθίας (αkn) της (αn) οπότε θα υπα΄ρχουν άπειροι όροι της υπακολουθίας σε μια περιοχή του γύρω από το s ( s-ε< ακn <s+ε ) oπότε και άπειροι όροι της ακολουθίας (αn)

Αλλά από το (2) στο (1) ακόμα δεν το έχω βγάλει,

σας ευχαριστώ πολύ για το χρόνο σας και συγνώμη που δεν έθεσα σωστά το πρόβλημα

[quote="spinos"]Ίσως έτσι που το έγραψα να δημιούργησε παρερμηνείες, τώρα που το ξαναδιάβασα κατάλαβα καλύτερα τι ρωτάει

Δίνεται μια ακολουθία {an} πραγματικών αριθμών. Συμβολίζουμε με S το σύνολο των σημείων συσσώρευσης της {an}. Δείξτε ότι τα ακολουθα (1) και (2) είναι ισοδύναμα.

1)  s=supS

2)  Για κάθε ε>ο ισχύουν τα παρακάτω
     α. υπάρχουν άπειρα n (φυσικοί) τέτοια ώστε an>s-ε (όπου s=supS και για κάθε ε>0)
      β. υπάρχουν πεπερασμένα n (φυσικοί) τέτοια ώστε an>s+ε (πάλι s=supS και για κάθε ε>0)
[/quote]

Ok, your statement leaves a bit to be desired but the idea is simple enough ;)
(Το γράψιμο είναι δυστυχώς λίγο terse, αλλά δεν έχω πολύ χρόνο :()

$ 1\Longrightarrow 2$
α. Αν ε>0, θα υπάρχει σημείο συσσώρευσης ανάμεσα στο $ s \equal{} \sup S$ και στο s-ε/2 οπότε θα υπάρχουν και άπειρα $ a_n > s \minus{} \varepsilon$.
β. Αν υπαρχουν άπειρα $ a_n > s \plus{} \varepsilon$ για κάποιο $ \varepsilon > 0$, έπεται ότι θα υπάρχει σημείο συσσώρευσης $ s'\geq s \plus{} \varepsilon > s \equal{} \sup S$, άτοπο.

$ 2\Longrightarrow 1$
Έστω ότι το $ s$ ικανοποιεί τις συνθήκες α,β και έστω $ s'$ σημείο συσσώρευσης με $ s' > s$. Τότε, για $ 0 < \varepsilon < \frac {s' \minus{} s}{2}$, θα έχεις άπειρα $ a_n > s \plus{} \frac {\varepsilon}{2}$, άτοπο (λόγω β). Άρα το $ s$ είναι άνω φράγμα του $ S$: $ s\geq\sup S$.
Τώρα, αν $ \sup S < s$ και $ 0<\varepsilon < \frac {s \minus{} \sup S}{2}$ θα υπάρχει πεπερασμένος αριθμός $ a_n > \sup S \plus{} \frac {\varepsilon}{2}$ (αλλιώς θα υπήρχε σημείο συσσώρευσης $ s' > \sup S$, άτοπο). Αυτό όμως αντιβαίνει την α και καταλήγουμε ότι $ s \equal{} \sup S$, QED.

Cheerio,

Durandal 1707

PS: Η λύση πάει με αγωνιστικούς χαιρετισμούς στον r_boris που με έμαθε ανάλυση ;)


If you have Xbox live just post your name and highest rank

Griff336: 23 in Team Slayer

Oh yah if you're in a clan higher than 15 send me an invite.

EliTe SuzUkI:   I think i messed up the capitals..........i havent played in a while, i'm generally into halo 2 and rs3

comeon noone else plays xbox live? you gotta b kidiin me

I really dont see why you are expecting a forum full of people deeply facsinated by mathematics, to also be intrigued by x-box, or x-box live to be more specific.  However, there are a few people, you and I among them, who are engaged in mathematics and have superb taste in video gaming :P .

i think a topic like this has already been made.

I am re-stating my forum so if anyone hears me just put down yer name and level.

Well, I play Call of Duty 2 and a bit of Full Auto online. Do you have a 360, or just an Xbox?

If you get a 360, just add me to your friends list. My Gamertag's in my signature.

What will your Gamertag be?


for real numbers $a,b,c > 0$ prove:
$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq \frac{a+b}{b+c}+\frac{b+c}{a+b}+1$

see here : http://www.mathlinks.ro/Forum/viewtopic.php?t=107534


What is the sum of the largest and smallest prime factors of 5445?

We can factor $ 5445$ into $ 3^25^1{11}^2$, so the sum we want is $ 3\plus{}11\equal{}\boxed{14}$


AMC
AIME
USA(J)MO
USAMO
AMC 12
AMC 12 B
AMC 12 A

i know a guy who had an index of 232.5

that just sucks :?

Naw, Danny Puller, some Florida kid that doesn't go on this forum.

10th grader, 8 on AIME I. But I'm geographically not eligible anyway (the only not eligible North American country. darn!)

My son missed USAMO by one AIME point. Again. I know one point is not THAT unusual, but I believe he did it three times already (not so sure about the floor in the previous years)...

9 grade: 129+6
10 grade: 136+6
11 grade: 144+8

I guess, he just is not good in the whole-numbers department. Can't count primes under 50 and all such stuff.  He might do much better at USAMO level, but alas...

tetrahedr0n, I feel your pain. Half-point...  :(  Well, I hope you'll have another chance next year.

[quote="DPopov"]Frank....?[/quote]

Me?  Nah.  I got 132.5 and 5.

But I know someone (besides tetrahedr0n) who got 132.5 and (I think)10.

I didn't exactly "just" miss (122+9 = 212) if you look at it score-wise.

However, had I not completely stalled out on a question on the AMC12, I'd have potentially gotten a full 11.5 more points on it. Add this to my miscounting somewhere along my brute-force attempt at #13 of the AIME, and I have 133.5+10=233.5, half point over the cut.

I mean, had a couple more things gone right, I'd have made it... but then again, I'm not at all disappointed with my effort, especially compared to my astonishing 143.5 combined index last year. I've also had a [rather unsuccessful] crack at the USAMO two years ago, so it isn't like I've always been off the cut.

As for you 232.5's out there, just remember that you outscored over 99.9% of the people who took the AMC. Granted, it's terribly disappointing to miss by such a small amount.

was 13 the (0,0) to (5,5) with no right angles? i spent about an hour counting by hand. kept getting lost and had to recount. then got an asymmectricity. doubt that's a word.

I was debating whether to take the AMC12B and pay for an extra bundle of 10 tests, since I am practically the only one at my school that takes AMC.  AMC12A=131, AMC12B=133.5, AIME1=10.  I would have missed had I not gained 2.5 points from A to B!

argh, on the AIME II, i was off by ONE on two of my answers... otherwise i would have made usamo....

when it asked for the sum of the numerator and denom. for the 2nd one, i said 70+9 = 078.
and i said the amount of numbers from 2-12 was 10... so i got 053 instead of 054 for one of my answers. arghhh.

These stories make me feel better about myself.

My problem was that I didn't think I had a chance to get more than 5 right.  I got 3 last year and I didn't study since then, so qualifying for the USAMO seemed far-fetched.

As a result, I checked the first 5 problems about 5 times.  I wasted like 40 minutes double checking answers.

I finally made this year..

However,
Sophomore year 144 on AMC 10, 7 on AIME (floor was 8!)
Junior year 126 on AMC 12, 7 on AIME (I thought I got 9 even after looking at the official answers - maybe misbubbled two)

Senior year 139.5 on AMC 12, 10 on AIME :)

Good luck to you all who had to go through this miserable moments ;)

yea it sux i also got an 8. but again i barely passed amc 10, so an 8 isn't that bad, for me at least.

I got an 8, but I still have 5 years to try.

I did horrible on the AMC 10. 112.  :blush:  I know kill me now!
But then i didn't prepare so i guess i did okay considering that....

[quote="mcalderbank"]For me, as a junior, this is the first year I am taking these math contests. I was nowhere near any cutoff. 122.5 on AMC12 and 7 on AIME. <twiddles thumbs.>[/quote]

You blew away the index needed in Wyoming  :lol:

Don't lose hope. I missed the usamo in 10th grade by one stupid aime mistake, and in 11th grade by two stupid aime mistakes. This year by some freak of nature my aime score jumped 6 points (maybe because I ordered all of the aimes that the company offers and practiced like none other), and now I'm completely screwed for the usamo.  :D

[quote="h_s_potter2002"][quote="mcalderbank"]For me, as a junior, this is the first year I am taking these math contests. I was nowhere near any cutoff. 122.5 on AMC12 and 7 on AIME. <twiddles thumbs.>[/quote]

You beat my index.  By about 40 points too.[/quote]

potter, did u make the usamo?? it seems like your index sounds not so good. My index this yr is barely over 200 but i made it through the floor value.

[quote="marma103"]He might do much better at USAMO level, but alas...
[/quote]

I know exactly what you mean. USAMO requires a different mind-set, which may not have a lot to do with AMC/AIME scores.


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Prove that the sequence $a_n =3^{n} - 2^{n}$ contains no three numbers in geometric progression.


[quote="Erken"]As far as i know problems from shortlist must be kept in secret until the next IMO :wink: .So please delete your post.[/quote]

This is indeed problem N1 from the ISL'2007.
It is really unfortunate that these problems are leaking into the MathLinks...

Of course, there is no point in removing the problem now.
It has been viewed by more than a hundred users.

By the way:
ISL'2007 contained two extremely ugly problems. The posted problem N1, and
problem C7. (Both were submitted by Austria.)  Let's hope for something better
for IMO'2008.

This is an easy problem.

It should not even be an ISL problem 1.

Who want me to provide alternative solution? :D

[quote]By the way, Dida Drogbier, are you from Chelsea or AC Milan? [/quote]

However, Today he must play in the match against Liverpool so he shouldn't mind this now  :)  Go ahead & Good luck  :P

[quote="test20"]

ISL'2007 contained two extremely ugly problems. [/quote]

Why do you say it's extremely ugly? I think it's quite beautiful.

[quote="csbiggy123"]
However, Today he must play in the match against Liverpool so he shouldn't mind this now  :)  Go ahead & Good luck  :P[/quote]

Yeah, maybe I'll be in the same team as him, if Avram Grant selects me.  :wink:  :P

Let's hope the best for Chelsea. Maybe we'll finally win the Champions League this year.  :first:

[quote="test20"][quote="Erken"]ISL 2007 contains much more ugly problem. And some of them the same time extremely hard.[/quote]

Are you talking about A5?[/quote]
About A5, С7, C4, N7

[quote="Erken"][quote="test20"]
Are you talking about A5?[/quote]
About A5, С7, C4, N7[/quote]

Aaah yes, I agree: N7 is also very ugly.
And I thought that C4 and G7 are on the borderline,
but still bearable.

Now I am curious:
Which are your favorite ISL'2007 problems? (by beauty)

As I can see from the above posts , there are many people , not team leaders , who have allready seen the shortlist 2007. After this it has no meaning not to send here the problems.The regulation must change ! Or the team leaders must keep the problems entirely sectret( too much difficult , since they use them) , or the short list must be given to the communities after the competition.I think that the second is better.The problems are made for the contestants and not for the others. Students must have the problems  during the whole previous year  to study them and to become better in problem solving.This makes IMO more fair to the students from all countries.

 Babis

I DO NOT agree anyone who open the ISL problems before that date comes.
Many country cannot make problems by themselves, they are necessarily using the ISL problems to select their contestant.
Opening these problems or posting their solutions make the selection corrupted.
So please do not open them or post their solution again, OK?  :mad:

P.S. N7 is very hard. I don't like it.  :|

[quote="Valentin Vornicu"][quote="m.essien"]Let's hope the best for Chelsea. Maybe we'll finally win the Champions League this year.  :first:[/quote]Go Liverpool! Red Devils all the way.

On the matter of the shortlist, however will post a shortlist problem until the deadline is not over will be [b]banned[/b]. Do not do that.

As for the countries using the ISL as a selection means, they have been advised at the IMO not to do that, as always ISL questions leak on the internet before (just because they are not known by everyone doesn't mean a few people don't have them which makes the selection even more flawed).[/quote]
Forca BARCA!!!
ISL 2007 problems were posted in internet,i saw them three or four months ago,but now i can not find them anymore.
So i think that there is still a reason to not post the problems here.As we saw many times, problems from ISL were used in different TST's

[quote="Erken"][quote="test20"]
Now I am curious:
Which are your favorite ISL'2007 problems? (by beauty)[/quote]
I am not a Combinatorics lover,so i think that G3, N6, A4 are pretty good. :wink:[/quote]

I agree that G3, N6, A4 are very nice.
A4 and G3 were eliminated at the beginning of the jury meeting,
since they were known from other sources.

I also liked A6, C3,  C6, N5.


The amount of state income tax that a Missouri family must pay is " $ 315 plus 6% of any Adjusted Gross Income over
$9,000 ". If a Missouri family's  Adjusted Gross Income is $ 36,400, the amount of Missouri state income tax they must pay is
A.$1644	B.$ 1959	C.$2184	  D.$ 2499 E.$16,755

[quote="sri340"]The amount of state income tax that a Missouri family must pay is " $ 315$ plus $ 6\%$ of any Adjusted Gross Income over $ 9,000$ ". If a Missouri family's  Adjusted Gross Income is $ 36,400$, the amount of Missouri state income tax they must pay is
$ \textbf{(A)}\ 1644\qquad \textbf{(B)}\ 1959\qquad \textbf{(C)}\ 2184\qquad \textbf{(D)}\ 2499\qquad \textbf{(E)}\ 16,755$[/quote]

Thanks 007math
[hide] The tax is $ 315\plus{}.06(36400\minus{}9000)\equal{}315\plus{}274(6)\equal{}\boxed{1959}$. Hence, the answer is B.[/hide]


calculus
derivative
geometry open
geometry

A piecewise C1 curve(Ck curve means k-th derivative of the curve exists and continuous) may have two tangents at each junction point. Define a piecewise C1 curve to be convex if it lies on one side of each tangent line.

Could you give an example to show that the theorem[A simple closed regular plane curve a(s) is convex iff curvature of a(s) has constant sign.] is false for piecewise C2 curves??

Assume that the C2 curve is a kind of cusp, and arcs between singular points are circular. Then the arcs are C2.

Obviously, between singular points curvarture is negative, which means curvature has constant sign, but the overall curve is not convex. Thus, for piecewise C2 curve, the theorem is false.


superior algebra
superior algebra unsolved

For $\sigma \in S_{n}$, let $f_{k}(\sigma)$ denote the number of $k$-cycles in $\sigma$. Show that
$\frac{1}{|S_{n}|}\sum_{\sigma \in S_{n}}f_{k}(\sigma) = \frac{1}{k}$.

Ok, I found a solution.  For every $k$-cycle $\tau \in S_{n}$, let $g(\tau)$ be the number of permutations $\sigma \in S_{n}$ which contain $\tau$ in their presentations as products of disjoint cycles.

Then
$\sum_{\sigma \in S_{n}}f_{k}(\sigma) = \sum_{\tau \in S_{n}\, \text{, k-cycle}}g(\tau)$.

It's not hard to realize that the amount of $k-cycles$ is $\binom{n}{k}(k-1)!$ and that for every k-cycle $g(\tau) = (n-k)!$. Plugging all of this in gives us the answer.


If $ x$ is a positive integer, what is the value of $ x$ for the equation $ (x!\minus{}(x\minus{}3)!) \div 23 \equal{} 1$?

Is there another way of finding the answer without using the method of guess-and-check?

[hide=Solution]Multiplying both sides by 23, we have : $ x!-(x-3)!=23$ Note that 23 is closest to 4!. Checking for 4 we see that 24-1=23 so 4 is our answer. [/hide]

[quote=5849206328x]We must have $ x!\minus{}(x\minus{}3)!\equal{}23$.  We find that $ x\equal{}\boxed{4}$ works and clearly no other positive integer does.[/quote]

[hide] dun't forget hide tags[/hide]


it might be very easy but i cant find it :(
 given  a triangle $ ABC$,points $ P,Q,R$ are taken on $ BC,CA,AB$ so that
$ BP: PC\equal{}CQ: QA\equal{}AR: RB\equal{}m: n.$
prove that

$ area(PQR): area(ABC)\equal{}m^3\plus{}n^3 : (m\plus{}n)^3$

pls help

$ \frac {[RAQ]}{[ABC]}\equal{}\frac {AR\cdot AQ}{AB\cdot AC}\equal{}\frac {mn}{(m\plus{}n)^2}$ a.s.o  $ \implies$ $ \frac {[PQR]}{[ABC]}\equal{}$ $ \frac {S\minus{}\sum [RAQ]}{S}\equal{}$

$ 1\minus{}\sum \frac {[RAQ]}{S}\equal{}$ $ 1\minus{}3\cdot\frac {mn}{(m\plus{}n)^2}\equal{}$ $ \frac {m^2\plus{}n^2\minus{}mn}{(m\plus{}n)^2}\equal{}$ $ \frac {m^3\plus{}n^3}{(m\plus{}n)^3}$


I have just come across a new online drawing program called [url=http://www.cumulatelabs.com/cumulatedraw/]Cumulate Draw[/url] where you can draw varous diagrams, including flowcharts then save in various formats - jpeg, png, pdf for including in LaTeX documents and MMD/FMD for re-editing later. It's all done in your browser so nothing to install.

Hmm...is there a way to include Greek symbols?

Yes, but. Click on the A in the tool to get Text mode. Then drag out a box. In that box you can paste symbols from another program which does accept Greek letters.

But there seems to be a fatal flaw - any text I use gives an empty final image. This is clearly a bug but you can copy the screen and use that for now.

Hello, I am the lead developer on cumulate draw. Thanks for using our site. Greek characters are now supported for image generation. Please see an example at:
[url]http://www.cumulatelabs.com[/url]

Great service! I can confirm that Greek letters can now be used - see attached.

Could I ask for more? Would it be possible to type symbols and non-Latin characters directly rather than copying and pasting them from elsewhere?

Also, no matter what font you use in the text box for the symbols, it reverts back to showing Arial when you click outside the box, but appears to save in the correct font. Attached png shows Times New Roman correctly.


inequalities
linear algebra
linear algebra theorems

Can somenone tell me some inequalities for the $ rank$ of a matrix.
For exemple the rank of sum and the rank of product.
And the inequality of $ Sylvester$.

Have you tried searching for them ?

http://www.mathlinks.ro/Forum/viewtopic.php?t=20558 (Sylvester)

Thank you. I have seen somewhere the inequality but I wasn't sure about the information


geometry
conics
hyperbola
incenter
geometry unsolved

In the configuration of quadrilateral ABCD with incircle O,
    diagonals AC and BD intersect at P.  A pair of opposite vertices
    could be thought of  as foci of a hyperbola which passes  thru the
    other pair of vertices (so there are two hyperbolas). Show that the circle
    thru the incenters of  APB,BPC,CPD,DPA  has it's center on the intersection
    of tangents to active branches of both hypebolas  at  apex points.

   Another way to tackle it is the Mediterranean  approach from the other end.

  A Philo line is a line thru a point inside an angle that has the shortest segment
  inside this angle.

  Explain why the center of this circle thru incenters is the mid-point between two Philo
  points on the diagonals of quadrilateral of incenters ( each diagonal is a shortest
  segment thru a Philo point on it ).

  Maj. Pestich

In Russian slang word " filonit' " means to avoid work, to do as little as possible.
I wonder if Philo's name was the origin, because the segment is the shortest.

In general I really like the sources where you find your problems, though quite uncommon. But are you sure about the source for this problem ?  :D


[quote="Kath"]Let $ \ a,b,c,k,l$ be positive real numbers and $ \ l>k>0$.Prove that:

$ \sum\sqrt{\frac{kx}{y+z}+l}\geq 3\sqrt{\frac{k+2l}{2}}$[/quote]
Let $ \frac{k}{l}=2m.$
Hence, $ \sum_{cyc}\sqrt{\frac{kx}{y+z}+l}\geq 3\sqrt{\frac{k+2l}{2}}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\sqrt{\frac{2mx}{y+z}+1}\geq 3\sqrt{m+1},$ where $ 0<m<\frac{1}{2}.$
It's true. See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=64811


I have posted a sprint round now. Please do not pm the answers. Please check your own answers. (I don't have time to check them.) The answers are conveniently at the back page. The format is a little messed up, but I think it is okay.

There is no due date, so you don't have to do this post haste. Again, I repeat, [b]DO NOT PM THE ANSWERS; THE ANSWERS ARE AT THE BACK!!![/b]

[quote="anirudh"]I could create one, but people have to do it themselves. If more than 10 people are interested, I'll make one.[/quote]
Nutz_for e is making a Mathcounts competition now.

[quote="anirudh"]where is it?[/quote]
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=112054[/url]

I believe it is conveniently titled "Mathcounts Competition."

And... You guessed it! It's in the Mathcounts forum.  :D

Okay. I have posted it. the format is a little messed up though  :(

are these problems original, the pictures are familiar

[quote="anirudh"]These are altered saab problems; they are based off of saab, but are otherwise original.[/quote]
Um original means completely unique...

I think what he's saying is that they are not original. You know, changing the numbers here and there does not change the substance of the problems. In addition, most guys here have already studied all the saab problems.

Also, I'm not sure about putting the answers in the back. That pretty much eliminates the point of a mock contest (esp when it's based off of saab in the first place).

No! It is for one to download and use as one's own math counts contest, and do it as many times they want etc.

[quote="Phelpedo"]1) It's from Saab. Putting saab problems on AoPS doesn't change the fact that they are saab problems and can be found at http://mathcounts.saab.org/mc.cgi.

2) Why would they want to do it "many times?"[/quote]
1) The link doesn't work.

@ anirudh - This is seriously starting to get annoying - random topics everywhere, spam, "original" problems, etc. Please stop. :wallbash:

ER....what are we supposed to do?

my mom actually was impressed by your layout of the questions, by the way.

Okey dokey. This is like what we do at school...but whatever. It's better to have new problems than some from like...1988...

yiou do know that thoe problems are a bit harder than now (at least thats what i heard)

Yeah, well, then it's better practice. Right?

Chatspeak is rather seriously frowned upon in the forums where we try to do math. Similarly chatting between just a few parties in a thread is frowned upon. If you can take a discussion to back-and-forth PMs between 2 people, do it and don't clutter the forums with the conversation nobody else cares about.

(Bluestalk, you're fairly new and you're basing your actions on the Bay Area forum, but be aware that in the Mathcounts forum we're a little more serious. moogra, you should know better.)

Billy

I just wanted to say, unless you make numbers REALLY ugly (17 vs. $\sqrt[5]{23^{8}+72\sqrt{3}+8\sqrt[4]{8}}$) the numbers don't affect the difficulty.

[quote="i_like_pie"]Just do the problems and check your answers with the answers at the back. Not very fun. :|[/quote]

It is still good practice  :wink:

[quote="lotrgreengrapes7926"]What's the pattern on number 15? Is it supposed to be alternating? :maybe:[/quote]
Yes, it's +, +, -, +, +, -...

[quote="lotrgreengrapes7926"]What's the pattern on number 15? Is it supposed to be alternating? :maybe:[/quote]

It is alternating with the format of $+,+,-$. If you need a hint,

[hide]Arithmetic series, looking for patterns[/hide]

[quote="anirudh"]It is alternating with the format of $+,+,-,-$.[/quote]

No, it's $+1+2-3+4+5-6...$

I am making a target one--original ones without answers at the back :wink:  :D

What level is this supposed to be? Chapter, State, National, etc.

I would say that it is an easy state competition.

[quote=mathwiz_aku]How do you solve number 7?[/quote]

Make a quadratic equation by adding sqrt(7x) to both sides and squaring. The left side becomes always positive now. Solve for x, and you should get 0 and 1/7. The arithmetic mean (basically just the mean) is (0+1/7)/2 = 1/14

[quote=CatDog76]Where can I find these[/quote]

I think this was the [url=http://mathcounts.saab.org/mc.cgi]source[/url]


limit
trigonometry
calculus
calculus computations

1. Given that $ \lim_{x \rightarrow a}f(x)\equal{}5, \lim_{x \rightarrow a}g(x)\equal{}\minus{}\sqrt{2}, \lim_{x \rightarrow a}h(x)\equal{}0$, evaluate $ \lim_{x \rightarrow a} \frac{7f(x)\minus{}g(x)}{3h(x)}$, or explain why it doesn't exist.

2. Evaluate:

a) $ \lim_{x \rightarrow 0}\frac{\sin 5x}{x}$
b) $ \lim_{x \rightarrow 0}\frac{1\minus{}\sec^2 2x}{4x^2}$
c) $ \lim_{x \rightarrow 0} \frac{\csc x \minus{} 1}{x \csc x}$

hello, for 2a) we have
$ \lim_{x \to 0}\frac{\sin(5x)}{x}\equal{}\lim_{x \to 0}5\cdot\frac{\sin(5x)}{5x}\equal{}5$ because
$ \lim_{u \to 0}\frac{\sin(u)}{u}\equal{}1$.
Sonnhard.

[hide="2"]
a)
$ \lim_{x\to0}\frac{\sin 5x}{x}=\lim_{x\to 0} 5\cos 5x=\boxed{5}$
(L'Hopital's rule). To do it without L'Hopital's, multiply and divide by 5 and then use the Squeeze Theorem.

b)
\begin{eqnarray*}
\lim_{x\to 0}\frac{1-\sec^22x}{4x^2} &=& \lim_{x\to 0}\frac{-\tan^22x}{4x^2}\\
&=&\lim_{x\to 0}-\left(\frac{\tan 2x}{2x}\right)^2\\
&=&\lim_{x\to 0}-\left(\frac{\sin 2x}{2x}\cdot\sec 2x\right)^2\\
&=&\boxed{-1}
\end{eqnarray*}

c)
\begin{eqnarray*}
\lim_{x\to 0}\frac{\csc x-1}{x\csc x}&=&\lim_{x\to 0}\frac{\frac{1-\sin x}{\sin x}}{\frac{x}{\sin x}}\\
&=&\lim_{x\to 0}\frac{1-\sin x}{x}\\
&=&\lim_{x\to 0}\frac{1}{x}-\frac{\sin x}{x}
\end{eqnarray*}
Doesn't exist.
[/hide]

hello for 2b) we get
$ \frac{1\minus{}\frac{1}{\cos^2(2x)}}{4x^2}\equal{}\minus{}\left(\frac{\sin(2x)}{2x}\right)^2\frac{1}{\cos^2(2x)}$
and from here follows that the searched limit is $ \minus{}1$ as $ x$ tends to zero.
Sonnhard.

hello for 2c) the limit
$ \lim_{x \to 0}\frac{\csc(x)\minus{}1}{x\csc(x)}\equal{}\lim_{x \to 0}\frac{1\minus{}\sin(x)}{x}$ doesn't exist.
Sonnhard.

$ L\equal{}\lim_{x \rightarrow a} \frac {7f(x) \minus{} g(x)}{3h(x)}$

$ L\equal{}\frac{\lim_{x \rightarrow a}7f(x)\minus{}\lim_{x \rightarrow a}g(x)}{\lim_{x \rightarrow a}3h(x)}$

$ L\equal{}\frac{35\plus{}\sqrt{2}}{0}$

Thus,L doesn't exist.


geometry
trapezoid
perpendicular bisector
geometry solved

In acute triangle $ABC$, $C_0$ is feet of perpendicular from $C$ to $AB$, and $C_1$ is midpoint of $AB$. Let $E$ and $F$ be points on $AB$ such that $\measuredangle ACE=\measuredangle FCB$. Feets of perpendiculars from $A$ and $B$ to $CE$ and $CF$ are denoted by $M$ and $N$, by that order. Prove that points $M,N,C_0$ and $C_1$ are cocyclic.

I sent you PM with solution for this problem (mrzi me da ga kucam na engleskom :) )...

pozz

@Dule_00: Does the "easy" mean that you have solved it? Then it should go into Geometry Proposed&Own Problems, not into Unsolved Problems. (Just for your info.)

Anyway, it's a very nice problem. As usual it is easier to prove more:

[color=blue][b]Theorem 1.[/b] We use directed angles modulo 180°. Let ABC be a triangle, let Z be the foot of its altitude from the vertex C, and let C' be the midpoint of its side AB. Let E and F be two points in the plane such that < ACE = < FCB. Let M and M' be the orthogonal projections of the points A and B on the line CE, and let N and N' be the orthogonal projections of the points B and A on the line CF.

[b](a)[/b] The points M, M', N, N' lie on one circle with center C'.
[b](b)[/b] The points M, N, Z, C' lie on one circle.
[b](c)[/b] The points M', N', Z, C' lie on one circle.[/color]

[i]Proof.[/i] First, note that the condition < ACE = < FCB is equivalent to < ACF = < ECB. This is because

  < ACF - < ECB = (< ACE + < ECF) - < ECB = < ACE - (< ECB - < ECF) = < ACE - < FCB.

Thus, the roles of the points E and F are symmetric, i. e. these points can be interchanged, with the points M and N being replaced by N' and M', respectively. As a consequence of this, once Theorem 1 [b](b)[/b] will be proven, Theorem 1 [b](c)[/b] will follow by analogy.

But let's prove [b](a)[/b] first. Since < BM'C = 90° and < BNC = 90°, the points M' and N lie on the circle with diameter BC. Thus, < CM'N = < CBN. On the other hand, $BN\perp CF$ yields < (CF; BN) = 90°. Thus,

  < MM'N = < CM'N = < CBN = < (CB; BN) = < (CB; CF) + < (CF; BN)
              = < BCF + 90°.

Similarly, < NN'M = < ACE + 90°. Thus,

  < MM'N = < BCF + 90° = - < FCB + 90° = - < ACE + 90°
              = - < ACE - 90°         (since 90° = -90° when working with directed angles modulo 180°)
              = - (< ACE + 90°) = - < NN'M = < MN'N.

Thus, the points M, M', N, N' lie on one circle.

Let $M_1$ be the midpoint of the segment MM'. The quadrilateral AMM'B is a trapezoid with its bases AM and BM' both being perpendicular to the line CE. Thus, the line $C^{\prime}M_1$, joining the midpoints C' and $M_1$ of its legs AB and MM', must be parallel to its bases, and thus is also perpendicular to the line CE. In other words, $C^{\prime}M_1\perp MM^{\prime}$. Since $M_1$ is the midpoint of the segment MM', this means that the point C' lies on the perpendicular bisector of the segment MM'. Similarly, the point C' lies on the perpendicular bisector of the segment NN'. On the other hand, the center of the circle through the points M, M', N, N' must lie on the perpendicular bisectors of the segments MM' and NN', too (because the center of a circle lies on the perpendicular bisector of any chord). Hence, both the point C' and the center of the circle through the points M, M', N, N' are common points of the perpendicular bisectors of the segments MM' and NN'. But since these two perpendicular bisectors have only one common point (in fact, if they would coincide, then the lines CE and CF would be parallel, thus also coincide, but then it would follow that M = N' and N = M' and Theorem 1 [b](a)[/b] would be already proved), this shows that C' is the center of the circle through the points M, M', N, N', and thus Theorem 1 [b](a)[/b] is proven.

Now, since the point C' is the center of the circle through the points M, M', N, N', the central angle theorem yields < C'MN = 90° -. < NM'M = 90° - < NM'C. Since < BZC = 90°, < BM'C = 90° and < BNC = 90°, the points Z, M' and N lie on the circle with diameter BC, so that < NM'C = < NZC. Thus, < C'MN = 90° - < NZC = < BZC - < NZC = < C'ZN. Thus, the points M, N, Z, C' lie on one circle, and Theorem 1 [b](b)[/b] is proven. As we already said, Theorem 1 [b](c)[/b] follows by analogy. This completes the proof of Theorem 1.

  Darij


[quote="b2"]Let $k$ is an integer not less than $2$
Find the best constant $M_{k}$ so that the following inequality hold for all $x_{1},x_{2},...,x_{n}>0$  ($\forall n\ge k$)
$M_{k}(\sum_{1\le i_{1}< i_{2}< ...<i_{k}\le n}x_{i_{1}}x_{i_{2}}...x_{i_{n}})<(\sum_{1}^{n}x_{i})^{k}$[/quote]

By Maclaurin's inequality, $(\frac{\sum x_{i}}{n})^{k}\geq \frac{\sum_{1\le i_{1}< i_{2}< ...<i_{k}\le n}x_{i_{1}}x_{i_{2}}...x_{i_{k}}}{ \binom{n}{k}}$. see http://www.mathlinks.ro/Forum/viewtopic.php?t=73460 .


[quote="Farenhajt"]Numbers $a^{2},b^{2},c^{2}$ are pairwise distinct and they form an arithmetic progression. Prove that the numbers $\frac{1}{b+c},\frac{1}{c+a},\frac{1}{a+b}$ also form an arithmetic progression.[/quote]

[hide]From the given,

$c^{2}-b^{2}=b^{2}-a^{2}$
$(c+b)(c-b)=(b+a)(b-a)$

Consider the sequence, $\frac{1}{b+c},\frac{1}{c+a},\frac{1}{a+b}$

The difference between the first two terms is:

$\frac{1}{c+a}-\frac{1}{b+c}$
$=\frac{b-a}{(b+c)(c+a)}$
$=\frac{(c+b)(c-b)}{(b+a)(b+c)(c+a)}$
$=\frac{c-b}{(a+b)(a+c)}$

Since the difference between the next two terms is also

$\frac{1}{a+b}-\frac{1}{c+a}=\frac{c-b}{(a+b)(a+c)}$,

this sequence is an arithmetic sequence.[/hide]

I guess it's harder to observe it than to prove it.
[hide]$\frac{1}{b+c}, \frac{1}{c+a}, \frac{1}{a+b}$ form an arithmethic progression iff \[\begin{aligned}\frac{1}{b+c}+\frac{1}{a+b}= \frac{2}{c+a}\ & \Longleftrightarrow\ (c+a)(a+b)+(b+c)(c+a) = 2(b+c)(a+b) \\ & \Longleftrightarrow\ a^{2}+c^{2}= 2b^{2}.\end{aligned}\] This is true iff $a^{2}, b^{2}, c^{2}$ form an arithmetic progression. [/hide]
So it's even an equivalence!  :P


linear algebra
matrix
linear algebra theorems

Given $ m$, how can one decompose a Hermitian $ n\times n$ matrix $ A$ into the product $ B B^\top$, where $ B$ is an $ n\times m$ matrix?

Do you mean $ BB^{T}$ or do you mean $ BB^{*},$ where $ B^{*}=\overline{B}^{T}\ ?$

If it's the latter, then such a decomposition is only possible for $ A$ positive semidefinite or positive definite. And the rank of $ B$ can't be smaller than the rank of $ A.$

Yes, I meant Hermitian positive-definite, and $ BB^{*}$.


analytic geometry
graphing lines
slope
calculus
geometry
LaTeX

A workman is trying to move an inflexible ladder around a corner at which two corridors meet at a right angle. If the widths of the two corridors are $a$ and $b$, find the maximum length of the ladder that can be successfully moved from one corridor to the other. (Assume the workman won't try anything tricky, like slanting the ladder.)  :P

This is a nice problem...aoother way to say it is...:

Suppose we have a line, L, that passes through the point (a,b). Provided the slope, m, of the line is $0\in(-\infty,0)$, find the minimum length of the segment formed by the intersections of L with the positive x and y axes.

I think we might bust out the calculus eventhough you probably can just complete the square or something...

[hide="Calculus"]
Assuming I didn't make a mistake..

Used co-ordinate geometry similar to above.
$d = a\sqrt{1+(\frac{b}{a})^\frac{2}{3}}+b\sqrt{1+(\frac{a}{b})^\frac{2}{3}}$
[/hide]

This problem was posted 3 years ago:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=12563[/url].
In that thread, I gave a solution using Holder's inequality.  (Back then, the forum didn't have LaTeX, and used some BBcodes that no longer work.)


Ok, since everybody's started seriously working for the IPhO, here's something from me :)

Two small equally charged balls, each of mass $m$, are hung from the same point by strings of length $\ell$. The distance between the balls is $x \ll \ell$. The charges of the balls change at a rate $\text dq/\text dt = \dot q$, so the speed of the balls changes as $v = a/\sqrt{x}$, where $a = \text{const.}$ Find $\dot q$.

well from what i understand from the problem is initially the two balls were $x$ distance apart and had a speed of $\frac{a}{\sqrt x}$  then due to repulsion each started moving away in circles of radius $l$

let at any point the angle made with the vertical by the strings is $\theta$ if we assume that at any time within our considerations seperation between balls $x(t)$ is negligibly small as compared to the $l$ then we can safely assume $\cos\theta = 1$ then the horizontal component of velocity is equal to
$|v_{h}|=\frac{a}{\sqrt x}$
let my origin(this allows me to comnsider only the case of  a single ball  due to symmetry) be m.p between the two point charges so since $\theta$ is negligibly small then we can assume  that to be $\frac{x}{2}$ from each charge
so their horizontal component of velocity $|v_{h}|$ $=$ $|\frac{dx_{or}}{dt}|$ $=$ $|\frac{dx}{2dt}|$
so
$|a_{h}|$ $=$ $\frac{a^{2}}{x^{2}}$

now $\sin\theta$ is negligibly small so the vertical component of velocity is and also the vertical component of accelaration
so we have
$T\cos\theta$ $=$ $mg$  where $T$ is the tension in the string
this gives $T$ $=$ $mg\sec\theta$
now let's consider horizontal component of accelaration
$\frac{q^{2}}{4\pi\epsilon_{0}x^{2}}-T\sin\theta$ $=$ $ma_{h}$ $=$ $m\frac{a^{2}}{x^{2}}$
this gives
$\frac{q^{2}}{4\pi\epsilon_{0}x^{2}}-mg\tan\theta$ $=$  $m\frac{a^{2}}{x^{2}}$
multiplying throught by $x^{2}$ gives

$\frac{q^{2}}{4\pi\epsilon_{0}}-mg\frac{x^{3}}{2l}$ $=$  $ma^{2}$
diffrentiating both sides with repect to time
$\frac{2qq^{.}}{4\pi\epsilon_{0}}-\frac{3amgx^{\frac{3}{2}}}{l}$ $=$  $0$

so

 $q^{.}$ $=$ $\frac{3amgx^{\frac{3}{2}}4\pi\epsilon_{0}}{2ql}$

where $x$ is the instantaneous distance and $q$ the instantaneous charge
but from the above eqyation we get
$q$ $=$ $\sqrt{4\pi\epsilon_{0}(ma^{2}+mg\frac{x^{3}}{2l})}$
putting this in the above equation we get
$q^{.}$ $=$ $\frac{3amgx^{\frac{3}{2}}4\pi\epsilon_{0}}{2\sqrt{4\pi\epsilon_{0}(ma^{2}+mg\frac{x^{3}}{2l})}l}$

Um, sorry, I forgot to specify that the balls are moving towards each other, but slowing down. This doesn't affect your solution much.

On the other hand, the solution itself is incorrect :( $\dot q$ should be constant...

well yes my solution atleast the answer doesn't change because of ur change ion question...but just now i proved that the rate can't be constant provided this is what the question is.....i think there's nothing wrong with the proof :)

I know, I got the same solution as you. Unfortunately I don't have the official solution, only the final answer, and since it's constant I'd like to see your proof :) (It is quite probable that there is a mistake in the problem itself.)

this problem is very common among indian authors.

and every 'good' physics book of problems carries the answer which is a constant as you think it to be as the official solution.
owing to non-acquiantance with latex i wont write their exact result here.

but their approach is plain rape of the problem.

their policy: the moment you see an equation with symbols start differentiating with respect to god knows what.

mr. djole the pathetic solutions of these books are:

suppose there isn't any leakage of charge. so at equilibrium you can easily derive the relation of charge(q) with the distance between them (x)

now differentiate with respect to time both sides.
and some manipulation will give you the result which i hope is the official answer.

but they dont realise that once you start moving in an accelerated manner you cant justuse your equations of equilibrium.its plain indifference towards newton.

Yes, you are right, I did get the specified solution.

Thanks for the help! I think you have a point -- at least, the problem could've been formulated better...

yes this is just another example of the so called 'good' problems ..with absolutely bull... solutions

I think that the problem is in the problem itself :lol:


linear algebra
matrix
algebra
polynomial
linear algebra unsolved

If $ A \in M_{n}(\mathbb{F})$ , and $ A$ satisfy: $ A' \equal{} A$  (where $ A'$ is the transposed matrix of $ A$). we say: $ A$ is a symmetric matrix :D

Ok, If $ A \equal{}$ $ \begin{bmatrix}a_{11} & ... & a_{1n} \\
... & ... & ... \\
a_{n1} & ... & a_{nn}\end{bmatrix}.$

we denote the principal minor of $ A$ of order $ r$ by $ det \begin{bmatrix}a_{i_{1}i_{1}} & ... & a_{i_{1}i_{r}} \\
... & ... & ... \\
a_{i_{r}i_{1}} & ... & a_{i_{r}i_{r}}\end{bmatrix}.$

where $ 1 \leq i_{1} < ... < i_{r} \leq n$

and my question is: If $ A$ is a symmetric matrix or a Hermite Matrix , then $ A$ has a nonzero principal minor which order equal $ rank(A)$.  :roll:

another question is: the Geometric meaning of Symmetric matrix is？ :huh:  :D

Geometric meaning- the best I can give you is the theorem that symmetric matrices are orthogonally diagonalizable over $ \mathbb{R}$, which means that we can find an orthogonal basis in which the action of $ A$ is entirely a combination of stretches along the axes. For the Hermitian version, it's a complex orthogonal basis (a unitary matrix), and real eigenvalues.

Your other question is tricky. I'll get an easy case over with now: it's true over $ \mathbb{R}$, by the diagonalizability. The geometric and algebraic multiplicities of the zero eigenvalue are the same, so the coefficient of $ x^{n\minus{}r}$ in the characteristic polynomial is nonzero. That coefficient is a linear combination of principal minors of order $ r$, and hence at least one of those minors is nonzero.


[i][b]find all functions that satisfy: x et y de R:
f(xy+f(x)) = xf(y)+f(x)

good luck[/b][/i]

First suppose that $ f(a) = f(b) = c$
Then $ x = a, y = b$ gives $ f(ab+c) = ac+c$ and $ x = b, y = a$ gives $ f(ab+c) = bc+c$. Then $ ac+c = bc+c$ and $ a = b$ or $ c = 0$.

Suppose that $ c = 0$ and $ a \neq b$. WLOG $ a \neq 0$. Then $ x = a, y = 1$ gives $ af(1) = 0$, so $ f(1) = 0$. Plug $ y = 1$ in to get $ f(x+f(x)) = f(x)$. Suppose $ f(x) \neq 0$ for some $ x$, then by the argument above $ x+f(x) = x$, so $ f(x) = 0$, contradiction. So $ f(x) = 0$ in this case.

Now suppose the opposite, that $ f$ is injective. Then $ x = 0$ gives $ f(f(0)) = f(0)$ so $ f(0) = 0$. $ y = 0$ then gives $ f(f(x)) = xf(0)+f(x) = f(x)$, and thus $ f(x) = x$.

So just two answers: $ f(x) = x$ and $ f(x) = 0$.

My solution:

Let $ f(0)=c$. Putting $ x=0$ gives $ f(c)=c$ and $ y=0$ gives $ f(f(x))=cx+f(x)$. Hence $ f(f(c))=c^{2}+f(c)\Rightarrow c=c^{2}+c\Rightarrow c=0$.

Thus $ f(f(x))=f(x)$. Now let $ f(a)=f(b)$ where $ a,b$ both aren't equal to $ 0$ at the same time. Then we have $ f(ab+f(a))=f(ab+f(b))\Rightarrow af(b)+f(a)=bf(a)+f(b)\Rightarrow f(a)=0$ or $ a=b$. If the latter holds then $ f$ is injective and from $ f(f(x))=f(x)$ we conclude that $ f(x)=x$. So assume that $ f(a)=f(b)=0$.

Putting $ y=1$ and $ x=a$ into our original equation we get $ f(a)=af(1)=0$. If $ a= 0$ we get by the same reasoning $ b=0$, which contradicts our assumption. So $ a,b\neq 0$. Thus we have $ f(1)=0$. So substituting $ y=1$ in our equation we have $ f(x+f(x))=f(x)$. Let $ x\mapsto f(x)$. Then this becomes $ f(f(x)+f(f(x)))=f(f(x))$ which in turn, using $ f(f(x))=f(x)$, is equivalent to $ f(x+f(x))=x$ thus $ f(x)=x$.

So the solutions are $ f(x)=0$ and $ f(x)=x$.

How did you conclude $ f(x) = 0$ so quickly? I thought all you got from the statement before is that $ f$ is injective on $ \mathbb{R}-\{0\}$.

Sorry for that. But I think it is correct now. Actually I'm not very good at functional equations, so it's not impossible that I make mistakes. :blush:

After posting my solution(correct version), I noticed that it's somewhat equivalent to MellowMelon's one. :oops:

My solution doesn't use injectivity explicitly.
 $ y \equal{} 0\Rightarrow f(f(x)) \equal{} xf(0)\plus{}f(x)\ (*)$. $ y \equal{} f(x)\Rightarrow f((x\plus{}1)f(x)) \equal{} x^{2}f(0)\plus{}(x\plus{}1)f(x)\ (**)$.

From (*) we have $ f(xf(y)\plus{}f(x)) \equal{} (xy\plus{}f(x))f(0)\plus{}xf(y)\plus{}f(x)$. Set y = x to get $ f((x\plus{}1)f(x)) \equal{} [x^{2}\plus{}f(x)]f(0)\plus{}(x\plus{}1)f(x) \equal{} (**)$, so either $ f(x) \equal{} 0$ or $ f(0) \equal{} 0$. If the latter is true then $ f(f(x))\equal{}f(x)$ and set $ x \equal{} f(x), y \equal{} x\Rightarrow f((x\plus{}1)f(x)) \equal{} [f(x)]^{2}\plus{}f(x) \equal{} (x\plus{}1)f(x)$ thus $ f(x) \equal{} x$.

Thus $ f(x) \equal{} 0,\ f(x) \equal{} x$ are the only solutions.


group theory
superior algebra
superior algebra solved

Hi.

Write down all cyclic groups that have exactly two generators.

A cyclic group is a group that can be generated by a single element..

A cyclic group of finit order $n$ has exactly $\phi (n)$ generators.
Now use the multiplicativity of $\phi$ and that $\phi (p^k)=p^{k-1}(p-1)$ for all primes $p$ and natural $k$ to get that all these groups are that with orders $3$, $4$ and $6$.

Now consider the infinit case, in which the group is isomorphic to $\mathbb{Z}$ and so has exactly two generators.

oops..I misunderstood the problem  :blush:  :blush:


number theory
relatively prime
greatest common divisor

If $ gcd(a,b) \equal{} 1$, prove that
$ (a) gcd(a\minus{}b,a\plus{}b)\leq 2$
$ (b) gcd(a\minus{}b,a\plus{}b,ab) \equal{}1$
$ (c)gcd(a^{2}\minus{}ab\plus{}b^{2}, a\plus{}b)\leq 3$

[hide="a"]Suppose there existed an odd prime that divided both $ a\plus{}b$ and $ a\minus{}b$. Then $ p\mid 2b$, so $ p\mid b$. But since $ p\mid a\plus{}b$, we also have $ p\mid a$, so $ a$ and $ b$ are not relatively prime.[/hide]

[hide="b"]The greatest common divisor of $ a\plus{}b$ and $ a\minus{}b$ is $ 2$ only if $ a$ and $ b$ have the same parity. Since they are relatively prime, both must be odd. But then $ ab$ is odd, and thus relatively prime to $ 2$.[/hide]

Blah, think that's all I have time for. This should probably be moved.


Is there a set of primes, such as $\mathbb{P}$ like there is of $\mathbb{N}, \text { natural numbers?}$

Also, in ACoPS, it is mentioned that the notation $: =$ is used to define something. So would it be acceptable to say, for instance, "$p$ must be in $\mathbb{P} : = \{ 2, 3, 5, 7, 11, \ldots \}$, the set of primes," on a proof-based test, i.e. USAMTS? :)

What are you asking in your first problem? I don't get the grammar  :blush:

He's asking if there $\exists$ a set such that $x\in\mathbb{P}$; basically if a set known as the Prime Numbers, denoted by $\mathbb{P}$ exists.  I don't think there is one that exists, however I believe you are allowed to say for example:
Let $\mathbb{P}=\{x:$ only $1,x|x\}$.

Masoud Zargar

Is there a set of primes? Yes. Are you asking if there is a symbol, like $\mathbf{R}$ for the reals, that is commonly used for it? Then I don't know, but you might as well just say "is prime."


Aratatzi ca daka a + b + c = 3 atunci este adevarata inegalitatea:

            (3-2a)(3-2b)(3-2c) <= a^2 * b ^ 2 * c ^ 2

RESPECT :|

notam $2x=b+c-a$, atunci $\sum x=3/2$, si $a=y+z$ etc...

inegalitatea devine $(\prod (x+y))^{2}\geq 8xyz$
omogenizam, amplificand in stanga cu $27/8$ si in dreapta cu $(\sum x)^{3}$, desfacem parantezele si obtinem inegalitatea echivalenta :
voi folosi notatia :
 $(u,v,w)=\sum_{sym} x^{u}y^{v}z^{w}$
$LHS=27(4,2,0)+27(3,3,0)+135(4,1,1)+54(3,2,1)+45(2,2,2)$
$RHS=32(4,1,1)+192(3,2,1)+64(2,2,2)$,
$LHS\geq RHS$ echivalent cu :
$27(4,2,0)+27(3,3,0)+103(4,1,1)\geq 138(3,2,1)+19(2,2,2)$. (*)
deoarece orice triplet din stanga majoreaza orice triplet din dreapta, si $27+27+103=138+19=157$, rezulta ca (*) este adevarata conform inegalitatii muirhead.

E deja postata la Resources la Romania Junior Balkan TST 2005..cu toate problemele

Nu vazusem...sorry :|...oricum iti dai seama ca era cunoscuta ca a fost data la OBJM 2005 deci recent :|...dar voiam sa subliniez ca doar "animalul" de stoican a facut-o :)

http://www.mathlinks.ro/Forum/viewtopic.php?p=457344#p457344
Sunt niste solutii misto aici

bravo lui :)
mie nu mi`a luat foarte mult sa gasesc solutia de mai sus, doar desfacutu parantezelor a durat asa vreo 20 de minute

In general daca x+y+z=s,
$s>0 \Rightarrow  (s-2x)(s-2y)(s-2z) \leq x^2y^2z^2 \cdot \frac{27}{s^3}$

a+b+c = 2p ...si se inmulteste cu 3/2 si in final se ajunge la faza cu sin A + sin B + sin C :)

Sau mai simplu,zic eu,se observa ca prin notatziile $a=\frac{3x}s,b=\frac{3y}s,c=\frac{3z}s$ se obtine problema din primul post,a+b+c=3 sa se arate ca $(3-2a)(3-2b)(3-2c)\leq a^2b^2c^2$ Deci faza cu x+y+z=s e mai mult vrajeala..cheia era sa se observe echivalentza dintre problema"generala" si aia cu a+b+c=3


limit
function
algebra
polynomial
complex analysis
superior algebra
superior algebra unsolved

I am familar with a Complex Analysis short proof that $ \mathbb{C}$ is algebraically closed. I was curious is there a [b]short[/b] algebraic proof of this?

If anybody is interested here is the one Im talking about.

[b]Proof:[/b] Say $ f(z)$ is non-constant and doesnt have a zero. Then $ \lim_{|z|\to \infty}\frac{1}{f(z)}=0$ so, $ |f(z)|\leq K$. And $ \frac{1}{f(z)}$ is an entire function because $ f(z)\not = 0$. So by Louiville's theorem $ f(z)=C$ a contradiction.

of course there's no 'pure algebraic' proof because the claim depends on the analytic structure of $ \mathbb{C}$ or rather $ \mathbb{R}$. look here:

http://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra#Algebraic_proofs

the mentioned proof also shows, that $ K[i]$ is algebraically closed for every real closed field $ K$, i.e. an ordered field whose polynomials satisfy the intermediate value theorem.


[quote="worthawholebean"]If $S=\sum_{k=1}^{\infty}k^{-2}$, find, in terms of S, the value of $1+\sum_{k=1}^{\infty}(2k)^{-2}$.[/quote]

[hide]Then
\[1+\sum_{k=1}^{\infty}(2k)^{-2}= 1+2^{-2}\cdot S = \boxed{1+\frac14 S}. \][/hide]

When I rewrote it in sigma notation it definitely got easier.

Also, I believe that the first one is equal to $\frac{\pi^{2}}{6}$ (famous Euler's BASEL problem), so it's possible to get an exact value:
If we have the second equal to $1+\frac{S}{4}$, the the value of the second is $1+\frac{\pi^{2}}{24}$


Egor has two parents, four grandparents, and so on.
Write an explicit formula and a recursive formula for the number of ancestors Egor has if we go back n generations.

This is very obviously a textbook problem.

very easy...

Let $ a_n$ be the number of ancestors $ n$ generations ago.  Then we have $ a_1 \equal{} 2$ (parents) and $ a_n \equal{} 2a_{n \minus{} 1}$; hence $ a_n \equal{} 2^n$.  If you want the number of ancestors going back $ n$ or less generations then we just have a partial sum of a geometric series.

Formulas for finding how many ancestors of Egor existed in the $ n^{th}$ generation:
[list]Explicit: $ a_{n} \equal{} 2^{n}$.
Recursive: $ a_{1} \equal{} 2$; $ a_{n} \equal{} 2 a_{n \minus{} 1}$.[/list]Formula for geometric series that goes back $ n$ generations (what [b]facis[/b] was suggesting): $ A_{n} \equal{} \sum_{i \equal{} 1}^{n} 2^{i}$.


there's this problem, find the square root of 9095625. i read all the directions but i still dont get it. can anyone give me a clear explanation to this problem

Try to prime factorize 9095625. If any exponents are even then that is a square.

how do u factorize a number that big? is it just plain luck or is there a technique to do it and if there is plz tell me

Well if you look at the unit's digit, you note that it's a 5. Therefore, try dividing it by $ 25$.

Primes such as 2 or 5 should be easy to factor out. 3 is quick to factor out as well. Make the number as small as possible by factoring out these primes. when the primes get larger than, say, 17, the going gets tough.

[hide="Discussion"] If you're familiar with the powers of $ 5$, you'll realize that because the number ends in $ 5^4 \equal{} 625$, we have

$ 9095625 \equal{} 9095000 \plus{} 625 \equal{} 5000 \cdot 1819 \plus{} 625 \equal{} 625 \cdot 8 \cdot 1819 \plus{} 625 \equal{} 625(8 \cdot 1819 \plus{} 1)$.

This is a good start.  $ 8 \cdot 1819 \equal{} 14552$, so we now know that

$ 9095625 \equal{} 5^4 \cdot 14553$.

Because $ 1 \plus{} 4 \plus{} 5 \plus{} 5 \plus{} 3 \equal{} 18$, we know that $ 9 | 14553$.  In fact,

$ 14553 \equal{} 13500 \plus{} 900 \plus{} 153 \equal{} 9 \left( 1500 \plus{} 100 \plus{} 17 \right) \equal{} 9 \cdot 1617$.

Because $ 1 \plus{} 6 \plus{} 1 \plus{} 7 \equal{} 15$, we know that $ 3 | 1617$.  In fact,

$ 1617 \equal{} 1500 \plus{} 117 \equal{} 3(500 \plus{} 39) \equal{} 3 \cdot 539$.  We have now determined that

$ 9095625 \equal{} 3^3 \cdot 5^4 \cdot 539$.

Now, $ 3, 5$ don't divide $ 539$, but $ 539 \equal{} 490 \plus{} 49 \equal{} 49 \cdot 11$, so our complete prime factorization is

$ 9095625 \equal{} \boxed{3^3 \cdot 5^4 \cdot 7^2 \cdot 11}$.

Can you simplify the square root from here?  :) [/hide]

Yes of course that is just $ 525\sqrt{33}$

t0rajir0u brings up a good point. Whenever you are stuck with prime factorization, separate the number by products of powers of ten. This can often guide you and help you remember some things that you couldn't (for instance, if you couldn't remember the divisibility rule for 11, you might be able to bring it to help you if you begin to separate the numbers).


A,B,C are three arbitrary points on a plain.Draw three lines $ A_1B_1$,$ A_2B_2$,$ A_3B_3$ pasing point P where two circles
($ O_1$);($ O_2$) intersect.Then draw $ \triangle C_1AB \sim \triangle CA_1B_1$;$ \triangle C_2AB\sim \triangle CA_2B_2$;
$ \triangle C_3AB\sim \triangle CA_3B_3$. Prove that $ C_1$,$ C_2$,$ C_3$ are Collinear

Sorry,I can't understand the statement of your problem.Can you re-post it more clearly? :?:


rotation algebra polynomial geometry circumcircle symmetry

given any regular polygon with n sides inscribed in the unit circle; show that:
$v_{1}+v_{2}+...+v_{n}=(0,0)$
where the $v_{i}$ are the vertices of the polygon.

[hide="hint"]consider $P(x)=x^{n}-1$[/hide]

This is equivalent to the $n$th roots of unity, since any polygon with no vertex at $(1,0)$ can be rotated to be so.
Consider the polynomial $P(x): x^{n}-1=0$
We know $1$ is a root, so factoring it out, we have for any root $a\neq 1$, we have:
$a^{n-1}+a^{n-2}+...+1=0$, and we are done.

[quote="Altheman"]the center of mass has to be the center of the circumcircle by symmetry...[/quote]

whats center of mass?

I dont understand davidyko's solution.
for example if you have a square, and you put one of its centers on the 1,0 then the other two will be on 0,1 0,-1 -1,0 not the 4th roots of unity.

this seems like a nice problem.

Er... $1, i,-1,-i$ are, in fact, the 4th roots of unity.

[quote="math227"][quote="Altheman"]the center of mass has to be the center of the circumcircle by symmetry...[/quote]

whats center of mass?[/quote]

The centroid of a collection of points $(x_{1}, y_{1}), ... (x_{n}, y_{n})$ is given by

$\left( \frac{x_{1}+...+x_{n}}{n}, \frac{y_{1}+...+y_{n}}{n}\right)$

A regular polygon inscribed in the unit circle has rotational symmetry through its centroid, which must therefore coincide with the center of the circle.  The center of the unit circle is the origin, hence the centroid of a regular polygon inscribed in the unit circle must also be the origin.


geometry
incenter
vector
cyclic quadrilateral
geometry unsolved

In triangle $ABC$ ($AB>AC$) let $S$ be incenter and $K,M$ are respecively midpoints of $AB$ and $AC$. Point $P$ is intersection point of lines $KM$ and $CS$, and point $Q$ is chosen such that $PQ \perp KM$ and $MQ ||BS$. Prove that $SQ \perp AC$.

P.S. I'm particularly interested in solution using vectors

[color=red][Moderator edit: See also http://www.mathlinks.ro/Forum/viewtopic.php?t=86460 .][/color]

[b] Lemma :[/b] Let $\triangle ABC$ . $(I)$ incribes $\triangle ABC$ and tangents $BC, CA , AB$ at $X, Y , Z$ . $M , N$ respectively are the midpoints of $AB , AC$ . Then , $MN , BI , XY$ are concurrent at $K$ and $\angle AKB = 90$ .
  [u][i]Proof :[/i] [/u]
 Let $BI$ intersect $XY$ at $K$ .
 We have : $(YX,YZ) = (IB,IZ) = (IY, IZ)$ . So $ZIKY$ is a cyclic quadrilateral .
 This infers $AZIK$ be also a cyclic quadrilateral . So $\angle AKB = 90$
 Then , $KM$ ia a median of the right-angle $AKB$ . This means $KM=MA=MB$
 So $\angle MKB= \angle MBK = \angle IBC$ . This infers $KM$ is parallel with $BC$ .
 Thus $K , M, N$ are collinear .

   Come back with this problem :
 Let $(S)$ tangent $BC, CA , AB$ at $X, Y , Z$ .
 By lemma : $X , P , Z$ are collinear and $YSPZ$ is a cyclic quadrilateral .
 So $\angle PYS = \angle SZP = \frac{\angle B}{2}$
 Let $YS$ intersect $MQ$ at $Q'$ .
 Because of $MYQ'P$ is a cyclic quadrilateral $(\angle PYS = \angle PMQ=\frac{\angle B}{2})$ , we gets $\angle MPQ' = 90$ . So $Q \equiv Q'$
  This means $SQ \perp AC$

We will prove the following equivalent statement.

[i]Let K and M be the midpoints of sides AB and AC of $\triangle ABC$ respectively. Denote the incenter of $\triangle ABC$ by I. Let $CI\cap KM=P$. Suppose the incircle is tangnet to AC at E. EI intersects the perpendicular of P to KM at Q. Then MQ//BI.[/i]

Throughout the proof, I will assume $a\ge c$, for the sake of clearness and preventing to be tedious.
Obviously P,Q,M,E are concyclic. Note that MQ//BI iff $\angle BIQ=\angle IQM=\frac{A-C}{2}$, hence the problem is reduced to showing $\angle EPM=\frac{A-C}{2}$. Now we look into $\triangle EPM$. First of all, $\angle EMP=C, MP=MC=\frac{c}{2}, EM=(s-a)-\frac{c}{2}=\frac{a-c}{2}$. In $\triangle ABC$, we construct a point X on side BC such that AB=BX. It can be easily seen that $\triangle CAX\sim\triangle MPE$, hence $\angle EPM=\angle XAC=\frac{A-C}{2}$.


Mot so ban du ky thi quoc gia nam nay:
[b]lqdschool[/b](Khanh Hoa),[b]namanhams[/b](Ha Noi ),[b]nttu,minhkhoa[/b](TP.Ho Chi Minh ),[b]nguyenquockhanh[/b](Nam Dinh ),[b]hungkhtn[/b](dai hoc khoa hoc tu nhien HaNoi),[b]K09[/b](HaiDuong)

I don't participate in it. But a question: Is it your first VMO ?  :)

[quote="orl"] But a question: Is it your first VMO ?  :)[/quote]
Yes it is the first time I take the VMO

more: [b]luuduchoan[/b]: VINH-team
[b]nttrung[/b]: NAMDINH-team

oh, sap thi rui-------->chuc moi nguoi thi tot,okie?
day la co cau 1 so doi tuyen:
[b]DHKHTN[/b]:15 thi sinh-7 thi sinh lop 11
[b]DHSP[/b]:10 thi sinh-6 thi sinh lop 11
[b]LEHONGPHONG-ND[/b]:8 thisinh-2 thi sinh lop 11
[b]LAMSON-THANHHOA[/b]:10 thi sinh-7 thi sinh lop 11
@pi[b]G[/b]fly[b]:are you NGUYENKHACMINH[/b]

Ở diễn đàn này thì phải Viết tiếng Việt chứ các bạn! ? nghĩa của nó là vậy mà.

[quote="benny"]pigfly khong phai thay Minh dau
Mac du danh gia khong cao nhung truong hn-ams cung co 9 nguoi trong do 3 nguoi la lop 11[/quote]

vây bạn là ai? và [b]pigfly[/b] là ai
cám ơn những thông tin chính xác.okie?

sao lai noi the
dan VIET post tren nay nhieu lam ma.okie?

M?i ngư?i thi thế nào ?
Tình hình năm nay ngày thứ hai dễ hơn ngày thứ nhất  :lol:  ?iểm vòng 2 không biết bao nhiêu ?

Các em trư?ng Ams thi thế nào, tốt chứ?? ;). Có bạn làm được 5 bài cơ đấy!! :) . Năm b?n anh thi không thằng nào làm đủ được 4 bài cả  :D

:?  :?  :?  :?  :?  co le de thi chi kho o cho ve duoc hinh bai 3 thoi
con lai la de.oh,. chan qua, minh khong ve duoc hinh

you sloved problem very well didn't you?[b]nguyenquockhanh,pigfly,lqdschool,namanhams,nttu,minhkhoa,hungkhtn [/b] and[b] K09[/b]?  :)

Tuyển Ams làm tốt lắm anh ạ ,có mỗi 3 ngư?i làm 5 bài ,còn lại thì ....làm 4 bài rưỡi  :D

thằng bạn em bảo nó làm hết cả 6 bài cơ đấy, mà hình như là đúng hết :D

Khoa cung vao day roi ah
To namanhams : anhnham rui, lop 11 khong phai ai cunglam tot thedau, hinh nhu ngoai Trung Anh ra
to Hoang Viet :ban anh truong nao vay
Moi nguoi post cai de len ho cai

Chào m?i ngư?i. Lâu rồi không tham gia diễn đàn, thấy m?i ngư?i đang bàn tán v? thi QG nên tui cũng cho vài thông tin mặc dù chưa chình thưc: ?ội tuyển Toán HP năm nay thi tất cả đ?u làm tốt, ít nhất là 5 bài(mặc dù tui cũng chưa kiểm chứng lại). Nói chung năm nay các đội đ?u làm tốt cả, bởi vì đ? thi vừa sức, lại ngắn g?n chứ không dài dòng như mấy năm trước.

benny:
thằng bạn tớ mới chỉ h?c lớp 11 thôi

Khi nào mới có kết quả thi toàn quốc thế các bạn . Có một người em dự thi mà đến bấy giờ chưa có kết quả cụ thế nữa. Nghe nó nói làm được 5.5 bài .

[quote="hardsoul"]Sorry,I don't know Vietnamese Language but does anyone know the cut-off's  for the medals.[/quote]

From non-official source: Gold: 38.5/40
                                     Silver: 31.00/40
                                     Bronze: 27.5/40
                                     Honorable: 24.5/40

[quote="K09"]Kiểu này thì ối giải nhất các bác nhỉ. Đội hd tụi mình ,đứa nào cũng 5 bài ,một vài 5,5 song chán quá luôn ,bọn tui ẩu lắm.Này bác nào ở KHTN>ĐHSP hay mấy nơi nào làm hết nhỉ.[/quote]

Giải nhất môn Toán bao giờ cũng rất khó. Thường là các bạn phải làm 5.5 -> 6 bài mới được.

5 den 6 bai nhung ma quan trong phai lam duoc het moi mong co giai nhat chu.

[quote="Namdung"][quote="nguyenquockhanh"]giai nhat: Pham kim hung, Nguyen Duc Tam: 39 diem, Tran Trong Dan, Do Quoc Khanh[/quote]

Do Quoc Khanh ( Truong Le Quy Don - Da Nang) duoc 39.5 diem[/quote]

 Vay la ban Do Quoc Khanh( Le Quy Don - Da Nang) xep hang nhat luon ah ? Tuyet voi ! Chuc mung !

 Con ban Nguyen Quoc Khanh ( Nam Dinh ) thi sao ? Truong PTNK TP HCM co duoc nhieu giai khong thay Nam Dung ?

 Co trang Web nao co the xem toan bo danh sach hoc sinh doat giai Toan quoc gia nam nay khong ?

[quote="dbkhoa"]Hi, em moi tham gia vao dien dan. Em dang hoc lop 10, nhung em dang o My nen khong thi duoc VMO, em vua moi duoc qualified USAMO nhung ma co le khong duoc thi vi khong phai la resident cua US :(.
Moi nguoi co ve it post bai len day nhi?  :D[/quote]

yea!! dbkhoa o My~ ma` tai sao khong phai? la` resident cua My~?? but anyways !! u were qualified as a USAMO. im proud of our vietnamese people. !! and im expected to learn more from ya!! by english and vnese.


if you have XP go to start, accessories, system tools, sytem information
then, go to components, display, and it'll be at the top

Yes. What's this game you're buying? Most games today should support a wide variety of graphics cards; however, whether the game can run [b]smoothly[/b] on your computer is another matter.

well, if you would like to upgrade ur graphics card, i sell them for very very cheap..


number theory
Diophantine equation
combinatorics proposed
combinatorics

let $ p,k$ be 2 positive integers with non-zero $ p$.
We have $ k\geq \kappa _1\geq \kappa _2\geq \ldots \geq \kappa _p\geq 0$
So the set formed by $ \sum _{i\equal{}1}^p \left(
\begin{array}{c}
 \kappa _i\minus{}i\plus{}p \\
 \kappa _i\minus{}1
\end{array}
\right)$ for all $ \kappa$ above have all numbers from $ 0$ to $ k\minus{}1$

I don't understand, what is there to prove here?

There were a little typo there. not $ k \minus{} 1$ but $ \frac {(k \plus{} p) \left( \begin{array}{c} k \plus{} p \minus{} 1 \\
k \minus{} 1 \end{array} \right)}{k}$

In binary, the numbers are:
$ \left( \begin{array}{ccc} 0 & 0 & 0 \\
0 & 0 & 1 \\
0 & 1 & 0 \\
0 & 1 & 1 \\
1 & 0 & 0 \\
1 & 0 & 1 \\
1 & 1 & 0 \\
1 & 1 & 1 \end{array} \right)$
etc.

And Can be Represented as $ a \plus{} b 2 \plus{} c 4$
Now, the system I said is:
$ \left( \begin{array}{cc} 0 & 0 \\
1 & 0 \\
1 & 1 \\
2 & 0 \\
2 & 1 \\
2 & 2 \\
3 & 0 \\
3 & 1 \\
3 & 2 \\
3 & 3 \end{array} \right)$,$ \left( \begin{array}{ccc} 0 & 0 & 0 \\
1 & 0 & 0 \\
1 & 1 & 0 \\
1 & 1 & 1 \\
2 & 0 & 0 \\
2 & 1 & 0 \\
2 & 1 & 1 \\
2 & 2 & 0 \\
2 & 2 & 1 \\
2 & 2 & 2 \\
3 & 0 & 0 \\
3 & 1 & 0 \\
3 & 1 & 1 \\
3 & 2 & 0 \\
3 & 2 & 1 \\
3 & 2 & 2 \\
3 & 3 & 0 \\
3 & 3 & 1 \\
3 & 3 & 2 \\
3 & 3 & 3 \end{array} \right)$

So is kind of a base system with a variable base, since there is no simple periodic pattern.
The problem were, For all $ p,n$ the $ \kappa$ can, by the formula, represent all numbers without repetition.
Or the problem itself can be to find this formula and proove it.

Or yet. the diophantine equation $ \sum_{i\equal{}1}^{p}\left(\begin{array}{c}\kappa_{i}\minus{}i\plus{}p\\ \kappa_{i}\minus{}1\end{array}\right) \equal{} n$ has only one solution in the positives integers for each $ n$


function
real analysis
real analysis unsolved

1)let f(x) be a differentiatable funtion twice at point a and it satisfies f''(a) !=0 (!= doesn't equal).
prove that for h small enough the t given by lagrange formula: f(a+h)=f(a)+hf'(a+th) is a function which is one to one function of h, and it satisfies:
lim t(h)=0.5
h->0

2)let f(x) be a differentiatable funtion twice at point a. prove that:
lim [f(a+h)-2f(a)+f(a-h)]/h^2=f''(a)
h->0

for the second question here what iv'e done:
all the lim are h->0
f'+(a)=lim[f(a+h)-f(a)]/h
f'-(a)=lim[f(a)-f(a-h)]/h

f''(a)=lim [f'+(a)-f'-(a)]/h
and we getting what we shoulv'e proven in the first place, is it correct.

p.s
about the first question i really don't have a clue, as far as i gone is to conclude that f'(a+th)=f'(a).

your help is appreciated.


Prove that for a given positive integer $d$ not equal to $2,5 or 13$
 $2d - 1 , 5d - 1$and$13d - 1$ cannot simultaeneously be perfect squares.
plz help

this looks mighty familiar... ah yes, IMO 1986, problem 1

$\alpha \beta\gamma hello$
[quote="rohitsingh0812"]I have got the solution using congruences modulo 16. I don't need the soln now.thank u[/quote]

ARGH, I have posted the following link in the wrong thread before...
http://www.mathlinks.ro/Forum/viewtopic.php?t=60769


Let $(x_{n})$ be a positive sequence and $a \in (0,1)$.
With every $n \in N$ , $k_{n}$ is the lowest integer such that: $\sum_{i=1}^{k_{n}}\frac{1}{i^{a}}\geq \sum_{i=1}^{n}x_{i}$
Prove that: $\exists$ $lim_{ n \to \infty}(k_{n+1}^{1-a}-k_{n}^{1-a})\leftrightarrow \exists$ $lim_{n \to \infty}x_{n}$


How is technology used and taught in your school? Does your school use SMART boards? have a laptop program? require and use graphing calculators in all science and math courses? have a wireless network? a computer fluency requirement?

How effective do you think these things are in enhancing the education and technological fluency of the students at your school? Do you think they are a good use of resources? What are the downsides to the technology used in your school?

Our school(a middle school), has lots of smart boards, I think they plan to put a SMARTboard in every room. For the classes that do not have a SMARTboard, they have a overhead projector. We have 4 computer labs and some labtops to be rented(for teachers to use in their classrooms). We also have new heating and cooling system installed. There is a recycling bin for paper in every classroom and a plastic and bottles one in the cafeteria. We just need a regular scientific calculator for math class, but the teacher has some graphing ones that he only lends to some students.

I think some of the electronics is helpful but sometimes it is an overdose. Like in US History, the teacher depends on technology and some of things don't even need his technology but he still uses it. I have no idea what my school uses for energy or resources.

Huh? I don't even know what a SMARTboard is...

Math: you shouldn't need anything but white boards and ONE TI-83/84/89/nSpire for the ENTIRE class

A SMARTboard is a board, obviously, that is smart, obviously. Anyways, it is linked to the computer and basically it is a projector that comes out of the ceiling and projects the image of the computer onto a board. With the board, you can use a color and draw stuff on the projected image. And you can also use an eraser. And your hand acts like the mouse. So you can tap the board, and that will be a left click on the mouse.

Oh that's neat... I don't think I've ever even heard of those... must be rather unnecessary...

Our high school just installed ELMOs to pretty much every science/math classroom. It allows the teacher to project a piece of paper (similar to a transparency but with better results and detail) so she can write, lecture, or post post a student's answers. It's very effective  :)

[hide="ELMO"]

[img]http://salestores.com/stores/images/images_747/ELMO9419.jpg[/img]

[/hide]

For science and math, we only require TI-83+ calculators. We might have Wi-fi, but all the computers are hooked via Ethernet cables.

I've seen commercials where there was a class that used laptops and hooked onto projector screens to present their projects. These seem more effective than standard classrooms (no more of those "I left my work at home" excuses).

Our school's use of technology varies based on the teacher.  We have teachers who have SMARTboards in their classrooms, use laptops, document projectors, etc, and those who stick to just plain chalk.

For math, each class has a class set of TI-83's that are available for use in class.

Well, we have a few SMARTboards, plus calculators and computers.

Pretty much the same as above, except our building is new (8 years old) so no chalk boards.

A neighboring school (OE) issues Palm TX to their students for note taking, reading text, and Internet access...

All math classrooms have a smart board or an ELMO, and we require some TI graphing calc. Umm yeah I think that's it...In terms or computers, we have a computer lab, a tech center, and a student center, each with about 25 computers. Then we have a language lab and a media center, so we have around 95-100 computers total, or about 1 comp for every 3 students.

Uhh our school as Wifi but its like locked so that we can't get on the internet during class.

We have smartboards in every room but I don't see teh reason why people can't just use whiteboards instead of smartboards. lol.

We have Smart boards in half the math rooms, overhead projectors and the erm digital projectors in every room, a million useless Mac labs, a shiny black mac whateveritscalled or 2 in every room for attendance or something? Wireless internet yes, and a bunch of shiny macbook pros for library or something.

Listing the particular pile of circuits available at your school is one of several things gauss asked.  It would be far more interesting to address his other questions:
[quote="gauss202"]How is technology used and taught in your school? ...  How effective do you think these things are in enhancing the education and technological fluency of the students at your school? Do you think they are a good use of resources? What are the downsides to the technology used in your school?[/quote]

I go to a school in fiji where technology in schools is not really advanced yet.
My school does not use smartboards but we do have projectors in the IT lab and the principal's office and the library i think taht projects stuff shown on the computer ... and students taking a certain course in the IB diploma are required to have TI-83/84 ...

I actually think its pretty unnecessary to use smartboards n stuff when you can use a whiteboard(or maybe i am just ignorant of the many advantages of the smartboard thinge because i havent ever seen one..). firstly it uses up too much energy(imagine every teacher in every class using one) and then secondly, i would not like to come to school just to learn from the computer .. i dont really know how to phrase this but I learn and take interest when the teacher him/herself is up there writing and stuff(i think its more engaging) and not just pointing out stuff on a screen with a mouse or pointer or whatever. Technology is important in the sense that computer labs provide internet which students can use for their research (i am ignoring teaching IT related stuff)and nowadays it is really important to know how all these things work.

Our school is about to get a wireless network so studentsd can get their laptops and work anywhere in school ... i think this is good maybe for universities but not school because everyone in school would just be hooked on to their laptops playing games or doing whatever random stuff and I think that is just a waste of school time not in the sense that they are not studying but the whole sense of being in a school with different people they can interact with.

Another downside to technology is that I feel it would kind of make the teacehrs lazy ... its great for teachers to get their research material from the internet but i think whatever they can do with smartboards n stuff ( except maybe showing some complex diagrams) they can do with a white board.

oh yeah and regarding calculators .. when i was back in my old school we were not allowed to use calculators in our exams but in my current school kids use it from like grade 6 i think and i this harms them in the future because 90% of my class cannot identify or do simple things like cross multiplying or cancellation in fractions and some other things i cannot rememebr easily( no they arent dumb). This i fugured was because of being too dependent on calculatiors.

My school uses ELMO and projector's that hook up to the computers, morning announcements and TV. We also use high powered laptops in our Project: Lead the Way(PLTW). The teachers also each recieve an interwrite pad which hooks up to their computer and works like an art pad. Only the art is projected by the projector for the students to see. It's very affective in math. Unfortunately, my school district has now fallen into a huge hole of debt (projected to be 36 million dollars by 2012) so I doubt our computers are going to be updated anytime soon.

By the way, does anyone else's school offer PLTW courses?


In a 30 days month, a basketball team plays at least one time a day. However they just play at most 45 match.
Prove that there are some consecutive days that the basketball team play 14 match in total.  :)

It's easy (if i'm not missing something )
Let the team plays $a_{i}$ games during the i-th day.And let $s_{1}=a_{1}$,
$s_{2}=a_{1}+a_{2}$... $s_{15}=a_{1}+\dots+a_{15}$.
Then the numbers $s_{1}\dots s_{15}$  are 15 distinct positive integets
by pigeon-hole principle there are two congruent modulo 14.
This means that there are some consecutive days(in the first 15) that the team has played 14.k games.
Similarly for the next 15 days (16th-30th) we get that there are some consecutive days that the team has played 14m games.
Notice that neither m nor k is zero.
now if $m\geq2$ and $k\geq2$ we get that the team has played more than 28+28=56 games which is not true .THis contradiction means that the team has played 14 games during first 15 or second 15.
Actually this shows that 45 can be changed to 55.


Seven distinct points are identified on the circumference of a circle. How many different triangles can be formed if each vertex must be one of these 7 points?

Suppose there are $ n$ points. If we fix a line segment, there are $ n\minus{}2$ triangles. There are $ (n\minus{}1)\plus{}(n\minus{}2)\plus{}\cdots\plus{}1$ line segments. In this way, each triangle is counted $ 3$ times. So there are $ \frac{(n\minus{}2)(1\plus{}2\plus{}\cdots\plus{}(n\minus{}1))}{3}\equal{}\frac{n(n\minus{}1)(n\minus{}2)}{6}$ triangles altogether.

or you can do $ \dbinom{7}{3}$.

wangzirui- can you explain your solution in a pm, i dont get.

[hide=Solution]
Another way to say this problem is: How many ways can you choose $3$ points out of $7$ distinct points on a circumference of a circle?

The answer is $\binom{7}{3} = \frac{7\cdot6\cdot5}{3!} = \frac{210}{6} = \boxed{35}$


A sequence is defined in the attachment below:

As n gets very large, the terms an converge to the value 0.5059 (to four decimal places). Use this fact to estimate 1/1 + 1/2 + 1/5 + 1/10 + ... + 1/1000001).

Thanks

[hide] Well we have that as $n$ gets large, $\arctan n$ approaches $\frac{\pi}{2}$. So,
$\sum_{k=0}^{\text{whatever}}\frac{1}{k^{2}+1}\approx \frac{\pi}{2}+.5059 \approx 2.0767$ [/hide]


number theory
prime numbers
number theory unsolved

I think you should state: "inspired from APMO 2004" :D

[quote="X-man"]Help me solve this problem:
Let $n$ be integer grater than $4$. Prove that: $n$ and $n+1$ are not prime numbers iff the closest integer to  $\frac{(n-1)!}{n^2+n}$ is even.[/quote]
If n and n+1 are not prime numbers:
$\frac{(n-1)!}{n(n+1)}$ is a integer  and is even
If n is a prime number(n=p)
$(p-1)!\equiv -1(mod \ p)$ ( Wilson's teorem)
$p+1\equiv 1(mod \ p)$
$\frac{(p-1)!}{p+1}\equiv -1(mod \ p)$ and is even
therefore $\frac{(p-1)!}{p+1}\equiv p-1(mod \ 2p)$
$\Rightarrow \frac{(p-1)!}{p+1}=2pk+p-1$
$\Rightarrow \frac{(p-1)!}{p(p+1)}=2k+1-\frac{1}{p}$; p=n
If n+1 is a prime number(n=p-1)
$(p-1)!\equiv -1(mod \ p)$ ( Wilson's teorem)
$\frac{(p-2)!}{(p-1)}=\frac{(p-1)!}{(p-1)^2}$
$(p-1)^2\equiv 1(mod \ p)$
therefore $\frac{(p-2)!}{(p-1)}\equiv -1(mod \ p)$ and is even
$\Rightarrow \frac{(p-2)!}{(p-1)}=2pk+p-1$
$\Rightarrow \frac{(p-2)!}{(p-1)p}=2k+1-\frac{1}{p}$; p=n+1

Im confused... :huh:
How are you using Wilson's theorem when n isn't a prime?

Opps :oops: nvm, I missed a line while reading it... :oops:


A rectangular box is 5 inches long, 4 inches wide, and 2 inches high. What is the number of inches in the length of the longest diagonal of the box? Express your answer in simplest radical form.

I assume by diagonal, the problem means a face diagonal.

We want the largest diagonal, so we need to longest sides, 5 and 4.

$ 5^2 \plus{} 4^2 \equal{} 41$

Thus our answer is $ \boxed{\sqrt{41}}$.

However, the longest diagonal is the space diagonal, which is calculated by $ \sqrt {l^2 \plus{} w^2 \plus{} h^2}$.  Thus, the answer is $ \sqrt {45}$, simplified to $ 3\sqrt {5}$.

Did the problem mean any diagonal or only face diagonals?

It didn't state, so I would presume a space diagonal.

Its probably 3rt5. It's hard to understand ambiguous questions.

Ernie was actually right. The longest diagonal in a rectangular prism is the space diagonal in 3D space, so it should be $ 3$$ \sqrt{5}$. Thus there really isn't any ambiguity in the problem...


Parallelogram $ ABCD$ has diagonals that intersect at point $ E$.  Suppose that $ AB \equal{} 6x\plus{}5$, $ BC\equal{}2x$, $ AE\equal{}4x\minus{}3$, and $ EC\equal{}3x\plus{}4$.  Is angle $ ABC$ an acute, right, or obtuse angle?

Since $ AE \equal{} EC$, $ 4x \minus{} 3 \equal{} 3x \plus{} 4\implies x \equal{} 7$, so $ (AB,BC,AC) \equal{} (47,14,50)$.

$ AB^2 \plus{} BC^2 \equal{} 47^2 \plus{} 17^2 \equal{} (50 \minus{} 3)^2 \plus{} 2894$

$ \equal{} 50^2 \minus{} 300 \plus{} 9 \plus{} 289 \equal{} 50^2 \minus{} 2 < 50^2 \equal{} AC^2$, so

$ \triangle{ABC}$ is obtuse, and since $ \angle ABC$ is the largest angle of the $ 3$, we conclude that it must be $ \boxed{\text{obtuse}}$.


Undamped oscillator with natural frequence w0 = .5 rad/s and a = 1 m/s^2. Calculate the response function for an impulse forcing function acting for tau = 2pi / w0

I know that the general solution for the response function is
x = b/w1 * e^ -B(t-t0) * sin w1(t-t0)

I was thinking that I can find the expression for acceleration from x and just multiply it by m to find the force, but without knowing B (beta) or w1, Im not sure how to proceed.

I would appreciate any help.

i think you need to do some fourier analysis.. which I have absolution no idea know how to do. :maybe:

try go to physicsforum.com. they have more experts (and i mean EXPERTS) over there.


trigonometry
calculus
integration
ratio
geometric series

Prove that $ \sin(\phi) \plus{} \sin(\phi \plus{} \alpha) \plus{} \sin(\phi \plus{} 2\alpha) \plus{} \dots \plus{} \sin(\phi \plus{} n\alpha) \equal{} \dfrac{\sin\dfrac{(n \plus{} 1)\alpha}{2}\sin\left(\phi \plus{} \dfrac{n\alpha}{2}\right)}{\sin\dfrac{\alpha}{2}}$.

By $ \boxed{\sin x\sin y \equal{} \frac {1}{2}\{\cos (x \minus{} y) \minus{} \cos (x \plus{} y)\}}$,

$ \sum_{k \equal{} 0}^n \sin (\phi \plus{} k\alpha)\sin \frac {\alpha}{2}$

$ \equal{} \frac {1}{2}\sum_{k \equal{} 0}^n \left[\cos \left\{\left(k \minus{} \frac {1}{2}\right)\alpha \plus{} \phi\right\} \minus{} \cos \left\{\left(k \plus{} \frac {1}{2}\right)\alpha \plus{} \phi\right\} \right]$

$ \equal{} \frac {1}{2}\left\{\cos \left(\phi \minus{} \frac {\alpha}{2}\right) \minus{} \cos \left(\phi \plus{} \frac {2n \plus{} 1}{2}\alpha\right)\right\}$

By $ \boxed{\cos x \minus{} \cos y \equal{} \minus{} 2\sin \frac {x \plus{} y}{2}\sin \frac {x \minus{} y}{2}}$.

$ \equal{} \frac {1}{2}\cdot 2\sin \left(\phi \plus{} \frac {n\alpha}{2}\right)\sin \frac {(n \plus{} 1)\alpha}{2}$.

Dividing both sides by $ \sin \frac {\alpha}{2}$ gives the result.

You should add the condition $ \alpha \neq 2n\pi$.

Hmmm I did this in the USSR book and it used a complex number argument that's kinda confusing :|

What's confusing you?  The complex number argument interprets this sum as the complex part of a complex geometric series, and geometric series are straightforward.

$ \sum_{k\equal{}0}^n\sin (\phi\plus{}k\alpha)\equal{}\Im\sum_{k\equal{}0}^ne^{i(\phi\plus{}k\alpha)}$.  Then, as t0rajir0u stated, you use geometric sums.  The fact that $ e^{i\theta}\equal{}\cos\theta\plus{}i\sin\theta$ can be proved using (a) integration, or (b) power series.

Since the angles are in the format of an A.P. , we multiply and divide with sin of half the common difference 'a' in this case.
Inside we would get the Difference of Two cos ratios , and for each sin , the 2nd term of the cos difference will get canceled with the 1st of the next sin.

And this would prove our result , these are standard results for Sum of cos's when angles are in AP and for sines and tans as well.

[quote="kunny"]See #2. :wink:[/quote]
Yea , same thing.
 Its just a standard result. how come it became a problem??


Prove or disprove that any integer can be represent by the string of number in $2005^n$ for $n\in\mathbb{N}$ . ( example : 123456 can be represent as 2005^k = ABC123456XYZ ...etc .... )

You can even claim that this string is at the beginning, which was (with slightly differences) posted before: http://www.mathlinks.ro/Forum/viewtopic.php?t=50019

Thanks Zetax . But my friends says that it only required some algebra to solve it and the solution from the link is somehow a bit complicated since I havent learn anything about " dense " yet ....  :blush:

I wouldn't call it purely algebraic, but the prove using density can easily be written by just using elementary terms.
(Note: to be dense doesn't mean more than that when you take any number (real here), then there are numbers from the desired set as close to that number as you want (technically: $D$ is dense in $\mathbb{R}$ iff $\forall x\in \mathbb{R}$ and $\forall \epsilon>0:$ $\exists s \in D$ with $|x-s| <\epsilon$; $\mathbb{R}$ can be changed to any other set); also no special theorems about dense sets is required, it's just used there to shorten the whole ;) )


geometry
perimeter
area of a triangle
Heron\u0027s formula

Is there a nice way of getting the area of a triangle given that one side is 10, the other is 5, and the median going to the third side is 6.5 inches?  I thought about working with the semi-perimeter, but I 'm afraid the problem will get nasty quickly.

Could someone post a hint before posting a solution?

Thanks!

     -- Pawel

well, knowing the Median formula (Or Stewart's Theorem), you could solve for the third side.

Solution:

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Now you can use Heron's formula to finish it up:

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If you do not know the median fomulu, you could draw a line from the midpoint paralel to one of the sides, then use Heren and the area ratios.