algebra
polynomial
Rational Root Theorem

Hi there!

Prove or give a counter-example:

"If x is a real number such that x*(x+1) is an integer, then x is an integer."

It isn't true.

Let's reformulate your hypothesis: if $ x(x \plus{} 1) \equal{} C$, where $ C$ is an integer, then $ x$ is an integer. But as we know,

$ x_{1,2} \equal{} \frac { \minus{} 1\pm \sqrt {1 \plus{} 4C}}{2}$. It is obvious that these numbers can be irrational, when $ 1 \plus{} 4C$ isn't a perfect square.

Proof by Contradiction (but only if $ x$ is a rational number; as hsiljak said, the claim is not true for irrational numbers)

[hide]Assume there is a rational number $ x$ such that $ x\cdot(x\plus{}1)$ is an integer.  This rational number can be expressed as a mixed number in the form $ a\frac{b}{c}$, where $ c$ is not $ 0$.  For $ a\frac{b}{c}\cdot(a\plus{}1)\frac{b}{c}$ to be an integer, $ \frac{b}{c}\cdot\frac{b}{c}$ must be an integer.  However, any proper fraction (except $ \frac{0}{c}$, which would make $ x$ an integer) multiplied by itself yields a value between $ 0$ and $ 1$, and thus cannot be an integer.  Therefore, we have a contradiction, and the original claim is proven.[/hide]

I hope that works (I haven't written very many proofs before).

[quote="cf249"]  For $ a\frac {b}{c}\cdot(a \plus{} 1)\frac {b}{c}$ to be an integer, $ \frac {b}{c}\cdot\frac {b}{c}$ must be an integer. [/quote]

Maybe you could elaborate this claim more thoroughly (although it may seem intuitively obvious) . The rest of the proof is OK.

The argument's not quite correct.  In particular, you ignore the fact that the $ 1$ in $ a \plus{} 1$ is also multiplied by $ \frac {b}{c}$.

What you're trying to prove is that if $ x$ is [i]rational[/i] and $ x(x \plus{} 1)$ is an integer, then $ x$ is an integer.

[hide="Short proof"] $ x^2 \plus{} x \minus{} a \equal{} 0$ has a rational root, so by [url=http://en.wikipedia.org/wiki/Rational_root_theorem]the Rational Root Theorem[/url] it has an integer root that divides $ a$. [/hide]
Another application of this idea:  show that the square root of an integer is either an integer or an irrational number.


[quote="Papadopoulos"]Το λογικό σφάλμα είναι το εξής: Επειδή η ισότητα (=) ως διμελής σχέση είναι μεταβατική, δεν επιτρέπεται να γράψουμε
    [color=red]limf(x) = k[/color] , [color=red]limf(x) = m[/color]
αν δεν αποδείξουμε πρώτα την μοναδικότητα του ορίου. Η γραφή και μόνο των δύο αυτών ισοτήτων είναι σαν να λέει εξ αρχής k=m. Αυτό που πρέπει να πούμε είναι κάτι της μορφής:
 Έστω το όριο της f(x) καθώς x----->a είναι k και επίσης m διαφορετικό του k.
Τότε για ε= l k-m l/2 > 0 έχουμε ..... και λοιπά ...[/quote]

+1

Από την άλλη... μακάρι να ήταν όλα τα λάθη τόσο αθώα :)

Καλή χρονιά gents,

Durandal 1707

[quote="Papadopoulos"][quote="τρικλινο"]Πολλα βιβλια στην αναλυση δινουν την ακολουθη αποδειξη για την μοναδικοτητα του οριου.

........................................ειναι σωστη??............................................

Υποθετουμε οτι η f στο σημειο a εχει δυο ορια k ,m με k$ \neq$ m.

Eστω ε= l k-m l/2 > 0

Επειδη [color=red]limf(x) = k[/color]  , x---->a ( x πηγαινει στο a ) υπαρχει r >0 τετοιο,ωστε για καθε x,με 0<l x-a l <r

να ισχυει; .......................................l f(x)-k l <ε..................................................................................(1)

Επειδη [color=red]limf(x) = m[/color] ,x----->a υπαρχει θ>0 τετοιο,ωστε για καθε x,με  0<l x-a l <θ

να ισχυει:........................................l f(x)-m l < ε...................................................................................(2)

Αν θεσουμε δ = mim{ r, θ },τοτε για καθε x,με 0 < l x-a l < δ ισχυουν ταυτοχρονα οι (1) και (2),οποτε εχουμε

0<l m-k l = l m-f(x)+f(x)-k l $ \le$ l m-f(x) l + l f(x)-k l < ε+ε = 2ε = l m-k l, δηλαδη

..............................................l m-k l < l m-k l.............................................................................

Αυτο ομως ειναι αδυνατο,που σημαινει οτι στο σημειο x, η f εχει μονο ενα οριο.[/quote]

Η απόδειξη έχει ένα λογικό σφάλμα στα σημεία που σημείωσα με κόκκινο.
Πέρα από αυτά, οι πράξεις με τις ανισώσεις και τα ε είναι σωστά.

Το λογικό σφάλμα είναι το εξής: Επειδή η ισότητα (=) ως διμελής σχέση είναι μεταβατική, δεν επιτρέπεται να γράψουμε
    [color=red]limf(x) = k[/color] , [color=red]limf(x) = m[/color]
αν δεν αποδείξουμε πρώτα την μοναδικότητα του ορίου. Η γραφή και μόνο των δύο αυτών ισοτήτων είναι σαν να λέει εξ αρχής k=m. Αυτό που πρέπει να πούμε είναι κάτι της μορφής:
 Έστω το όριο της f(x) καθώς x----->a είναι k και επίσης m διαφορετικό του k.
Τότε για ε= l k-m l/2 > 0 έχουμε ..... και λοιπά ... Άτοπο.
Το άτοπο είναι στο "k και επίσης m διαφορετικό του k".
Άρα το όριο είναι ένα.

Τάκης Π.[/quote].

Κυριε Τακη η αποδειξη στην 1η γραμμη γραφει.....................................

......................" Υποθετουμε οτι η f στο σημειο a εχει δυο ορια k,m με $ k\neq m$"........... επομενως μπορουμε να πουμε,ανορθοδοξα ισως

..............limf(x)=k  kai limf(x) = m     x------->a ,που ειναι περιπου το ιδιο με αυτο που λετε εσεις :......" Έστω το όριο της f(x) καθώς x----->a είναι k και επίσης m διαφορετικό του k".

Θα μπορουσαμε ισως ,για να αποφυγουμε,εμπλοκες με την μεταβατικοτητα της ισοτητας να πουμε:

.......................$ limf(x)\equiv m$  kai $ limf(x)\equiv k$ x----->a.......................................................

Tελος παντων δεν ειναι εκει ,νομιζω το μεγαλο λαθος

Tην αποδειξη αυτη εχω παρει απο το βιβλιο¨.

.......................................Μαθηματικα ................................................

.....................................................γ' λυκειου.......................................

.................................................αναλυση......................................................

ΕΚΔΟΣΗ Β 1993 και η συγγραφικη ομαδα ειναι:

ΚΑΤΣΑΡΓΥΡΗΣ ΒΑΣ........................κ.τ.λ

η αποδειξη ειναι στη σελιδα 301

Είναι πολύ λεπτό το σημείο.

Με λίγα λογια πάντως η φράση

" Υποθετουμε οτι η f στο σημειο a εχει δυο ορια k,m με $ k\neq m$"

είναι σωστή αρχή της απόδειξης ενώ η

" .............limf(x)=k  kai limf(x) = m     x------->a"

που της μοιάζει, δεν είναι.

Μόνο [b]αφού[/b] αποδείξουμε την μοναδικότητα του ορίου μια ακολουθίας $ a_n$, επιτρέπεται να του δώσουμε όνομα. Το συμβολίζουμε, ως γνωστόν, lim$ a_n$.

Τ.Π.

[quote="Papadopoulos"]Μόνο [b]αφού[/b] αποδείξουμε την μοναδικότητα του ορίου μια ακολουθίας $ a_n$, επιτρέπεται να του δώσουμε όνομα. Το συμβολίζουμε, ως γνωστόν, lim$ a_n$.[/quote]

Και πάλι, +1. Κανονικά, ο συμβολισμός $ \lim a_n$ θα έπρεπε να εισάγεται μόνο μετά το θεώρημα απόδειξης μοναδικότητας. Αυτό που κάνουν πολλά (καλά) βιβλία ανάλυσης είναι να λένε ότι "αν υπάρχει, το όριο το συμβολίζουμε έτσι, κάτι που δικαιολογείται από το παρακάτω θεώρημα μοναδικότητας blah, blah, blah..."

Κανονικά, ίσως θα έπρεπε να λέμε ότι "ο αριθμός $ \ell$ καλείται όριο της $ f$ στο $ x_0$ όταν..." και το θεώρημα μοναδικότητας να λέει "αν $ \ell, m$ όρια της $ f$ στο $ x_0$, τότε $ \ell \equal{} m$", και μετά να εισαγάγουμε τον συμβολισμό για το όριο που είναι μοναδικό (όταν υπάρχει τουλάχιστον ένα όριο), αλλά... meeeeeeeeh ;)

Τρίκλινο, τί σε απασχολεί στην παραπάνω απόδειξη? Προφανώς όχι nitpicking issues όπως το παραπάνω, υποθέτω έχεις κάποιο σημαντικότερο πρόβλημα...

Cheerio,

Durandal 1707

[quote="Durandal"][quote="τρικλινο"][b]ΠΡΩΤΑ το θεωρημα της υπαρξης και κατοπιν της μοναδικοτητας[/b][/quote]

Θεώρημα ύπαρξης? :huh:[/quote]

Για παραδειγμα .

Εστω a(n) ακολουθια αυξουσα και φραγμενη απο πανω,τοτε δια του αξιωματος της συνεχειας των πραγματικων αριθμων μπορουμε να αποδειξουμε οτι:

Υπαρχει ενας πραγματικος αριθμος m ,μοναδικος ετσι ωστε:

Για ολα τα ε>0 υπαρχει φυσικος αριθμος κ τετοιος ωστε : $ n\geq k\rightarrow$ l a(n)-m l<ε

.........$ \forall \epsilon [ \epsilon>0\rightarrow\exists k ( k\epsilon N\wedge\forall n ( n\geq k\rightarrow$ la(n)-m l<e))].........................σε συμβολα.

Βλεπουμε λοιπον οτι εχουμε ενα θεωρημα υπαρξεως του οριου ,του οποιου την μοναδικοτητα αποδεικνυουμε κατοπιν

Τρικλινο, πρώτον [i]σίγουρα[/i] δεν καταλαβαίνω και πάλι τι εννοείς με το θ. ύπαρξης, από τη στιγμή που προφανώς δεν ισχύει για κάθε σημείο κάθε συνάρτησης :huh: Αν απλώς θέλεις να δείξεις ότι ο ορισμός είναι non-vacuous, η σταθερή συνάρτηση αρκεί.

Δεύτερον, πες μου αν έχεις πρόβλημα με την παρακάτω διατύπωση:

Ορσ: το $ \ell\in\mathbb{R}$ θα λέγεται όριο της $ f$ στο $ x_0$ όταν...

Θεώρημα: έστω ότι $ \ell, m\in\mathbb{R}$ είναι όρια της $ f$ στο $ x_0$. Τότε $ \ell\equal{}m$.

Απόδειξη: αφού $ \ell,m\in\mathbb{R}$, θα ισχύει είτε $ \ell < m$ είτε $ \ell\equal{}m$ είτε $ \ell>m$.
1. Aν $ \ell>m\implies\ldots 0>0$, άτοπο. Άρα δεν μπορεί να ισχύει $ \ell>m$.
2. Aν $ \ell < m\implies\ldots$ αναλόγως, άτοπο.
3. Συνεπώς, θα πρέπει να έχουμε $ \ell \equal{} m$.

Sorry, αν θέλεις νόμους λογικής θα πρέπει να περιμένεις να απαντήσει ο Demetres... λογική μια φορά έκανα και μετά κρατάω μία απόσταση σεβασμού και ασφαλείας (in the sense: I get it, people did it, now things work and we know what we're talking about, let's go solve some real problems now :) ).

Προς τί το τόσο nitpicking? Αν σε αυτά τα "λάθη" αναφέρεσαι, ελπίζω να συμφωνείς ότι βιβλία γραμμένα όπου λέγεται σε κάθε βήμα ποιός νόμος λογικής χρησιμοποιείται... θα ήταν μάλλον unwieldy από εκπαιδευτικής άποψης ;)

Cheerio,

Durandal 1707

[quote="Durandal"]Τρικλινο, πρώτον [i]σίγουρα[/i] δεν καταλαβαίνω και πάλι τι εννοείς με το θ. ύπαρξης, από τη στιγμή που προφανώς δεν ισχύει για κάθε σημείο κάθε συνάρτησης :huh: Αν απλώς θέλεις να δείξεις ότι ο ορισμός είναι non-vacuous, η σταθερή συνάρτηση αρκεί.

Δεύτερον, πες μου αν έχεις πρόβλημα με την παρακάτω διατύπωση:

Ορσ: το $ \ell\in\mathbb{R}$ θα λέγεται όριο της $ f$ στο $ x_0$ όταν...

Θεώρημα: έστω ότι $ \ell, m\in\mathbb{R}$ είναι όρια της $ f$ στο $ x_0$. Τότε $ \ell \equal{} m$.

Απόδειξη: αφού $ \ell,m\in\mathbb{R}$, θα ισχύει είτε $ \ell < m$ είτε $ \ell \equal{} m$ είτε $ \ell > m$.
1. Aν $ \ell > m\implies\ldots 0 > 0$, άτοπο. Άρα δεν μπορεί να ισχύει $ \ell > m$.
2. Aν $ \ell < m\implies\ldots$ αναλόγως, άτοπο.
3. Συνεπώς, θα πρέπει να έχουμε $ \ell \equal{} m$.

Sorry, αν θέλεις νόμους λογικής θα πρέπει να περιμένεις να απαντήσει ο Demetres... λογική μια φορά έκανα και μετά κρατάω μία απόσταση σεβασμού και ασφαλείας (in the sense: I get it, people did it, now things work and we know what we're talking about, let's go solve some real problems now :) ).

Προς τί το τόσο nitpicking? Αν σε αυτά τα "λάθη" αναφέρεσαι, ελπίζω να συμφωνείς ότι βιβλία γραμμένα όπου λέγεται σε κάθε βήμα ποιός νόμος λογικής χρησιμοποιείται... θα ήταν μάλλον unwieldy από εκπαιδευτικής άποψης ;)

Cheerio,

Durandal 1707[/quote]

Durandal το θεωρημα υπαρξεως ηταν για ακολουθειες και οχι γενικα για συναρτησεις και οπως σωστα εγραψες το χρησιμ. για non vacuous ορισμους.

Πες μου Durandal θεωρεις nitpicking την [b]ΛΑΝΘΑΣΜΕΝΗ χρησιμοποιηση των νομων της λογικης στην μαθηματικη αποδειξη??[/b]

Στο προηγουμενο post      μου εδειξα την παραβιαση του κανονος της υποθετικης αποδειξεως η αλλως θεωρηματος παραγωγης (deduction theorem),εαν δεχτουμε οτι η αποδειξη ειναι σωστη.

Και να ηταν αυτη η αποδειξη η μονη λανθασμενη στην αναλυση θα ημαστε ευτυχισμενοι.

[b]Ετσι λοιπον στο ονομα της κακοηθους συντομιας στις μαθηματικες αποδειξεις πρεπει να πεταξουμε τους νομους της λογικης στον τραφο [/b]

Δεν θα αναφερθω στα εκατομμυρια των μαθηματικων αποδειξεων που οι μαθηματικοι δεν μπορουν να εξηγησουν στους μαθητες τους επειδη αλλο απο αυτη την κακοηθη συντομια δεν εχουν τιποτα αλλο να προσφερουν.

[b]Μπορουν να αναλυσουν μια μαθηματικη αποδειξη σε αξιωματα θεωρηματα ορισμους και κανονες λογικης[/b]

και ετσι να ανακαλυψουν που ο μαθητης κουτσαινει??

Ξερεις ποσοι P .Hντιδες αγνοουν βασικους κανονες λογικης,ξεχωρα που δεν μπορουν να τους χρησιμοποιησουν.

Στην παραπανω αποδειξη [b] εχρησ. μονο τον σκελετο της αποδειξεως[/b] για να δειξω οτι η αποδειξη ηταν εσφαλμενη [b] και οχι πληρη φορμαλιστικη αποδειξη[/b]

Τωρα θα μου πεις ,οπως σωστα ανεφερες, τι θα κανουμε θα γραψουμε ολες τις μαθηματικες αποδειξεις σε φορμαλιστικη μορφη ?? [b] οχι βεβαια[/b],αλλα πρεπει να εχουμε ενα τροπο να ελενχουμε τις μαθηματικες αποδειξεις.

Ο Ευκλιδης παρεα με την γεωματρια εγραψε και το βιβλιο ψευδαρια (Pseudaria) στο οποιο αναφερει τροπους ελενχου και αποφυγης λανθασμενων αποδειξεων

Οσον αφορα την παραπανω αποδειξη επιμενω οτι ειναι λαθος και περιμενω και του Δημητρη την γνωμη.

Επι τη ευκαιρια ο νομος της τριχοτομιας στους πραγματικους αριθμους που ανεφερες στην πληρη σωστη διατυπωση του ειναι:

 ........... (l<m v l=m v l>m ) & ~( l<m & l=m) & ~( l<m & l>m) & ~( l=m & l>m).............l,m εR..........................

 WHERE "~" means not end  & και....................

................................................................η..................................................................................

........l<m or l=m or l>m..............end:

......l<m-------> $ l\neq m$. end l<m------->~l>m end l=m-------->~ l>m.............................

[quote="τρικλινο"]Πες μου Durandal θεωρεις nitpicking την [b]ΛΑΝΘΑΣΜΕΝΗ χρησιμοποιηση των νομων της λογικης στην μαθηματικη αποδειξη??[/b][/quote]

$ \text{mathematics}\neq\text{logic}$

Short answer? Ναι :P

Longer answer? Ναι, θεωρώ nitpicking το να κάνεις spell out πλήρως τους νόμους της λογικής που χρησιμοποιείς έναν προς έναν, ιδίως όταν έτσι καταλήγεις να επαναλαμβάνεις το ίδιο και το ίδιο χιλιάδες φορές, χωρίς να προσθέτεις κάτι σε μία απόδειξη εκτός από εντελώς τετριμμένα λογικά συμπράγκαλα τα οποία σε απομακρύνουν από την ουσία του μαθηματικού επιχειρήματος και γράφονται μόνο και μόνο για να είσαι φορμαλιστικά σωστός. Αν γράφαμε τα πάντα, τότε θα κατέληγαν οι μαθηματικές αποδείξεις να είναι τρις χειρότερες από τα νομικά συμβόλαια: για να καλύψουμε τετριμμένες και μη ενδιαφέρουσες λογικές υποθέσεις, θα έπρεπε να γράφουμε σεντόνια.

Έλεος βρε συ :) Τα μαθηματικά διαβάζονται αρκετά δύσκολα as is, δε χρειάζεται να τα φορτώσουμε με 1337 σύμβολα παραπάνω τα οποία θα κάνουν ακόμα πιο δυσνόητο το μαθηματικό επιχείρημα αυτό καθαυτό.

Σε κάθε περίπτωση, για να μην παρεξηγηθώ, in principle, συμφωνώ απολύτως με το λογικό έλεγχο κάθε απόδειξης. Ωστόσο, δε συμφωνώ ότι είναι δουλειά των μαθηματικών να το κάνουν αυτό, ιδίως τώρα που έχουμε υπολογιστές. Για να σιγουρευτούμε ότι δεν έχει ξεφύγει κάτι, περνάμε σιγά σιγά όλα τα μαθηματικά θεωρήματα από το κόσκινο της formal logic με τη βοήθεια υπολογιστών που ελέγχουν το κάθε βήμα. Αυτό είναι το automated theorem proving (ή κάπως έτσι); σκέψου ότι μέχρι και το θ/μα Hahn-Banach έχει ελεγχθει (που είναι πολύ πιο tricky από το παραπάνω λημματάκι)!

Οπότε, από τη στιγμή που έχουν ελεγχθεί πλήρως λογικά (από υπολογιστές ακόμα!) τα θεωρήματα που χρησιμοποιούμε, και γνωρίζουμε ότι είναι σωστά, ποιος ο λόγος να τα φορτώσουμε με περαιτέρω πράγματα? Αυτά που λες ανήκουν σε ένα εξειδικευμένο μάθημα λογικής, όχι ανάλυσης.

Over and out,

Durandal 1707

Durandal απο την μια πλευρα εχουμε αυτα που εγραψες εσυ και παραθετω¨.

"Σε κάθε περίπτωση, για να μην παρεξηγηθώ, in principle, συμφωνώ απολύτως με το λογικό έλεγχο κάθε απόδειξης. Ωστόσο, δε συμφωνώ ότι είναι δουλειά των μαθηματικών να το κάνουν αυτό, ιδίως τώρα που έχουμε υπολογιστές. Για να σιγουρευτούμε ότι δεν έχει ξεφύγει κάτι, περνάμε σιγά σιγά όλα τα μαθηματικά θεωρήματα από το κόσκινο της formal logic με τη βοήθεια υπολογιστών που ελέγχουν το κάθε βήμα. Αυτό είναι το automated theorem proving (ή κάπως έτσι); σκέψου ότι μέχρι και το θ/μα Hahn-Banach έχει ελεγχθει (που είναι πολύ πιο tricky από το παραπάνω λημματάκι)!

Οπότε, από τη στιγμή που έχουν ελεγχθεί πλήρως λογικά (από υπολογιστές ακόμα!) τα θεωρήματα που χρησιμοποιούμε, και γνωρίζουμε ότι είναι σωστά, ποιος ο λόγος να τα φορτώσουμε με περαιτέρω πράγματα? Αυτά που λες ανήκουν σε ένα εξειδικευμένο μάθημα λογικής, όχι ανάλυσης"

Και απο την αλλη ειναι αυτα που βρηκα στο βιβλιο του PATRICK SUPPES,Introduction to LOGIC,page 69  :wink:

και παραθετω:

" [b]  IT was rigorously proved in 1936 by the contemporary American logician Alonzo Church that there is no decision procedure,that is,no mechanical test,for the validity of arbritary formulas in first-order predicate logic.Since all of mathematics may be formalised within first-order predicate logic, the existence of such a decision procedure would have startling consequences: a machine could be build to answer any mathematical problem or to decide on the validity or ivalidity of any mathematical argument.But Church's theorem ruins at a stroke all such daydreams of students of logic and mathematics .Not only is there no known decision procedure:his theorem establishes that there never will be any [/b]"

Δηλαδη οπως καταλαβαινεις (αφου γνωριζεις Αγγλικα) λεει χοντρικα : μηχανη που να αποφασιζει για οποιοδηποτε μαθηματικο επιχειρημα,εαν ειναι σωστο η λαθος [b]ουτε υπαρχει ουτε ποτε θα υπαρξη[/b].

Βεβαια τα κομπουτερς [b]μπορουν να λυσουν ολα τα προβληματα της προτασιακης λογικης[/b]

Oooook...

[url=http://en.wikipedia.org/wiki/Automated_theorem_proving]Automated Theorem Proving[/url]

Έλεος, σε μία 1st-order logic που περιέχει τους φυσικούς αριθμούς, προφανώς και υπάρχουν προτάσεις που δεν είναι machine-provable (or even provable :D). Αυτό δε σημαίνει ότι και τα θεωρήματα που συζητάμε δεν είναι machine-provable.

Cheerio,

Durandal 1707


After spending some hours with trig. i see only $\pi\,\pi ... \cos, \tan $  :wacko:  :what?:

Here are some so you can have fun like i did :) :play_ball:

(1) Express as a product the following expresion $E=\tan x+\tan 2x -\tan 3x$

(2) Evaluate $S=\cos\frac{2\pi}{9}+\cos\frac{8\pi}{9}+\cos\frac{14\pi}{9}$

and a nice one

(3) Compute $\sin\frac{\pi}{36}\cos\frac{7\pi}{36}\sin\frac{13\pi}{36}$. this one is  :dry:

So it should be like this: :P
$-\tan{x}\tan{2x}\tan{3x}$

Yes 2 is correct. And shobber 's answer as well :P

For #3, this is as far as I got it...in degrees btw

[hide]$\frac{\sin{5}}{4}(2\cos^2{5}+\cos{10})$[/hide]

[quote="gotztahbeazn"]wow how do u kno that?[/quote]

$\tan A + \tan B + \tan C = \tan A \tan B \tan C$

then $\tan x + \tan 2x - \tan 3x = \tan x + \tan 2x + (- \tan 3x) = -\tan x \tan 2x \tan 3x$

however, the formula works only if $A+B+C = 180$, so unless if that is one of the condition for this problem, we cannot say that our answer is true.

can someone post how they got 1 and/or 3?

Note for 2 and 3:
The measure of the angles form an arithmetic sequence.

the third one was rather fun :P

[hide="3."]

$\displaystyle \sin\frac{\pi}{36}\cos\frac{7\pi}{36}\sin\frac{13\pi}{36}$

$=\displaystyle \frac12(\cos\frac{\pi}{3}-\cos\frac{14\pi}{36})\cos\frac{7\pi}{36}$

$=\frac14\cos\frac{7\pi}{36}-\frac14(\cos\frac{7\pi}{36}+\cos\frac{21\pi}{36})$

$=-\frac14\cos\frac{7\pi}{12}$

$=-\frac14\cos(\frac\pi4+\frac\pi3)$

$=-\frac14(\cos\frac\pi4\cos\frac\pi3+\sin\frac\pi4\sin\frac\pi3)$

$=\frac{\sqrt6-\sqrt2}{16}$
[/hide]

:winner_first: very good :). I spent a lot of time with it :P

edited my numerous latex errors  :blush:  :blush:


abstract algebra
superior algebra
superior algebra unsolved

Let $ G$ be finitely generated module on a commutative ring $ R$ show that:

$ \sqrt{Ann_R(G/IG)}\equal{} \sqrt{ I\plus{}Ann_R(G)}$

Can somebody post a more detailed solution? Thanks in advance.

Inclusion $"\supset"$ is immediate as $\text{Ann}_{R}(G/IG)\supset I+\text{Ann}_{R}(G)$

For $"\subset"$ take $f\in \text{Ann}_{R}(G/IG)$.
Let $\phi\in \text{Hom}_{R}(G,G)$ be the multiplication by $f$.
Then $\phi(G)\subset IG$.
By appropriate version of NAK we have
$\phi^{n}+\sum_{i=1}^{n}a_{i}\phi^{n-i} = 0$
for some $n$ and $a_{i}\in I$.
Hence $f^{n} + \sum_{i=1}^{n}a_{i}f^{n-i}\in \text{Ann}_{R}(G)$ and so $f^{n}\in I + \text{Ann}_{R}(G)$.


Let a,b,c>0 such that       ab+bc+ca=3                            Prove that
3$ \sqrt{(a\plus{}b)(b\plus{}c)(c\plus{}a)}>\equal{}2(\sqrt{a\plus{}b}\plus{}\sqrt{b\plus{}c}\plus{}\sqrt{c\plus{}a})>\equal{}6\sqrt{2}$

$ \bigstar$First , we prove the left ineq .
Setting :$ x\equal{}ab;y\equal{}bc;z\equal{}ca$=>$ x\plus{}y\plus{}z\equal{}3$
By cauchy-schwarz , we need to prove :
$ 9(x\plus{}y)(y\plus{}z)(z\plus{}x) \ge 8(x\plus{}y\plus{}z)(xy\plus{}yz\plus{}zx)$(Right)
$ \bigstar$, Next, we prove the right ineq :
By Am-Gm , we need to prove :
$ (a\plus{}b)(b\plus{}c)(c\plus{}a) \ge 8$
<=>$ 3(a\plus{}b\plus{}c)\minus{}abc \ge 8$(Right because $ ab\plus{}bc\plus{}ca\equal{}3$)


Well, testing out this forum, and hopefully i can get some help before next class (two days from now):

the chapter introduces Euclidean n-space, so we're talking about inner products and basic properties of n-tupled vectors; the previous chapters had been about matrices and determinants (if that helps at all). Anywho, here's the skinny -

Let [b]u1[/b]=(-1,3,2,0), [b]u2[/b]=(2,0,4,-1), [b]u3[/b]=(7,1,1,4) and [b]u4[/b]= (6,3,1,2). Find the scalars, c1, c2, c3, and c4, such that

c1[b]u1[/b]+c2[b]u2[/b]+c3[b]u3[/b]+c4[b]u4[/b]=(0,5,6,-3)

any help is appreciated. Thanks much.

One equation in terms of 4-dimensional vectors becomes 4 equations when written in components: \[\begin{cases}-1c_{1}+2c_{2}+7c_{3}+6c_{4}&=0,\\ \dots&=5\\ \dots&=6\\ \dots&=-3 \end{cases}\] Now you can apply your knowledge of matrices and determinants to solve this system of 4 equations with 4 unknowns.

Thank you very much. Now, i have another problem >.< Again, i'm glad you gave me the set up and not necessarily the answer, so i'm hoping something along those lines for the next one.

Solve the linear system for x1,x2 and x3.

(1,-1,4)*(x1,x2,x3) = 10
(3,2,0)* (x1,x2,x3)  = 1
(4,-5,-1)*(x1,x2,x3)= 7

I tried putting the problem back into a linear system, and solving by row-reduction; but i wound up with pretty wicked numbers. The solutions manual doesn't have anything for this particular problem, either, so i'm at a standstill.

Again, any help is appreciated :D

You should expect 38 (or some factor of 38) to show up in the denominator. This probably wasn't rigged to generate nice answers.


Many thanks to everyone who participated in the contest!  It was more successful than we expected.  Hopefully you all will continue to use the AoPSWiki and add content :)  Here are your top 3:

:winner_first: b-flat
:winner_second:  lotrgreengrapes7926
:winner_third: Xantos C. Guin
 :ninja: (honorable mention) bubka

What about i_like_pie? I saw him have a ton...

I thought he just added a bunch of problem pages with no solutions or something

The problem pages where already there, I just added the no solution template to all of them and added links to the next and previous problems. If I got an honorable mention, I would be amazed. :ninja:

W00T!!!  I am surprised that I won, especially because I didn't have enough time to finish the 2004 AMC 10A problems.

Congrats b-flat.

I finished the 2003 AMC 10A Problems, but my solutions probably were pitiful in comparison to yours.

You guys are helping people too!!!!
*Funcia gives funcia-land community service badges*


If $ f(x)\equal{}\cos(x)$ and $ g(x)\equal{}\log_{10}{x}$, how many points of intersection are there for the graphs of these two equations?

Hint: all intersection points must be in the interval $ [1,10]$.  What is the period of cosine?


O is a point outside the triangle ABC. And OHI || BC, GOE|| AB, OJF||AC. If $S_{OEF}=m^2$, $S_{OGH}=n^2$, $S_{OIJ}=k^2$. Find $S_{ABC}$.

[hide="hint"]this is a very easy application of similar triangles...[/hide]

[hide]
The bases of each of those triangles is proportional to the square root of their areas, or $m, n,$ and $k$. Let $B_{\triangle ABC}$ denote the base of $\triangle ABC$. Notice that $B_{\triangle ABC} = B_{\triangle OEF}+B_{\triangle OGH}-B_{\triangle OIJ}$. Thus $B_{\triangle ABC}$ is proportional to $m-n+k$ and the area $S_{\triangle ABC} = (m+n-k)^2$.[/hide]

[quote="shobber"]Shouldn't that be $(m+n-k)^2$?[/quote]

Yes. Sorry, I don't know why I wrote $(m-n+k)^2$ when all my work showed $(m+n-k)^2$... :blush:


AMC USA(J)MO USAMO email MATHCOUNTS ARML

I hope this isn't too irrelevant/pushy/spammy, but will a full list of MOsPpers be posted at some point?

[quote="jb05"]Even people who made it don't have one yet......[/quote]

Yeah I know, the reason I was asking was because I made it and want to see who will also be there...

Has there been one in the past? I was accepted last year, but I didn't go, so I didn't pay much attention to what they sent out.

Yes, they don't send it out but they usually post it online.

It tends to be somewhat guessable from the e-mail addresses, if they send out an e-mail notification.

The list of 2008 MOSP Participants is now up on our website:
http://www.unl.edu/amc/a-activities/a6-mosp/a6-1-mosparchives/2008-ma/08mosppart.shtml

The list is sorted by state, then by last name.

Steve Dunbar
MAA Director, American Mathematics Competitions

There is a Beijing, CT? :rotfl:

Yay no 8th grade blue moppers... That means I didn't totally get beasted  :D

According to the list of USAMO qualifiers, Illinois Math & Science Academy is in five different cities, namely Chicago, Aurora, Evanston, Palatine and Vernon Hills. The list of MOPpers adds Madison, WI to this list.

I think for schools, it's the person's hometown, not the school's town. For example, Elizabeth Synge lives in Lexington, MA, but her school, BU Academy, is in Boston. However for the Exeter kids, I don't know what's up, because it's a boarding school...and as far as I know, Inyoung Cho does not live in New Hampshire.

[quote="mcsquared"]Wait, does this mean that Eric Larson didn't make MOP?
Holy crap.[/quote]

He would have, being an 11th grade HMer, but if I'm not mistaken, he chose RSI.

Yes!  I'm the only person on the list who takes up two lines!!!

[quote]I believe there is a small: two people are listed as Phillips Exeter Academy, but in places other than Exeter, NH (Minson Shin, David Rush).[/quote]

I think that they're probably boarders, yet their mailing addressess/hometowns are listed as their cities.

Huh...so Joy Zheng is going to CGMO but not MOP?

If I am interpreting your comment correctly, Joy Zheng and Shijie Zheng are the same person.

Same person.  :D  I tend to write Shijie (my legal name) on all the AMC things so that there aren't a bunch of issues when they try to deal with anything along the lines of passports, plane tickets, etc.

Huh. jhredsox said he was going, but his name doesn't seem to be on there.

[quote="mcsquared"]Wait, does this mean that Eric Larson didn't make MOP?
Holy crap.[/quote]
He [url=http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2008-ua/2008usamotop12.shtml]qualified[/url] but chose not to go. My guess would be that he's going to RSI instead.

He knows he can win next year.. He probably thinks this year was a failure, being the beast that he is...

[quote="the future"]^huh i hope you were joking.  i think mostly anyone would choose rsi over mop.[/quote]
huh
i wouldnt

[quote="the future"]^huh i hope you were joking.  i think mostly anyone would choose rsi over mop.[/quote]It definitely depends on what you are trying to accomplish with math in high school.

its on the inside cover of the blue handbook in the package

:blush: Probably should have realized that, but it's been quite a while since I looked at the USAMO qualifiers list

[quote="CatalystOfNostalgia"][quote="piggyinc"] jhredsox said he was going, but his name doesn't seem to be on there.[/quote]

You never know with jhredsox...I mean it does kind of give it away that he didn't make the USAMO, but he'll probably come to MOP anyway. I sure look forward to it![/quote]
lollllz
that's only slightly mean

Oh, I see. So unless I completely fail at counting, we're going to have 11 girls, not 10. Hmm. I wonder which of them is the one CGMO team member whose email doesn't give their name away.

[quote="mathemonster"]MOP's way harder to get into for sure.[/quote]
That doesn't necessarily make it better for certain people.

[quote="mathemonster"]why would u choose rsi over MOP; MOP's way harder to get into for sure.[/quote]

Would you say so?  Apparently not the case for me...last year, I both was rejected from RSI and got into MOP.  Then again, that's for if you apply for math...

[quote="mathemonster"]why would u choose rsi over MOP; MOP's way harder to get into for sure.[/quote]

I'm with K8 -> I'm not sure this is true, particularly if you are from a highly populated state.

I'll also note that RSI and MOP have *very* different missions, very different structures, and  considerably different student populations, so there are lots of reasons you might choose one over the other.

yea it seems way easier to get in if you dont do math....do u know of any interesting finance math related things i could do?

ps. how good was your application?

My impression is that MOP feels more like...an exclusive sort of elite. It's like running a race: if you're fast enough, you move on; if not, you don't. RSI is a bit more complicated. The selection isn't quite as concrete, which is why you sometimes see "random" people beat out USAMO winners. There are some forces beyond your control, e.g. how good are your recs? What state are you from?

I think making MOP is more of an achievement than getting into RSI, but that's not to say it's better. The former is a great distinction in itself; the latter provides you with an opportunity for greater accomplishments in scientific research.

Some got into MOP but not RSI.

Some got into RSI but not MOP.

Some went to MOP over RSI.

Some went to RSI over MOP.

Some, even with good chance of reaching IMO for the first time, went to RSI over MOP.

And then of course some went to MOP, for the chance of reaching IMO for the first time, over RSI (I think most in that situation would choose to do so).

As mentioned already, a capable student has reasonable control towards getting into MOP, but has no control towards getting into RSI.

[quote="Boy Soprano II"]Well, keep in mind that if you would make the IMO team this year, chances are you will make it next year, which is a reason to choose RSI instead.[/quote]
Reasonable, but it had not always turned out that way.

[quote="janecw"]No mopper from TX this year?No Kevin Chen?[/quote]

Texas isn't known for being that great in olympiad math (looking at MOP lists and qualifier numbers...), which is surprising for a state that pwns Mathcounts and does well at ARML

Also, I agree it's what your goal for math is. Eric probably is more interested in the further research in math and science he may want to go into as a career instead of just another IMO medal. Plus, he may see Blue MOP as a waste of time since he is already an IMO winner. This was my main point.

[quote="Anaxerzia"][quote="janecw"]No mopper from TX this year?No Kevin Chen?[/quote]

Texas isn't known for being that great in olympiad math (looking at MOP lists and qualifier numbers...), which is surprising for a state that pwns Mathcounts and does well at ARML
[/quote]
I know at least one person who got a 19 and I'm somewhat sure Kevin got a 9... but yeah we've had low numbers at mop lately... (dk about earlier)


$ 2*2$ typed marices $ A,\ B$ satisfy $ AB\minus{}BA\equal{}A$. Prove that $ A^2\equal{}\mathbb{O}$.

I will prove that for every two $ n\times n$ matrices $ A$ and $ B$ satisfying $ AB\minus{}BA \equal{} A$, we have $ A^n \equal{} 0$.

Note that for every $ k>0$
\[ tr\,  A^k \equal{}  tr(AB\minus{}BA)A^{k\minus{}1} \equal{} tr(ABA^{k\minus{}1}) \minus{} tr(BA^k) \equal{} tr(BA^k) \minus{} tr(BA^k)\equal{}0.\]
Here we used the identity $ tr\, XY \equal{} tr\, YX$. Denote the eigenvalues of $ A$ by $ \lambda_1,\dots,\lambda_n$. Then for every $ k>0$,
\[ \sum_{i\equal{}1}^n \lambda_i^k \equal{} 0.\]
Hence all eigenvalues $ \lambda_i$ are equal to 0. So $ A$ is nilpotent, and $ A^n \equal{} 0$. (Or, in other words, the characteristic polynomial is $ t^n$; by the Cayley–Hamilton theorem, $ A^n \equal{} 0$).


logarithms function inequalities calculus derivative

Find $ x$

$ \left( 10^{\frac{1}{\log 4}}\right) ^{\log x}\minus{}4^{\log \left( x\plus{}6\right) }\minus{}6\equal{}0$

Was that just brute force, or what? I think bambang wanted the solution, not just the answer.

Write the equation as $ 10^{\log _4 x} \minus{} 4^{\log_{10} (x \plus{} 6)} \equal{} 6$

What's the one number that's going to give you a "nice" value for the LHS? The problem looks like it was written with the answer in mind, rather than with some particular method in mind. The only method I can think of is as follows:

[hide]Define the function $ \textbf{D}[f(x)]$ to be the rate of change of $ f$ at $ x$.

It is intuitively clear that $ \textbf{D}[10^{\log _4 x}] > \textbf{D}[4^{\log _4 x}] > \textbf{D}[4^{\log _{10} x}] > \textbf{D}[4^{\log _{10} (x \plus{} 6)}]$

The first inequality is true because exponentials with higher bases increase more quickly. The second is true because logs with lower bases increase more quickly. The third is true because log "flattens out" for higher x; in other words, it is concave.

Then, $ \textbf{D}[10^{\log _4 x}] \minus{} \textbf{D}[4^{\log _{10} (x \plus{} 6)}] > 0 \implies \textbf{D}[10^{\log _4 x} \minus{} 4^{\log _{10} (x \plus{} 6)}] > 0$

Therefore, the LHS is increasing for all $ x > 0$, and there can be at most one solution.

Notice that $ x\equal{}4$ works, and therefore this must be the uniqe solution.

[b]Note:[/b] The point is to show that you don't need to know how to take a derivative to use what you know about the shapes of function graphs and apply them to an increasing/decreasing argument.[/hide]

I agree that the solution seemed obvious.

However, it's still better to post the solution.

And TZF, even if you just wanted to graph, so you wouldn't have to work with derivatives, logarithms are not the easiest things to graph, and twice as hard when in an equation like this. Just saying.


number theory
greatest common divisor
number theory unsolved

PRove 2^2k + 1 is relatively prime to 2^2n + 1 if k and n are distinct positive integers.

Use mods p or use factorisation of squares and result follows.
General gcd can apply for a general no.

Bomb

[quote="kiddos"]PRove $2^{2k} + 1$ is relatively prime to $2^{2n} + 1$ if k and n are distinct positive integers.[/quote]
 :huh:
I can' understand the question since $k=3$ and $n=5$ are distinct though

$2^{2k}+1=2^6+1=65$ and
$2^{2n}+1=2^{10}+1=1025$

But $(65,1025)=5$

PRove 2^2k - 1 is relatively prime to 2^2n- 1 if k and n are distinct positive integers.

Now they're both divisible by $2^2-1=3$ :).


geometry incenter perimeter trigonometry

Let $I$ be the incenter of the triangle $ABC$. Let $A_1,A_2$ be two distinct points on the line $BC$, let $B_1,B_2$ be two distinct points on the line $CA$, and let $C_1,C_2$ be two distinct points on the line $BA$ such that $AI = A_1I = A_2I$ and $BI = B_1I = B_2I$ and $CI = C_1I = C_2I$.
Prove that $A_1A_2+B_1B_2+C_1C_2 = p$ where $p$ denotes the perimeter of $ABC.$

Pierre.

We'll use the well-known fact:
In any triangle ABC we have that the projection of O(circumcenter of ABC)to BC is equal to RcosA.

Using for triangle A1AA2 we have r=AIcosA1AA2 hence cosA1AA2=2*sin 2(A/2)-1. Using cosine theorem in A1AA2: A1A2=2(p-a).
Adding similar conclusions we get the desired result.

Let $D$ and $E$ be the feet of the perpendiculars from $I$ to $AB$ and $BC$ respectively.  Then it's obvious that $\triangle ADI \equiv \triangle A_1EI$, so $AD = A_1E$.  Similarly for sides $CA$ and $AB$.

i thought it was quite a cute cutting and pasting of the triangle.

kennyworm

when i solved it, it didn't occur to me. but now its obvious (= i just used pythagoras without noticing the congruent triangles.

3 consecutive posts by the Singapore team! ;) Wow! :D


Okay so I have been fiddling around with this problem for half a week and my take home quiz is due really soon. I've gotten really desperate and decided to post it on here if anyone can point me in the right direction and help me out. This is a classic 2 column proof. So here it goes:

Given: E is the midpoint of segment DA, F is the midpoint of segment DB. Segment FC is congruent to segment EF.
Prove: EABC is a parallelogram.
[Hint: consider what you know about segment EF in triangle ADB? and consider quadrilateral BCDE.]

here is the picture of what it looks like:
[img]http://i6.photobucket.com/albums/y247/jimmywashere/geom.jpg[/img]

MOD: Moved to pre-olympiad.

If this is a take home quiz, it compromises your academic integrity to receive hints from us.
Sorry, but that's not really morally acceptable in my book.

Well i tried and tried and i can't think anything of it. Judge me however you like, but i'm in search of some help. Why are these forums are made if you're going to be testing my intelligence. Let me make something clear to you, i am NOT in search of "answers" i need to be pointed in the RIGHT direction. Reading > you. So if there's anyone out there who is more helpful and would kindly help me out, please do so. If you're going to post something stupid and not help out, then just click backspace becuase this thread will not be for you and neither will it be appreciated. So for the people who do help, thank you in advance.

[quote="tobies"]Why are these forums are made if you're going to be testing my intelligence.[/quote]

As difficult as it might be for you to grasp the concept, some people actually enjoy mathematics and work math problems because they find them interesting.  We don't do them just because we want good grades, and even if we did, we'd have the foresight, maturity, and responsibility to learn, study, and understand the concepts before we're assigned a test.

[quote]If you're going to post something stupid and not help out, then just click backspace becuase this thread will not be for you and neither will it be appreciated. So for the people who do help, thank you in advance.[/quote]

What is not appreciated is your attitude, which I find immature, disrespectful, and offensive.  You post as if you own this forum, which you clearly do not, or else you would already know how to solve such an elementary question.  You have the arrogance to tell others not to comment; and if they do, that their posts are stupid, when it's you who has nothing to offer.  You solicit help but you do it with a sneer, so although I could have provided all the help you could possibly ask for on any mathematical subject you could imagine, I've decided that such efforts would be lost upon you.  Had you expressed a more humble attitude, perhaps I would have been more inclined toward leniency.

I didn't mean to come off with an "attitude". I came here for help because I didn't understand. I study hard and get good grades, but some geometry concepts I can't "grasp" as you say. Never did I intend to intrude the forums, especially with an attitude and acting as I "own the forums", no, I didn't come here for that. I came here to seek help from someone who is willing to help. I posted up my problem... and the very next post I get insulted? If you're going to put me in the wrong, then okay, I'll apologize, sorry for coming in here with an attitude. But I still need help on this, so if anyone can help, please do.

You should have posted this problem on a more basic level. EF is a midline of $\triangle ABD.$ What can you say about this midline ? Diagonals of the quadrilateral BCDE cut each other in half at F. What can you say about this quadrilateral ?

[quote] EF is a midline of  What can you say about this midline ?[/quote I guess that would mean that EF is 1/2 line AB?
[quote]Diagonals of the quadrilateral BCDE cut each other in half at F. What can you say about this quadrilateral ?[/quote][/quote]
DE and CB are parallel to each other. Does that mean that DE and EA are congruent to each other. therefore EA and CB are parallel.
Thus, EABC is a parallelogram? Maybe this isn't going through my head right...

DE parallel to CB means also EA parallel to CB. How about that midline EF of $\triangle ABD$ ?

[color=red]Mods can move the topic to the apropriate section.[/color]

[quote="tobies"]
Given: E is the midpoint of segment DA, F is the midpoint of segment DB. Segment FC is congruent to segment EF.
Prove: EABC is a parallelogram.

[/quote]

 Well , to clear this topic and because every long  journey needs the first step,(following Yetti's hint) , we observe that:

[b]a)[/b] In triangle $DAB$ segment $EF$ connects two midpoints , so $EF \parallel AB$, that is $EC \parallel AB$.
[b]b)[/b] In quadrilateral $DEBC$ diagonals $EC,BD$ bisect each other, so it is a parallelogram. Hence $DEA \parallel BC$.But since $EC\parallel , AE\parallel BC$ , we take that $DEBC$ is a parallelogram.

  [b]NOTE[/b]

 As allready has been said above , send these questions to the high school section - basics.


I need to show if these "vectors" are linearly independent...
y1(x)=x, y2(x)=sin(x), y3(x)=e^x.
Besides Wronskian, how can I solve this?

Consider the equation $ c_1x \plus{} c_2e^x \plus{} c_3\sin x \equal{} 0.$

We would like to show that if that is true for all $ x,$ then $ c_1 \equal{} c_2 \equal{} c_3 \equal{} 0.$ One thing we could do is to find three useful points and plug them into this equation. I'd try $ 0,\pi,$ and $ \frac{\pi}2.$ That would give you three linear equations for the three unknowns $ c_1,c_2,$ and $ c_3.$


I only need help with Sheet 2 What do you call a message printed on a lion with chickenpox?

Please do a problem or two. I can figure it out if you do just show me how to do a problem.

1 cancel 2xy^3 to get 4x^2y=H
2 cancel, f, 4, x^2 to get 2x^2y^3/3y5=x^2/3y^2=D
5 cancel x-1 to get -6x^2y/4xy^3=-3x/2y^2=G

I'll do more later (or someone else will).

thanks. feel free to do some more if you would.


Dar mosallase ABC ,parekhathayi be toole AM=CN=p (p=nesfe mohit)ruye nimkhat haye AB va CB joda shode and.(B beine A va M, va beine C va N gharar darad.)farz konid K noghteyi ruye dayereye mohitie ABC va moghabel ghotrie B bashad.sabet konid amud vared az K bar MN az markaze dayereye mohatie mosallas migzarad.

khob kafie sabet konim $ KI\perp MN$,pas kafie sabet konim $ KM^2 - KN^2 = IM^2 - IN^2$.
hala biaid dotarafe ebarate akharo hesab konim,vaseye samte chapesh,deghat konid ke mosalase $ \triangle AKM$ ghaemozavias pas:

$ KM^2 = AK^2 + AM^2 = AK^2 + p^2$

hamchenin mosalase $ \triangle CKN$ ham ghaemozavias pas darim:

$ KN^2 = CK^2 + CN^2 = CK^2 + p^2$

pas darim:

$ KM^2 - KN^2 = AK^2 - CK^2 = \left((2R)^2 - AB^2\right) - \left((2R)^2 - BC^2\right) = a^2 - c^2$

khob hala biaid samte rastesho hesab konim,az $ I$ bar $ BC,AB$ amoodhaye $ IH,IH'$ ro rasm konid,hala too mosalase ghaemozavie $ \triangle IHN$ darim:

\begin{eqnarray*}IN^2 = IH^2 + HN^2 = r^2 + HN^2 = r^2 + \left(HB + BN\right)^2 = r^2 + \left(p - b + p - a\right)^2 = r^2 + c^2\end{eqnarray*}

be tarighe moshabeh natije mishe:

$ IM^2 = r^2 + a^2$

pas darim:

$ IM^2 - IN^2 = a^2 - c^2$

$ \Rightarrow KM^2 - KN^2 = IM^2 - IN^2$.


geometry
incenter
projective geometry
radical axis
power of a point
complex numbers
geometry proposed

Although this is trivial by pole-polars, let us try something different.

Let $Q$ be the projection of $O$ on $BE;$ second intesection of the circle $\odot(OMBK)$ with diameter $OB$ and the circle with diameter $OE.$ Now, $AC,MK,OQ$ are pairwise radical axes of $(O),\odot(OMBK)$ and $\odot(OQE)$ concurring at their radical center $P$ $\Longrightarrow$ $O,P,Q$ are collinear, or $OP \perp BE,$ as desired.

pole and polar property
PJ is the tangent to the incircle

This is my first problem I actually solved with polars, without help  :lol:

MK is the polar of B wrt O since BM and BK are tangents to the circle, and AP is the polar of E wrt O, since E lies on the circle and AP is tangent to it at E, so P is the intersection of the two polars. Thus, BE is the polar of P, as Luis and plane geometry said.

EDIT: Here is another proof
Let Q be the perpendicular from O to BE. Since $ \angle BMO \equal{} \angle BKO \equal{} \angle BQO \equal{} 90$, BKQOM is cyclic. Also, note that the circle with diamter OE passes through Q and is tangent to AC, since $ OE \perp AC$. Thus, the radical axis of these two circles and the incircle must be concurrent, so MK, AC, and OQ are concurrent, so $ PO \perp BE$.

[quote="SUPERMAN2"]The circle with the center $ O$ is inscribed in a triangle $ ABC$ and touches the sides $ AB,BC,CA$ at $ M,K,E$ respectively.Denote by $ P$ the intersection of $ MK$ and $ AC$.Prove that $ OP$ is perpendicular to $ BE$.

Can we solve this problem by another method such as harmonic division?[/quote]

Easy :D
Let $ AK\cap (O)\equal{}\{J\}$
$ JMKE$ is a harmonic quadrilateral then JP is the tangency of $ (O) \rightarrow$ QED

Thank you very much for your nice solutions.Complex number can also kill this problem easily but I still prefer the geometric method.

[quote="SUPERMAN2"]Let $ w \equal{} C(I,r)$ be the incircle of $ \triangle ABC$ and $ w$ touches the sides $ AB,BC,CA$ at $ M,K,E$

respectively . Denote by $ P$ the intersection of $ MK$ and $ AC$ . Prove that $ PI\perp BE$ .[/quote]

[u]Very  nice the [b]Dgreenb801[/b]'s proof[/u] !

I"ll present a simple proof without complex numbers, radical axis, harmonical division (pole, polar) :

$ PM\perp BI\Longleftrightarrow PB^2 \minus{} PI^2 \equal{} MB^2 \minus{} r^2\Longleftrightarrow$ $ PB^2 \minus{} \left(PE^2 \plus{} r^2\right) \equal{} MB^2 \minus{} r^2\Longleftrightarrow$

$ PB^2 \minus{} PE^2 \equal{} MB^2$ $ \Longleftrightarrow PB^2 \minus{} PE^2 \equal{} IB^2 \minus{} IE^2\Longleftrightarrow PI\perp BE$ .

See [url=http://www.mathlinks.ro/viewtopic.php?t=284253][color=red][b]here[/b][/color][/url] a nice extension of this problem.


trigonometry
geometry
circumcircle
parallelogram
geometry proposed

Hi !
 I've found a solution for this problem:

If in a triangle $ABC$, median $BD$ is such that $\widehat{A}=\widehat{DBC}$ and $\widehat{ADB}=45$ then $\widehat{A}= 30$.

A very nice figure : [url]http://agutie.homestead.com/files[/url] problem proposed n° 1.
My solution has a little bit of trigonometry and I would like to know if there is a more synthetic one ! :maybe:

Here $CB$ is tangent to circumcircle of $ADB, <DBC=<A$, denote $AD=DC=m$, then $BC^{2}=CD.CA=m(2m), BC=m\sqrt{2}$.Draw the altitude $CE$ of $BDC$, now $<EDC=<ECD=45^{o}, CE=ED= m \frac{\sqrt{2}}{2}$, further in the right triangle BEC we have $CE=\frac{BC}{2}$, hence $<EBC=<A=30^{o}$.

Muchas gracias juancarlos !

Here is mine solution :
I have prolonged $BD$ to make a parallelogram $ABCE$. Then I found that $\widehat{BCE}=45^{\circ}$ and I use the [b]sine law[/b] in the triangles $BCE$ and $ABD$. If we note $\widehat{BAC}=x$, $BD=DE=a$ and $AB=CE=b$ I have the relations : \[\frac{2a}{sin45^{\circ}}=\frac{b}{sinx}\] and \[\frac{a}{sinx}=\frac{b}{sin45^{\circ}}\] Eliminating $a$ and $b$ from these two relations (I've divided them), I've got $sinx=0.5$ so $x=30^{\circ}$. :roll:

[color=darkblue][b]Proof 1 (trigonometric).[/b] $m(\widehat{CBD})=A\ ,\ m(\widehat{ABD})=135-A\ ,\ C=45-A$ $\Longrightarrow$ $c\cdot\sin\widehat{ABD}=a\cdot\sin \widehat{CBD}$ $\Longleftrightarrow$
$\sin (45-A)\cdot\sin (45+A)=\sin^{2}A$ $\Longleftrightarrow$ $\cos 2A=2\sin^{2}A$ $\Longleftrightarrow$ $\cos 2A=\frac{1}{2}$ $\Longleftrightarrow$ $A=30\ .$

[b]An easy extension.[/b] Let $D$ be the middlepoint of the side $[AC]$ in the triangle $ABC$. Prove that $\boxed{\ m(\widehat{BDC})=B\Longleftrightarrow\ A< 90\ \ \wedge\ \ \sqrt 2\cdot\sin A=\sin B\ }\ .$
[b]Proof.[/b] Show easily that $A<B$ and $A<180-B$, i.e. $A<90$. Thus, $c\sin (B-A)=a\sin A$ $\Longleftrightarrow$ $\sin (B+A))\sin (B-A)=\sin^{2}A$ $\cos 2A-\cos 2B=1-\cos 2A$ $\Longleftrightarrow$ $\cos 2A=\cos^{2}B$ $\Longleftrightarrow$ $2\sin^{2}A=\sin^{2}B$ $\Longleftrightarrow$ $\sqrt 2\cdot\sin A=\sin B\ .$
[b]Remark.[/b] Similarly, $m({\widehat{BDA})=B}$ $\Longleftrightarrow$ $C<90\ \ \wedge\ \ \sqrt 2\cdot\sin C=\sin B\ .$[/color]


calculus
integration
calculus computations

It doesnt look tough but I simply cant find it
Can u help me out

$\int{x^{x}}$

Maybe this integral can't  be expressed in the form of basic function.

I wonder this calculation will be solved by the expansion of a series.

I was thinking the same thing
I proved every single method i know and I still cant sove it.

And it looks relatively simple for it not to have a solution

i used mathematica on my friend's computer and it was unable to solve this. so i am pretty sure it does not have an antiderivative that is an elementary function.

See also : http://mathworld.wolfram.com/LambertW-Function.html


[quote="maskman"]Let $x,y,z \ge 0$

Show that :

$(x^{\frac{1}{2}}+y^{\frac{1}{2}}+z^{\frac{1}{2}})(x+y+z) \le \sqrt 2 (x+y+z)^{\frac{3}{2}}+2(xyz)^{\frac{1}{2}}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=32399


Find all n integer that can divisor of  5.(n^11)-2.(n^5)-3n

5.(n^11)-2.(n^5)-3n

*factors*

n(5n^10 - 2n^5 - 3)
n(5n^5 + 3)(n^5 - 1)

That should help, unless I misread the problem.

Also, n^5 - 1 factors into (n - 1)(n^4 + n^3 + n^2 + n + 1), if that helps, hehe..
Not quite sure what the question is asking though.

Sorry...My problem is:How many integer can be find that divisor of
5n^11-2n^5-3n for all positive integer value of n?

I think what you really mean is,

"What is the set of positive integers which divide $5n^{11} - 2n^5 + 3n$ for all positive integers n?"

JBL: you've got wrong sign, it's -3n (if it's as you write then it's quite easy, cause for n=1 this expression has value 6 and it's jsut about checking 2,3,6) So problem is:

"What is the set of positive integers which divide $5n^{11} - 2n^5 - 3n $ for all positive integers n?"

Is this  supposed to be a serious question? Even if you interpret as the one given above, Isn't the answer    trivial ?

Is there someting I am missing?

The  expression is (n-1)n(n+1)(  rest)
so 1,2,3,6 will always  divide, and just looking at GCD of values of rest  when  n=2, and 3,   one can see that there are no more such  universal divisors... except  another 3 or 5

So you   the answer would be somethng like 1,2,3,6,9,18 . all factors of 90 . unless I missed something..

Gyan, you seem to confuse "trivial" and "solvable with a method that I know."  Perhaps this should be in an easier forum, but it's not "trivial," it just uses a well-known method.

But JBL you shouldn’t always trust the definition  given by Webster. :) According to  dictionary of  mathematicians,  anything that can be proven is “trivial? and mathematician can ONLY prove trivial theorems.   :)

 A conversation between two math graduate students:  (noted by    Feynman ) :

[quote]“ I still remember a guy sitting on the couch, thinking very hard, and another guy standing in front of him, saying, "'nd therefore such-and-such is true."
"'Why is that?" the guy on the couch asks.
"It's trivial! It's trivial!" the standing guy says, and he rapidly reels off a  series of logical steps:
"First you assume thus-and-so, then we have Kerchoff's this -and-that; then there's Waffenstoffer's Theorem, and we
substitute this and construct that. Now you put the vector which goes  around here and then thus-and-so..." The guy on the couch is struggling to understand all this stuff, which goes on at high speed for about fifteen minutes!
Then the second guy agrees and says,
[size=150]"yeah, yeah, it is trivial[/size]" ?[/quote] :)  :)

Okay Okay .. Here is the short reasoning:

[hide] n* (5(n^6-1) + 3 (n^4-1)
or
  Mod 2:
Eiter  n == 0 or   n^6 == n^4 == 1
Mod   5, eiterr n == 0 or n^4 == 1
Mod  9  , either n == 0 or n^6 == 1

In all cases the expression is divsible by 90,.
(And direct  calculation, as pointed out in the previous message,  shows  that we are not missing anythig else)[/hide]

Now all should agee that its trivial .. :D


23. Melissa is driving a sports utility vehicle along the interstate at a constant speed of 55mph. A sports car one-half mile behind her that is moving at a constant speed passes her in 60 seconds. How many miles per hour is the speed of the sports car?

18. Given that $ 9a^2\minus{}8b^2\equal{}1800$ and $ \frac{a}{b}\equal{}\frac{4}{3}$, what is the value of the product $ ab$?

[quote="Bomboyage"]23. Melissa is driving a sports utility vehicle along the interstate at a constant speed of 55mph. A sports car one-half mile behind her that is moving at a constant speed passes her in 60 seconds. How many miles per hour is the speed of the sports car?

18. Given that $ 9a^2 \minus{} 8b^2 \equal{} 1800$ and $ \frac {a}{b} \equal{} \frac {4}{3}$, what is the value of the product $ ab$?[/quote]

[hide]
23.

The sports car gains 1/2 a mile on Melissa in 60 seconds. So, 1/2 a mile in one minute. multiply both sides by 60 u get 30 miles in an hour.

So the car is 30 mph faster so the sports car goes 85 mph.[/hide]

[hide]

with 2nd equation, a/b=4/3

so 3a=4b so a= (4/3)b

take first equation

9a^2-8b^2=1800

substitute.

9(4/3)b)^2-8b^2=1800

9(16b^2/9)-8b^2=1800

(144b^2)/9 - 8b^2=1800

16b^2-8b^2=1800
8b^2=1800

b^2=225
b=15

substitute into first equation:
a/15=4/3

60=3a

a=20

so 20*15=300[/hide]


geometry
convex quadrilateral
circle
parallel
diameter
IMO
IMO 1984

Let $ABCD$ be a convex quadrilateral with the line $CD$ being tangent to the circle on diameter $AB$. Prove that the line $AB$ is tangent to the circle on diameter $CD$ if and only if the lines $BC$ and $AD$ are parallel.

Let $M,N$ be the midpoints of $AB,CD$ respectively, and let $P,Q$ be the feet of the perpendiculars from $M,N$ to $CD,AB$.

We know that $MP=MA$, and $AB$ is tangent to the circle with diameter $CD$ iff $NQ=NC$, which is equivalent to the triangles $AMN,MNC$ having equal areas $(*)$. Since $AM=MB,\ NAM,NMB$ always have equal areas, so $(*)$ is, in turn, equivalent to $NMC,NMB$ having equal areas, i.e. $MN\|BC$ (because $B,C$ lie on the same side of the line $MN$), and this is just another way of saying that $BC\|AD$.

[quote="grobber"]
We know that $MP=MA$, and $AB$ is tangent to the circle with diameter $CD$ iff $NQ=NC$, which is equivalent to the triangles $AMN,MNC$ having equal areas $(*)$.[/quote]

Sorry for reviving such an old thread, but I don't understand why that is equivalent to the triangles $AMN,MNC$ having equal areas.

Thank you.

[hide="My solution"]
suppose that $O$ and $O'$ are the midpoints of $AB$ and $CD$. Let the circles with the diameter $AB$ and $CD$ be $w_1$ and $w_2$.
And $CD$ and $AB$ be tangent to $w_1$ and $w_2$ at points $P$ and $Q$.
Obviously quadrilateral $OO'PQ$ is concylic; So $\hat{PO'Q}=\hat{POQ}$ but: $\hat{PO'Q}=2\hat{PCQ}$ and $\hat{POQ}=2\hat{PBQ}$
Hence, quadrilateral $BQPC$ is concylic and by the same way quadrilateral $ADPQ$ is also concylic. it follows that: $\hat {BCD}=\hat{AQP}=\pi -\hat{ADC}$ and the conclusion follows immediately.
[/hide]

[hide="for the the other side"] its enough to consider the point $Q$ on line $AB$ such that $ADPQ$ is concyclic.[/hide]

Assume that line $AB$ is tangent to the circle on diameter $CD$ and let the intersection point be $E$; similarly, define the intersection of line $CD$ and the circle on diameter $AB$ to the $F$.  Also, let $O_1$ be the midpoint of $AB$ and $O_2$ be the midpoint of $CD$.  By the definition of tangent lines, we have $\angle CFO_1=90^{\circ}$ and $\angle BEO_2=90^{\circ}.$

[b]Claim 1:[/b] $EFO_2O_1$ is cyclic.
[i]Proof:[/i] Note that $\angle O_2FO_1=90^{\circ}=\angle O_2EO_1$. $\Box$

By Claim 1, we have $\angle FO_2E=\angle FO_1E$, which implies that the supplement angles are equal or $\angle CO_2E=\angle BO_1F. $

[b]Claim 2:[/b] $ADFE$ is cyclic.
[i]Proof:[/i] Because a central angle equals twice an inscribed angle, we have $\angle CO_2E=2\angle CDE$ and $\angle BO_1F=2\angle BAF.$  But, $\angle CO_2E=\angle BO_1F $ so $$\angle FDE=\angle CDE=\angle BAF=\angle EAF,$$ proving the desired. $\Box$

[b]Claim 3:[/b] $BCFE$ is cyclic.
[i]Proof:[/i] This is proven similarly like Claim 2.$\Box$

By Claim 2 and 3, we have $$\angle DAB=\angle DAE=180^{\circ}-\angle DFE=\angle CFE=180^{\circ}-\angle CBE=180^{\circ}-\angle CBA,$$ proving $AD\parallel BC$. $\blacksquare$

Now, assume that $AD\parallel BC$.  Let $O_1, O_2$ be the midpoint of $AB, CD$ respectively and let $E$ be the perpendicular from $O_2$ to line $AB$ and $F$ be the point of tangency of the circle with diameter $AB$ and line $CD$.

[b]Claim 4:[/b] $EFO_2O_1$ is cyclic.
[i]Proof:[/i] By construction, $\angle O_2FO_1=90^{\circ}=\angle O_2EP_1$. $\Box$

[b]Claim 5:[/b] $ADFE$ is cyclic.
[i]Proof:[/i] Since $AD\parallel BC$, $ABCD$ is a trapezoid with midline $O_1O_2$. Thus, $\angle O_2O_1B=\angle DAE.$  By Claim 4, $\angle EFO_2=\angle O_2O_1B.$  Hence, $$\angle DAE=\angle O_2O_1B=\angle O_2FE,$$ proving the desired. $\Box$

By Claim 5, $\angle FAO_1=\angle O_2DE.$  By Claim 4, $FO_1A=\angle DO_2E$.  By AA similarity, $\triangle O_2DE\sim \triangle O_1AF$.  As $\triangle O_1AF$ is an isosceles triangle with base $AF$, $\triangle O_2DE$ must be an isosceles triangle with base $DE$.  Hence, $DO_2=EO_2$ so $E$ lies on the circle with diameter $CD$.  Thus, the line $AB$ is tangent to the circle with diameter $CD$ as $O_2E\perp AB$. $\blacksquare$


Αν $ f: N^*\to N^*, f(f(n))\le an\plus{}bf(n)$ για κάθε $ n\in N^*,a>0,b>0,a\plus{}b\equal{}1$βρείτε τον τύπο της $ f(n)$
Λύση σε λίγες μέρες

[quote="r_boris"]Αν $ f: N^*\to N^*, f(f(n))\le an \plus{} bf(n)$ για κάθε $ n\in N^*,a > 0,b > 0,a \plus{} b \equal{} 1$βρείτε τον τύπο της $ f(n)$
Λύση σε λίγες μέρες[/quote]
Συγνώμη ξέχασα να πω ότι η $ f$ είναι $ 1\minus{}1$ στην προηγούμενη άσκηση .

[quote="r_boris"]Αν $ f: N^*\to N^*$ συνάρτηση $ 1\minus{}1$, $ f(f(n))\le an \plus{} bf(n)$ για κάθε $ n\in N^*,a > 0,b > 0,a \plus{} b \equal{} 1$βρείτε τον τύπο της $ f(n)$ [/quote]

[b][u]ΛΥΣΗ[/u][/b]

Από την υπόθεση έχουμε: $ f(f(n))\minus{}f(n)\leq \minus{}a(f(n)\minus{}n)$

Αν $ f(n)>n$ $ \forall n \in \mathbb{N}$ τότε $ LHS>0$ και $ RHS<0$, [u]ΆΤΟΠΟ.[/u]

Συνεπώς $ f(n)\leq n$ $ \forall n\in \mathbb{N}$ και αφού η $ f$ είναι $ 1\minus{}1$, συμπεραίνουμε ότι $ f(n)\equal{}n,$  $ \forall n \in \mathbb{N}$

Nίκος Ράπανος

[quote="r_boris"][quote="Nick Rapanos"][quote="r_boris"]Αν $ f: N^*\to N^*$ συνάρτηση $ 1 \minus{} 1$, $ f(f(n))\le an \plus{} bf(n)$ για κάθε $ n\in N^*,a > 0,b > 0,a \plus{} b \equal{} 1$βρείτε τον τύπο της $ f(n)$ [/quote]

[b][u]ΛΥΣΗ[/u][/b]

Από την υπόθεση έχουμε: $ f(f(n)) \minus{} f(n)\leq \minus{} a(f(n) \minus{} n)$

Αν $ f(n) > n$ $ \forall n \in \mathbb{N}$ τότε $ LHS > 0$ και $ RHS < 0$, [u]ΆΤΟΠΟ.[/u]

Συνεπώς $ f(n)\leq n$ $ \forall n\in \mathbb{N}$ και αφού η $ f$ είναι $ 1 \minus{} 1$, συμπεραίνουμε ότι $ f(n) \equal{} n,$  $ \forall n \in \mathbb{N}$

Nίκος Ράπανος[/quote]
NIck μήπως μετά το ΑΤΟΠΟ πρέπει να πεις ότι : υπάρχει φυσικός $ k$ τέτοιος ώστε $ f(k)\le k$και όχι $ \forall n \in \mathbb{N}$?[/quote]

Nai exete dikio- auta pathainei kaneis otan grafei apeutheias sto pc kai de to skeftetai ligo me xarti kai molyvi.

Έστω ότι υπάρχει $ n_0$ ώστε $ f(n_0) < n_0$. Θεωρούμε την ακολουθία $ x_{n}$ με $ x_0 \equal{} n_0, x_{n \plus{} 1} \equal{} f(x_{n})$. Τότε από τη δοθείσα ανισότητα έχουμε $ x_{n \plus{} 2} \leq ax_{n} \plus{} bx_{n \plus{} 1}$. Tώρα αφού $ x_1 \equal{} f(n_0) < n_0$ παίρνουμε επαγωγικά $ x_{n} < n_0, \forall n \in N$. Όμως οι όροι της ακολουθίας είναι διαφορετικοί μεταξύ τους. Πράγματι αν $ x_k \equal{} x_l$ με $ k > l$ τότε $ f(x_{k \minus{} 1}) \equal{} f(x_{l \minus{} 1})$, άρα $ x_{k \minus{} 1} \equal{} x_{l \minus{} 1}$ και τελικά $ x_{k \minus{} l} \equal{} x_0 \equal{} n_0$, που είναι άτοπο αφού $ x_{k \minus{} l} < n_0$. Συνεπώς υπάρχουν άπειροι διαφορετικοί θετικοί ακέραιοι (οι όροι της ακολουθίας $ x_n$) μικρότεροι από $ n_0$, που είναι αδύνατο. Άρα $ f(n) \geq n, \forall n \in N$, απ' όπου εύκολα παίρνουμε ότι $ f(n) \equal{} n$.

Η απόδειξη που σκέφτηκα είναι παρόμοια με του Savva ,αν και λίγο διαφέρει ως προς την παρουσίαση της και είναι στο attachment

r_boris σχετικά με το θέμα που μου είχες πει για τον ελάχιστο όγκο του,που πρέπει να πάρω την ανισότητα του αριθμητικού και γεωμετρικού μέσου το έκανα και μου βγήκε ό ελάχιστος όγκος V=9/2αbc
όμως πρέπει να προσδιορίσω και τα α, b,c συναρτήσει του σημείου (x1,ψ1,z1) που διέρχεται το επίπεδο
μήπως υπάρχει τρόπος να τα προσδιορίσω
εγώ το έκανα με δεσμευμένα ακρότατα, έβγαλα τον ίδιο όγκο
και τα α=3χ1, b=3ψ1 και c=3z1
αλλά ήθελα να δω αν υπάρχει πιο εύκολος τρόπος,αν δεν σου είναι κόπος κοίταξε το.
Σ΄ευχαριστώ!


algebra
polynomial
parameterization
function
complex numbers
Galois Theory

Hi, I don't know which section this problem should go, it looks like algebra, although it also has complex numbers. Here it is:

I have a polynomial $ p(z,n,r)$ that contains a parameters $ n$ and $ r$. Here, $ n$ is a natural number, $ r$ is a positive real number. For each $ n$, as $ r$ changes, the roots (including complex roots) of $ p$ would change continuously.

In my particular $ p(z,n,r)$, the roots begin on the left half of the complex plane, and some of them cross the imaginary axis as $ r$ increase. For each $ n$, I call the value $ r$ that causes the root to sit on the imaginary axis $ r_{max}(n)$, such that whenever $ 0 < r < r_{max}$, the roots of $ p(z,n,r)$ is still on the OLHP.

My goal is to analytically solve for $ r_{max}(n)$ as a function of $ n$. I say analytically because I have already experimentally obtained the values, and have analytically solved for $ r_{max}(n)$ if $ n$ is known. (i.e. If you ask me what is $ r_{max}(197)$, I could run some equations and return a correct answer, without iteratively trying different $ r$ until a root first hit the imaginary axis.)

What I have not been able to do, is to obtain the function $ r_{max}(n)$. I could plot the existing data, and it shows a rather smooth curve, but I have problem solving for the equation of that curve.

I was wondering how this kind of problem is commonly solved. I am assuming that I am not aware of a whole set of math tricks that would solve it.

The first equation I was stuck with:

$ p(z,n,r) \equal{} \frac {1}{z}((z \plus{} 1)^n \minus{} 1) \plus{} r$

$ r_{max}(n) \equal{}$ ?

At the moment, for each $ n$, I can solve for $ r_{max}$ because I know that at $ r_{max}$, $ z \equal{} ib$ is a solution, and I know that the real and imaginary parts of $ p(z,n,r)$ are both 0. I observe that $ r$ is only involved in the real part, therefore, I use the imaginary part to solve for $ b$, then plug in that $ b$ to solve for $ r_{max}$.

The problem of this method is that when I solve for $ b$, $ b$ is solved by solving roots of the coefficient polynomial of the imaginary part of $ p(z,n,r)$. I don't know how to express that subroutine as an algebraic operation. That is why although I could solve for $ r_{max}(n)$ for each $ n$, I cannot solve for the formula that is the curve $ r_{max}(m)$, where $ m$ is a real number and when $ m\equal{}n$, I get the meaningful values.

Any ideas?

- Thanks

[quote="Twinbee"]The problem of this method is that when I solve for $ b$, $ b$ is solved by solving roots of the coefficient polynomial of the imaginary part of $ p(z,n,r)$. I don't know how to express that subroutine as an algebraic operation. [/quote]
That's because, in general, this is impossible (depending on what you mean by "algebraic operation").

What are some ways to tell whether a solution has an algebraic expression? What are some examples where it can be proven that a relation has no closed form solution?

The keyword here is [url=http://en.wikipedia.org/wiki/Galois_theory]Galois theory[/url].  If you can compute the Galois group of a polynomial, then the roots of the polynomial can be expressed using radicals and arithmetic operations if and only if the group satisfies a technical condition called [url=http://en.wikipedia.org/wiki/Solvable_group]solvability[/url].

What this really tells you is that the whole program of using radicals to solve is hopeless beyond degree four equations.  (Note that there are other ways to express the solutions of certain polynomials besides radicals.)


Kindly see if my solution is valid.

Question:  Prove that $D_{x}x= 1$
$D_{x}x= lim_{h \rightarrow 0}= \frac{f(x+h)-f(x)}{h}$
$= lim_{h \rightarrow 0}= \frac{x+h-x}{h}$
$= 1$

Am I correct?


You are given a very large two dimensional grid where points are marked at intervals of $1$ centimeter.
You are asked to choose $3$ points on this grid at random.

What is the probability that the $3$ points form an equilateral triangle?

[hide="Answer"]Zero. An equilateral triangle can't exist on an integer lattice.[/hide]

[hide="my answer"]An equilateral triangle with width $x$ will have a height of $x*rt3/2$. Because of this, there cannot be an equilateral triangle with all integer vertices, so the probability is 0[/hide]


probability
function
geometry
real analysis
algebra unsolved
algebra

let $f: [0,1] \longrightarrow[0,1]$ be a dense but [i]not[/i] continuous function(not even in a single point) in $[0,1]$x$[0,1]$

Is the prabobality that a random point in $[0,1]$x$[0,1]$ belongs to $f$ Zero?

Something like this :

If $mn$ is a non-vertical segment in $[0,1]$x$[0,1]$ then

$f$ cuts the $mn$ in at least one point.

So, it automaticaly means that $f$ is discontinuous everywhere.

If I understand it right, only one point out of $a \times [0,1]$ is in your set, so the probability will be $0$.

But I already misunderstood the other problem...

What is the probability space in this problem? We can't speak about probability without it.

I was wondering if such a probablity is suppose to be bigger than Zero then $f$ must form something like AREA .Is it true ?

Let me try to say it in one dimenstional argument :

let $I=[0,1]$ and let $A \subset I$ be a set such that :

1.$A$ is dense in $I$ ,ie between any two distinct $a,b,\in I$ there is an element of $A$.
2.There is [i]NO[/i] interval like $t \subset I$, such that $t \subset A$.

Is the probabilty of random point in $I$ belonging to $A$, Zero?

Mr.Myth What do you mean ? a point either belongs to $f$ or not So :fool:
I tried to read this but
http://en.wikipedia.org/wiki/Measure_%28mathematics%29

EDIT: ah , never mind Its wrong , Let $A=[0,1]-Q$ and probabilty is one.

[quote="lomos_lupin"]let $I=[0,1]$ and let $A \subset I$ be a set such that :

1.$A$ is dense in $I$ ,ie between any two distinct $a,b,\in I$ there is an element of $A$.
2.There is [i]NO[/i] interval like $t \subset I$, such that $t \subset A$.

Is the probabilty of random point in $I$ belonging to $A$, Zero?
[/quote]
It depends on measure you define for the probability space. For example, take $A=R\setminus Q$ and define Lebesgue measure. Then the probability to choose point from $A$ on $[0,1]$ is 1.

It is not that simple to define probability ;)


Let  $ A,B,C\in M_{n}{\mathbb{(C)}}$  are Hermitian matrices such that  the system equation matrix :
$ \{\begin{array}{c}{1}A^3 = 9C^2 - 27(C - I_n) \
B^3 = 9A^2 - 27(A - I_n) \
C^3 = 9B^2 - 27(B - I_n)\end{array}$
Prove that system equation matrix have unique root $ A = B = C = 3I_n$
 Here $ I_n$ is $ nxn$ identity matrix

1)Let $ A,B,C\in M_{n}{\mathbb{(C)}}$ are Hermitian matrices such that the system equation matrix :
$ \left\{\begin{array}{c}{1}A^{5}- A^{4}+2A^2B=2I_n\ B^{5}- B^{4}+2B^2A=2I_n\C^{5}- C^{4}+2C^2A=2I_n \end{array}$
 Prove that $ A=B=C=I_n$
2)Let $ A,B,C\in M_{n}{\mathbb{(C)}}$ are Hermitian matrices and positive define such that the system equation matrix :
$ \left\{\begin{array}{c}{1}A^2+2009AB=2010B\B^2+2009BC=2010C \C^2+2009CA=2010A \end{array}$
 Prove that $ A=B=C=I_n$
...................... :P .............


sure...i totally agree, but this was an extremely pointless post...

you guys don't rule, I rule. I rule the whole universe.


solve these equations:
1-$Arc tg\frac{x+1}{x-1}+Arc tg\frac{x-1}{x}=k\pi+Arc tg(-7)$
2-$sin x.cos^2 x=1$ :)

does this equation have solution?
$*$)$-x^3+x-1=0$ :)
the second equation can solve with this equation $*$. :)

Normally any ecuation of the $n$ order has $n$ solutions. I`m guessing that you want to know if it has any real solutions.
Well, it has because it`s order is $3$ which is odd, so your ecuation has at least one real solution.

$Arc tg\frac{x+1}{x-1}+Arc tg\frac{x-1}{x}=k\pi+Arc tg(-7) \Longrightarrow \frac{x(x+1)+(x-1)^2}{1-x}=-7 \Longrightarrow x=2$.
$sinxcos^2x=1 \Longrightarrow sin(2x)cosx=2$ have not real solution.

thank you Rust. :)
thank you ZetaX, :)
How did you calculate the root of this equation?

Using Maple (one could do it by hand with Cardano's formula but I'm to foolish for this...).

[quote="ZetaX"]Using Maple (one could do it by hand with Cardano's formula...).[/quote]
What is cardano's formula? :)

It's how the formula to solve cubic equations sometimes is called: http://mathworld.wolfram.com/CubicFormula.html


number theory
prime numbers
combinatorics proposed
combinatorics

I'm going to assume taht you meant the product of two numbers equal the third one, since if it's not equal , then the problem is ridiculously trivial...

[hide]
Yes.

Let the set of positive integers colored by the ith color be A_i.

Define f(n):N->N to be:

1 if n = 1
and $ e_1 \plus{} e_2 \plus{} \ldots \plus{} e_k$ if $ n \equal{} p_1^{e_1}p_2^{e_2}\ldots p_k^{e_k}.$

Define g(n):N->N to be the exponent of the largest power of 2 that divides f(n).

Let A_1 be the set of positive integers n with g(n) = 0, let A_2 be the set of positive integers n with g(n) = 1, etc. (Except let A_2009 be the set of points with g(n) >= 2008.)

Note that if m,n are non-one positive integers then f(mn) = f(m)+f(n).

Assume that there are 3 numbers a,b,c with different colors such that ab=c. Then, clearly a,b,c are all non 1. Then f(c) = f(a)+f(b). a,b are different colors, so g(a) doesn't equal g(b). Then, wlog g(a) < g(b). Then, 2^g(a) divides both f(a) and f(b), so 2^g(a) divides f(c) = f(a)+f(b). Since g(a), g(b) are integers, g(a)+1 <= g(b). Thus, $ 2^{g(a) \plus{} 1}$ divides f(b). However, since 2^g(a) is the largest power of 2 that divides f(a), $ 2^{g(a) \plus{} 1}$ doesn't divide f(a) but divides f(b), so it doesn't divide f(a)+f(b) = f(c), so the largest power of 2 that divides f(c) is 2^f(a), so c is colored the same as a, a contradiction.[/hide]

Another solution.

First of all, we paint all prime numbers in arbitrary colors, such that each color must be used at least one (infinity) time(s). Finally, if $ n\equal{}p_1^{k_1}p_2^{k_2}...p_s^{k_s}$ and $ p_1<p_2<...<p_k$ are primes then $ n$ is painted in color of $ p_1$.

First consider the sets $S_1,S_2,......,S_{2009}$ such as all elements of a set $S_i$ are colored with a same color and are divisible by a fixed prime,where no element from a set $S_i$ is divisible by that prime.$WLOG$ consider $S_1$, suppose $S_1=\{kp_1\}$ for all $k\in\mathbb N$ where $p_1$ is a prime.Now so we don't have any problem with those triples whose at at least one number is divisible by $p_1$. Now similarly considering $p_2>p_1$ for set $S_2$ we don't have problem with triples such that among them at least one is divisible by $p_2$. Similarly carrying on this process upto $S_{2009}$ and taking prime $p_{2009}$ as maximum we don't have any problem with those primes factors. Now note no two number can't go to same set.Now fro other numbers doesn't belong to those $S_i$'s , just put all of them in $S_{2009}$.Clearly now also no two numbers go to same set. So we're done with $2009$ colors.

Let S1,S2...S2009 be the subsets of the colors.Let S1 be the subset with the smallest prime 2,S2 with the smallest prime 3,...S2008 with the smallest prime p2008 and S2009 with the smallest prime that is larger than p2008.This coloring obviosly satisfaies the conditions of the problem,so we are finished.

[quote=April]Can be colored the positive integers with 2009 colors if we know that each color paints infinitive integers and that we can not find three numbers colored by three different colors for which the product of two numbers equal to the third one?[/quote]

Let $c_i, p_i, 1\leq i \leq 2008,$ be the $i$'th color and prime, respectively, and $c_0$ be the other color.
Color $n = p_i^{k}, 1\leq i \leq 2008,$ with $c_i$ and all the rest $n$ with $c_0.$

@April, do you mean:

Can the positive integers be colored with 2009 colors? Conditions for the coloring are:
 + there are an infinite amount of integers with each color, and
 + for all positive integers $a$ and $b$, $a$, $b$, and $ab$ are colored with 3 different colors


[quote="hollandman"]For positive reals $ a_1,a_2,...,a_n$, prove that

$ (1 \plus{} \frac {a_1^2}{a_2})(1 \plus{} \frac {a_2^2}{a_3})...(1 \plus{} \frac {a_n^2}{a_1})\geq (1 \plus{} a_1)(1 \plus{} a_2)...(1 \plus{} a_n)$[/quote]

It's obvious true since $ \left( {1 \plus{} \frac{{a_1^2 }}{{a_2 }}} \right)\left( {1 \plus{} a_2 } \right) \ge \left( {1 \plus{} a_1 } \right)^2$. :)


could sumbody explain to me the power of a point theorem (of cyclic quads) ??

Given a point P and a circle, pass two lines through P that intersect the circle in points A and D and, respectively, B and C. Then AP·DP = BP·CP. The point P may lie either inside or outside the circle. The line through A and D (or that through B and C or both) may be tangent to the circle, in which case A and D coalesce into a single point. In all the cases, the theorem holds and is known as the Power of a Point Theorem.

When the point P is inside the circle, the theorem is also known as the Theorem of Intersecting Chords (or the Intersecting Chords Theorem) and has a beautiful interpretation. When the point P is outside the circle, the theorem becomes the Theorem of Intersecting Secants (or the Intersecting Secants Theorem.)

The proof is exactly the same in all three cases mentioned above. Since triangles ABP and CDP are similar, the following equality holds:

  AP/CP = BP/DP,

which is equivalent to the statement of the theorem: AP·DP = BP·CP.

The common value of the products is then depends only on P and the circle and is known as the Power of Point P with Respect to the Circle. Note that, when P lies outside the circle, its power equals the length of the square of the tangent from P to the circle.

Sometimes it is useful to employ signed segments. The convenience is in that it now it is possible to tell points inside the circle from the points outside the circle. The poser of a point inside the circle is negative, whereas that of a point outside the circle is positive. This is exactly what one obtains from the algebraic definition of the power of a point.

The theorem is reversible: Assume points A, B, C, and D are not collinear. Let P be the intersection of AD and BC such that AP·DP = BP·CP. Then the four points A, B, C, and D are concyclic. To see that draw a circle through, say, A, B, and C. Assume it intersects AP at D'. Then, as was shown above, AP·D'P = BP·CP, from which D = D'. (If, say, B and C coincide, draw the circle through A tangent to PB at B.)


1.How many polynomials are there of the form x^3-8x^2+cx+d such that c and d are real numbers and the 3 roots of the polynomial are distinct positive integers?

2. The sides of a triangle satisfy  a^2-a-2b-2c=0 and a+2b-2c+3=0 . Find the measure of the largest angle of the triangle.

[hide="1"]
Recall that the sum of the three roots must be $ 8$ from Vieta's formulas or expanding out a polynomial with three roots.  Let $ j,k,l$ be the three roots offered.  Since they are distinct, we can assume WLOG that $ j<k<l$.  Then the solutions for the roots are $ (1,2,5)$ and $ (1,3,4)$.  Each yields a distinct $ d$ value and a distinct $ c$ value so there are $ 2$ solutions.[/hide]


inequalities
geometry
ratio
Triangle
area of a triangle
IMO Shortlist

In a triangle $ ABC,$ choose any points $ K \in BC, L \in AC, M \in AB, N \in LM, R \in MK$ and $ F \in KL.$ If $ E_1, E_2, E_3, E_4, E_5, E_6$ and $ E$ denote the areas of the triangles $ AMR, CKR, BKF, ALF, BNM, CLN$ and $ ABC$ respectively, show that
\[ E \geq 8 \cdot \sqrt [6]{E_1 E_2 E_3 E_4 E_5 E_6}.
\]

This is very easy once you compute area ratios through side length ratios.  For example, we have $ [AMR]\equal{}\frac{MR}{MK}\cdot\frac{AM}{AB}\cdot\frac{BK}{BC}[ABC]$.  We will find the $ \frac{CK}{BC}$ factor in the expression for $ [ALF]$.  By AM-GM, $ \frac{1}{4}\ge\frac{BK}{BC}\cdot\frac{CK}{BC}$.  We similarly maximize the other factors, and the answer follows.

For ease of notation, let the points $D\in LM, E\in MK, F\in KL$

We have \[\frac{[AME]}{[ABC]}=\frac{[AME]}{[AMK]}\times\frac{[AMK]}{[ABK]}\times\frac{[ABK]}{[ABC]}=\frac{ME}{MK}\times\frac{AM}{AB}\times\frac{BK}{BC}.\] Hence, by AM-GM we have \[\sqrt[3]{\frac{[AME]}{[ABC]}}\le\frac{ME/MK+AM/AB+BK/BC}{3}.\]

Similarly \[\frac{[CKE]}{[ABC]}=\frac{KE}{MK}\times\frac{KC}{BC}\times\frac{MB}{AB}\] and by AM-GM \[\sqrt[3]{\frac{[CKE]}{[ABC]}}\le\frac{KE/MK+MB/AB+KC/BC}{3}.\]

Adding, we get \[\sqrt[3]{[AME]}+\sqrt[3]{[CKE]}\le\sqrt[3]{ABC}.\] We repeat the same procedure for the other pairs. If we denote the areas of the six smaller triangles as $A_1, A_2, \ldots, A_6$ we have $\sum \sqrt[3]{A_i}\le 3\sqrt[3]{[ABC]}.$

Applying AM-GM again we get \[\sum\sqrt[3]{A_i}\ge 6\left(\prod\sqrt[3]{A_i}\right)^{\frac{1}{6}}.\]

Hence $\sqrt[3]{[ABC]}\le 2\left(\prod\sqrt[3]{A_i}\right)^{1/6}.$ Cubing gives \[\left(\prod A_i\right)^{\frac{1}{6}}\le\frac{1}{8}[ABC].\] We are done.


hypochlorous acid and the OH- ion

i need this so i can figure out delta G, which is easy

do u have a link or do u know it

Why do you need to calculate the variation if free energy for that reaction?


A baseball league has nine teams. During the season, each of the nine
teams plays exactly three games with each of the other teams.
What is the total number of games played?

Each of 9 teams must play 8 games 3 times, which is 72*3=216. Each game had been counted twice, so dividing by 2 gives you $ \boxed{108}$


how many lines can be drawn from a given point in space such that the angles between any two fo them be equal

[hide]as said by professor Sharygin in the article, this was one of the intersting facts that he knows[/hide]

the answer

[hide]only $6$ lines are possible to draw[/hide]

What do you mean by space? In mathematics there are many different spaces and the solution depends on what space do you mean.

(I hope your space is $R^3 $;).
This problem was at the final stage at the Polish Olympiad year ago, it's equivalent to ask (easier question) how many (maximal number) two-sided, distinct cones in $R^3$ can be pairwise tangential. It occurs that the answer is : 6 longest diagonals in the dodecahedron.

sorry for the delay

i was on a trip

yes, as a space, i mean the usuall one and that means $R^3$

and usually when someone just say [b]space[/b] and there  is no definatin

it is $R^3$, isnt it(:?:) (sorry for not clarifing the pace)


Vorood Professor Farhat ro be MATHLINKS tabrik migam.
 :lol:  :)

[quote="Ashegh"]man niz be nobeye khod vorood e in professor e geran ghadr va geran vazn ro be in mahfel tabrik arz minamayam.

toooooooooooooooooooooooooooooooooooooooooppppppp.[/quote]
ashegh jan kheili mamnoon.

[quote="samanfarhat"][quote="jensen"]bah,che khoob.in site hame chish kamel bood vali hamvare ye chizi kam dasht amma hes mikonam dige kamel shode. :D[/quote]
kheili mokhlesim jensen.
khejalat nade. :blush:[/quote]
man fahghat haghighat ro goftam.


I would appreciate some help with this one.

[url=http://a.imagehost.org/view/0825/Sp_a_Sida_179_Geometri_4138][img]http://a.imagehost.org/t/0825/Sp_a_Sida_179_Geometri_4138.jpg[/img][/url]

or

http://a.imagehost.org/0825/Sp_a_Sida_179_Geometri_4138.jpg

That picture is a bit large for my screen.... Can you shrink it??

Edit: Yeah thanks, the url helps a lot...

[hide]
i think the area is a circle. try that.
[/hide]

[hide="2nd hint"]The area is a circle minus that 45º arc + a tiny circle around A.[/hide]


Your post is nonsense.  In particular, the second inequality isn't true.

:?: take a=1, b=c=0 and it's clearly it's a non-sense... maybe you should add some conditions...

Well, I don't know if it's true or not, but your counterexample doesn't work (the RHS is 0?)

:blush: obsviously, i'm wrong. i should've been more regardful...  :blush:

In fact,
\[ (a+b+c)^6-27(a^2+b^2+c^2)(ab+bc+ca)^2=(a^2+b^2+c^2+8(ab+bc+ca))(a^2+b^2+c^2-ab-bc-ca)^2  \]
The latter expresion is positive whenever $ab+bc+ca\geq 0$.

Nice.  Any motivation for that factorization?

Well, there's nothing that says the inequality applies only to positive reals, so we should look to factorize.

I tried factors of $ab+bc+ca$ and $a^2+b^2+c^2-ab-bc-ca$; the second one worked.

blahblahblah, cool! :ninja:
Try it :
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=78337&sid=88950a55260cf404dab00c6bf39b9331
This is for you. ;)

hi

i have a quite short soln through substitution

a+b+c=p
ab+bc+ca=q>=0

so lhs = p^6

rhs=27(p^2-2q)(q^2)

so the ineq reduces to
p^6 + 54q^3 >= 27 p^2 * q^2

p^6>=0, q^3>=0
applying AM>=GM to p^6,27q^3,27q^3  we get the result :)

[quote="arpit goel"][quote="arqady"]arpit goel, hi!
I meant it in accuracy! :lol:[/quote]
what do u mean??  :roll:[/quote]
I mean to,  that my proof exactly as your.

[quote="arqady"]$ ab \plus{} ac \plus{} bc\geq0.$ Prove that
     $ (a \plus{} b \plus{} c)^6\geq27(a^2 \plus{} b^2 \plus{} c^2)(ab \plus{} ac \plus{} bc)^2.$[/quote]

It obviously true by AM-GM

[quote="Mirus"][quote="arqady"]$ab+ac+bc\geq0.$ Prove that
     $(a+b+c)^6\geq27(a^2+b^2+c^2)(ab+ac+bc)^2.$[/quote]
I think it is obvious after expanding
Please ,someone,  expand it by computer, I am too lazy to expand it by hand   :D[/quote]

Surely , the OP meant an elegant way of doing that


calculus
integration
logarithms
real analysis
real analysis unsolved

Prove

$ \int_u^v \frac{\ln x}{(x\plus{}u)(x\plus{}v)}\, dx\equal{}\frac{\ln(uv)}{2(v\minus{}u)}\, \ln\left(\frac{(u\plus{}v)^2}{4uv}\right)$

Using the substitution $ x\equal{}\frac{uv}{t}$ we get that

$ I\equal{}\ln(uv)\int\limits_{u}^{v}\frac{dt}{(v\plus{}t)(t\plus{}u)}\minus{}I$,

and the result follows by a simple computation.

Yes, it works. You can also show $ \int_0^\infty \frac{\ln x}{(x\plus{}u)(x\plus{}v)}\, dx\equal{}\frac{\ln^2(u)\minus{}\ln^2(v)}{2\, (u\minus{}v)}$ with this substitution.

Next task: Show $ \int_0^1 \left(\frac{1}{a\plus{}1\minus{}x}\minus{}\frac{1}{a\plus{}x}\right)\, \ln x\, dx\equal{}\frac12 \ln^2\left(\frac{a\plus{}1}{a}\right)$

I have found a solution today. See this [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=185149]link[/url].


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $ G$ is a group of order $ n$ and let $ H$ be a unique subgroup of order $ p$, where $ p$ is a prime factor of $ n$. Show that $ H$ is contained in the center of $ G$. The center of $ G$ is the set of all elements in $ G$ which commutes with all elements in $ G$. Thank you for the solution before.

I think this is not true if p\neq 2.  The group S_3 is a mimimal order conterexample if we choose p=3.

For general p>2, D_p is a counterexample.


algebra
polynomial
calculus
derivative
Vieta
combinatorial geometry
real analysis

Ok, I will post the problem without indications and if nobody manages (I really think it will be the case) I will post the hints given by the examinator. The problem is actually a conjecture of Popoviciu.
     Suppose that $f$ is a complex polynomial of degree at most $n$ and with at least 2 distinct roots, then the polynomial $F(x)=f(x)f'(x)...f^{(n-1)}(x)$ has at least n+1 distinct roots.
     First hint: let $z_{1},...,z_{k}$ the distinct roots of $F$  and $m_{i}(l)$ the multiplicity of $z_{i}$ as root of $f^{(l)}$. Prove that there are polynomials of degree at most $k-1$ called $g_{1},...,g_{k}$ such that $g_{i}(l)=m_{i}(l)$ for $l$ between 0 and $n-1$ and also $g_{i}(n)=0$.

Very nice problem :) and a very strong hint :(. Here is my [hide="sketch of a solution"]
The sum of $r$-th powers of the roots of the $\ell$-th derivative of a polynomial of degree $n$ is a polynomial of degree $r$ in $\ell$ according to Vieta's formulae  ($r=1,\dots,n$). So, taking the first $k-1$ equations of this kind together with the trivial equation $\sum_{j}m_{j}(\ell)=n-\ell$, and solving them with respect to $m_{j}(\ell)$, we get the desired polynomials. (The determinant of the system is the Vandermonde determinant, so it is not $0$).
Now, if $z_{1}$ is a vertex of the convex hull of the roots of the original polynomial, we get the values $m,m-1,\dots,1,0,0,\dots,0$ for the multiplicity of the root $z_{1}$ with some $m<n$. It remains to note that you need a polynomial of degree at least $n$ to interpolate such a sequence.
[/hide]

oh i guess one month ago i post this problem in algebra section ,
http://www.mathlinks.ro/Forum/viewtopic.php?t=88050
I saw it in a really precious material due to Prasolov,there were many uncanny problem about polynomial in that book


geometry function algebra polynomial induction calculus integration

[quote="JSteinhardt"]On the other hand, that would have some people doing nothing.

4 sections seemed decent, as this allowed (at least TJ) to split into groups of 2-3 and work on each section in that manner.[/quote]

TJ pwned the power round at PUMaC.  I heard through the grapevine that they got something like 60 points, while the next team got around 30.

How did you guys do so well on this?  Any tips?  I think perhaps a problem for us was letting the stronger students immediately start working on the harder problems (they immediately understood all definitions).  Then we missed some points on the harder problems, because the were hard, and then missed points on the easier problems, because none of our best people were working on them.

We had people split into groups based on which sections appealed to them.

I believe our two Freshmen worked on part A, Mark and I worked on part C, Sarah, Bruce, and Jack worked on part D, and Freddy, Haitao, and Jonathan worked on part B.

This allowed us to gain more intuition specifically about the portion of the problem that we were asked about.

Haitao and I were consulted if people wanted to make sure that their proofs were rigorous, and in general if someone thought they were missing something then they asked someone from another section to help (in hindsight I should've looked at section D in the last 5 minutes when I was asked to instead of trying to write some random stuff for C3, as the techniques I was looking at for part C could be used to quite easily prove that $\ln(x)+c$ is not refineable).

Part of our team selection required people to take a $2$-hour test with $3$ olympiad-style problems, which not only selected people who were better at proofs but encouraged our top people to practice them more. We also hold what we refer to as "TJUSAMO" (Olympiad preparation) after school every Thursday, and have three members (me, Haitao, and Dan Li) enrolled in WOOT, though I would say that that is a relatively small part of our success as we have seen in the past that three people do not do too much to make up for an overall lack of experience on the team.

We also took into account team mandelbrot performance for team selection, and we held two practice power round contests with the 20 people we were sending to Princeton. Haitao and I gave feedback to our B team on how they could write better proofs, though I guess that doesn't have anything to do with our A team's performance. Also, the second time we did this, I stayed with the B team, and the topic was one of Haitao's weakest areas, so the team as a whole was forced to work on the problems without relying on me and Haitao.

However, to be completely honest, I am not sure why we did so much better than everyone else. I was fully expecting us to get owned by Saiem's team (with Alex and Tony) team on the Power Test and then hopefully make up for it on the Relay and Individual tests, but apparently we got owned on the Relays and at the same time owned on the Power. We certainly do not have anywhere near the best nor the highest quantity of extremely strong people on our team. Haitao, our best person, is still far below Alex, Arnav, Jeremy, or Tony, and I am the only other MOPper on the team. I think that we had just practiced so much that when we got the problems we naturally split up into effective groups, and the amount of time spent messing around was minimal (though there was a bit of it near the end). I think A was the only section we solved completely, but we got the first two parts in each other section, which allowed us to rack up a huge number of points (potentially 80ish though apparently some people's proofs were incorrect or something).

By the way, if you want the problems we used to help our team practice, it is all up at http://activities.tjhsst.edu/vmt/pages/tjusamo/tjusamo2k7/. I have been trying to publicize this site as I believe that it should not be limited to TJ students, though I am not sure how effective I have been, especially with the preponderance of other resources out there.

Hopefully that helped.

Actually we didn't [i]quite[/i] solve part A completely...We had some small errors in A3 and didn't get to fixing them.

I can remember clearly that we spent the first 5 minutes just staring at the definition. "A function is finitely 2-refinable if ..." and we were all like, "what?...?" The biggest part of power round is understanding the definition at the beginning of the paper. Obviously, if you don't know what the question asks, you can't solve it.

Second, you should have some conversation between groups and try to minimize redundency. Having all 10 people work on D3 does you 0.0001 good unless it's your last problem. However, at the same time, you shouldn't have only 1 person working on D3 because he/she might need help.

As one of the two freshman who took on section A, I'd say that the biggest thing to make sure you don't lose too many points on the easy problems is make sure that the weaker people are at least capable of doing part A. If you're unsure of this, it could be very beneficial to send one experienced person to section A to rack up those ~20 points.

Thirdly, make sure that your proofs are correct! Obviously, don't worry about this when you have 1 minute left as you should be organizing your papers, but you should have someone read each of the proofs before it gets marked as done.

And last, USE THE BLACKBOARD! I know that us two freshmen used the blackboard as scratch paper just because it's big, so Dan Li can write big and because there's like 6 blackboards there, so you can step back and look at what you've done and get the big picture in a more literal sense. The blackboard also duals as a way to keep track of who's done what. Its uses aren't limited to writing definitions and "WTF?" underneath.

Oh yeah organizing your papers with like 5 minutes left instead of 1 is a good idea as we just sort of randomly threw papers together when we had 5 seconds left before time was called.

We made the mistake of letting our great people work on section D first. They realized the problem with that in the middle, and began to work on other parts. For part A, we had LaRue and another pretty competent guy write the proofs. I hope we didn't get like a 30something. That'd be pretty lame. I thought we did pretty decent to be honest. I wasn't surprised when we got 2nd in Power to TJ. No amount of talent in a single person can upset a balanced team on power round. I think the power round says alot for the balance of the rest of a team.

Congrats to TJ for beating everyone by a mile in power. We wasted alot of points in Section B because we only had one competent guy on it, so eh. Some organization and better team selection is in order for our team. We probably had better top individuals than most teams, but some of our weaker members weren't exactly up to par.

[quote="Pakman2012"]Congrats to TJ for beating everyone by a mile in power. We wasted alot of points in Section B because we only had one competent guy on it, so eh. Some organization and better team selection is in order for our team. We probably had better top individuals than most teams, but some of our weaker members weren't exactly up to par.[/quote]

fortunately they didn't

it wouldn't be good to embarrass all the MOPpers like that

As Jacob mentioned, it really works a lot better to focus on one problem or type of problem the whole time. If I had been working on C the whole time, I would have had time to complete C3 and maybe C4. I also probably should have worked on the last B problem, since I figured it wouldn't be a particularly tricky linear independence thing, but the people working on it were like DJFSALK what are you talking about? :)

DJFSALK, huh

well that is certainly pretty tricky

Thanks for the great post, Steinhardt.

Vestavia is going as an actual school to HMMT, so we'll be MOPless, to say the least.

[quote="Xevarion"][quote="Elemennop"]B wasn't htat bad, they just wnated you to know about vandermonde matrices in order to show uniqueness of solutions which definitely warrants DJKFLJ.[/quote]
Actually I was hoping that somebody would prove it for polynomials with only real roots using induction on the degree, but I was unable to find such a proof myself (and I guess no one else did either). I thought maybe it would be possible to prove it for some reasonably large class of polynomials using integration, too, but again I couldn't get this to work myself...[/quote]

yea we were actually gonna do vandermonde matrices cause Saiem said so but by the time we got around to that part of the proof we didn't really have time so we had someone (jia?) just write up some BS induction in degree proof.

Well we were told before results were out that the 2nd highest score was a 30. Apparently that was wrong?

That's what I heard too, but I guess it was just a false rumor.

I think the most important single tip for relatively hard power rounds such as this one is to soak up all the easy points - and do so in as efficient a manner as possible. Sure, you should let the less experienced people do the easier problems, but make sure the solutions are good enough.

[quote="Altheman"][quote="mysmartmouth"]
TJ pwned the power round at PUMaC.  I heard through the grapevine that they got something like 60 points, while the next team got around 30.
[/quote]

wtf are you talking about... TJ got 60, then we, florida a, got 55, and carolina got 50....that is not that large of a difference...[/quote]

LOL, the guy who gave us our packet must have lied (not Nathan but one of the other tall guys...


I'm complete chemistry dummy, but I need organic formula (visual representation) of some element which will cause dehydration of a living cell. Basically I need organic substance that will initiate dehydration of the cell. Also i need something that will cause cell to gain amount of water. Well it sound more like bio question but i really need the formula and preferably organic one (non organic would be NaCl and we would be done) Also i don’t care od formula is only hypothetical. I just need it for my business management class. Im sure my professor has no idea about chemistry anyway.

Thanks a lot for help!

For something to be put into solution outside a cell that would dehydrate the cell, you simply need anything that can be dissolved to a high solute concentration. With that in mind, how about sugar? Sucrose, plain old table sugar. You can dissolve quite a prodigious amount of it in water. Of course, there are plenty of other highly soluble organic compounds that could be used to the same effect: ethanol, for instance. Or glycerol. Or ethylene glycol. Lots of ways to pickle something.

To go the other way (with something external to the cell), you just have to lower the solute strength. So, make that distilled water.

I like the idea of glycerol, if that would take of the water it will still give energy to the cell to keep it functional on low metabolism level.

But for reversing the process the cell will intake all required water to get enough for its operations, but what if i want to force it to take say 10% extra of required. so we would have 110ng instead if 100ng?

Thanks for help!