Let $ p$ be a prime number,greater than $ 3$ and $ k\equal{}[\frac{2p}{3}]$(the notation $ [x]$ denote the largest integer wich is not exceeding $ x$).prove that $ \sum^{k}_{i\equal{}1}$ is a multiple of $ p^2$

[quote="hanamvn"]Let $ p$ be a prime number,greater than $ 3$ and $ k \equal{} [\frac {2p}{3}]$(the notation $ [x]$ denote the largest integer wich is not exceeding $ x$).prove that $ \sum^{k}_{i \equal{} 1}$ is a multiple of $ p^2$[/quote]

$ \sum^{k}_{i \equal{} 1}$ of what ?

Let $ p$ be a prime number,greater than $ 3$ and $ k \equal{} [\frac {2p}{3}]$(the notation $ [x]$ denote the largest integer wich is not exceeding $ x$).prove that $ \sum^{k}_{i \equal{} 1}C^i_p$ is a multiple of $ p^2$

This problem is from PEN problems number A24 :wink:

Hello, Erken and all!
I have known that this is PEN A24. But in the PEN forum, we haven't any exact and simple solutions, example: the Darij's solution! I think that this isn't a hard problem and we can solve it more simply! I cann't understand the termes, example: "p-ardic"... Help me, please!
PEN A24 is in that link:
http://www.mathlinks.ro/viewtopic.php?t=150392


topology
abstract algebra
advanced fields
advanced fields unsolved

Show that $ \mathbb{R}P^3$ and $ SO(3)$ are diffeomorphic (using the quaternions)

If you can use the statement: Any homeomorphic three dimensional manifolds are also diffeomorphic, then I can do it quite easily.

We have that $ S^3$ is a double cover of $ O(3)$ via the unit quaternions. So use this map and mod out by the kernel to get $ S^3/\{\pm 1\}\cong RP^3$ (this is quotient is just identifying antipodal points) is homeomorphic to $ O(3)$. Thus they are diffeormorphic.

But how can we prove that $ S^3$ is a double cover of $ O(3)$?

I don't have time right now to go into depth, but the basic idea is to extend the idea that $ S^1\equal{}\{v\in\mathbb{C} : |v|\equal{}1\}$ to $ S^3\equal{}\{v\in\mathbb{H} : |v|\equal{}1\}$. Then just use the "inclusion" map from $ S^3\to O(3)$ and you see that $ v$ and $ \minus{}v$ are identified. That was my intuition when I wrote the first post, I haven't worked through the details, though.


[i]Την είδα σε ένα pdf πρόσφατα και μου άρεσε![/i]

Έστω δύο σημεία $ A,B$ στο επίπεδο σε απόσταση $ 1$. Βρείτε το μέγιστο μήκος μιας διαδρομής από το $ A$ στο $ B$, χρησιμοποιώντας το πολύ $ n$ τμήματα, με την ιδιότητα ότι κατά μήκος της διαδρομής , η απόσταση από το $ B$ διαρκώς μειώνεται.

Νίκο δεν είμαι σίγουρος, γι'αυτό αν είναι λάθος πες μου!

Το μεγαλύτερο τμήμα σε κάθε βήμα είναι πάντα $ <2$, και το μέγιστο είναι αυτό που ενώνει τα $ A$ και $ B$, αφού προεκταθεί κατά τμήμα ίσο και λίγο μικρότερο.
Στη συνέχεια, για κάθε βήμα φέρουμε τμήματα, λίγο μικρότερα από διαμέτρους του κύκλου με κέντρο το $ B$ και ακτίνα $ 1$.
Οπότε για τα πρώτα $ n\minus{}1$ η μέγιστη απόσταση θα είναι $ <2(n\minus{}1)$. Στο τελικό βήμα ενώνουμε το τελικό σημείο με το $ B$, που έχουν απόσταση $ <1$

Οπότε το μέγιστο μήκος της διαδρομής θα είναι: $ L_{max} < 2n \minus{} 1$

orfikoss den einai sosto, alla den boro na katalavo 100% ti skepsi sou gia na pos me sigouria pou einai to lathos. mallon ekei pou les gia proektaseis kai gia ligo mikrotero, mallon ekei  xalaei to pragma.

Έχει το πνεύμα της λύσης του Orfikos.....

Είναι εύκολο να καταλάβουμε ότι για να κατασκευάσουμε διαδρομή ($ n$ τμημάτων) θα έχει μάλλον τη μορφή $ 2n \minus{} 1$.

Από την περίπτωση $ n \equal{} 2$ καταλαβαίνουμε πολλά. Στη διαδρομή $ APB$ το σημείο $ P$ το πέρνουμε πάνω στον κύκλο κέντρου $ B$ και ακτίνας 1 , τέτοιο ώστε
$ AP \plus{} PB$ μέγιστο. Για $ P$ αυστηρά μέσα στον κύκλο ορίζουμε το σημείο τομής $ D$ μεταξύ κύκλου και $ AP$ .Από τριγωνική ισχύει πάντα $ AP \plus{} PB < AD \plus{} DB$. Έτσι το $ D$ τίνει στον ίδιο τον κύκλο. Προφανώς $ DB \equal{} 1$ οπότε και μέγιστη απόσταση όταν $ AD \equal{} 2$. Έτσι κατασκευάζουμε διαδρομή που να τίνει τη μέγιστη απόσταση $ 3$.

Έστω τώρα η  $ a_i$  η  $ i \minus{}$η  απόσταση από το $ B$. Από τον περιορισμό πρέπει να έχουμε μια θετική μονοτονική μειωμένος διαδοχή. Είναι προφανές ότι κάθε βήμα δν ξεπερνά το μήκος $ 2$ , ενώ μάλιστα το τελευταίο θα είναι αυστηρά μικρότερο του $ 1$. Τότε η μέγιστη απόσταση των αποστάσεων δν υπερβαίνει το
$ 2(n \minus{} 1) \plus{} 1 \equal{} 2n \minus{} 1$ , αλλά το ίδιο αποτελεί το μέγιστο αφού μπορούμε να κατασκευάσουμε τέτοια διαδρομή.
Πράγματι, $ AD \plus{} DA \plus{} AD \plus{} DA \plus{} .... \plus{} (AB or DB) \equal{} 2n \minus{} 1$.

Από νόμο συνημητόνων γινόταν τπτ?

[quote="wallace"]
Πράγματι, $ AD \plus{} DA \plus{} AD \plus{} DA \plus{} .... \plus{} (AB or DB) \equal{} 2n \minus{} 1$.
[/quote]

Απ'τα δεδομένα του προβλήματος "η απόσταση απ' το Β συνεχώς μειώσεται (δεν λέει μπορεί να είναι και ίσο). Αρα αυτή η διαδρομή δεν είναι δεκτή...

[quote="silouan"]Ρε παιδιά αυτό που ζητάει ο Nick είναι [u]μέγιστο [/u]  . Εσείς βρίσκεται απλά ένα upper bound  :wink:[/quote]

akrivos..

[quote="gpapargi"]Νικο να σου κανω μια διευκριστικη ερωτηση
Εννοεις ότι στo τελος κάθε βηματος πρεπει να ειμαστε πιο κοντα από ότι ειμαστε στο τελος του προηγουμενου βηματος (και αδιαφορουμε για τα ενδιαμεσα σημεια) η οτι  κάθε σημειο της διαδρομης πρεπει να είναι πιο κοντα απο κάθε προηγουμενο σημειο της διαδρομης;

Στην πρωτη περιπτωση μου βγαινει ότι τα συνολο των πιθανων διαδρομων είναι το [1, 2ν-1) που δεν εχει μεγιστο, ενώ στη δευτερη περιπτωση κατεληξα καπως βιαστικα σε τετραγωνικη ριζα του ν.

Μπορεις να δωσεις αυτή τη διευκρινηση για να μη μπλεκω τζαμπα;[/quote]

το δευτερο.... :P

[quote="gpapargi"]Οσοι εφτασαν στο 2ν-1 μαλλον καταλαβαν το πρωτο  :)

Το ριζα ν ειναι σωστο η υπαρχει καλυτερο;[/quote]

sosto giorgo! grapse ti lysh an theleis!

Λοιπόν το πρόβλημα είναι πολύ όμορφο και θα προσπαθήσω να παρουσιάσω τη λύση μου . Για να εξηγηθεί καλά αυτή η άσκηση θέλει αρκετό μπλα μπλα , το οποίο θα παραλείψω . Λοιπόν θα σας δώσω μια γεύση από τη $ n\equal{}3$ και μετά θα το γενικεύσω .
Έστω $ a_1,a_2,a_3$ τα τρία τμήματα  και έστω $ a_1\equal{}AP$ . Για να ισχύουν τα δεδομένα , η βέλτιστη θέση για το $ P$ είναι το να βλέπει την $ AB$ υπό ορθή γωνία (γιατί? :wink: ) .Αν τώρα $ a_2\equal{}PK$ και $ a_3\equal{}KB$ όπως και πριν η βέλτιστη θέση για το $ K$ είναι να βλέπει την $ PB$ υπό ορθή γωνία . Επομένως στη βέλτιστη θέση θα ισχύει $ a_1^2\plus{}a_2^2\plus{}a_3^2\equal{}1$ (1) (Πυθαγόρειο θεώρημα ). Τώρα για να πάμε στα 4 τμήματα αφαιρούμε ένα τμήμα (το τελευταίο) και προσθέτουμε δύο για τα οποιά το άθροισμα των αποστάσεςων μεγιστοποιείται δηλαδή θα βλέπει και το νέο σημείο υπό ορθή γωνία . Άρα βγάζουμε το $ a_3$ και βάζουμε τμήματα $ a_3',a_4$ για τα οποία ισχύει $ a_3'^2\plus{}a_4^2\equal{}a_3^2$ επομένως η (1) δίνει $ a_1^2\plus{}a_2^2\plus{}a_3'^2\plus{}a_4^2\equal{}1$
Οπότε για $ n$ τμήματα επαγωγικά θα ισχύει $ a_1^2\plus{}...\plus{}a_n^2\equal{}1$ . Το θέμα είναι τώρα πως πρέπει να τοποθετηθούν αυτά μεταξύ τους ώστε η διαδρομή που είναι η $ a_1\plus{}...\plus{}a_n$ να είναι μέγιστη καθ'ως οι ορθές γωνίες <<παίζουν>> .
Αυτό όμως είναι εύκολο γιατί από BCS $ n(a_1^2\plus{}...\plus{}a_n^2)\geq (a_1\plus{}...\plus{}a_n)^2$ από όπου
$ a_1\plus{}...\plus{}a_n\leq \sqrt{n}$  άρα η μέγιστη διαδρομή είναι $ \sqrt{n}$ και επιτυγχάνεται με την παραπάνω κατασκευή και επιπλέον να ισχύει ότι $ a_1\equal{}a_2\equal{}...\equal{}a_n$   qed  :)

[quote="gpapargi"]Το σχήμα στο οποίο κατέληξα είναι σα σαλιγκάρι και είναι αυτό που φτιάχνουμε για να κατασκευάσουμε τη ρίζα του 2, 3 …ν, μόνο που έχει φάει μια σμίκρυνση κατά παράγοντα ν. Έτσι κάθε ευθύγραμμο τμήμα είναι ρίζα(ν)/ν και όλα μαζί ρίζα(ν).

Την απόδειξη για το ότι πληρείται η προϋπόθεση, «όσο πάει πλησιάζουμε» και ότι τα μήκος είναι  μέγιστο μήκος θα τη στείλω δευτέρα.

Με τη ευκαιρία πως βάζεις latex μέσα; Τώρα που έμπλεξα μαζί σας λέω να το μάθω.  :)
Επίσης τι πρόγραμμα μπορώ να βρω εύκολα για σχήματα;[/quote]

giorgo gia to latex grafeis tis math mode entoles mesa se duo dollaria $ \$$. boreis na vreis ena topic pou uparxei sxetika me to latex sta stickies (de thymamai pou akrivos) i kai na gogolareis ligo (px. latex math symbols etc). gia ta sxhmata boreis na xrisimopoihseis to sketchpad i to geogebra.

silouan oraios...stin episimi lysh vevaia parousiazetai enas polu omorfos kai formal tropos...isos tin anevaso tis apomenes meres ego i kai o dimitris me ton opoio ti syzhthsame sto msn.

Νίκο σου γράφω τη σκέψη μου

Προτιμώ να βλέπω το πρόβλημα λίγο διαφορετικά γιατί κάτι τέτοιο δίνει καλύτερη εποπτεία.

Αντί να ξεκινήσω από το Α και να πάω στο Β έτσι ώστε κάθε στιγμή να βρίσκομαι πιο κοντά από ότι την προηγούμενη, μπορώ να θεωρήσω ότι ξεκινάω από το Β και πάω στο Α έτσι ώστε κάθε στιγμή να είμαι πιο μακριά από το Β από ότι την προηγούμενη.

Επίσης μπορεί να σκεφτεί κανείς ότι δε χρειάζεται να φτάσεις στο Β. Μπορείς να φτάσεις σε οποιοδήποτε σημείο απέχει 1 και στη συνέχεια να περιστρέψεις το σχήμα γύρω από το Β έτσι ώστε το τέλος της διαδρομής (που απέχει 1 από το Β) να πέσει πάνω στο Α.

Μια ακόμα σκέψη απλοποίησης που μπορείς να κάνεις είναι ότι δε χρειάζεται να φτάσεις σε απόσταση 1. Αν βηματίζοντας φτάσεις σε απόσταση L από το Β τότε μπορείς πολύ απλά να κάνεις μια σμίκρυνση στο σχήμα τόσο όσο το L να γίνει 1. Δηλαδή να βρεις το όμοιο του σε κλίμακα 1/L.

[b]Αυτό που είναι σημαντικό είναι το να βηματίζεις κάθε φορά με τέτοιο τρόπο ώστε να μεγιστοποιείς το λόγο διανυόμενο μήκος (άθροισμα βημάτων) προς απόσταση από του τελικού σημείου από το Β. Αυτή είναι και η σκέψη κλειδί της άσκησης. Ο λόγος που ισχύει αυτό είναι ότι σε όποιο απόσταση από το Β και αν φτάσουμε τελικά θα διαιρέσουμε με αυτή για να αναχθούμε σε απόσταση 1 και νε καταλήξουμε στο όμοιο σχήμα με απόσταση 1 από το Β).[/b]

Θα αποδείξουμε λοιπόν ότι το σχήμα που θα προκύψει είναι ο σαλίγκαρος που φτιάχνουμε για να σχεδιάσουμε τη ρίζα του ν. Η απόδειξη θα γίνει επαγωγικά.

Έστω ότι κάνουμε το πρώτο βήμα απόστασης κ από το Β και καταλήγουμε στο Β1. Θέλουμε να βρούμε το σημείο Β2 ώστε να μεγιστοποιεί το λόγο άθροισμα βημάτων προς το τμήμα ΒΒ2.
Για να εντοπιστεί το βέλτιστο Β καταρχήν πρέπει να αποφασίσουμε προς ποια διεύθυνση θα κινηθούμε δηλαδή ποια είναι η γωνία Β1. Μπορεί να αποδειχτεί ότι θα είναι ορθή. Αν είναι οξεία τότε φέρνοντας το ύψος ΒΛ τότε κατά τη απόσταση Β1Λ δεν αποκρινόμαστε από το Β άρα  δεν επαληθεύεται η βασική προϋπόθεση της άσκησης. Αν η Β1 είναι αμβλεία τότε δεν μεγιστοποιούμε την καλυπτόμενη διαδρομή (φέρνεις και το ορθογώνιο με την τρίτη κορυφή να απέχει το ίδιο από το Β και έχεις 2 τρίγωνα με ίδιες 2 πλευρές, οπότε τη μεγαλύτερη τρίτη πλευρά την έχει αυτό που έχει τη μεγαλύτερη περιεχόμενη γωνία).

Άρα η σωστή κατεύθυνση είναι προς την ορθή γωνία. Μένει να βρούμε την κατάλληλη απόσταση που μεγιστοποιεί το λόγο άθροισμα βημάτων προς απόσταση από το Β.
Έστω x η απόσταση. Άρα θέλουμε να βρούμε το x που μεγιστοποιεί τη συνάρτηση:
f(x) = (κ+x)/ρίζα (κ^2+x^2)

Μετά από παραγώγιση προκύπτει ότι το ζητούμενο x είναι το x=κ.

Πάμε τώρα στο επαγωγικό βήμα. Μετά από ν βήματα έχουμε διανύσει απόσταση ν*x ενώ η απόσταση από το Β είναι κ* ρίζα(ν). Βλέπουμε ταυτόχρονα και το σχήμα με τον σαλίγκαρο. Εύκολα αποδεικνύουμε ότι το καλύτερο βήμα (που μεγιστοποιεί το λόγο) είναι πάλι απόσταση κ στην κάθετο και πάμε σε απόσταση ρίζα(ν+1).

Την αναλυτική απόδειξη του επαγωγικού βήματος την παραλείπω, αλλά αν κάποιος τη θέλει είναι διαθέσιμη (είναι ίδιο σκεπτικό με τη περίπτωση για ν=2).

Για να κανονικοποιήσουμε διαιρούμε όλες τις αποστάσεις του σχήματος με κ*ριζα(ν). Έτσι στο νέο σχήμα η απόσταση από το Β είναι 1 και κάθε βήμα μας έχει μήκος 1/ρίζα(ν).

Κάνουμε και μια περιστροφή γύρω από το Β έτσι ώστε το τελικό σημείο μετά από ν βήματα να είναι το Α και βρήκαμε τη λύση.

Ζητάω συγνώμη για τη μη χρήση latex (υπόσχομαι να είναι η τελευταία φορά) αλλά για την ώρα είμαι λίγο πνιγμένος.

Νίκο ρίξε μια ματιά και πες μου αν είναι όλα κατανοητά. Η σκέψη κλειδί ήταν να βρεθεί ο λόγος που πρέπει να μεγιστοποιηθεί ώστε μετά που θα καταλήξουμε στο όμοιο σχήμα υπό κλίμακα να έχουμε το ζητούμενο.

Πολύ ωραία Γιώργο!


This is the first day of the Iberoamerican Mathemathical Olympiad, later I'll translate the second.

Daniel


Consideramos los $2004$ números enteros $n$, desde $1$ hasta $2004$. Determinar para cuántos de estos valores de $n$ se verifica que el número $n^3+3^n$ es múltiplo de $5$.

disculpa que ahora no tenga latex es que ando de viaje, pero se me ocurrio una idea no se si esta bien pero me parece que obsevando en mod 5, vemos que los residuos de n^3 en mod 5 son {1, 3, 2, 4, 0} para los valores que tome n bien sea desde la forma 5k+1, 5k+2, 5k+3, 5k+4 o 5k... Ahora bien los 3^n, dejan en mod 5 los siguientes restos y se repiten {3, 4, 2, 1}, bueno para que n^3+3^n sea congruente con 0 mod 5, entonces podemos ver cuales residuos sumen 5 o multiplo de 5 es decir ver las listas

1 3 2 [color=red]4[/color] 0 [color=red]1[/color] [color=red]3[/color] 2 4 0 1 3 [color=red]2[/color] 4 0 1 3 2 4 0 ...
3 4 2 [color=red]1[/color] 3 [color=red]4[/color] [color=red]2[/color] 1 3 4 2 1 [color=red]3[/color] 4 2 1 3 4 2 1 ...

y ver en vertical cuales sumas son 5, como [4,5] es 20 seria ver cuantos n hay en los primeros 20 que lo cumplan y multiplicarlo por 2004/20.., en los primeros 20 hay 4 ns (ver los ke estan en rojo) que lo cumplen y la lista se sigue repitiendo de 20 en 20....entonces hay 404 n que lo cumplen. si no me ekivoco..

cercita cercita.... fijate de nuevo porque estas ahi

:blush:
creo que me equivoque contando, hay 4 por cada 20, lo que significa que hay 400 en los primeros 2000...y me quedan 4 lo que suma uno mas ..asi que habrian 401 n's con tal propiedad.


To clarify any misunderstandings caused by yaofan's post...

Order of operations.
First, you solve (-1)^11 because it is in parenthesis, plus it's an exponent, which are the two first operations.
(-1)^11=-1

Then, you solve the contents of the braces, which are these [].
1--1=1+1=2

Finally, you finish with the exponent on the very outside of the problem.
Evaluate and there you go.
2^2=4

4 is your answer.


LaTeX
trigonometry
real analysis
real analysis unsolved

Prove that $\displaystyle \sum_{k=1}^{n}\frac{sin(2k-1)x}{2k-1}>0$ for every $x\in(0,\pi)$

That's a $\LaTeX$ mistake you're reading. The $n$ is intended to be the upper limit of the sum.

This is a partial sum of the Fourier series of a square wave. Fixing what Moubinool wrote in #2,

$\sum_{k=1}^{\infty}\frac{\sin((2k-1)x)}{2k-1}=\frac{\pi}4$ for $0 < x < \pi$, extended to be odd and periodic of period $2\pi$. (So that the value is $-\frac{\pi}4$ for $\pi<x<2\pi.)$

With a summation by parts argument, I can prove that

$\left|\sum_{k=n}^{\infty}\frac{\sin((2k-1)x)}{2k-1}\right|\le\frac1{(2n+1)|\sin x|}$

Since the sum of the whole series is $\frac{\pi}4$, this is enough to prove the partial sum is positive whenever $\sin x > \frac4{\pi(2n+1)}$.

We should be able to argue that when this isn't true (for $x$ near enough to $0$ or $\pi$), then all or most of the terms of the series are positive.


Solve the systems of equation:$ \begin {cases} x^4\minus{}x^3y\plus{}x^2y^2\equal{}1\\x^3y\minus{}x^2\plus{}xy\equal{}\minus{}1\end {cases}$

[quote="phuong"]Solve the systems of equation:$ \begin {cases} x^4 \minus{} x^3y \plus{} x^2y^2 \equal{} 1 \\
x^3y \minus{} x^2 \plus{} xy \equal{} \minus{} 1\end {cases}$[/quote]

We can call $ a\equal{}\minus{}x^2,b\equal{}xy$ then we have:
$ \begin {cases} x^4 \minus{} x^3y \plus{} x^2y^2 \equal{} 1 \\
x^3y \minus{} x^2 \plus{} xy \equal{} \minus{} 1\end {cases}\Leftrightarrow \begin {cases} a^2 \plus{}ab\plus{}b^2 \equal{} 1 \\
a \plus{} b\minus{}ab \equal{} \minus{} 1\end {cases}$
Now it is easy with $ S\equal{}a\plus{}b, P\equal{}ab$!


Let $ \omega$ be the circle and $ AB$ is diameter of $ \omega$.Point $ C$ lies on the continius of $ AB$ . Let $ O$ be the centre of $ \omega$ and $ l$ be the line through $ C$ that intersects $ \omega$ at points $ D$ and $ E$.Let $ \omega_{1}$ be the circumcircle of $ \triangle{BOD}$ and $ OF$ is the diameter of $ \omega_{1}$.If $ G$ is the intersections of $ CF$ and $ \omega_{1}$ then prove that quadrilateral $ OAEG$ is inscribed in circle.


Could anybody solve the [b]SECOND[/b] part of this theorem?

Let be a complete quadrylateral inscrite in a circle, its diagonal points (the intersection points of polygonal diagonals) form a autopolar triangle. More over, the tangents though the poligonal vertex to the circle form a circunscript quadrylateral such that its diagonals are the sides of the above autopolar triangle.

Thanks a lot

José Carlos

hmm...it is very strange that you should ask to prove the second part if you already know the first part. One usually proves the first part by noting that two of these "diagonal points" are collinear with one of these intersection-of-tangencies by Pascal and then that the same two diagonal points are collinear with another one of these intersection-of-tangencies (symmetry) - this suffices to prove the second part of the above theorem - one then usually finishes off the first part by applying a polar map or two.

:blush:  I've already prove the second part... thanks anyway

José Carlos :oops:


A 6-digit security code is made up of four 0s and two 1s. How many security codes of this type can be formed?

[hide="um"]We have six places to put the ones.  $ \binom{6}{2}=\boxed{15}$[/hide]

[quote="buzzer11"][hide="EXTENSION"]What if it was a 10-digit security code made up of 4 ones, 3 twos, and 3 threes?[/hide][/quote]
[hide="Not much harder"]For an $ n$ digit security code made up of $ m_{1}$ 1's, $ m_{2}$ 2's, etc... up to $ m_{i}$ i's, we have:
$ \frac{n!}{m_{1}!m_{2}!\cdots m_{i}!}$

In your case,
$ \frac{10!}{4!3!3!}= \boxed{4200}$[/hide]

Hader extension: How many security codes of 10 consisting of 4 zeros, 3 ones, and 3 twos do not start with 0 and have no two consecutive zeros

[hide]There are three possible ways you can lay out the zeros into the code.

Thus there are 6 slots left for the 3 ones and 3 twos.

$ 3\cdot \frac{6!}{3!\cdot 3!}= \boxed{60}$[/hide]

That's quite simple.

Actually, buzzer11, unless I'm mistaken, there are 15 ways.

[hide]Using the 15 ways to place the zeroes (I counted them out by hand :blush: ), we know that there are 6!/3!/3!=20 ways to place the remaining digits.  Thus, the answer should be 300 ways. :lol: [/hide]

spell it out 11 got it right, but here's the work behind his answer.

Here's a slightly impossible extension:

How many security codes with 60 digits do not contain any of the even digits in the even places (the 2nd, 4th, etc.) and none of the odd in the odd places? (Count 0 as even)

It shouldn't be that hard, actually.  We just place the odd digits in the even places and the even digits in the odd places.  Thus, the answer is:

[hide]4x5^29x5^30.  I'm too lazy to calculate that.  It's something like 1.676380634x10^21.  PWN!!!! :lol: [/hide]
Tell me if I'm wrong.  If it's slightly impossible, I must have made a grave mistake.  Oh well, I wouldn't know. :maybe:

Errmmm.  :maybe:

I think it would be higher.


geometry
circumcircle
combinatorics proposed
combinatorics

For each 2 points A and B on one side of the line Ab we wil,l have at least n+1 points. From these we choose C such that angle <ABC is the n+1-th as value in decreasing order. Then the circumcircle of AABC contain exactly n points in interior. Each of these 'good' circles is counted three times for all the sides of the triangle. Therefore we have at least $\frac{1}{3}\binom{2n+3}{2}$ such circles.

[quote="nhat"]Let $2n+3$ points be given in the plane in such a way that no three lie on a line and no four lie on a circle.Prove that the number of circles that pass through three of these points and contain exactly $n$ interior points is not less than $\frac{1}{3}\binom{2n+3}{2}$[/quote]
Prove that we can choose three of the points and draw a circle through these points, so that exactly $n$ of the remaining of $2n$ points lie inside this circle and $n$ outside.


On day 1 of a series of days, the probability of an event happening is 100%. On day 2, the probability of it happening is 99%, on day 3, it is 98%, and so on.
1. What is the probability that out of the first 20 days the event happened atleast 15 days?
2. What is the probability that after the 50th day, exactly 37 events have happened?
3. What is the expected number of days of the event happening after 80 days?

[quote="easyas3.14159..."]I don't quite understand your third question-the number of days the event happened over 80 days could range anywhere from1-80.[/quote]
He means that how many times should it be expected to have happened. I agree though the wording is rather subglacious.

sorry :oops: i was in a hurry. u know what it means though. what's the expected number of days the event will occur? ill edit it

i don't seem to be getting answers. ill post it in the high school section.


The state meet results are online.

Team (WOCOMAL teams highlighted):
http://www.wocomal.org/statistics/maml/2006/2006m_team.html
Individual (WOCOMAL students highlighted):
http://www.wocomal.org/statistics/maml/2006/2006m_individual.html

Team (MML teams highlighted):
http://www.massmathleague.org/statistics/maml/2006/2006m_team.html
Individual (MML students highlighted):
http://www.massmathleague.org/statistics/maml/2006/2006m_individual.html

I wanted to get the stats online last night, so I apologize for the delay.  Unfortunately the people running the computers had had trouble reading and/or transcribing student names from scoresheets; I spent some time trying to fix the mistakes.  In the top 100, misspelled names listed included Sergi [sic] Bernstein, Anod [sic] Swawantham [sic], Feigi [sic] Jiang, Rahnl [sic] Banerjee, Alex Kazberonk [sic], Nathaniel Rahich [sic], Anton Liu [sic], Anil Vassireddi [sic], George Hons [sic], Shengzhi Lian [sic], and Russel [sic] Wolf.  Those were the errors I caught.  If your name is misspelled, let me know and I'll fix it for you.  This offer also applies to the 2005 and 2004 results for MAML and NEAML.

If you're curious, the hardest rounds (based on average student score) were, in order:

Round 5 (average score 0.797)
Round 6 (average score 1.051)
Round 4 (average score 1.173)
Round 1 (average score 1.604)
Round 3 (average score 2.179)
Round 2 (average score 2.224)

Congratulations to the high scorers.

Rick Yanco


trigonometry
ARML
geometry
complex numbers

what are the roots of the equation $x^n+x^{n-1}+x^{n-2}+\cdots+x+1=0$?

is it always on a circle in the complex plane?  thanks.

\[x^n+x^{n-1}+x^{n-2}+\cdots+x+1=\frac{x^{n+1}-1}{x-1}=0.\]

So the roots are the $n+1$ roots of unity excluding $x=1$.

I think I've posted this question before in another thread, but I can't find it so, what exactly are roots of unity?

First, unity means one.  So saying roots of one might be a little easier to understand (but roots of unity is the common term).  Now, the $n$ roots of unity are the complex solutions to the equation:

\[x^n=1.\]

We can actually go a step further and find the value of these roots.  It is easiest to use the exponential or polar form of complex numbers to find them.  The exponential form of a complex number is generally $e^{i\theta}$.  This is known to also equal $\cos \theta +i\sin \theta$ (this is the polar form).  That second expression is often times expressed with the shorthand $\text{cis} \theta.$

Ok, so let's say we wanted to find the $n$ roots of unity.  In other words, we want to find $x$ such that $x^n=1$.  Now using the polar form we are looking at:

\[(\cos \theta + i\sin \theta)^n=\cos n\theta+i\sin n\theta=1.\]

(We used de Moivre's theorem when going from the first to the second. de Moivre says that $(\text{cis}\theta)^n=\text{cis}n\theta.$  If you use the exponential form this becomes obvious.)

This is obviously satisfied when $\sin n\theta=0$ and $\cos n\theta=1.$  If you ponder this a little bit you can see that

\[\theta=\frac{2\pi}{n} \ , \ 2\left(\frac{2\pi}{n}\right),\ldots,(n-1)\left(\frac{2\pi}{n}\right)\]

will be the angles we want (you can keep going up but in reality those will just give you coterminal angles to the ones you already have...and thats not very interesting ;) ).

Ok, so the $n$th roots of unity are:

\[\text{cis}\frac{2k\pi}{n} \ \ \text{for} \ \ k=1,2,\ldots, n-1.\]

Or in exponential form:

\[e^{2ki\pi/n} \ \ \text{for} \ \ k=1,2,\ldots, n-1.\]

Exercise 1:  Find the roots of the equation $x^5-1=0.$

Exercise 2 (ARML 2004):  Compute the area of the polygon whose vertices are the solutions in the complex plane to $x^7+x^6+x^5+x^4+x^3+x^2+x+1=0$.

I've seen the following problem on a few tests (in various forms):

Exercise 3:  How many real solutions are there to the equation:

\[x^5+x^4+x^3+x^2+x+1=0?\]

Exercise 4:  Convine yourself that if one of the $n$ th roots of unity is $\omega$ where $\omega=\text{cis}\frac{2\pi}{n}$ then the $n$ roots of unity are:

\[1,\omega,\omega^2,\ldots,\omega^{n-1}.\]

Exercise 5:  Prove that

\[1+\omega+\omega^2+\cdots+\omega^{n-1}=0.\]

Hope that helps!  :)

O, and if you want a much better and in-depth exploration of roots of unity and complex numbers in general I highly suggest taking the Intermediate Trigonometry/Complex Numbers Class here on AoPS.  There are also plenty of sources to read which contain material on complex numbers.  Usually your basic precal book wil have a section or so on them.  AoPS Volume 2 and the Art and Craft of Problem Solving would be more preferable if you wanted to gain a better understanding of them.

Just a tiny we typo in your post, should be $\displaystyle (\cos{x}+i \sin{x})^n=\cos{nx}+i\sin{nx}$

We would be missing the second 'i'... :)


integration
calculus
calculus computations

Dear everyone.

please read the following

find the value of

$ 1 \minus{} \frac {1}{5} \plus{} \frac {1}{7} \minus{} \frac {1}{11} \plus{} \cdots \plus{} \frac {1}{6n \minus{} 5} \minus{} \frac {1}{6n \minus{} 1} \plus{} \cdots$

thanks for reading. :)

Let $ P(x)\equal{}\sum_{n\equal{}1}^{\infty }\left( \frac{x^{6n\minus{}5}}{6n\minus{}5}\minus{}\frac{x^{6n\minus{}1}}{6n\minus{}1}\right)$

Desired sum $ \equal{}P(1)$

$ P^{\prime }(x)\equal{}\sum_{n\equal{}1}^{\infty }\left( x^{6n\minus{}6}\minus{}x^{6n\minus{}2}\right) \vspace{1pt}\equal{}\allowbreak \frac{x^{2}\plus{}1}{x^{4}\plus{}x^{2}\plus{}1}$

Hence $ \int_{0}^{1}P^{\prime }(x)dx\equal{}P(1)\equal{}\int_{0}^{1}\frac{x^{2}\plus{}1}{x^{4}\plus{}x^{2}\plus{}1}dx\equal{}\allowbreak \frac{\pi }{2\sqrt{3}}$


ratio
quadratics
geometric sequence
geometric series

Excellent :)

But since I see these recursion topics as a sort of crash course for everyone interested, perhaps you could bring forward some details on how to find those transformations you used.

All right, I have just given a supplementary explanation.

Note that LHS and numerator in those equations have the same form, except that LHS is in $n+1$ and the numerator is in $n$. Hence, when he divides the two equations, he'll get a geometric recursion (check the next row).

Consider the recurrence $a_{n+1}=\frac{pa_{n}+q}{a_{n}+r}.$

For distinct $\alpha ,\beta$, we are to rewrite this into the form of $\frac{a_{n+1}-\beta}{a_{n+1}-\alpha}=\frac{p-\beta}{p-\alpha}\cdot \frac{a_{n}-\beta}{a_{n}-\alpha}.$ Now for the above recurrence, let $\lambda$ be the root of the equation $x=\frac{px+q}{x+r}\Longleftrightarrow x^{2}-(p-r)x-q=0,$ we have $\lambda^{2}-(p-r)\lambda-q=0\Longleftrightarrow \lambda^{2}-p\lambda =q-r\lambda .$

Thus  $a_{n+1}-\lambda =\frac{pa_{n}+q}{a_{n}+r}-\lambda =\frac{(p-\lambda)a_{n}+q-r\lambda}{a_{n}+r}=\frac{(p-\lambda)a_{n}+\lambda^{2}-p\lambda }{a_{n}+r}$

$=\frac{(p-\lambda)(a_{n}-\lambda)}{a_{n}+r}.$

Then for distinct $\lambda=\alpha,\ \beta ,$ we obtain $\frac{a_{n+1}-\beta}{a_{n+1}-\alpha}=\frac{p-\beta}{p-\alpha}\cdot \frac{a_{n}-\beta}{a_{n}-\alpha}.$

How about for the case $\alpha =\beta ?$

I still am not quite sure about the role of solving for the zeroes... is this what enqures that the denominators are nice?

By the way, how do you solve the recurrence $a_{n+1}=2a_{n}+1,\ a_{1}=1?$

to kunny, Substitute $a_{n}=b_{n}+C$, and solve for a C so you can get rid of the constant term.

[b]Remark.[/b] $a_{1}=a\ ,\ \left(\forall\right)n\in N^{*}\ ,\ a_{n+1}=\frac{pa_{n}+q}{a_{n}+r}$ $\Longrightarrow$ $\left(\forall\right)n\in N^{*}\ ,\ a_{n}=\frac{ax_{n}+y_{n}}{az_{n}+t_{n}}$ , where $\left(\begin{array}{cc}x_{n}& y_{n}\\ z_{n}& t_{n}\end{array}\right)=$ $\left(\begin{array}{cc}p & q\\ 1 & r\end{array}\right)^{n-1}\ .$
[u]Particular case :[/u] $a=1$ $\Longrightarrow$ $a_{n}=\frac{x_{n}+y_{n}}{z_{n}+t_{n}}$, where ... .

[quote="kunny"]Yes, that's right. :)

This is the standard solution for Japanese high school students.

$a_{n+1}=2a_{n}+1.$
$\alpha =2\alpha+1.$
Subtracting both sides, $a_{n+1}-\alpha =2(a_{n}-\alpha ).$

Can you understand this transformation?[/quote]

Sorry I'm really bad at this recursive stuff.. :oops: what does that do to solving for $a_{n}$? How does it help...?

[quote="kunny"]$a_{n+1}-\alpha =2(a_{n}-\alpha ).$

From this, can you understand that the sequence $\{a_{n}-\alpha\}$ is the geometric sequence with the common ratio 2?  :)[/quote]

OH. Oops. Ok, thanks.  :jump:

[quote="me@home"]ok so we are solving equation in such a way that if we subtract a zero from each side we get something nice, then have a geometric series?[/quote]

Subtracting zero often does nothing; on the other hand, subtracting a constant does quite well.

I mean solving for a zero of a quadratic (giving a constant) and subtracting that...


a)

Show that $ \sin^{\minus{}1}\left(\frac{x}{3}\minus{}1\right) \equal{}\frac{\pi}{2}\minus{}2\sin^{\minus{}1}\sqrt{ 1\minus{}\frac{x}{6}}$

b)

Let $ a$,$ b$ and $ c$ satisfy $ bc \equal{} 1\plus{}a^{2}$. Show that:

$ \tan^{\minus{}1}\left(\frac{1}{a\plus{}b}\right)\plus{}\tan^{\minus{}1}\left(\frac{1}{a\plus{}c}\right) \equal{}\tan^{\minus{}1}\left(\frac{1}{a}\right)$

[hide="hint for b"]take the tan of each side and plug in the formula $ \tan(\alpha+\beta)=\frac{\tan{\alpha}+\tan{\beta}}{1-\tan{\alpha}\tan{\beta}}$[/hide]


[quote="Totally Zealous"]Sarah pours four ounces of coffee into an eight ounce cup and four ounces of cream into a second cup of the same size. She then transfers half the coffee from the first cup to the second and, after stirring thoroughly, transfers half the liquid in the second cup back to the first. What fraction of the liquid in the first cup is now cream?
(A) $\frac{1}{4}$
(B) $\frac{1}{3}$
(C) $\frac{3}{8}$
(D) $\frac{2}{5}$
(E) $\frac{1}{2}$[/quote]

[hide]We have 4 oz coffee in cup 1, 4 oz cream in cup 2.
After the transfer,
we have 2 oz coffee in cup 1, 2 oz coffee and 4 oz cream in cup 2.
After the 2nd transfer,
we have 3 oz coffee and 2 oz cream in cup 1, 1 oz coffee and 2 oz cream in cup 2.

So the answer is D.[/hide]

In the first cup, there are 4 ounces coffee in 1st cup 4 ounces cream 2nd cup originally.

After mixing (and stirring), there are 2 ounces coffee in 1st cup 2 ounces coffee, 4 ounces cream 2nd cup.

After mixing again (and stirring again), there are 3 ounces coffee, 2 ounces cream 1st cup, 1 ounces coffee, 2 ounces cream 2nd cup.

1st cup: 2 ounces cream / 5 ounces total = 2/5 = D

A pretty simple mixing problem :P

[hide]
It wasn' that hard-
We began with 4 ounces of coffee and 4 ounces of Cream
Half of the coffee was mixed into the creamer- That means we have 2 ounces of coffee in cup 1 and 2 ounces of coffee and 4 ounces of cream in cup 2.
Then, half of cup 2 (which was mixed) was added to cup 1- That means 1 ounce of coffee and 2 ounces of ream were adde to cup giving cup one a total of 5 ounces. 3 ounces bing coffee and 2 ounces being cream.

Our final answer is 2/5!
[/hide]


In the row of Pascal's triangle that starts 1, 11, what is the ninth number?

the row has the number $ \binom{11}{0}, \binom{11}{1}, \binom{11}{2}...$ has ninth number $ \dbinom{11}{9\minus{}1}\equal{}\binom{11}{8}\equal{}\binom{11}{3}\equal{}\frac{11\times 10\times 9}{6}\equal{}11\times 5\times 3 \equal{} 11\times 15\equal{} \boxed{165}$


group theory
abstract algebra
superior algebra
superior algebra unsolved

Both planetmath and mathworld say that if we have a group $G$ and a subgroup $H$, then the normalizer $N_{G}(H)$ is a subgroup of $G$. It seems as if they are saying that $N_{G}(S)$ is not necessarily a subgroup for any non-empty subset $S \subset G$.

However, I think that $N_{G}(S)$ is always a subgroup. Am I correct?

I have another related problem. If I have a subgroup $H \le G$ and an element $C \in G$ such that $i_{C}(H) \subseteq H$ (inner automorphism of $H$ by $C$), can I conclude that $i_{C}(H) = H$ even if $H$ is not finite?  Or must I also show that $i_{C^{-1}}(H) \subseteq H$?

Thanks for help


[quote="humblemastermind"]To find LCMs in general, you want to take the prime factors of each number and remove any extras.

Here, $13$ prime factors into $13 \times 1$, $26$ factors into $13 \times 2$, and $39$ factors into $13 \times 3$. You can get rid of the two extra $13$s, leaving you with, as goldendomer said, $13 \times 1 \times 2 \times 3 = 78$[/quote]

Yeah that's what I did. Just didn't explain it though.


We can assume that

$ 0 \leq a \leq b \leq c$ with $ ab\plus{}bc\plus{}ca \neq 0$

$ f(c)\equal{}\sqrt{9a^2\plus{}16bc}\plus{}\sqrt{9b^2\plus{}16ca}\plus{}\sqrt{9c^2\plus{}16ab}\minus{}5(a\plus{}b\plus{}c)$

It is easy to show that $ f''(c)<0$ and so

$ f(c) \geq min\{f(b),f(a\plus{}b)\}$ since $ b \leq c \leq a\plus{}b$

And the rest is trivial.  :wink:

[quote="crazyfehmy"]We can assume that

$ 0 \leq a \leq b \leq c$ with $ ab \plus{} bc \plus{} ca \neq 0$

$ f(c) \equal{} \sqrt {9a^2 \plus{} 16bc} \plus{} \sqrt {9b^2 \plus{} 16ca} \plus{} \sqrt {9c^2 \plus{} 16ab} \minus{} 5(a \plus{} b \plus{} c)$

It is easy to show that $ f''(c) < 0$ and so

$ f(c) \geq min\{f(b),f(a \plus{} b)\}$ since $ b \leq c \leq a \plus{} b$

And the rest is trivial.  :wink:[/quote]
How to show that $ f''(c)<0$? Can you show it? Thanks.

I think there is a stupid mistake :blush:

When $ f''(c)>0$ when $ a\equal{}b\equal{}c$  :P

[quote="Guest"][quote="can_hang2007"]Let $ a,$ $ b,$ $ c$ be the side-lengths of a triangle. Prove that
$ \sqrt {9a^2 \plus{} 16bc} \plus{} \sqrt {9b^2 \plus{} 16ca} \plus{} \sqrt {9c^2 \plus{} 16ab} \ge 5(a \plus{} b \plus{} c).$[/quote]
This is not true. In fact, the inequality inversely holds, i.e.,

$ \sqrt {9a^2 \plus{} 16bc} \plus{} \sqrt {9b^2 \plus{} 16ca} \plus{} \sqrt {9c^2 \plus{} 16ab} \le 5(a \plus{} b \plus{} c).$[/quote]
No, my inequality is not reversed. Try $ a\equal{}b\equal{}1,$ $ c\equal{}\frac{1}{2}.$

[quote="can_hang2007"][quote="Guest"][quote="can_hang2007"]Let $ a,$ $ b,$ $ c$ be the side-lengths of a triangle. Prove that
$ \sqrt {9a^2 \plus{} 16bc} \plus{} \sqrt {9b^2 \plus{} 16ca} \plus{} \sqrt {9c^2 \plus{} 16ab} \ge 5(a \plus{} b \plus{} c).$[/quote]
This is not true. In fact, the inequality inversely holds, i.e.,

$ \sqrt {9a^2 \plus{} 16bc} \plus{} \sqrt {9b^2 \plus{} 16ca} \plus{} \sqrt {9c^2 \plus{} 16ab} \le 5(a \plus{} b \plus{} c).$[/quote]
No, my inequality is not reversed. Try $ a \equal{} b \equal{} 1,$ $ c \equal{} \frac {1}{2}.$[/quote]
You are correct; I confused this one with another.

[quote="can_hang2007"]Let $ a,$ $ b,$ $ c$ be the side-lengths of a triangle. Prove that
$ \sqrt {9a^2 \plus{} 16bc} \plus{} \sqrt {9b^2 \plus{} 16ca} \plus{} \sqrt {9c^2 \plus{} 16ab} \ge 5(a \plus{} b \plus{} c).$[/quote]
See here: http://can-hang2007.blogspot.com/2010/02/inequality-69-v-q-b-can.html

yes,holds in triangle,but for positive numbers $ a,b,c$,prove that:

$ \sqrt {9a^2 \plus{} 14bc \plus{} b^2 \plus{} c^2} \plus{} \sqrt {9b^2 \plus{} 14ca \plus{} c^2 \plus{} a^2} \plus{} \sqrt {9c^2 \plus{} 14ab \plus{} a^2 \plus{} b^2} \geq 5(a \plus{} b \plus{} c).$
BQ

[quote=mudok][quote=can_hang2007]Let $ a,$ $ b,$ $ c$ be the side-lengths of a triangle. Prove that
$ \sqrt{9a^2\plus{}16bc}\plus{}\sqrt{9b^2\plus{}16ca}\plus{}\sqrt{9c^2\plus{}16ab} \ge 5(a\plus{}b\plus{}c).$[/quote]

Not only triangle sides, it is true for all nonnegative $a,b,c$ such that

$17(a+b+c)^2(ab+bc+ca)\ge 4(a+b+c)^4+25abc(a+b+c)+4(ab+bc+ca)^2$[/quote]
What do you think about an ugly condition? I think it make the original inequality is not nice  ;)

[quote=dm_edogawasonichi]
What do you think about an ugly condition? I think it make the original inequality is not nice [/quote]

I agree, it is not nice condition. But it is more general. And it is interesting that If we reverse the condition, the inequality also will be reversed. See here:

http://artofproblemsolving.com/community/c34h1249035_reverse_inequality_of__my_previous

[quote=mudok]

I agree, it is not nice condition. But it is more general. And it is interesting that If we reverse the condition, the inequality also will be reversed. See here:

http://artofproblemsolving.com/community/c34h1249035_reverse_inequality_of__my_previous[/quote]
You can see that some inequality masters on AoPS, their inequalities look very nice with a simple condition. I think it is the pinnacle of inequality and people like these inequalities. Also, for a year recently, I see that Inequalities on Mathlink look very bad with some nicks from China. Now, it is horrible if we compare with AoPs few years ago.

[quote=arqady][quote=can_hang2007]Let $ a,$ $ b,$ $ c$ be the side-lengths of a triangle. Prove that
$ \sqrt{9a^2\plus{}16bc}\plus{}\sqrt{9b^2\plus{}16ca}\plus{}\sqrt{9c^2\plus{}16ab} \ge 5(a\plus{}b\plus{}c).$[/quote]
By Holder
$$\left(\sum_{cyc}\sqrt{9a^2+16bc}\right)^2\sum_{cyc}\frac{(3a^2+b^2+c^2+6bc+2ab+2ac)^3}{9a^2+16bc}\geq\left(\sum_{cyc}(3a^2+b^2+c^2+6bc+2ab+2ac)\right)^3=125(a+b+c)^6$$
Thus, it remains to prove that $5(a+b+c)^4\geq\sum_{cyc}\frac{(3a^2+b^2+c^2+6bc+2ab+2ac)^3}{9a^2+16bc}$,
which is true and not so hard, but my proof still  is not nice.[/quote]

it is a simple one


geometry
circumcircle
trigonometry
trapezoid
ratio
geometric transformation

Using ceva threom we obtain that BG=CG
By angle chasing we acquire that ⊿BQD∽⊿BNC  ⊿CQD∽⊿CMB
=> DH*BC=DC*BM=CN*BD
Notice AB/AC=BM/CN   =>  DC*AB=AC*BD
ABCD is a harmonious quadrilateral
This completes our proof

[b]Here is my solution[/b] (Using [b]spiral similarity[/b], no [b]calculation[/b])
Because $ (\overline {BMA},\overline {BPN},\overline {CMP},\overline {CNA})$ is a complete quadrilateral, therefore, $ Q$ is the [b]Miquel[/b] point wrt $ (\overline {BMA},\overline {BPN},\overline {CMP},\overline {CNA})$. Therefore, $ (A,N,Q,B)$, and $ (A,M,Q,C)$ are the sets of concyclic points. Hence, $ \angle BAQ \equal{} \angle BNQ\equiv \angle PNQ \equal{} \angle QCP$. Therefore, we just need to prove that $ \angle QCP \equal{} \angle PAN$. Indeed, it is easy to notice that $ \triangle QAB\sim \triangle QCP$, further, these $ 2$ triangles have the same direction, hence, through the spiral similarity with center $ Q$, denote this $ f$, $ f: \triangle QAB\mapsto \triangle QCP$, therefore, $ f: A\mapsto C$, $ B\mapsto P$ $ \Longrightarrow f: A\mapsto B$, $ C\mapsto P$, which implies that $ f: \triangle QAC \mapsto \triangle QBP \Longrightarrow \frac {QC}{QP} \equal{} \frac {AC}{BP}$, $ (1)$. Now, by [b]Thales theorem[/b], we notice that: $ \frac {AC}{AN} \equal{} \frac {BC}{MN} \equal{} \frac {PB}{PN} \Longrightarrow \frac {AC}{PB} \equal{} \frac {AN}{PN}$, $ (2)$. From $ (1),(2)$, we obtain, $ \frac {NA}{NP} \equal{} \frac {QC}{QP}$, combine with $ \angle ANP \equal{} \angle CQP\Longrightarrow \triangle ANP\sim \triangle CQP\Longrightarrow \angle PAN \equal{} \angle PCQ$, which implies to the result of the problem.
Our proof is completed :lol:

We have:
$ \frac {AN\sin{\widehat{QAN}}}{AM\sin{\widehat{QAM}}} \equal{} \frac {S_{AQN}}{S_{AQM}} \equal{} \frac {AN}{AM}.\frac {QN}{QM}.\frac {\sin{\widehat{QNC}}}{\sin{\widehat{QMB}}}$

$ \Rightarrow \frac {\sin{\widehat{QAN}}}{\sin{\widehat{QAM}}} \equal{} \frac {QN}{QM}.\frac {\sin{\widehat{QNC}}}{\sin{\widehat{QMB}}} \equal{} \frac {QM}{QB}.\frac {QN}{QM} \equal{} \frac {QN}{QB} \equal{} \frac {AN}{AM} (1)$

It is because of the fact that $ \triangle QCN \sim \triangle QMB$ and since, $ \frac {QN}{QB} \equal{} \frac {AN}{AM}$.

From $ (1)$, it is followed that $ AQ$ is the symmedian of $ \triangle AMN$, which implies the result of the problem.

[b]P/S:[/b] My $ 200^{\text{th}}$ post :lol:

[quote="Ahiles"] [color=darkred]Let $ MN$ be a line parallel to the side $ BC$ of a triangle $ ABC$, with $ M$ on the side $ AB$ and $ N$ on the side $ AC$ . The lines $ BN$ and $ CM$

meet at point $ P$ . The circumcircles of triangles $ BMP$ and $ CNP$ meet at two distinct points $ P$ and $ Q$. Prove that $ \angle BAQ \equal{} \angle CAP$ .[/color][/quote]

[color=darkblue][b][u]Proof[/u].[/b] $ \left\|\begin{array}{c} \widehat {BMQ}\equiv\widehat {BPQ}\equiv\widehat {NCQ} \\
\ \widehat {MBQ}\equiv\widehat {CPQ}\equiv\widehat {CNQ}\end{array}\right\|$ $ \implies$ $ \left\|\begin{array}{c}\widehat {BMQ}\equiv\widehat {NCQ} \\
\ \widehat {MBQ}\equiv\widehat {CNQ}\end{array}\right\|$ $ \implies$ $ \triangle BMQ\sim\triangle NCQ\implies$

$ \frac {AB}{AC} \equal{} \frac {BM}{NC} \equal{} \frac {\delta (Q,BM)}{\delta (Q,NC)}$ $ \implies$ $ \frac {\delta (Q,AB)}{\delta (Q,AC)} \equal{} \frac {AB}{AC}$ , i.e. the point $ Q$ belongs to the $ A$-symmedian.

I used the notation $ \delta (X,d)$ - [u]the distance of the point[/u] $ X$ [u]to the line[/u] $ d$ . I"ll present below two simple and well-known properties :[/color]

[color=darkred][b]$ 1\blacktriangleright$ The geometrical locus of an interior  point $ L$ of the angle $ \widehat {BAC}$ for which $ \frac {\delta (L,AB)}{\delta (L,AC)} \equal{} \frac {AC}{AB}$ is the $ A$-median.[/b]

[b]$ 2\blacktriangleright$ The geometrical locus of an interior point $ L$ of the angle $ \widehat {BAC}$ for which $ \frac {\delta (L,AB)}{\delta (L,AC)} \equal{} \frac {AB}{AC}$ is the $ A$-symmedian.[/b][/color]

[quote="Virgil Nicula"][color=darkred][b][u]An easy extension[/u].[/b]

[b]Consider in $ \triangle ABC$ two mobile points  point $ M$ , $ N$ which belong to the sidelines $ AB$ , $ AC$ respectively so that $ \frac {MB}{NC} \equal{} k$ (constant)

 and the line $ BC$ doesn't separates the points $ M$ , $ N$ . Denote $ P\in MC\cap NB$ and the second intersection $ Q$ between the circumcircles

 of the triangles $ BMP$ , $ CNP$ . Prove that $ \frac {\delta (Q,AB)}{\delta (Q,AC)} \equal{} k$ , i.e. the geometrical locus of the point $ L$ is a line which passes through $ A$ .[/b][/color][/quote]

a Turkish student solve it from four trigonometric ceva :lol:

Let $ H$ be the midpoint of $ BC$.
Applying the lemma about trapezium we have $ A,P,H$ are collinear then $ AP$ is the median of triangle $ ABC$.
We need to prove $ AQ$ is a symmedian.
Let $ K, J$ be the projection of $ Q$ on $ AB, AC$.
$ \angle BMQ \equal{} \angle BPQ \equal{} \angle NCQ, \angle MBQ \equal{} \angle MPB \equal{} \angle NQC$
$ \Rightarrow \Delta MBQ\sim \Delta CNQ \Rightarrow \frac {AB}{AC} \equal{} \frac {BM}{CN} \equal{} \frac {QK}{QJ}$
$ \Rightarrow$ QED

Complete quadrilateral - need symmedian - spiral similarity - done:

http://tlovering.wordpress.com/2009/05/05/balkan-mo-2009-q2-a-geometers-geometry/

Note that the problem was posted 2 years ago from the competition in that forum.
See here
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=175434

You can see
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=358752
Sincerely
Jean-Louis

I solved this using lots of ratios and sine rules.
Let $AP$ meet $BC$ at $X$. Let $\angle BMQ = \angle ACQ = x$, $\angle QBA = \angle QNC = y$ and $\angle MQB = \angle NQC = z$.  Here we make the observation that $Q$ is the Miquel Point of the compelte quadrilateral with vertices $B, M, A, N, C, P$.
$\frac{AQ}{\sin x} = \frac{MQ}{\sin \angle BAQ}$ and $\frac{AQ}{\sin y} = \frac{NQ}{\sin \angle CAQ}$ so $\frac{\sin x}{\sin y} = \frac{\sin \angle BAQ}{\sin \angle CAQ} \cdot \frac{NQ}{MQ}$ (1).
$\frac{NQ}{\sin x} = \frac{NC}{\sin z}$ and $\frac{MQ}{\sin y} = \frac{MB}{\sin z}$ so $\frac{NQ}{MQ} = \frac{\sin x}{\sin y} \cdot \frac{AC}{AB}$ (2).
(1), (2) $\implies \frac{\sin \angle BAQ}{\sin \angle CAQ} = \frac{AB}{AC}$, and it is quite well known that this implies that $AQ$ is a symmedian of $\triangle ABC$.
Using Ceva, we get $\frac{BM}{MA} \cdot \frac{AN}{NC} \cdot \frac{CX}{XB} = 1 \implies CX = BX$ as $MN || BC$. Thus $AX$ is a median, so $\angle BAQ = \angle CAQ$, as required.

Thus $AX$ is a median, so $\angle BAQ = \angle CAQ$, as required.[/quote]
BAP=CAQ

A simple proof which highlights the configuration:

We proceed using directed angle $\pmod{\pi}$.

Let $AP\cap BC=X$, by Ceva $AP$ is a median of $\triangle AMN$.  Now note that
\[\measuredangle ABQ=\measuredangle MBQ=\measuredangle MPQ =-\measuredangle QPM =-\measuredangle QPC =\]\[=-\measuredangle QNC=-\measuredangle QNA =\measuredangle ANQ.\]Hence, $ABNQ$ is cyclic.  Similarly, $ACMQ$ is cyclic.

Let $F=(CNPQ)\cap MQ$.  We claim that $NF\parallel AB$.  Now $Q$ is the center of the unique spiral similarity sending $BM$ to $NC$.  Then
\[\measuredangle QMB =\measuredangle QCN =\measuredangle QFN =-\measuredangle NFQ =-\measuredangle NFM=\measuredangle MFN.\]So $NF\parallel AB$.  Similarly, if $G=BMPQ\cap NQ$, then $MG\parallel AC$.

Let $H$ be the point such that $AMHN$ is a parallelogram.  Then $H$ lies on $MG,NF,AP$.

Note that $\measuredangle PGN =\measuredangle PGQ=\measuredangle PMQ=\measuredangle PMF.$  Similarly, $\measuredangle PFM=\measuredangle PNG.$  So $P$ is the center of the unique spiral similarity sending $MF$ to $GN$, so $P=(MFH)\cap (NGH)$ and hence $MFHP,NGHP$ are cyclic.

We are close! Note that $\measuredangle ANP =\measuredangle ANB=\measuredangle AQB$ and
\[\measuredangle NPA=\measuredangle NPH =\measuredangle NGH=\measuredangle NGM=-\measuredangle MGN =-\measuredangle MGQ=\]\[=-\measuredangle MBQ=\measuredangle QBM=\measuredangle QBA.\]So $\triangle PAN\sim \triangle BAQ$, and $\angle PAN=\angle PAC=\angle BAQ$, as desired. $\square$

[hide="Motivation"]First get $AP$ median and cyclic quads.  Now $H$ is probably a good point, draw it in, ooh it looks like $F,G$ are good.  Now we sense it suffices to show that $MFHP,NGHP$ are cyclic.  Go back and prove both of these things, then ooh similar triangles and done.

Spiral similarities are nice.  Also, I find it easier to do the whole thing for a particular configuration in terms of regular angles, then it's easy enough to rewrite in terms of directed angles.[/hide]

My solution : By Miquel's theorem $AMQC,BANQ$ are cyclic quads. Apply $\sqrt bc$ inversion with centre $A$.
We get that
1.) $B$ goes to $C$, $C$ to $B$
2.) Let $M$ go to $M'$ and $N$ go to $N'$ and $Q$ go to $Q'$
3.)Since a circle through the centre of inversion becomes a line we get that $ Q' $ is the intersection of $BM'$ and $CN'$. So $AQ'$ is a median in triangle $ABC$. So $AQ$ is its isogonal and we are done .

My solution :

From the problem [url=http://www.artofproblemsolving.com/community/c6h282732]Three concurrent radical axes[/url] $ \Longrightarrow AQ $ pass through the isogonal conjugate of the complement of $ P $ ,
so $ MN \parallel BC \Longrightarrow AP $ is A-median of $ \triangle ABC \Longrightarrow AP $ pass through the complement of $ P \Longrightarrow \angle BAQ=\angle CAP $ .

Q.E.D

[quote=buratinogigle]An extension

Let $ABC$ be a triangle with $E,F$ are arbitrary points on $CA,AB$. Circle $(ABE),(ACF)$ intersect again at $D$. $M,N$ are midpoints of $BC,EF$. $G$ lies on $BC$ such that $AG\parallel MN$. Prove that $\angle DAB=\angle GAC$.[/quote]

My solution :

Let $ T \equiv BE \cap CF $ and $ S $ be the reflection of $ T $ in $ M $ .

Since $ MN $ passes through the midpoint of $ AT $ ( Newton line theorem ) ,
so $ A, G, S $ are collinear $ \Longrightarrow AG $ passes through the complement of $ T $ WRT $ \triangle ABC $ ,
hence from the problem [url=http://www.artofproblemsolving.com/community/c6h282732]Three concurrent radical axes[/url] $ \Longrightarrow AD, AG $ are isogonal conjugate WRT $ \angle BAC $ .

Q.E.D

[quote=Mr.Buratino]An extension

Let $ABC$ be a triangle with $E,F$ are arbitrary points on $CA,AB$. Circle $(ABE),(ACF)$ intersect again at $D$. $M,N$ are midpoints of $BC,EF$. $G$ lies on $BC$ such that $AG\parallel MN$. Prove that $\angle DAB=\angle GAC$.[/quote]
My solution:
$MN\cap{CA, AB} = U, V$; $H$ is the midpoint of $CF$
$X, Y$ are the projections of $D$ on $AB, AC$
$K, L$ are the projections of $G$ on $CA, AB$
We have: $\triangle{DBF}$$\sim$$\triangle{DEC}$
$\Rightarrow$ $\frac{DX}{DY} = \frac{BF}{EC}$ (1)
On the other hand: $\triangle{HMN}$$\sim$$\triangle{AVU}$
$\Rightarrow$ $\frac{BF}{EC} = \frac{HM}{HN} = \frac{AV}{AU} = \frac{sin\angle{AUV}}{sin\angle{AVU}} = \frac{sin\angle{GAK}}{sin\angle{GAL}} = \frac{GK}{GL}$ (2)
(1), (2) $\Rightarrow$ $\frac{DX}{DY} = \frac{GK}{GL}$ and the conclusion follows
Q.E.D

Nice solution dear tranquanghuy7198, a little change without sine

Let $AZ\perp UV, Z\in UV$ then $\triangle AZU\sim\triangle AKG$ and $\triangle AZV\sim\triangle ALG$ so $\frac{BF}{EC} = \frac{HM}{HN} = \frac{AV}{AU} =\frac{AV}{AZ}.\frac{AZ}{AU} =\frac{AG}{GL}.\frac{GK}{AG}= \frac{GK}{GL}$.

Note that by a standard Yufei/spiral similarity configuration, $\triangle QMB\sim\triangle QCN$.  This means that $\angle BMQ=\angle NCQ$, so quadrilateral $AMQC$ is cyclic.  Similarly, $ANQB$ is also cyclic.

Now perform a $\sqrt{bc}$ inversion $\Phi$ about $A$.  Note that $\Phi$ sends $B$ to $C$ and vice versa.  Furthermore, $\Phi$ sends $M$ to the point $M'$ on ray $\overrightarrow{AC}$ such that \[AM\cdot AM'=AB\cdot AC\quad\implies\quad AM'=2AC.\] Similarly, this inversion sends $N$ to the point $N'$ such that $AN'=2AB$.  As a result, $\Phi$ sends $(ANB)$ to $CN'$ and $(AMC)$ to $BM'$, meaning that the image of $Q$ about $\Phi$ is the centroid of the triangle homothetic to $\triangle ABC$ with scale factor $2$ (i.e. $\triangle AN'M'$).

Hence $Q'$ lies on the $A$-median of $\triangle ABC$, meaning that $Q$ lies on the $A$-symmedian, so by the definition of symmedian we have $\angle BAQ=\angle CAP$ as desired.  $\blacksquare$

First, I claim that $AP$ is the $A$-median of triangle $ABC$. Consider the reflection of $P$ about the midpoint $D$ of $BC$, say $D'$. Then we have $\triangle MPN \sim PD'C$ as $PBD'C$ is a parallelogram, hence by considering the homothety taking $P$ to $D'$ centered at $A$ we know $A, P, D$ are collinear.

Next, I claim $AQ$ is the $A$-symmedian of triangle $ABC$. If we prove this then $AP$ is isogonal to $AQ$ and we are done. Denote the projections of $Q$ onto $AB$ and $AC$ as $M'$ and $N'$ respectively. It's well-known that $\triangle BQM \sim \triangle NQC$, so we have $\frac{QM'}{QN'}=\frac{MB}{NC}=\frac{AB}{AC}$. This completes the proof.

Firstly by Ceva's theorem; $AP$ is the median.
Note that $Q$ is miquel point of complete quadrilateral $MBPC$. Therefore; $AMCQ$ and $ANBQ$ are cyclic.
Then note that $\sqrt{bc}$ inversion sends $MN$ to a line parallel to $BC$, and $Q$ to $M^*C^* \cap N^*Q^*$, if $X^*$ is inverted image of any point $X$ under this inversion. Therefore $AQ^*$ is $A-$median of $\triangle ABC$. But since $\sqrt{bc}$ inversion compounds of reflection around $\angle A$- bisector, $AP$ and $AQ$ are isogonal; as desired. $\square$

[quote=AMN300] So we have $\frac{QM'}{QN'}=\frac{MB}{NC}=\frac{AB}{AC}$. This completes the proof.[/quote]
How it's complete the proof? Is it a property of symmedian? please explain.

Here's a sketch of a bary solution: Observe that $Q$ is the Miquel point of the quadrilateral $AMPN$, hence the two circles $\omega_1 = \omega(A,M,Q,C)$, $\omega_2 = \omega(A,N, Q, B)$ exists. Also by Ceva's theorem, we have $P$ on the $A$-median of $\Delta ABC$, hence let $P = (p:1:1)$ and thus $M = (p:1:0)$ and $N = (p:0:1)$.

It can be easily verified that the equation of $\omega_1$ is $-\sum_{\sigma}a^2yz + \frac{c^2py}{(p+1)}(x+y+z) = 0$, and the equation of $\omega_2$ is $-\sum_{\sigma}a^2yz + \frac{b^2pz}{(p+1)}(x+y+z) = 0$. I claim that if you let $q = \frac{p(b^2+c^2) - a^2(p+1)}{(p+2)}$, then $Q = (q:b^2:c^2)$. Due to the symmetry in swapping $b \leftrightarrow c, y \leftrightarrow z$, it suffices to show that $Q \in \omega_1$. Now, $$ Q \in \omega_1 $$ $$\Leftrightarrow -a^2b^2c^2 - b^2c^2q - b^2c^2q + \frac{c^2b^2q}{p+1}(q+b^2+c^2) = 0$$ $$\Leftrightarrow a^2+2q = \frac{p(q+b^2+c^2)}{p+1}$$ $$\Leftrightarrow (p+1)(a^2+2q) = p(q+b^2+c^2)$$ $$\Leftrightarrow q = \frac{p(b^2+c^2) - a^2(p+1)}{(p+2)}$$, which is true, hence the point is $Q$.

Now this finishes the point since $AQ$ is the $A$-symmedian line which is the isogonal conjugate of the $A$-median line.

[hide=Same as #26, Posted for Storage]
Note that $Q=(ANB)\cap(AMC)$ by Miquel's theorem. Let $\phi$ be inversion at $A$ with power $AN\cdot AB=AM\cdot AC$ composed with reflection over the $\angle A$ bisector. We see that
\[\phi(Q)=\phi((ANB)\cap(AMC))=MC\cap NB=P,\]
so $AP$ and $AQ$ are isogonal, as desired.
[/hide]

It's easy to see that: $P$ lies on $A$ - median of $\triangle$ $ABC$
We have: $\widehat{BQM}$ = $\widehat{BPM}$ = $\widehat{CPN}$ = $\widehat{CQN}$, $\widehat{MBQ}$ = $\widehat{QPC}$ = $\widehat{CNQ}$
So: $\triangle$ $BMQ$ $\sim$ $\triangle$ $NCQ$
Then: $\dfrac{d (Q; BM)}{d (Q; NC)}$ = $\dfrac{BM}{CN}$ = $\dfrac{AB}{AC}$ (With $d (X; YZ)$ is distance from point $X$ to line $YZ$)
Hence: $Q$ lies on $A$ - symmedian of $\triangle$ $ABC$ or $\widehat{BAQ}$ = $\widehat{PAC}$

From Ceva's Theorem clearly $AP$ is the $A$-median. Now consider an inversion $\Phi$ with $r=\sqrt{AN.AB}$ followed by the reflection of the angle bisector. We heve $$\Phi : M \mapsto C, N \mapsto B$$.  Note that $Q$ is the center of spiral similarity sending $BM \mapsto NC$. Hence it also maps $BN \mapsto MC$. Hence it lies on $\odot(ANQB)$ and $\odot(ANQC)$. So $Q'=(ANB)^* \cap (AMQ)^*=BN \cap MC=P \implies AQ $ is the $A$ symmedian $\implies $ $\angle BAQ =\angle PAC$.

[quote=Ahiles]Let $ MN$ be a line parallel to the side $ BC$ of a triangle $ ABC$, with $ M$ on the side $ AB$ and $ N$ on the side $ AC$. The lines $ BN$ and $ CM$ meet at point $ P$. The circumcircles of triangles $ BMP$ and $ CNP$ meet at two distinct points $ P$ and $ Q$. Prove that $ \angle BAQ = \angle CAP$.

[i]Liubomir Chiriac, Moldova[/i][/quote]

The problem becoms easy when you use inversion.

Here it is an easy solution;
We can prove easly by angle chasing that quadrilateral (ANQB) and (AMQC) are cyclic.

we now that $MN$ is parallel to $BC$ and it is known that $AM$ $AC$=$AN$ $AB$
Lets get an inversion at center A whith radius $\sqrt{AM AC}$ and also this inversion is reflection on angle bisector of A(Almost the same thing like $\sqrt{bc}$) Lets name this inverzion $\Phi$.  $AM$ $\Phi$(AM)=$AM$ $AC$.From this we can se that $\Phi$ sends M to C and N to B from this we now that after $\Phi$ (BQN) becoms line BN and $\Phi$(Q) is somewhere in BM,Same after $\Phi$  (MQC) becoms line CM and $\Phi$(Q) is somewhere in CM...
From all this we can see that P=$\Phi$(Q) .Now looking back how we define $\Phi$ we cay that AP and AQ are isogonal $\blacksquare$

Note that $Q$ is a Miquel Point of $MNBC$, so $(ABQN)$ and $(ACQM)$ are cyclic. Now performing a $\sqrt{bc}$ inversion followed by a reflection over the angle bisector of $\angle BAC$, everything stays fixed and $P$ inverts to $Q$. Therefore, $P$ and $Q$ are reflections across the angle bisector, done.

$Q$ - Miquel point of $AMPN$, so $Q\in (AMC), Q\in (ANB)$.
With flip-inversion $\sqrt{AM\cdot AC}$ we have $BN\cap CM\mapsto (AMC)\cap (ANB)=P\mapsto Q$ $\blacksquare$

[color=#600][b]Remark[/b][/color]: $M, N$ - isogonal conjugates in $\triangle APQ$.
We can get a solution from this fact without inversion. I'll sketch it: By Miquel we already know that $QM,QN$ are isogonal in $\angle AQP$, so we need to prove that $AP,PQ$ are isogonal in $\angle BPC$ , or $PQ$ - symmedian in $BPC$.
Let $MN$ cut $(BPM),(CPN)$ at $K,L$.
By Reim $BK,CL$ are tangent to $(BPC)$.
$BCKL$ - cyclic trapezoid.
By Radical Axis $BK\cap CL\in PQ$ $\blacksquare$

It can be seen by homothety that $P$ is on the $A$-median, so the problem reduces to proving that $AQ$ is the $A$-symmedian. Notice that by spiral similarity, we know that $\triangle QBM\sim \triangle QNC,$ so $\frac{d(Q,AB)}{d(Q, AC)}=\frac{BM}{NC}=\frac{AB}{AC},$ and by one of the properties of the symmedian, we are done.

[b][color=#f00]Claim:[/color][/b] $ABQN$ and $ACQM$ are cyclic.

[i]Proof.[/i] By the Spiral Center Lemma, we know $Q$ is the center of the spiral similarity taking $BM$ to $NC$. Thus, $Q$ is also the spiral center between $BN$ and $MC$. But $BM \cap NC = A$, so $A$ lies on $(QBN)$ and $(QMC)$ as desired. $\square$

Now, consider the inversion about $A$ with radius $\sqrt{AM \cdot AC} = \sqrt{AN \cdot AB}$ (equal since $AMN \sim ABC$) followed by a reflection over the $A$-bisector. Notice $B$ swaps with $N$, and $C$ swaps with $M$. Thus, $$P^* = \overline{B^*N^*} \cap \overline{C^*M^*} = (ABN) \cap (ACM) = Q$$ (as $P^* \ne A$). The result follows easily. $\blacksquare$

[b]Remark:[/b] The inversion I performed was motivated by numerous "potential/desired" spiral similarities, namely $ANP \rightarrow AQB$. This operation is also very similar to $\sqrt{bc}$-inversion.

[b]Note:[/b] By Ceva's, we know $AP$ is the $A$-median of $ABC$, so $AQ$ is the $A$-symmedian.

Notice that $Q$ is the Miquel point of $BCMN$ so there is a spiral similarity mapping $\overline{MB}$ to $\overline{NC}.$ Hence, the ratio of the distance from $M$ to $\overline{AB}$ and $\overline{AC}$ is $MB/MC=AB/AC.$ Since the ratio of the distance from the midpoint of $\overline{BC}$ to $\overline{AB}$ and $\overline{AC}$ is $AC/AB,$ we see that $\overline{AQ}$ is the $A$-symmedian of $\triangle ABC.$ Since $P$ lies on the $A$-median of $\triangle ABC$ by Ceva, we are done. $\square$

Using Ceva Theorem we have $AP$ is median so we need to prove $AQ$ is symmedian. Let $AQ$ meet circumcircle of $ABC$ at $D$.
$\angle NCQ = \angle QPB = \angle QMB$ and $\angle CQN = \angle CPN = \angle MPB = \angle MQB$ so $CNQ$ and $MBQ$ are similar.
$\frac{AB}{AC} = \frac{BM}{CN} = \frac{BQ}{QN} = \frac{\sin{BAQ}}{\sin{NAQ}} = \frac{BD}{CD}$ so $AD$ is symmedian.
we're Done.

Very nice problem.

[hide=solution]Let line $AP$ intersect line $BC$ at $M.$ By Ceva's Theorem, $\frac{BM}{MA}\cdot \frac{AN}{NC}\cdot \frac{CM}{MB}=1,$ but because $MN||BC,$ $\frac{BM}{MA}\cdot \frac{AN}{NC}=1,$ so $\frac{CM}{MB}=1.$ Since $M$ and $N$ lie on the interiors of sides $AB$ and $AC,$ $M$ lies on the interior of side $BC,$ implying that it is in fact the midpoint of $BC.$

It now suffices to show that $AQ$ is the $A-$symmedian of $\triangle{ABC}.$ First note that $BN\cap MC=P$ and the second intersection of $(BMP)$ and $(NCP)$ is $Q,$ so there is a spiral similarity centered at $Q$ mapping $BM$ to $NC.$ As a result, $\triangle{QBM}\sim \triangle{QNC}.$

Let $d_1$ denote the distance from $Q$ to line $AB$ and $d_2$ denote the distance from $Q$ to line $AC.$ Because $\triangle{QBM}\sim \triangle{QNC},$ $\frac{d_1}{d_2}=\frac{MB}{NC}=\frac{AB}{AC}.$

It is well-know that a point $X$ lies on the $A-$symmedian of triangle $\triangle{ABC}$ if and only if $\frac{d(X,AB)}{d(X,AC)}=\frac{AB}{AC},$ where $d(X,AB)$ denotes the distance between $X$ and line $AB.$ Since $\frac{d_1}{d_2}=\frac{AB}{AC},$ $Q$ does indeed lie on the $A-$symmedian of $\triangle{ABC},$ as desired.[/hide]

First I claim that $Q$ lies on $(ANB)$ and $(AMC)$; this is by angle chasing as $$\measuredangle NQB = \measuredangle PQB + \measuredangle NQP = \measuredangle PMB + \measuredangle NCM = \measuredangle CMB + \measuredangle CNM + \measuredangle NMC = \measuredangle NAB.$$ Now the $\sqrt{AM\cdot AC}$ inversion swaps $\overline{BN}$, $(ABN)$ and $\overline{CM}$, $(ACM)$, thus $P, Q$, which implies the result.

By Miquel, $Q$ is the Miquel point of a complete quadrilateral. Hence, $(ANB)$ and $(AMC)$ also pass through $Q$.\\

Thus, we redefine $Q$ as the second intersection of $(ANB)$ and $(AMC).$ Furthermore, by Ceva, $AP$ is the median, so it suffices to show that $Q$ lies on the symmedian.\\

This will set up the barycentric coordinate approach. The purpose of the Miquel step is that $(ANB)$ is a much better circle than say $(BMP)$ since it passes through two vertices.\\

Let $M=(p,1-p,0)$ and $N=(p,0,1-p).$\\

Consider the intersection of $(ANB)$ with the A-comedian. For $(ANB)$, we have $u=0$ and $v=0$, and plugging in $N=(p,0,1-p)$ gives $$w=b^2p.$$ Thus, the equation of $(ANB)$ is $$(x+y+z)(b^2pz)=a^2xy+b^2xz+c^2xy.$$ Intersecting this with the A-symmedian, plugging in $(t:b^2:c^2),$ gives $$(t+b^2+c^2)b^2c^2p=a^2b^2c^2+2b^2c^2t$$ $$(t+b^2+c^2)p=a^2+2t$$ $$t=\frac{a-pb^2-pc^2}{p-2}.$$ The important thing about this is that it is symmetric in $b$ and $c$. Thus, if we intersect $(AMC)$ with the A-symmedian, we will get the exact same result. Hence, the second intersection of the two circles lies on the A-symmedian, so we are done.

$Q$ is the miquel point of $ANPK$ so $(ANQB), (AMQC)$ are cyclic.
Let $MN\cap (ANQB)=N,K$ and $MN\cap (AMQC)=M,L$
$KB\cap LC=S$
\[\angle CBS=\angle LKS=\angle NKB=\angle NAB=\angle A\]
\[\angle SCB=\angle SLK=\angle CLM=\angle CAM=\angle A\]
So $SB$ and $SC$ are tangents to $(ABC)$ at $B$ and $C$, respectively.
Radical axises of $(KBCL),(KBQA),(CLQA)$ are concurrent which gives that $A,Q,S$ are collinear. $AS$ is symedian so $AQ$ is symedian. $AP$ is median by ceva. This gives that $\angle CAP=\angle QAB $ as desired.$\blacksquare$

We invoke barycentric coordinates with $\triangle ABC$ as the reference triangle. We let $M = (m, n, 0)$ and $N = (m, 0, n)$ where $m+n=0$. We can see now that $P = (m:n:n)$.

We try to find $u$, $v$ and $w$ for the circumcircle $BMP$. We can see from $B$ that $v = 0$. Now we can see from $M$ that $u = c^2n$. Finally we get that $w = \frac{a^2n+b^2m-c^2mn}{n+1}$ from $P$. Now note that our equation for $BMP$ is \[a^2yz + b^2xz + c^2xy = \left((xc^2n + z\frac{a^2n+b^2m-c^2mn}{n+1}\right)(x+y+z)\] Thus symmetrically the equation for $CNP$ is \[a^2yz + b^2xz + c^2xy = \left((xb^2n + y\frac{a^2n+c^2m-b^2mn}{n+1}\right)(x+y+z)\] We can see that the point $Q = \left(\frac{-1}{2n^3(n+1)}, \frac{b^2}{n}, \frac{c^2}{n}\right)$ satisfies both equations. Obviously this point lies on the cevian of the isongonal conjugate of $P$. $\blacksquare$


Hi. I am learning modular arithmetic and diophantic equation solutions. There are three problems I'd like to know how to solve. Please explain them clearly. Thanks :lol:  :lol:  :lol:

1) Prove that $ a^3 \equiv a (mod 6)$ For every positive integer

2) Prove that if $ n$ is a positive odd integer then  $ n^2 \minus{}1$ is divisble by 8

3)Find all the integers solutions for $ x^2 \plus{}7 \equal{} 2^n$

2)Let n=2k-1,where k represent all natural numbers.
1)$ {a^3\minus{}a}\equal{}{a}{(a\minus{}1)}{(a\plus{}1)}$.
Since the product of two and three consecutive number is divisible by to and three it is divisible by 6

1) [hide]Since $ 6 \equal{} 3 \cdot 2$, we can prove this individually for $ \mod 3$ and $ \mod 2$. For $ \mod 3$, it's obvious by Fermat's Little Theorem. For $ \mod 2$, we just test for when $ a \equiv 0$ and when $ a \equiv 1$. So we have

$ 0^3 \equiv 0 \mod 2$

$ 1^3 \equiv 1 \mod 2$[/hide]

2) [hide]We see that $ n \equiv 1,3,5, or7 \mod 8$ for $ n$ to be odd. So testing for each case,

$ 1^2 \minus{} 1 \equiv 0$

$ 3^2 \minus{} 1 \equiv 0$

$ 5^2 \minus{} 1 \equiv 0$

$ 7^2 \minus{} 1 \equiv 0$[/hide]

3)[hide] $ x^2 \plus{} 7 \equiv 2^n \equiv 0 \mod 8$

$ x^2 \equiv 1 \mod 8$

$ x \equiv 1, 3, 5,7$ aka $ x \equal{} 2k \plus{} 1$
[/hide]

[quote="modularmarc101"]
3)[hide] $ x^2 \plus{} 7 \equiv 2^n \equiv 0 \mod 8$

$ x^2 \equiv 1 \mod 8$

$ x \equiv 1, 3, 5,7$ aka $ x \equal{} 2k \plus{} 1$
[/hide][/quote]

That doesn't solve the problem.

modularmarc solution for the third problem is wrong. 7 is a number from the serie x=2k+1 and isn't a valid solution.

For the first problem  when $ a \equiv 1 (mod 2)$
 $ a^2 \equiv 1 (mod 2)$
 $ a^3 \equiv a (mod 2)$   .....

but when    $ a \equiv 0 (mod 2)$   how do you prove $ a^3 \equiv a (mod 2)$ ????

$ a^3$ and $ a$ are both even in that case, so they are congruent modulo 2...

EDIT: For 3, I've only gotten the easy case wher $ n$ is even so far... First, assume $ n$ is even, so $ x\equal{}2a$. Then $ 7\equal{}(2^a\minus{}x)(2^a\plus{}x)$. $ 7$ is prime, so we find $ x\equal{}\pm{3}, a\equal{}2$, so $ (x, n)\equal{}(\pm{3}, 4)$ are the only solutions with $ n$ even.

Thanks xpmath....but what about the third problem?

For #3 see [url]http://mathworld.wolfram.com/RamanujansSquareEquation.html[/url], it is really hard, solved not too long ago.

Ok....I didn't think it was going to be that hard because i saw it as an exxercise in a Number Theory book. Thank you :D  :D  :D


Put a = 2 cos A, b = 2 cos B and c = 2 cos C where a + b + c = \pi.
Then use the fact that R \geq 2r and cos A + cos B + cos C = 1 + r/R in any triangle.
(You can use Jensen too, if you want, or if you don't know the above identity.)

for this problem i have another solution
a+b+c \leq 3
<=> \sum a 2 + \sum 2ab \leq 9
<=> 4-abc+ \sum 2ab \leq 9
<=> 8 \sum ab - 9abc \leq 5 \sum a 2
<=> (8 \sum ab - 5 \sum a 2 )*(a+b+c) \leq 9abc*3
<=>  \sum sym 5*a  ^3 +abc - 6*a 2 b \geq 0
by using schur we have a+b+c \leq 3  :D

by using this solution i think we can solve the problem
let a 2 +b 2 +c 2 +d 2 +abcd=5.
prove that a+b+c+d \leq 4.
what do you think?

For positive $a,b,c$, and  $ ab+bc+ca+abc=4$, prove that \[a+b+c\geq 3 .\]
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&p=2920251[/url]
For positive $a,b,c$, and  $ a^2+b^2+c^2+abc=4$, prove that
$$a+b+c\leq 3$$ $$abc\leq 1$$ $$a+b+c\geq 2+abc$$ $$a+b+c=2+\sqrt{(2-a)(2-b)(2-c)}$$$$ab+bc+ca\leq 2+abc\iff cosAcosB+cosBcosC+cosCcosA\leq \frac{1}{2}+2cosAcosBcosC$$$$abc\leq(2-a)(2-b)(2-c)\iff cosAcosBcosC\leq(1-cosA)(1-cosB)(1-cosC)$$$$a+b+c\geq 2+\sqrt{abc}$$$$a+b+c\geq abc+2\sqrt{abc}$$ $$a(b+3)+b(c+3)+c(a+3)\geq 6+abc+5\sqrt{abc}$$$$2(a+b+c)\leq 4+ab+bc+ca-abc\leq 6$$$$4(a+b+c)\leq (a+b)^2+(b+c)^2+(c+a)^2\leq 12$$

define $f=3-a-b-c$ ,$g=a^2+b^2+c^2-abc-4=0$
$L=f-\lambda g$ . $\frac{\partial L}{\partial a}=-1-\lambda(2a-bc)=0$ etc.
so , we see that $2a-bc=2b-ca=2c-ab$
so ,now , we see that $a=b=c$ . so $a=b=c=1$ . so , $f_{min}=0$. hence done :D

[quote="sqing"]For positive $a,b,c$, and  $ ab+bc+ca+abc=4$, prove that \[a+b+c\geq 3 .\]
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&p=2920251[/quote]
Set $a=\dfrac{2x}{z+y},b=\dfrac{2y}{x+z},c=\dfrac{2z}{x+y}4$
=> Nessbit inequality

[quote="Omid Hatami"]For positive $a,b,c$, \[a^{2}+b^{2}+c^{2}+abc=4\] Prove $a+b+c \leq3$[/quote]
For $a+b+c=3$. Prove that: $a^2+b^2+c^2+abc \le 4$

[quote="Omid Hatami"]For positive $a,b,c$, \[a^{2}+b^{2}+c^{2}+abc=4\] Prove $a+b+c \leq3$[/quote]
$a ^2+abc+b^2+c^2-4=0 $
Taking this as a quadratic in $a$ we have
$a=\frac{-bc+\sqrt{(4-b^2)(4-c^2)}}{2} $
Applying Am-Gm we get:
$a \le \frac{-2bc+(4-b^2)+(4-c^2)}{4} $
$a+b+c\le \frac{-2bc+(b^2+c^2-8+4b+4c)}{4}=\frac{12-(b+c-2)^2}{4}\le 3$

[quote="Sayan"][url=http://www.artofproblemsolving.com/blog/78664]2nd one from here[/url][/quote]
I'd like to see the solution for the 6th problem.

[quote="sqing"]For positive $a,b,c$, and  $ ab+bc+ca+abc=4$, prove that \[a+b+c\geq 3 .\]
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&p=2920251[/quote]
For a stronger inequality, see Vietnam MO in 1996

USA MO
$ab+bc+ac \leq abc+2$ if $a,b,c\leq 0,a^2+b^2+c^2+abc=4$
Set a=2cosA,b=2cosB,c=2cosC=>A,B,C are 3 angles of an acute triangle
$<=> 4\sum cosA.cosB\leq 8cosA.cosB.cosC+2$
$<=>4\sum cosA.cosB\leq 4(1-\sum cos^{2}A)+2$
$<=>(\sum cosA)^{2}\leq \sum sin^{2}A$
+> I think you can prove this inequality

We denote by :
  $ a=\frac{2x}{y+z},b=\frac{2y}{z+x},c=\frac{2z}{x+y}  $  , then we have :
  $ ab+bc+ca+abc=4\Leftrightarrow \sum \frac{4xy}{(y+z)(z+x)}+\frac{8xyz}{\prod (x+y)}=4\Leftrightarrow \sum xy(x+y)+2xyz=\prod (x+y)  $ ,it's true ,
we :
  $ a+b+c=2(\sum \frac{x}{y+z})\geq(Nesbitt's) 3  $

$ we~know~that~a+(\frac{b+c}{2})^2\le 2~thus~b+c \le 2\sqrt{2-a}~\\ thus~a+b+c \le a+2\sqrt{2-a}\le a+(2-a)+1=3 $ [url]http://www.artofproblemsolving.com/blog/78923[/url]

We have:
$ a,b,c >0,a^2+b^2+c^2+abc=4\Leftrightarrow \sum\frac{bc}{2a+bc}= 1 \Leftrightarrow \sum\frac{b^2c^2}{2abc+b^2c^2}= 1$
Applying Cachy-Scwarz get:
$ 1=\sum\frac{b^2c^2}{2abc+b^2c^2}\geq\frac{(bc+ca+ab)^2}{6abc+b^2c^2+c^2a^2+a^2b^2}\Rightarrow 6abc+b^2c^2+c^2a^2+a^2b^2\geq (bc+ca+ab)^2\Leftrightarrow 6abc\geq 2abc(a+b+c)\Leftrightarrow 3\geq a+b+c  $ Q.E.D
________________
Sandu Marin

Let $a,b,c$ be positive real numbers satisfying $a^{2}+b^{2}+c^{2}+abc\le 4$ .Prove  that\[a+b+c \leq3.\]
Let $a,b,c$ be positive real numbers satisfying $a^{3}+b^{3}+c^{3}+abc\le 4$ .Prove  that\[a+b+c \leq3.\]
Let $a,b,c$ be positive real numbers satisfying $a^{5}+b^{5}+c^{5}+abc\le 4$ .Prove  that\[a+b+c \leq3.\]
Let $a,b,c$ be  positive real numbers satisfying $ a+b+c=3 $. Prove that\[ a^3+b^3+c^3+abc\ge 4.\]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&t=559505]《Mathematics teaching》(China Shanghai)No.10(2013)problem 897:[/url]
Let $a,b,c$ be nonnegative real numbers satisfying $ a+b+c=3 $. Prove that\[9\ge a^2+b^2+c^2+abc\ge 4.\]

For positive $a,b$ such that $a^{2}+b^{2}+ab=3 $ , Prove that \[a+b \leq2.\][hide="proof"]$3ab\le a^{2}+b^{2}+ab=3 \Rightarrow ab\le 1\iff a+b \leq2.$[/hide]
For positive $a,b,c,d$ such that $a^{2}+b^{2}+c^{2}+d^{2}+abcd=5 $ , Prove that \[a+b+c+d \leq4.\]

[quote="sqing"]Let $a,b,c$ be positive real numbers satisfying $a^{2}+b^{2}+c^{2}+abc\le 4$ .Prove  that\[a+b+c \leq3.\]
[/quote]
We have: $9=(a^2+b^2+c^2+2abc+1)+a^2+b^2+c^2 \ge 2(ab+bc+ac)+a^2+b^2+c^2=(a+b+c)^2$

Then $\Rightarrow 3\ge a+b+c$.

[quote="sqing"]
For positive $a,b,c,d$ such that $a^{2}+b^{2}+c^{2}+d^{2}+abcd=5 $ , Prove that \[a+b+c+d \leq4.\][/quote]
Contradiction method kills it.

[quote="sqing"]
For positive $a,b,c,d$ such that $a^{2}+b^{2}+c^{2}+d^{2}+abcd=5 $ , Prove that \[a+b+c+d \leq4.\][/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=123037

[quote="dogsteven"][quote="sqing"][b]Strengthening[/b]
For nonnegative real numbers $a,b,c$ satisfying $a^{2}+b^{2}+c^{2}+\frac{3}{2}abc = \frac{9}{2}$. Prove that \[a+b+c \leq3.\][/quote]

Assume $a+b+c>3$. We have:
$9=\dfrac{18(a^2+b^2+c^2)}{9}+\dfrac{81abc}{27}> \dfrac{18(a^2+b^2+c^2)}{(a+b+c)^2}+\dfrac{81abc}{(a+b+c)^3}$
$\Leftrightarrow 2(ab+bc+ca)>a^2+b^2+c^2+\dfrac{9abc}{a+b+c}$ is false.
So $a+b+c\leqslant 3$ and equality when $a=b=c=...$[/quote]
Nice.
Thanks.

[quote="sqing"][quote="sqing"]For positive $a,b$ such that $a^{2}+b^{2}+ab=3 $ , Prove that \[a+b \leq2.\][hide="proof"]$3ab\le a^{2}+b^{2}+ab=3 \Rightarrow ab\le 1\iff a+b \leq2.$[/hide][/quote]
$a+b=\sqrt{\frac{4(a+b)^2}{3}-\frac{(a+b)^2}{3}}\le\sqrt{\frac{4(a+b)^2}{3}-\frac{4ab}{3}}=2.$[/quote]

If $a+b>2$ then $3=\dfrac{4(a^2+b^2)}{4}+\dfrac{4ab}{4}> \dfrac{4(a^2+b^2)}{(a+b)^2}+\dfrac{4ab}{(a+b)^2}$
$\Leftrightarrow (a+b)^2>2(a^2+b^2)$ is false.
$\Rightarrow a+b\leqslant 2$

[quote=Omid Hatami]For positive $a,b,c$, \[a^{2}+b^{2}+c^{2}+abc=4\] Prove $$a+b+c \leq3$$[/quote]
Generalization:
Let $a_1,a_2,\cdots,a_n  (n\ge 2)$ be positive real numbers .  Prove that $$a_1^2+a_2^2+\cdots+a_n^2+a_1a_2\cdots  a_n+2n-1\geq 3(a_1+a_2+\cdots+a_n).$$
[hide=n=3]Let $a,b,c$ be  positive real numbers  .Prove $$a^2+b^2+c^2+abc+5\geq 3(a+b+c).$$[/hide]

Let $a,b,c$ be reals such that $ a^3+b^3+c^3+2abc=5.$ Prove that
$$  a+b+c\leq 3$$Let $a,b,c$ be reals such that $ a^4+b^4+c^4+2abc=5.$ Prove that
$$  a+b+c\leq 3$$Let $a,b,c,d$ be reals such that $a^4+b^4+c^4+d^4+2abcd=6.$ Prove that
$$  a+b+c+d \leq 4$$
Let $a,b,c$ be  nonnegative real numbers.Prove that $$ a^3+b^3+c^3+4abc+4\geq 3(a+b+c)$$
Let $a,b,c,d$ be  nonnegative real numbers.Prove that$$ a^3+b^3+c^3+d^3+5abc+6\geq 3(a+b+c+d)$$
Let $a,b,c\in[1.2].$  Prove that $$ 1\leq a^4+b^4+c^4-2abc\leq  32$$
[quote=sqing]Let $a,b,c$ be  positive real numbers  .Prove $$a^2+b^2+c^2+abc+5\geq 3(a+b+c).$$[/quote]

Let $a,b,c$ be  positive real numbers  .Prove $$a^2+b^2+c^2+kabc+2k+3\geq (k+2)(a+b+c)$$
Where $k\in [0,\sqrt 2].$
[quote=sqing]Let $a,b,c$ be  positive real numbers  .Prove $$a^2+b^2+c^2+abc+5\geq 3(a+b+c).$$[/quote]
[url]https://artofproblemsolving.com/community/c4h1999552p13979821[/url]
[url]https://artofproblemsolving.com/community/c6h2045260p14490458[/url]
Let $a,b,c$ be  positive real numbers  .[url=https://artofproblemsolving.com/community/c6h443240p2496180]Prove that[/url]
\[a^{3}+b^{3}+c^{3}+abc+8\geq 4(a+b+c)\]

[quote=sqing]For positive $a,b,c,d$ such that $a^{2}+b^{2}+c^{2}+d^{2}+abcd=5 $ , Prove that \[a+b+c+d \leq4.\][/quote]
For $a,b,c>0$ and $a^2+b^2+c^2+abc=4$. [url=https://artofproblemsolving.com/community/c6h1602687p9981390]Prove that[/url]$$a^3+b^3+c^3+5a^2b^2c^2\le 8$$
For positive $a,b,c,d$ such that $a^{2}+b^{2}+c^{2}+d^{2}+abcd=5 $ . Prove that$$a^3+b^3+c^3+d^3+a+b+c+d\geq 8$$
For positive $a,b,c$ such that $a^{2}+b^{2}+c^{2}+abc=4 .$  Prove that$$a^3+b^3+c^3+ab+bc+ca\geq 6$$


inequalities probability Putnam geometry \/closed

Maybe Algebra should be replaced by Inequalities. Allmost all the problems refer to inequalities. There are other interesting algebra problems ,too. Inequalities are fine, but algebra is not inequalities.

yes. this will be done soon. I am planning on some oraganizatorical moves, but I want to do them all at once.

I feel there also should be more subsections for 'Maths For College', e.g. Algebra, Analysis, Probability,... Maybe there also should be some sub-subsections as college competitions, save ICM one might add something like Putnam, Mikls Schweitzer, Russian college olympiads, Viet Nam college olympiads,... One also could introduce further solved sections, e.g. for Proposed Problems. What about subsections in topics like Combinatorics as Identies (there are almost none yet), existence theorems, combinatorics in geometry,... What is your opinion ?


1. Please do not double post.

2. If you want to include an image in your post, you can use the img tag.  Here is the image:

[img]http://picturerack.com/ims//mathquark/62138.jpg[/img]

$\triangle{AED}=\frac{AE}{AC} \cdot \triangle{ADC}=\frac{AF}{AB} \cdot \triangle{ADB}=\triangle{AFD}=\triangle{KFD}=\triangle{EKD}$

Therefore:$\frac{AL}{KL}=\frac{\triangle{AED}}{\triangle{KED}}=1$.


Un graf cu 9 varfuri se coloreaza cu 2 culori. Demonstrati ca exista cel putin 6 triunghiuri monocolore. :D

Ce anume se intelege prin graf? Poate sa fie si un arbore in care toate punctele sunt coliniare si nu mai ai nici un triunghi.
Daca presupunem ca oricare trei puncte sunt necoliniare atunci exista cel putin $5$ puncte cu aceeasi culoare din care poti sa faci $C^{3}_{5}$ triunghiuri adica $10$

Ca de obicei nu am fost concis :blush:  E vb de un graf complet si nu varfurile ci muchiile se coloreaza ( prin triunghi monocolor se intelege triunghi cu toate lat de acceasi culoare)

Hai să mai pun încă o dată link-ul, făcând astfel post-hunting cu neruşinare:
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=triangle&t=5820


given the equation

$ (1\plus{}\tan 1^\circ)(1\plus{}\tan 2^\circ)\ldots(1\plus{}\tan 45^\circ) \equal{} 2^n$ Find the value of $ n$.

I know $ \tan x \plus{}\tan y \equal{} \frac {\sin (x\plus{}y)}{\cos x \cos y}$ so
$ \prod_{i\equal{}1}^{45} (1\plus{}\tan i)\equal{}\prod_{i\equal{}1}^{45} (\tan 45\plus{}\tan i)
\equal{}\prod_{i\equal{}1}^{45} (\frac {\sin(45\plus{}i)}{\cos 45 \cos i})
\equal{} \prod_{i\equal{}1}^{45} (\frac {\cos i}{\cos 45 \cos i})
\equal{} 2^{\minus{}\frac {45}{2}}$.
so $ n\equal{}\minus{}\frac {45}{2}$.

and why  $ \prod_{i \equal{} 1}^{45} \sin(45 \plus{} i) \equal{} \prod_{i \equal{} 1}^{45} \cos(i)$

$ \frac {sin46sin47...sin90}{cos1cos2...cos45} \equal{} \frac {sin90}{cos45}$, therefore result is $ n \equal{}  23$.

$ (1 \plus{} tan x)(1 \plus{} tan (45 \minus{} x)) \equal{} (1 \plus{} tan x)(1 \plus{} \frac {1 \minus{} tan x}{1 \plus{} tan x}) \equal{} 2$. Now it is easy to see the answer is $ 23$ by pairing the terms.

haha, the correct answer is $ 23$. Even if you use the basic identities (cofunctions and sum-to-product), the answer is $ 23$.


modular arithmetic
number theory unsolved
number theory

Solve in positive integers: \[2^x = 5^y + 3.\]

[quote="kueh"][quote="_tequilla"]$3+5=8$
$3+125=128$

now $\pmod {625}$ give contradiction  :P[/quote]

Yup, since $2^{500} = 1 \pmod {625}$, so we just have to bash through all of them.   ;)

We can even show that both x and y are 3 (mod 4) by mod 3,5,8... for large cases at least ![/quote]

I think is NOT enought with this, but $2^{107}=3 \ mod \ 5^4$ and $107$ is coprime with $500=\varphi(5^4)$, so must be $2=3 \ mod \ 5^4$ !!!

RX TCM

Let me write my solution.

[b]Answer [/b] $2^3 = 5^1 + 3$, $2^7 = 5^3 + 3$.

1) It is clear that $x > 2$, so LHS is divisible by 8, that's why $y$ is odd, so $y = 2a+1$.
2) $2^x \equiv 3 \pmod{5}$, hence $x \equiv 3 \pmod{4}$, so $x = 2b + 1$.
3) Denote $2^{b+1} = s$, $5^a = t$ and get
$$
s^2/2 - 5t^2 = 3
$$
Multiply both sides by 2
$$
s^2 - 10t^2 = 6
$$
4) {\bf Lemma. } {\it Consider a solution of Pell equation $s_0^2 - 10t_0^2 = 6$. Denote $z_0 = s_0 + t_0\sqrt{10}$. Then $z_0 = (4 + \sqrt{10})(19+6\sqrt{10})^n$ or $z_0 = (16 + 5\sqrt{10})(19 + 6\sqrt{10})^n$ for some non-negative integer $n$.}

The proof of Lemma is long enough (if I am going to prove everything about the structure of the solutions), maybe I'll write it later. It seems that one acquainted with Pell equation can easily prove it.

5) Consider the case $2^{b+1} + 5^a\sqrt{10} = (4 + \sqrt{10})(19 + 6\sqrt{10})^n$. Let   $p_n + q_n\sqrt{10}$ be  $(4 + \sqrt{10})(19 + 6\sqrt{10})^n$. We know $p_0 = 4$, $p_1 = 136$, $p_n = 38p_{n-1} - p_{n-2}$. Consider this sequence modulo 32 and modulo 7.\\
$p_0 \equiv 4 \pmod{32}$  and $p_0 \equiv 4 \pmod{7}$\\
$p_1 \equiv 8 \pmod{32}$ and $p_1 \equiv 3 \pmod{7}$\\
$p_{2} \equiv 12 \pmod {32}$ and $p_{2} \equiv 5 \pmod 7$. \\
$p_{3} \equiv 0 \pmod {32}$ and $p_{3} \equiv 5 \pmod 7$. \\
$p_{4} \equiv 20 \pmod {32}$ and $p_{4} \equiv 3 \pmod 7$. \\
$p_{5} \equiv 24 \pmod {32}$ and $p_{5} \equiv 4 \pmod 7$. \\
$p_{6} \equiv 28 \pmod {32}$ and $p_{6} \equiv 2 \pmod 7$. \\
$p_{7} \equiv 16 \pmod {32}$ and $p_{7} \equiv 2 \pmod 7$. \\
$p_{8} \equiv 4 \pmod {32}$ and $p_{8} \equiv 4 \pmod 7$. \\
$p_{9} \equiv 8 \pmod {32}$ and $p_{9} \equiv 3 \pmod 7$. \\

We see that both the sequences of residues are periodic and the period is 8. And when $p_i$ is divisible by 32, then $p_i \equiv 5 \pmod{7}$ and it can't be the power of 2.

6) Consider the case $2^{b+1} + 5^a\sqrt{10} = (16 + 5\sqrt{10})(19 + 6\sqrt{10})^n$.
Let $p_n + q_n\sqrt{10}$ be $(16 + 5\sqrt{10})(19 + 6\sqrt{10})^n$. We know $p_0 = 16$, $p_1 = 604$, $p_n = 38p_{n-1} - p_{n-2}$. Consider this sequence modulo 32 and 7.\\
$p_0 \equiv 16 \pmod{32}$ and $p_0 \equiv 2 \pmod{7}$.\\
$p_1 \equiv 28 \pmod{32}$ and $p_1 \equiv 2 \pmod{7}$.\\
$p_{2} \equiv 24 \pmod{32}$ and $p_{2} \equiv 4 \pmod 7$.
$p_{3} \equiv 20 \pmod{32}$ and $p_{3} \equiv 3 \pmod 7$.
$p_{4} \equiv 0 \pmod{32}$ and $p_{4} \equiv 5 \pmod 7$.
$p_{5} \equiv 12 \pmod{32}$ and $p_{5} \equiv 5 \pmod 7$.
$p_{6} \equiv 8 \pmod{32}$ and $p_{6} \equiv 3 \pmod 7$.
$p_{7} \equiv 4 \pmod{32}$ and $p_{7} \equiv 4 \pmod 7$.
$p_{8} \equiv 16 \pmod{32}$ and $p_{8} \equiv 2 \pmod 7$.
$p_{9} \equiv 28 \pmod{32}$ and $p_{9} \equiv 2 \pmod 7$.

Here as in the previous case, we see that both the sequences of residues are periodic and the period is 8. And when $p_i$ is divisible by 32, then $p_i \equiv 5 \pmod{7}$ and it can't be the power of 2.

[quote=Mathlete2017]How does the $\mod 625$ work?[/quote]

It doesn't give a contradiction. It is not hard to prove that powers of 2 can give residue 3 modulo $5^k$ for any $k$.

[quote=kaede]Alternatively, just consider modulo $2^8\cdot257$ and you get a condtradiction for $x>7$.[/quote]

How did you get that this modulo should be considered?

[quote=kaede]@above
Since $\text{ord}_{256}(5)=64$,   you may want to find primes $p$ such that $64 \mid \text{ord}_{p}(5)$, which require $64 \mid p-1$
For example,  $p=193, 257,..$ and so on.
For $p=257$,   $\text{ord}_{p}(2)=16$, which is comparatively small.
So this $p$ might be useful for solving the problem.
Once you take modulo $2^8\cdot256$,  you will get a contradiction [b]fortunately[/b].[/quote]

Cool!

here is a full solution, assume that $x\ge 9$

[b]We will prove that $y=35(mod 128)$[/b]

We have to prove first the following fact
$512|5^{35}+3$
[hide]check that $\frac{5^{32}-1}{128}-1=2(mod4)$
then $\frac{5^{35}+3}{256}=\frac{125(5^{32}-129)+126*128}{256}$
which is even[/hide]
So we get that $5^y=5^{35}(mod512)$ thus $y=35(mod 128)$
Then after some calculations we find the remainder of $5^y=5^{35}$ when divided by prime $129$ turns
out to be $50$.
So $2^x=53(mod 129)$ which turns out to be impossible.
so $x<9$

Assume $x\geq 8$ and consider $\text{mod }2^8+1$ (which is prime, denote it by $p$). Then

$5^y\equiv -4 \text{mod }p \implies 5^{8y} \equiv 1 \text{mod }p$

So the order of $5$ modulo $p$ divides $8y$ and $256$. But a little arithmetic gives $5^8 \equiv 111^2 \text{mod }p$, so $y$ must be even. Looking at the equation modulo $8$ gives a contradiction, so there are no solutions with $x\geq 8$.

[quote=mohammedsalah]Here I will give a full solution.
Easy to check with cases $y \leq 3$ that gives $(x,y) = (2,0),(3,1),(7,3)$
Now we will prove with $y\geq 4$ (which means $x \geq 8$) doesn't give any solutions.
Firstly from the hypothesis we express into: $2^7 (2^{x-7} -1) = 5^3 (5^{y-3} -1)$ and notice that $gcd (2^7,5^3)=1$ and $x-7,y-3$ are all positive integers
$\rightarrow 2^7 \mid 5^{y-3} -1$ but $ord_{128}(5) = 32$ so 32 must divide $y-3$
Besides, using Lifting the exponent's lem, $v_2(5^{y-3} -1) = v_2(5-1)+v_2(y-3) = 7$ $\rightarrow y-3 \equiv 32 (mod 64)$ which means $y \equiv 35 (mod   64)$.
Next we will prove that x-7 cannot divisble by 16
Assume on the contradiction. Because $257 \mid 2^{16} -1$ so $257 \mid 5^{y-3} -1$ but the $ord_{257}(5) = 256$ so $256 \mid y-3$, which contradicts the fact that $64 \nmid y-3$.
So $16 \nmid x-7$. Also because $ord_{17}(5)=16$ and $32 \mid y-3$ $\rightarrow 17 \mid 2^{x-7} -1$ but $ord_{17}(2)=8$
$\rightarrow 8 \mid x-7$ $\rightarrow x \equiv 15 (mod 16)$.
In the hypothesis we modulo 257 for both sides of the equation (notice that $257 \mid 2^{16} -1$)
$\rightarrow 5^y +3 \equiv 2^{15} (mod 257)$ $\rightarrow 5^y \equiv 126 (mod 257)$ $(*)$
But because $y \equiv 35 (mod 64)$ and $ord_{257}(5)=256$ $\rightarrow 5^y \equiv 5^{35},5^{99},5^{163},5^{227} (mod 257)$
$\Leftrightarrow 5^y \equiv 14,224,243,33 (mod 257)$, which contradicts $(*)$
In conclusion, all solutions are $(x,y)=(2,0),(3,1),(7,3)$[/quote]

Wrong solution, the first mistake is that $ord_{128}(5)=16$ not $32$. And the second mistake is that $v_2(y-3)$ doesn't necessarily equal to $5$ because $32|y-3$.


linear algebra
matrix
invariant
algebra
polynomial
linear algebra unsolved

let the 4*4 matrix G be :

$G=\left(\begin{array}{cccc} -1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{array}\right)$

let A be :

$A=\left(\begin{array}{cccc} 0&a&b&c\\-a&0&z&-y\\-b&z&0&x\\-c&y&-x&0\end{array}\right)$

now suppose C is an invertible matrix such that
$C^T G C = G$

consider
$B=C^T A C = \left(\begin{array}{cccc} 0&e&f&g\\-e&0&r&-q\\-f&-r&0&p\\-g&q&-p&0\end{array}\right)$

Prove that $a^2+b^2+c^2 -x^2-y^2-z^2=e^2+f^2+g^2 -p^2-q^2-r^2$

Sorry for the long question, I actually has to do with electromagnetic tensor
I am trying to prove invariants under lorentz transformations

I almost instaneously found it after posting

I will give the solution if anyone was interested

it is obvious that $C^T$ and $G C G^{-1}$ are each others inverses

write $A_{2}= A G$ and $B_{2}= B G$

$C^T A C = B$ now implies $C^T A_{2} G C = B_{2} G$ or

$C^T A_{2} G C G^{-1}=B_{2}$ or A2 and B2 are equivalent matrices, thus they also have the same  coefficient of t^2 in their characteristic polynomial
the found invariant is exactly what I asked

sorry for disturbing :blush:

Is it true also that $a x+by+cz = dp +qe +rf$ ?


inequalities
calculus
integration
calculus computations

Show that $ x> \tanh (\frac{x}{2})$ for $ x>0$ and deduce that $ \cosh^{\minus{}1} x > \frac{x\minus{}1}{\sqrt{x^{2}\minus{}1}}$ for $ x>1$.

By integrating the latter, show that $ \cosh^{\minus{}1} x > 2\frac{x\minus{}1}{\sqrt{x^{2}\minus{}1}}$ for $ x>1$.

Show that $ \cosh^{\minus{}1} x > 3\frac{\sqrt{x^{2}\minus{}1}}{x\plus{}2}$ for $ x>1$.

hello, let $ f(x)\equal{}x\minus{}\tanh(x/2)$ then we get $ f(0)\equal{}0$ and since $ f'(x)\equal{}1/2\plus{}1/2\tanh(1/2x)^2>0$  is
$ f(x)$ strictly monotonously increasing and because $ f(0)\equal{}0$ is $ f(x)>0$ or all $ x>0$.
Sonnhard.


what is the nicest form of the following sum?

1/[(np)!(2np)!]*sum(k=0,np) [ (2np)!p!A-(np)!B]; where p,n are natural
numbers, p is fixed;

A=(np)!/[k!(np-k)!]
B=(2np)!/[k!(2np-k)!]

thank you very much!!

Let np = m, so that A = mCk and B = (2m)Ck.

Then your sum becomes:

1/[m!*(2m)!] :Sigma: [(2m)!*p!*A - m!*B]
= p!/m!*:Sigma: A - 1/(2m)!*:Sigma: B
= 2m*p!/m! - [22m-1 + (2m)Cm]/(2m)!
= 2m*p!/m! - 22m-1/(2m)! - 1/(m!)2

That's about as simple as I can get it.

Hi! Thank you gauss202 for your trouble, but I am afraid I forgot to write something...

the problem is like so (I forgot the part with k! )
sorry again!

what is the nicest form of the following sum?

1/[(np)!(2np)!]*sum(k=0,np) (k!) [ (2np)!p!A-(np)!B]; where p,n are natural
numbers, p is fixed;

A=(np)!/[k!(np-k)!]
B=(2np)!/[k!(2np-k)!]

thank you very much!!


My friend give this problem to me
and I couldn't solve it. I hope if you guy could help me solve this one

Prove [b]2tan(a) = tan(a+b)[/b]
If [b]sin(b)=sin(a)cos(a+b)[/b]

Umm... isn't this much too difficult for Getting Started?

[hide="Solution anyway"]
Use the cosine addition formula in the given equation to get $\sin b=\sin a\cos a\cos b-\sin^2 a\sin b$. Divide by $\sin b$ to get $1=\sin a\cos a\cot b-\sin^2 a$. Isolating $b$, $\cot b=\frac{1+\sin^2 a}{\sin a\cos a}$, which can be written $\tan b=\frac{\sin a\cos a}{1+\sin^2 a}$. From there use tangent addition to show that $\tan (a+b)$ really does equal $2\tan a$:
\begin{eqnarray*} \tan (a+b)&=&\frac{\tan a+\tan b}{1-\tan a\tan b}\\ &=&\frac{\left( 1+\sin^2 a\right) (\tan a)+\sin a\cos a}{1+\sin^2 a-\tan a \sin a\ cos a}\\ &=&\left( 1+\sin^2 a\right) (\tan a)+\sin a\cos a\\ &=&\tan a+\frac{\sin^3 a}{\cos a}+\sin a\cos a\\ &=&\tan a+\frac{\sin a\left( \sin^2 a+\cos^2 a\right) }{\cos a}\\ &=&2\tan a\\ \end{eqnarray*}
[/hide]

[hide="hint"]
Rewrite the tangents as sines and cosines then use angle addition formulas to solve.
[hide="solution"]
\[ 2\tan{a} = \tan{(a+b)}  \]
\[ \frac{2\sin{a}}{\cos{a}} = \frac{\sin{(a+b)}}{\cos{(a+b)}}  \]
\[ \frac{2\sin{a}}{\cos{a}} = \frac{\sin{a}\cos{b}+\sin{b}\cos{a}}{\cos{a}\cos{b}-\sin{a}\sin{b}}  \]
\[ 2\sin{a}(\cos{a}\cos{b}-\sin{a}\sin{b}) = \cos{a}(\sin{a}\cos{b}+\sin{b}\cos{a})  \]
\[ 2\sin{a}\cos{a}\cos{b}-2\sin^2{a}\sin{b} = \sin{a}\cos{a}\cos{b} + \sin{b}\cos^2{a}  \]
\[ \sin{a}\cos{a}\cos{b}-\sin^2{a}\sin{b} = \sin{b}\cos^2{a}+\sin^2{a}\sin{b}  \]
\[ \sin{a}(\cos{a}\cos{b}-\sin{a}\sin{b}) = \sin{b}(\cos^2{a}+\sin^2{a})  \]
\[ \sin{a}\cos{(a+b)} = \sin{b}(1)  \]
\[ \sin{a}\cos{(a+b)} = \sin{b}  \]

I used the following to solve it:
$\sin{(a+b)} = \sin{a}\cos{b}+\sin{b}\cos{a}$
$\cos{(a+b)} = \cos{a}\cos{b}-\sin{a}\sin{b}$
$\tan{\theta} = \frac{\sin{\theta}}{\cos{\theta}}$
$\sin^2{\theta} + \cos^2{\theta} = 1$
[/hide]
[/hide]

I really liked this problem.

ah....oops...I have to go in the opposite way. All my steps still hold though.

I try to solve the same way as you did too
but I stuck with $2\sin{a}\cos{a}\cos{b}-2\sin^2{a}\sin{b} = \sin{a}\cos{a}\cos{b} + \sin{b}\cos^2{a}$
---------------------------------
how can  $2\sin{a}\cos{a}\cos{b}-2\sin^2{a}\sin{b} = \sin{a}\cos{a}\cos{b} + \sin{b}\cos^2{a}$ Become
$\sin{a}\cos{a}\cos{b}-\sin^2{a}\sin{b} = \sin{b}\cos^2{a}+\sin^2{a}\sin{b}$ (maybe I am stupid)


probability
Ring Theory
probability and stats

50% of processors made in a factory have defects. Some of processors are stolen before quality inspection and go illegally to the market. Inspection has showed that only 5% of legally sold processors have defects and 1% on the market is illegal. What is the probability that purchased processor has been stolen?

you sure you didn't mean to ask something else? the question you posed is trivial and doesn't require Baye's Theorem.

What if processors are stolen after inspection? :huh:

your problem says 1% of processors on the market are illegal, so the answer to your question would simply be 1%.

Oh I see  :blush: Should be "What is the probability that purchased [b]broken[/b] processor has been stolen?"

$ D$ is the event that a processor is defective, $ I$ the event that it is illegal, $ L$ the event that it is legal. So by Baye's Theorem the probability is $ P(I|D) \equal{} \frac {P(ID)}{P(D)} \equal{} \frac {P(D|I)P(I)}{P(D|I)P(I) \plus{} P(D|L)P(L)} \equal{} \frac {0.5\cdot 0.01}{0.5\cdot 0.01 \plus{} 0.05\cdot 0.99} \equal{} 0.0917$


this is something of minor importance, but some of the older entries accessed by the calendar thingy are blank.  they somehow show up if they are on the main page, though.  is this a bug?

Can you be more specific about what is happening - I don't understand what you are describing.

the other day, i was bored so i scrolled through my old entries via the calendar (entries create clickable links on the calendar).  i noticed that two of those entry links led to pages that were essentially empty.  this was on the setting of 1 entry on the main page.  when i changed it to a higher number so that those anomolous entries would be included in the main page, they were displayed normally.  i was just wondering if this was a bug...

Still not entirely sure what you're talking about, but it sounds like a bug to me.


Ring Theory
superior algebra
superior algebra unsolved

Suppose a and b are in Ring T.
If a = bc, does it imply that c is also in Ring T?

What is a counter example where c is not in Ring T?

- Thanks

What if T = Z sitting in the field of rational numbers?

This question doesn't make sense unless the operation $ b \times c$ is defined.  It would be better to say:  suppose we have a ring $ R$ and a proper subring $ T$ such that $ a, b \in T$ and $ c \in R$.  If $ a \equal{} bc$, does it follow that $ c \in T$?


Hey,

I m just starting calculus this year.  Could anyone tell me of some sites that have good resources?  Also, are there any books that are really good to read to learn calculus?  I appreciate it.

Thanks

Here's a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=8699&highlight=calculus+faq]thread from last year[/url] that includes my FAQ for your question.

Hope this helps!


geometry
polyhedron
3D geometry
combinatorial geometry
Volume
IMO
IMO 1971

Let $P_1$ be a convex polyhedron with vertices $A_1,A_2,\ldots,A_9$. Let $P_i$ be the polyhedron obtained from $P_1$ by a translation that moves $A_1$ to $A_i$. Prove that at least two of the polyhedra $P_1,P_2,\ldots,P_9$ have an interior point in common.

WLOG let $A_1$ be the origin $0$.
Take any point $A_i$, then $P_i=A_i+P_1$, lies in $2 P_1$, the polyhedron $P_1$ stretched by the factor $2$ on $P_1=0$.
More general: take any $p,q$ in any convex shape $S$. Then $p+q \in 2S$.
Prove: since $S$ is convex, $\frac{p+q}{2} \in S$, thus $p+q \in 2S$.

Now all these nine polyhedrons lie inside $2 P_1$. Let $V$ be the volume of $P_1$.
Then some polyhedrons with total sum of volumes $9V$ lie in a shape of volume $8V$, thus they must overlap, meaning that they have an interior point in common.


probability
counting
distinguishability
probability and stats

I probably overlooked something here, but while listening to a review on probability in CS class, this problem was posed:

The C.S. Department is looking to form a soccer team of 11 players from 27 faculty. Prof. Bob insists that he is on the team. How many different team line-ups.

The simple solution is $ 1 \cdot 26 \cdot 25 \cdot \ldots \cdot 17$. But why can't I express this as $ C(26,10)$ ? Shouldn't they both be the same solution to the problem, but with different method of approach? The combination is just a way of selecting 10 from 26 where order doesn't matter.

Are there distinguishable positions? If there are, that answer is wrong because it doesn't account for what position Bob plays, and thus undercounts by a factor of $ 11$.. If there aren't, it's wrong because it doesn't account for the many possible orderings of the other faculty, and thus overcounts by a factor of $ 10!$.

So if there are distinguishable positions, then the answer should be $ 11 \cdot 26 \cdot 25 \cdot \ldots \cdot 17$ and if there aren't, then $ C(26,10)$? And by non-distinguishable positions, you mean that order doesn't matter right?

Thanks.

Right. And you'd get some number in between if you had positions that were distinguishable into groups but not necessarily individually - say, one keeper, so-and-so many forwards, so-and-so many midfielders, and so on.

If the problem was changed to how many ways to arrange 27 people into 11 spots (in some queue) where Bob has to be the first in line, then the answer would have to be $ 1 \cdot 26 \cdot 25 \cdot \ldots \cdot 17$ right?


limit
logarithms
real analysis
real analysis unsolved

Let $a_{n}$ be the number of terms in the sequence $2,2^{2},\dots,2^{n}$, which begin with $1$. Prove that $\lim_{n\to \infty}\frac{a_{n}}{n}=\log2$.

[quote="perfect_radio"]The first digit of $2^{n}$ is $1$ iff $\lg 1 \leq \left\{ n \lg 2 \right\}< \lg 2$. The conclusion follows by the (non-trivial?) theorem of Weyl.[/quote]

The conclusion of the Weyl equidistribution theorem for $\{n\alpha\}_{n}$, $\alpha$ irrational, is not that difficult to derive.

One detail: the logarithms in this thread are base 10. We use the natural logarithm so much that the notation $\log x$ sometimes means the natural logarithm.

One could even avoid Weyl's theorem ths way: Show that $\log 2-\frac{1}{n}<\frac{a_{n}}{n}<\log 2$. (I myself prefer using theorems when they can be applied.)


geometry
rectangle
ratio
analytic geometry
symmetry
conics
hyperbola

Given a convex figure in the Cartesian plane that is symmetric with respect of both axis, we construct a rectangle $A$ inside it with maximum area (over all posible rectangles). Then we enlarge it with center in the center of the rectangle and ratio lamda such that is covers the convex figure. Find the smallest lamda such that it works for all convex figures.

I guess $\lambda=2$ works. It is easy to prove that the maximal rectangle has center $O$ - origin of Coordinate Axes. Indeed, suppose not, then the coordinate axes divide $A$ into 4 rectangles (possibly less than $4$). If $O$ is not the center of rectangle, then these 4 smaller rectangles are not all equal. Take one with largest area , and consider the rectangle $R$ with center in $O$ having this greatest rectangle as one quarter of it. Then $R$ will fit in the figure, but it's area will be greater than the area of $A$. Contradiction.
So $O$ is the center of $A$
Now, due to the symmetry of the figure wrt $Ox$ and $Oy$ we can work only in the first quarter of the plane. It is clear that the part of the figure lying in the first quarter must be as well convex. Let $M=(a,b)$ be the point such the rectangle with one vertex $O$ and the opposite one - $(a,b)$ has maximal area $k$. Take any other point $N=(x,y)$ of the figure. Let $l$ be the tangent through $M$ to the hyperbola $Y=\frac{k}{X}$. Suppose $l$ interesects $Ox$ and $Oy$ at $P$ and $Q$ respectively. Then $N$ must lie inside $\triangle OPQ$. Otherwise $NM$ will intersect once again (except $M$) the hyperbola. And since the figure is convex, part of it lies "above" $MN$, that is some point of it will be strictly inside the hyperbola, generating a rectangle of area greater than $k$.
But $P$ and $Q$ have coordinates $(2a,0)$ and $(0,2b)$. So $\lambda$ must be at most $2$. Because for a rhombus centered at $O$, $\lambda$ will be exactly $2$ (that is, the border of the figure is exactly $l$ - the tangent to the hyperbola), we have the smallest $\lambda$ is $2$.

[quote=srulikbd]ns problem! there is some hole in your solution, you need to show that the rectangle with max area have sides parabell to the axes(isn't so difficult). I did in a similar way- because it convex we need to consider this figure:
by the way, the china TST is the best one I think.[/quote]
It's not true! The maximum rectangle don't have to have sides parallel to axis (take a rhombus)


Find the number of ways of using diagonals to divide a convex (n+2)-sided polygon into triangles such that at least one side of each triangle is a side of the original polygon and such that each vertex of each triangle is a vertex of the original polygon.

Bump. Haven't tried it but [hide]induction[/hide] looks like one way.

ComplexZeta wrote:[hide]I don't think problems with Catalan numbers should be in Intermediate Topics.[/hide]

[hide]Catalan numbers? I don't know, but this was stated as being obvious in a solution to a HMMT guts round problem, but I couldn't find a way to prove it.[/hide]


logarithms
probability
combinatorics unsolved
combinatorics

assume $V(G)$ as vertices of the graph.  We say $A$ is a "prevailed " subset of $V(G)$ if for each $v \in V$, $v$ is in $A$ or has a neighbor in $A$. if $G(V,E)$ is a graph with $n$ vertices and for each vertices $\deg v \geq  \gamma$. Prove G has "prevailed" set with at most $\frac{n(1+\ln(\gamma +1))}{ \gamma +1}$ members.

[i]Edited by Myth[/i]

To Myth :
Yes, I know how to prove that. It uses a probabilistic approach.
I will post it later.

Anyway, the good terminology is not a prevailed set but a dominating set.

Pierre.

yes probablitistic method works well :)
if you assume that $S$ be random set of vertices by including each vertex independently with the probablity $p=\frac{\log(d+1)}{d+1}$ .and let $T$ be the set of vertices outside of $S$ having no neighbor in $S$ ;adding $T$ to $S$ yields prevailed set or as pierre said dominating set. it is clear that $E(|S|)=np$and it is easy to check that $E(|T|)\leq n(1-p)^{d+1}$ and if you use this ineq $1-p\leq e^{-p}$then we have $E(|S \bigcup T|)\leq n\cdot \frac{1+\log(d+1)}{d+1}$.
that's all.

That's exactly what I was willing to write now... :D

Pierre.

[quote="pbornsztein"]That's exactly what I was willing to write now... :D [/quote]
I don't believe you! I were waiting till sam-n write it!
 :lol:  :lol:  :lol:

No, no. I was waiting to have some time to write it. But it is cool that sam-n did it before  :D
Just a remark :
I think this bound is due to Alon and Spencer.

Pierre.

I also think it is classical result. It is really nice!  :D


An average person can walk a mile in 15 minutes. At this rate, it takes 18 minutes to walk around the outside of the Pentagon in Washington, DC. What is the number of feet in the perimeter of the Pentagon?

Set up a ratio.

$ \frac{1 \text{mile}}{15 \text{minutes}}\equal{}\frac{x \text{feet}}{18 \text{minutes}}$

$ \frac{1 \text{mile}}{15 \text{minutes}}\equal{}\frac{\frac{x}{5280} \text{miles}}{18 \text{minutes}}$ (since $ 1 \text{mile}\equal{}5280 \text{feet}$)

$ \frac{1}{15}\equal{}\frac{x}{95040}$

$ x\equal{}\frac{95040}{15}$

$ x\equal{}6336$

Thus, the perimeter of the Pentagon is $ \fbox{6336}$ feet.

1 mile=5280

1 mile can be run in 15 minutes

$\frac{5280}{15}$ $=$ 352
352=for 1 minute

352*19 =6688

so the answer is $\boxed{6688}$

[size=50]im really sorry if this is SPAM [/size]

Excuse me, but I'm pretty sure the answer is 352 times [b]18[/b], not 352 times 19. So rs15 is right.