Δίνεται ένα τρίγωνο $ABC$.Οι πλευρές του ορθικού του τριγώνου τέμνουν τις πλευρές του (αρχικού τριγώνου) σε 3 σημεία που είναι συνευθειακά (εαν δε το γνωρίζετε, τότε να το αποδείξετε).Να αποδειχθεί ότι αυτή η ευθεία είναι κάθετη στην ευθεία του Euler του τριγώνου $ABC$.

[i]Υ.Γ. Η άσκηση έχει προταθεί πολλές φορές και στο olympiad section, αλλά πιστεύω πως καλό είναι να υπάρχει και στο δικό μας forum με δικές μας λύσεις (γραμμένες στα Ελληνικά για να τις καταλαβαίνουν όλοι).Ελπίζω να συμφωνείτε...[/i]

[quote="silouan"]Για το ότι είναι συνευθειακά έχω μια καλή απόδειξη για την καθετότητα δε θα θέλατε να δείτε τη λύση μου  :D .
Λοιπόν έστω $AA_{1},\ BB_{1},\ CC_{1}$ τα ύψη .Από το θεώρημα Desarques τα ζεύγη των πλευρών $(AB,\ A_{1}B_{1})$, $(BC,\ B_{1}C_{1})$, $(CA,\ C_{1}A_{1})$ συνευθειακά .[/quote]

Mία από τις ωραιότερες λύσεις.Ακόμη μία πολύ ωραία χρησιμοποιεί το πλήρες τετράπλευρο του Miquel (πως?).Και οι δύο λύσεις πάντως οδηγούν στην ίδια γενίκευση (βρείτε την).
Βέβαια η άσκηση ξεκινάει από εδώ και πέρα...

[i]Υ.Γ.Σιλουανέ υπάρχουν κάμποσες λύσεις για την καθετότητα-όλες λίγο πολύ- δεν είναι και ιδιαίτερα κομψές (εκτός ίσως από μία που χρησιμοποιεί ορθολογικά τρίγωνα και ριζικούς άξονες), οπότε δεν πειράζει βάλε τη λύση σου όσο περίεργη κι αν είναι.
Εξ'άλλου την άσκηση τη βάζω για να προβληθούν όλες οι απόψεις και  για κανέναν άλλο λόγο.[/i]

Νίκο η λύση μου χρησιμοποιεί κυρίως τριγονομετρία . :(
Αν δεν βρεθεί λύση θα τη βάλλω .Η γενίκευση είναι η εξής (για το πρώτο ερώτημα):

Έστω τρεις σεβιανές $AA_{1},\ BB_{1},\ CC_{1}$  σε τρίγωνο ABC .Δείξτε ότι τα σημεία τομής των ευθειών των πλευρών του τριγώνου με τις αντίστοιχες πλευρές του $A_{1}B_{1}C_{1}$  είναι συνευθειακά σημεία .

Απόδειξη
Και πάλι Desarques  :D

[quote="silouan"]Νίκο μπορείς να βάλλεις τη λύση με τα ορθολογικά τρίγωνα και ριζικούς άξονες ?
Λοιπόν μια άλλη προσέγγιση είναι η εξής . Θεωρούμε ορθογώνιο σύστημα συντεταγμένων και αρχή το ίχνος Δ του ύψους ΑΔ .
Βάζουμε στις κορυφές συντεταγμένες ας πούμε Α(0,5) , Β(-3,0)
Γ(7,0) βρίσκουμε τον συντελεστή διεύθυνσης της ευθείας και μετά τον συντελεστή διεύθυνσης της ΟΗ . Αρκεί να έχουν γινόμενο -1 . :wink:  . Τέλος......[/quote]

Σιλουανέ, η λύση αυτή δεν είναι δικιά μου!Μου την είχε δείξει ο Darij στη Σλοβενία και νομίζω ότι ήδη υπάρχει στο advanced.Θα το βρω και θα σας δώσω το link αλλιώς θα στείλω ένα pm στον Darij.

Τη θυμάσαι ?Αν ναι γράψτην ή στείλτη σε pm γιατί για τον Darij το βλέπω δύσκολο αφού έχει πολύ δουλειά με το πανεπιστήμιο του .Τέλος πάντων θα του πω και αν μου δώσει λύση θα τη βάλλω και εδώ . :)

Βρήκα μια πραγματικά ωραία λύση και τη μοιράζομαι μαζί σας .
Έστω ABC το τρίγωνο και AD, BE, CF τα ύψη του .
Έστω $DF\cap AC=P$  ,  $FE\cap BC=M$  και  $DE\cap AB=Q$ .Το τετράπλευρο AFDC είναι εγγράψιμο άρα από το θεώρημα της δύναμης σημείου έχουμε $PA\cdot PC=PF\cdot PD$ . Όμως τα γινόμενα   $PA\cdot PC$ kai $PF\cdot PD$ είναι οι δυνάμεις του P ως προς τον περιγεγραμμένο κύκλο του ABC και τον κύκλο του Euler του ΑBC .Αφού είναι ίσες το P βρίσκεται στον ριζικό άξονα των δύο κύκλων .Όμοια τα Μ, Q ανήκουν στον ριζικό άξονα των δύο κύκλων . Η διακεντρική ευθεία αυτών των κύλων είναι η ευθεία Euler και η ευθεία που οριζεται απο τα P,M,Q είναι ο ριζικός άξονας των δύο κύκλων άρα το ζητούμενο  :wink:

P.S . Νίκο ευχαριστώ (ευχαριστούμε ) για την πολύ ωραία άσκηση . Ελπίζω η λύση να ήταν καλή .

Μπράβο, Σιλουανέ! Πολύ ωραία λύση :)

Silouan ,

 αυτό που έκανες είναι αυτό ακριβώς που λέμε
  '' θαυμάσια λύση''.
 Ο Erdoes θα έλεγε ότι είναι από το βιβλίο( from the book!) !
                     Congats !!!

Το σοφό είναι και απλό, όπως έλεγε ο Θεόφιλος Χρυσοστομίδης σε κάποιο σχόλιό του και απορώ πως μπορεί κανείς, διαβάζοντας μια απόδειξη σαν κι' αυτή, να σκεφτεί την καταστροφή της Γεωμετρίας.

Σιλουανέ σε ευχαριστώ για τη χαρά που μου έδωσες και σου εύχομαι κι' άλλες εμπνευσμένες αποδείξεις, για να μαθαίνουμε και μεις οι μεγαλύτεροι.

Κώστας Βήττας.

Σας ευχαριστώ όλους .... :)

[quote="vittasko"] Σιλουανέ σε ευχαριστώ για τη χαρά που μου έδωσες και σου εύχομαι κι' άλλες εμπνευσμένες αποδείξεις, για να μαθαίνουμε και μεις οι μεγαλύτεροι.

Κώστας Βήττας.[/quote]

Νομίζω ότι εμείς μαθαίνουμε από εσάς .... :wink:

Μπράβο Σιλουανέ!Αυτή είναι η <<ωραία>> απόδειξη στην οποία αναφερόμουν.
Πως λοιπόν μπορείς μετά από μια τέτοια άσκηση να απορρίψεις την Ευκλείδεια Γεωμετρία?

[i]Τώρα μένει να ζητήσω από τον Darij τα ενδιαφέροντα αποτελέσματα που έχει βγάλει χρησιμοποιώντας ορθολογικά τρίγωνα.[/i]

είναι πραγματικά Α-Ψ-Ο-Γ-Η

Χθες δέχτηκα  ένα e-mail (ξεκίνησαν πάλι τα όργανα φαίνεται (και πάλι τέτοιο καιρό τα ίδια ))(πέρσυ όμως ήταν πολύ χειρότερα δυστυχώς )(ευτυχώς φέτος είναι πιο ήπια)
το οποίο είχε τίτλο
"δέχεσαι συγχαρητήρια ? "
Και το μήνυμα έγραφε
"Γιατί δέχεσαι τα συγχαρητήρια των μελών του φόρουμ για τη λύση μιας άσκησης που
δεν έλυσες εσύ ?

Σε παρακαλώ να βγεις στο φορουμ και να πεις ότι δεν είναι δική σου η λύση .

Γ.Μ
"
Απλά να πω σε αυτόν τον ανώνυμο Κύριο ότι χθες το βράδυ το μύνημά του δε με άφησε να κοιμηθώ .Αν ήταν αυτός ο σκοπός του τον πέτυχε

Φίλε μου Σιλουανέ είναι απίστευτο κι όμως ελληνικό δυστυχώς ΝΑ ΕΙΜΑΣΤΕ ΚΟΜΠΛΕΞΙΚΟΙ ΜΕ ΚΑΠΟΙΟΝ ΠΟΥ ΠΡΟΟΔΕΥΕΙ όποιος κομπλέξας έστειλε αυτό το μήνυμα είναι πραγματικά άθλιος βλέπουμε μια πολύ ωραία απόδειξη στο πρόβλημα του Νίκου από τον Σιλουανό ο οποίος από όσο ξέρω είναι ένα πολύ καλό παιδί με θέληση να πετύχει κι με αξία μαθηματική κι έχουν πέσει οι καρχαρίες κι τα σκυλόψαρα που θέλουν τον τζερτζελε τον κακό να τον φάνε επειδή "δέχεται μια απόδειξη που κάποιος άλλος έλυσε ή που είδε από βιβλίο" ακόμα κι έτσι να έιναι τί ακριβώς τον ενοχλεί???? ακόμα χειρότερα για αυτόν που δεν την βρήκε να την δημοσιεύσει οκ??? Φϊλε μου Σιλουανέ ΔΕΝ ΧΡΕΙΑΖΕΤΑΙ ΝΑ ΣΤΕΝΑΧΩΡΙΕΣΑΙ ΜΕ ΚΟΜΠΛΕΞΙΚΟΥΣ (θυμίσου εμένα κι τον ανώνυμο)
ΣΥΝΕΧΙΣΕ ΕΤΣΙ ΚΙ ΑΠΛΑ ΘΑ ΠΕ΅ΤΥΧΕΙΣ ΚΙ ΟΙ ΑΛΛΟΙ ΘΑ ΦΑΝΕ ΤΑ ΛΥΣΑΚΑ ΤΟΥΣ
ΝΑ ΣΑΙ ΚΑΛΑ ΜΕ ΑΓΩΝΙΣΤΙΚΟΥΣ ΧΑΙΡΕΤΙΣΜΟΥΣ

Σιλουανέ,

το ότι έλυσες την άσκηση την έλυσες πρώτα για 'σένα! Εσύ ξέρεις την ευχαρίστηση που πρωτοπήρες όταν βρήκες αυτή την πανέμορφη λύση! Τώρα από τη στιγμή που τα λόγια αυτού του τύπου δεν ισχύουν, δεν έχεις κανένα λόγο να κάτσεις να ασχοληθείς περαιτέρω με το θέμα. Ξέρεις είναι πολλές φορές αυτός ο φθόνος και ο εγωισμός που μας κάνει να μην υποκλιθούμε σε κάτι ανώτερο, που εμείς προφανώς δεν ήμασταν ικανοί να το πετύχουμε. Αυτό όμως δεν μας πάει μπροστά! Άλλωστε όπως είπε και ο Cooley: [i][b]"Acknowledge your limitations or they will tyrannize over you"[/b][/i]. Το ότι το e-mail που έλαβες ήταν και ανώνυμο λέει πολλά επίσης! Με κάποιον που κρύβεται πίσω από το δάχτυλό του και δεν θέλει να υποστεί τις "συνέπειες" των πράξεών του δεν έχεις παρά να τον αγνοήσεις και να συνεχίσεις απτόητος να μας προσφέρεις αυτές τις υπέροχες λύσεις!

Στέλιος.

Σιλουανέ, τα είπαμε και μέσω ιμ. Πάντα θα υπάρχουν άνθρωποι που θα μας χαλάν τη διάθεση, το καλύτερο που έχουμε εμείς να κάνουμε είναι απλά να τους αγνοούμε.  :wink:
Βέβαια επειδή μ'αρέσει να ακούω όλες τις διαφορετικές απόψεις πάνω σε ένα θέμα πριν πάρω θέση, αν θέλει αυτός που σου έστειλε το e-mail ας γράψει εδώ στο forum τί τον ώθησε να το κάνει αυτό, ώστε να ακούσουμε και την άποψή του. Αν βέβαια δεν το κάνει, πάει να πει ότι έχει άδικο. :wink:

Στέλιος

Προσυπογράφοντας τα προηγούμενα μηνύματα των Αμβρόσιου, Στέλιου (Mostel) και Στέλιου (Stedes), θέλω να προσθέσω ότι δεν είναι κάτι ασυνήθιστο, το να (ξανα) ανακαλύπτει κάποιος ένα θεώρημα ή μία απόδειξη. Με τα χρόνια έχω πειστεί, πως αυτό μάλλον είναι ο κανόνας. Το πιθανότερο είναι κάποιος άλλος να έχει προηγηθεί και ιδιαίτερα ως προς την ξένη βιβλιογραφία.

Όπως έχουμε πει, η γνώση είναι μία διαδικασία μάθησης κυρίως από τους άλλους και αυτό δεν μειώνει καθόλου τη χαρά που νοιώθουμε όταν πετυχαίνουμε μία (γνωστή ίσως) ωραία απόδειξη. Η απόδειξη που έδωσε ο Σιλουανός, γνωστή όπως μας πληροφορεί ο Νίκος, μπορεί εύκολα να τη σκεφτεί ο καθένας μας, αλλά εξίσου εύκολα μπορεί να μας ξεφεύγει, όταν τίθεται κάποιο πρόβλημα.

Όμως ως μία όμορφη απόδειξη, μπορούμε να την απολαύσουμε τη στιγμή που εμφανίζεται και ας παραμερίσουμε τις όποιες κακεντρέχειες των ανώνυμων επιστολογράφων, γιατί το σίγουρο είναι ότι οφείλονται σε προσωπική τους ανεπάρκεια.

Κώστας Βήττας.

Συμφωνώ με τον vittasko που λέει :

"η γνώση είναι μία διαδικασία μάθησης κυρίως από τους άλλους και αυτό δεν μειώνει καθόλου τη χαρά που νοιώθουμε όταν πετυχαίνουμε μία (γνωστή ίσως) ωραία απόδειξη".

Σιλουανέ

είσαι η ψυχή του forum . Μην πτοείσαι !
Οι αποδείξεις σου είναι εξαιρετικές .

Νίκος Ρ

Σιλουανέ τα είπαμε και από το τηλέφωνο...Δεν αξίζει να ασχολήσε με τέτοια υποκείμενα...
Επίσης θα ήθελα να πω το εξής: H απόδειξη είναι του Σιλουανού  δεν είναι κλεμένη!Απλώς έτυχε να συμπίπτει με τη γνωστή.
Όποιος ξέρει έστω και λίγα Μαθηματικά μπορεί να ξεχωρίσει μια κλεμένη απόδειξη από μία αυθεντική!Ως εκ τούτου αυτός που έστειλε το e-mail δεν ξέρει Μαθηματικά (ή προσπαθεί χωρίς αποτέλεσμα, να μάθει), άρα δεν υπάρχει λόγος να ασχολούμαστε μαζί του.

Υ.Γ. Τα αρχικά Γ.Μ. παραπέμπουν κάπου (?)...

Σιλουανέ, καταλαβαίνω ότι σε ενόχλησε, αλλά δε χρειάζεται να δίνεις σημασία σε τέτοια άτομα. Καλά έκανες και το μετέφερες στο φόρουμ για να ξέρουμε με τι άτομα έχουμε να κάνουμε, αλλά όχι και να χάσεις τον ύπνο σου. Να ξέρεις ότι ποτέ μικροί άνθρωποι δεν μπορούν να επηρεάζουν μεγάλους. Παλιά κι εμένα με ενοχλούσαν τέτοιες μικρότητες, αλλά όταν έμαθα ποια είναι εκείνα τα άτομα που φτάνουν σε τέτοιο σημείο, έπαψε να με ενοχλεί.

Μία απορία: [u]Γιατί μερικοί ζηλεύουν???[/u]
Πρώτα απ όλα επειδή υπάρχει λόγος. Για να ζηλέψεις κάποιον, ξέρεις σίγουρα ότι τα καταφέρνει καλύτερα από σένα.
Αν όμως δεν το παραδέχεσαι, τότε έχεις πρόβλημα!

Καλό είναι να υπάρχει λίγη ζήλεια προς τους άλλους (θα την έλεγα θαυμασμό), τόση όση είναι απαραίτητη ώστε να μας κάνει να θέλουμε να τους μοιάσουμε. Όποιος μπορεί να μπει σ'αυτήν τη διαδικασία, βγαίνει κερδισμένος. Όποιος δεν μπορεί, φτάνει μέχρι εκεί που η νοημοσύνη του του επιτρέπει. Για να αμφισβητήσεις κάποιον δε χρειάζεται ιδιαίτερη νοημοσύνη.
Είναι πάγια τακτική των [u]αμόρφωτων[/u] ανθρώπων και δεν εννοώ την ακαδημαϊκή μόρφωση. Μπορέι κάποιος που δεν έχει τελειώσει ούτε το δημοτικό να είναι πολύ ανώτερος από καθηγητή πανεπιστημίου. Γιατί όταν η γνώση μετατρέπεται σε αλαζονεία χάνει την ουσία της και είναι σαν να μην υπάρχει. Αποδεικνύει απλώς τον κατώτερο χαρακτήρα.

Ο πολιτισμένος άνθρωπος έχει την ικανότητα να αποδέχεται την ανωτερότητα των άλλων. Όποιος δεν μπορεί να το κάνει αυτό είναι επειδή κουβαλάει πολλλλλλλά κόμπλεξ. Έχω δει πολλούς τέτοιους σε διάφορα φόρουμ, περισσότερο σε ελληνικά και λιγότερο σε ξένα.
Εγώ δεν τον κατηγορώ. Μέχρι εκεί μπορεί να φτάσει, οπότε αυτό κάνει. Εξάλλου, αν είχε την ελάχιστη νοημοσύνη να καταλάβει τι υποδηλώνει μία τέτοια ενέργεια, είμαι σίγουρος ότι δε θα την έκανε. Ίσως μάλιστα να είναι αυτός ο λόγος που έστειλε το μήνυμα ανώνυμα, για να μη διασυρθεί το όνομά του. Αλλά αυτό δεν αλλάζει τα πράγματα. Ο ίδιος ξέρει πολύ καλά ποιος είναι...

[color=brown]Για πες μου λοιπόν φίλε Γ.Μ. ή όπως αλλιώς σε λένε (γιατί είμαι σίγουρος ότι παρακολουθείς τη σελίδα)
Αισθάνεσαι ότι είσαι ο καλύτερος;
Ότι τα ξέρεις όλα;
Ότι όλοι οι υπόλοιποι δεν πιάνουν μία μπροστά σου;
Δικαίωμά σου να το πιστεύεις, αν αυτό σε κάνει να αισθάνεσαι καλύτερα. Αλλά να ξέρεις, έτσι δεν προοδεύεις. Γιατί επαναπαύεσαι σ' αυτήν την ιδέα και δεν προσπαθείς να βελτιωθείς.
Όταν βλέπεις ότι κάποιος έλυσε μία άσκηση που εσύ δεν μπορούσες, μη θυμώνεις. Δε φταίει εκείνος που τα κατάφερε καλύτερα από σένα! Δες το σαν μια ευκαιρία να μάθεις κάτι καινούριο. Γιατί όχι, σαν μία πρόκληση για να τον ξεπεράσεις. Στα πλαίσια του συναγωνισμού πάντα.
Κι εγώ δοκίμασα να λύσω την άσκηση με δύναμη σημείου και αντιστροφή ως προς κύκλο, χωρίς να καταφέρω κάτι ουσιαστικό. Ο Σιλουανός αποδείχτηκε καλύτερός μου. Τι θα 'πρεπε δηλαδή, να βγω και να τον κατηγορήσω; Για ποιο λόγο, για να πάρω τη ρεβάνς; Είναι τρόπος σκέψης λογικού ανθρώπου αυτός;
Επαναλαμβάνω, δε σε κατηγορώ. Σκοπός μου είναι να καταλάβεις το λάθος σου. Και αν μπορείς να αποβάλεις αυτόν τον τρόπο σκέψης, να ρθεις στην παρέα μας. Αν όχι, τότε το φόρουμ δε σου ταιριάζει γιατί εδώ ασχολούμαστε με ανώτερα πράγματα.[/color]

Δε θέλω να επεκταθώ περισσότερο.
Έχουμε ξαναπεί ότι εδώ δεν κάνουμε πρωταθλητισμό.
Προσπαθούμε να μοιραστούμε ιδέες, κι έτσι να βγούμε όλοι κερδισμένοι. Όσες περισσότερες λύσεις βρεθούν σε μία άσκηση, τόσο το καλύτερο.

Με την ευκαιρία να πω και το εξής:
Πριν βρω το mathlinks πίστευα ότι γνώριζα μαθηματικά. Μόλις το ανακάλυψα, ανακάλυψα ταυτόχρονα ότι "εν οίδα ότι ουδέν οίδα". Μαθαίνω απ' αυτό το φόρουμ, καθημερινά. Και μάλιστα μαθαίνω από άτομα που στην πλειοψηφία τους είναι μικρότερα σε ηλικία από μένα. Επομένως τώρα ταιριάζει ένα άλλο αρχαίο ρητό, "γηράσκω αεί διδασκόμενος".


Choose three points $ A_1;B_1;C_1$ on three sides of a triangle $ ABC(A_1 \in BC;B_1 \in CA;C_1 \in AB$) such that $ AA_1;BB_1;CC_1$ meet at a common point.Again,choose three points $ A_2;B_2;C_2$  on the sides of triangle $ A_1B_1C_1(A_2 \in B_1C_1;B_2 \in C_1A_1;C_2 \in A_1B_1)$
Prove that $ AA_2;BB_2;CC_2$ meet at a common point if and only if so do $ A_1A_2;B_1B_2;C_1C_2$

Denote by $ A_3\equal{}AA_2 \cap BC$. Now, denote by $ k_{A_2}\equal{}\frac{C_1A_2}{A_2B_1}$ and $ k_{A_3}\equal{}\frac{BA_3}{A_3C}$.

$ \overrightarrow{AA_3}\equal{}\overrightarrow{AB} \cdot \frac{1}{1\plus{}k_{A_2}}\plus{}\overrightarrow{AC}\cdot\frac{k_{A_2}}{1\plus{}k_{A_2}}$

$ \overrightarrow{AA_2}\equal{}\overrightarrow{AC_1} \cdot \frac{1}{1\plus{}k_{A_3}}\plus{}\overrightarrow{AB_1}\cdot \frac{k_{A_3}}{1\plus{}k_{A_3}}$

$ \overrightarrow{AA_2}\equal{}\overrightarrow{AB}\cdot \frac{AC_1}{AB} \cdot \frac{1}{1\plus{}k_{A_3}}\plus{}\overrightarrow{AC}\cdot \frac{AB_1}{AC}\cdot \frac{k_{A_3}}{1\plus{}k_{A_3}}$

$ \Rightarrow \frac{\frac{1}{1\plus{}k_{A_2}}}{\frac{AC_1}{AB} \cdot \frac{1}{1\plus{}k_{A_3}}}\equal{}\frac{\frac{k_{A_2}}{1\plus{}k_{A_2}}}{\frac{AB_1}{AC}\cdot \frac{k_{A_3}}{1\plus{}k_{A_3}}}$

$ \Rightarrow \frac{k_{A_2}}{k_{A_3}}\equal{}\frac{AB}{AC} \cdot \frac{AB_1}{AC_1} \Rightarrow \prod\frac{k_{A_2}}{k_{A_3}}\equal{}\prod\frac{AB}{AC} \cdot \frac{AB_1}{AC_1}$

The conclusion results easily, applying Ceva's theorem.


\[ \{\begin{array}{c}{1}
x^3-mx-y=0\
y^3-my+x=0
\end{array} (*)

\]
Find $ m$ to $ (*)$ have five root of systems of equations

(1) $ y\equal{}x^3\minus{}mx$
(2) $ (x^3\minus{}mx)^3\minus{}m(x^3\minus{}mx)\plus{}x\equal{}0$ has five different roots.
Obviously $ x\equal{}0$ is one root. Then divide LHS with $ x$ and we get
$ y^4\minus{}3my^3\plus{}3m^2y^2\minus{}(m^3\plus{}m)y\plus{}m^2\plus{}1\equal{}0$, where $ y\equal{}x^2$.

i) $ m^2\plus{}1\equal{}0$ or $ m\equal{}\pm i$
$ y^2\minus{}3my\plus{}3m^3\equal{}0$ has to have two nonzero roots, which is easy to verify.

ii) $ m^2\plus{}1\ne0$
$ y^4\minus{}3my^3\plus{}3m^2y^2\minus{}(m^3\plus{}m)y\plus{}m^2\plus{}1\equal{}0$ has to have exactly two different nonzero roots.
Let $ z\equal{}y\minus{}m$, we get
$ z^4\plus{}mz^3\minus{}mz\plus{}1\equal{}0$
Let the two roots be $ a,b$, we either have  $ LHS\equal{}(z\minus{}a)(z\minus{}b)^3$ or $ LHS\equal{}(z\minus{}a)^2(z\minus{}b)^2$. Then we can use Vieta's theorem.
For the first case, $ ab\plus{}b^2\equal{}0\Rightarrow b\equal{}\minus{}a$. Then it is not hard to see there is no solution for $ m$.
For the second case, it is not hard to see $ m^2\equal{}8$.

Conclusion, $ m\equal{}\pm 2\sqrt{2},\pm i$


If   $ e^^(sin^2x\plus{}sin^4x\plus{}sin^6x\plus{}....\infty)log_e3$ satisfies the equation $ x^2\minus{}28x\plus{}27\equal{}0$,find the value of $ (cosx\plus{}sinx)^^\minus{}1$,when $ 0<x<\frac{\pi}{2}$ :?:  :?:

[hide]$ \sin^2x \plus{} \sin^4x \plus{} \sin^6x \plus{} ... \equal{} \sum_{n \equal{} 1}^{\infty}(\sin^2x)^n \equal{} \frac {\sin^2x}{1 \minus{} \sin^2x} \equal{} \frac {1}{\csc^2x \minus{} 1} \equal{} \tan^2x$, for $ 0 < x < \frac {\pi}{2}$, and so $ 3^{\tan^2x}$ satisfies that equation. The solutions of $ x^2 \minus{} 28x \plus{} 27 \equal{} 0$ are $ x \equal{} 27$ and $ x \equal{} 1$. So we have $ 3^{\tan^2x} \equal{} 27 \equal{} 3^3 \equal{}> \tan^2x \equal{} 3 \equal{}> \tan x \equal{} \sqrt{3} [\equal{}> x \equal{} \frac{\pi}{3} \in (0,\frac{\pi}{2})]$ and $ 3^{\tan^2x} \equal{} 1 \equal{}> \tan^2x \equal{} 0 \equal{}> \tan x \equal{} 0 \equal{}> x \equal{} 0 \ni (0,\frac{\pi}{2})$. So we have $ \sqrt{3}$. But

$ \frac {1}{\sin x \plus{} \cos x} \equal{} \frac {\sqrt {1 \plus{} \tan^2x}}{1 \plus{} \tan x}$. Therefore, on this case, $ \frac {1}{\sin x \plus{} \cos x} \equal{} \frac{\sqrt{1 \plus{} 3}}{1 \plus{} \sqrt{3}} \equal{} \frac{2(1 \minus{} \sqrt{3})}{1\minus{}3} \equal{} \sqrt{3} \minus{} 1$.[/hide]


geometry
Harvard
college
Columbia
topology
number theory

Hi!

I would like to study algebraic number theory in the future, after finishing my bachelor in pure mathematics in Bergen. Which colleges are the best in this area??

[quote="blahblahblah"]I believe that Harvard is particularly strong in this area.  I could be wrong though, so take it with a grain of salt.[/quote]I thought that Antilopen was talking about Norway or the suroundings, that is why I didn't answer, I really don't know much about education in that part in northern Europe, but I have a few friends in Sweden, I could ask them. If it is in America, well, probably we could find out.

Hi!

I am talking about both Europe and the USA.
I recently talked with one of my professor (he is professor in algebraic geometry)
which took his Ph.D (or doctor) in Colombia University.
He recommended me to study in the USA, England or Germany.
Here in Bergen they are doing most research in algebraic geometry, topology and analysis when it comes to pure mathematics. The applied math is connected to the oil-industry, which I am not so interested in, although I won a competition in the area three years ago.

OK, Enlgand, I don't know much, basically what I have heard is about Cambridge and Oxford, and they do A LOT OF STUFF. A friend of mine was in Cambridge for a year as an Interchange student, just after he finished his Bachelor, exactly studying Algebraic Number Theory, but it is difficult university to get in.

Here in the US, well, there are many. I know that Berkeley use to have a good program there, I have to friends who got (one is still in progress to get it) their Ph.D from there in Number Theory, and both say that the place is great. I think also Texas A&M has a good research team in the area, but the only person that I know there is in Differential Geometry. I heard that Urbana Champain also have a strong research team, but the two people that I know from there study something nearest to Algebra than to Number Theory (things like p-regularity and those kind of things).

Germany, I don't know any german really interested on Number Theory, then, I don't know.


just wondering if any of you use Open Office

http://www.openoffice.org/

if so, how is it

I use openoffice...but then again I have no choice since I use Linux. I can't really say how it is, because I almost always LaTeX everything. It is slightly heavy for open source software (which is known for distributing the bare necessities and allowing you to add on what you want). I haven't run into any problems in my brief encounters. It is also nice because you can export as .doc if you really need to and import many many different file formats that something like word would just yell at you for.

I use open office too since i'm too cheap to by office 07  :oops:  I find open office pretty good for open source. It also allows you to export to many formats including pdf and is unusually compatible with ms office. If you nead anything specific like accent characters, special templates, or latex modules you can always download the add-ons. Since open office is open source, the software is developing pretty rapidly and may someday surpass ms office.


how to solve $k^{2}\equiv 36\pmod{100}$?

[hide="Solution"] We solve $\bmod 4, \bmod 25$.  Of course, $\bmod 4$ gives $k \equiv 0, 2 \bmod 4$.

$\bmod 25$ gives $k \equiv 6, 19 \bmod 25$.  We know there are only two solutions because the two factors $k+6, k-6$ cannot both be $\equiv 0 \bmod 5$.

By CRT this gives $k \equiv \boxed{6, 56, 44, 94}\bmod 100$. [/hide]

[quote="yeppyyep"]sorry I know very little number theorem. Could you explain why compute mod 4 and mod 25? and what calculation you did with CRT to get answers?[/quote]

4 and 25 are the prime-power factors of 100, and prime powers are good.

CRT isn't that complicated here, since $k$ even and $k\equiv \pm 6\pmod{25}$ occur when $k\equiv \pm 6\pmod{50}$. (If $k$ even and $k\equiv 6\pmod{25}$, then $k-6$ is divisible by 2 and 25, so it is divisible by 50. Same with the other case.)


What is the ratio of $ 0.1\overline6$ to $ 0.8\overline3$? Express your answer as a common fraction.

The first one is $ 1/6$ and the second is $ 5/6$.  So the ration is $ 1/5$.


do they mean a rectangular lattice? then it cannot be. maybe they mean equilateral triangular lattice? or they don't mean lattice at all...?

Can't you just type out the problem? That takes about 10-15 seconds to type it up  :maybe: ? It's much harder to read from a jpg and a pain for people using dial-ups [u]and[/u] you are using your limited upload quota.

[quote="boxedexe"]1) The problem says nothing about lattice points. It says that every point $(a,b,c)\in\mathbb{R}^{3}$ has a color.

2) Contradiction and Circles.[/quote]

They said in the plane though.

Sorry about that.  I didn't observe the problem carefully.  So it should be $(a,b)\in\mathbb{R}^{2}$.

[quote="Tim_Lou"]I don't know if this is correct
[hide]
draw a hexagon with a dot in the center, label the center 0, and the corners 1-6.

WLOG,
let 0 be white, since 1,3,5 or 2,4,6 cannot be all black, two of the corners must be white. let them be {1,3} and {1,4}

case I: {1,3} are white
2,6 must be black, and 4 must be black, contradiction, 2,4,6 forms a triangle.

case II: {1,4} are white
all the others must be black. draw one more dot by creating equilateral triangles using {2,3}, label the dot created as 7 so that {2,3,7} forms an equilateral triangle. 7 cannot be black since 2 and 3 are black.

7 is white but that implies {1,4,7} are white, a contradiction
[/hide][/quote]
correct  :)  I just considered triangular lattice. boxedexe your way is more elegant I assume..? can  you write it?


Hi all,

Not really topology, but this question came straight from Munkres. It's one of the first few questions of the book. I felt that I needed to work on my set theory so I dug up these questions.

I'm having trouble proving that $ A\backslash (A\backslash B)\subseteq B$ using rigorous logical notation. This is what I have down.

Let $ x\in A\backslash (A\backslash B)$. Then $ x\in A\wedge x\notin A\backslash B$. Then $ x\in A\wedge (x\notin A\wedge x\in B)\equal{}(x\in A\wedge x\notin A)\wedge x\in B$.

I am then stuck because it seems like I have a contradiction on my hands. (Note the $ x\in A \wedge x\notin A$ part.) Where did I go wrong?

Thanks folks.

[quote="polya.szego"]Hi all,

Not really topology, but this question came straight from Munkres. It's one of the first few questions of the book. I felt that I needed to work on my set theory so I dug up these questions.

I'm having trouble proving that $ A\backslash (A\backslash B)\subseteq B$ using rigorous logical notation. This is what I have down.

Let $ x\in A\backslash (A\backslash B)$. Then $ x\in A\wedge x\notin A\backslash B$. Then[/quote]

this step not correct

[quote] $ x\in A\wedge (x\notin A\wedge x\in B) \equal{} (x\in A\wedge x\notin A)\wedge x\in B$.

I am then stuck because it seems like I have a contradiction on my hands. (Note the $ x\in A \wedge x\notin A$ part.) Where did I go wrong?

Thanks folks.[/quote]

Hmm...which part? The logical connective should be associative, shouldn't it?

Look again at your condition for $ x \notin A \backslash B$.

A typical de Morgan violation...

[quote]... $ x\notin A\backslash B$. Then ... $ x\notin A\wedge x\in B$...[/quote]

is false; in fact

... $ x\notin A\backslash B$. Then ... $ x\notin A\vee x\in B$..., since this is the negation of
... $ x\in A\backslash B$. Then ... $ x\in A\wedge x\notin B$..., i.e. ... $ x\notin A\vee x\in B$...

Then everything works. But why not just use that $ A\setminus (A\setminus B) \equal{} A \cap B$ ?

Notation has nothing to do with rigor.  You can be perfectly rigorous in plaintext.

Thanks so much! I can't believe what a silly but pedagogical mistake that was.

Also, I agree with the last poster. But I want to type in symbols because that saves me lots of time, and also being the autodidact that I am, I feel that I need the symbols to be there to be absolutely confident in my answers.

Thanks all once again.


Gujurati

And not very well either.

I can speak Sanskrit to a limit as well.

Marathi......
Its not there :mad:  :mad:  :mad:

i'm not really fluent at any indian language since i was born and raised in america, but i can understand kannada (my parents r both from bangalore) and a bits and pieces of hindi...

Tamil ,hindi and Sanskrit(to quite a decent extent,though it is not a desi language in true sense..and one cant go about communicating in sanskrit)

May be of interest here:
[url=http://www.techtree.com/India/News/Google_Launches_Services_in_Hindi/551-79811-643.html]google transliteration for  Hindi[/url]
[quote]Google today announced the launch of Hindi transliteration on Blogger, its Web publishing service. The company also announced its News service in Hindi.

The company believes these launches will enable a majority of people, who are fluent in Hindi, to have quicker, easier access to lots more information, paving the way for creation/consumption of Indian language content.

oogle's transliteration technology converts English to phonetically equivalent text in Indian languages. Hindi transliteration on Blogger allows users publish content in Hindi, while using English keyboards for text entry.

For instance, when users type 'mera bharat', the technology will convert this into corresponding Hindi script. Besides, users will be able to switch freely between English and Hindi.

Google's News service in Hindi is based on its innovative clustering technology. The service gathers news stories from various Hindi news sources, and presents a ranked one page view with links to users' favorite news sources.

For all Indian language Web sites, Google supports adoption of Unicode
UTF-8 standard. With the launch of Google News in Hindi, users will be able to reach content from even those news sources which are not yet in this format. And, as these news sites transition, Google will internally convert their content to the Unicode UTF-8 standard.

Speaking on the occasion, Prasad Ram, Centre Head of Google India R&D, said, "Google is committed to promoting the creation and consumption of content in Indian languages, and we're proud to launch our first Indian language edition for Google News and Blogger. These launches will enable our Hindi-speaking users to express themselves better, and access information in a language they are comfortable with." [/quote]

i understand both hindi and kannada quite well, but i don't speak either of them very fluenty.

ola.. im tamil all the way... i wuz born in america, but i can speak it fluently. mostly because i watch alot of movies and listen to music. no hindi for me though, but my best friend is from rajistan. anyone here that does mu alpha theta?

[quote="bpms"]Does anyone know how an NRI who was born in the US and has lived there all their life (i've been to India a good number of times) can get better at Hindi?
I already know it decently, but I don't know it as well as people who live in India, and I really want to.

Any suggestions?[/quote]

even though i have visited indian numerous times, watching hindi movies with the subtitles on helps you learn a few hindi words and its entertaining to depending on the movie.

[quote="bpms"]Does anyone know how an NRI who was born in the US and has lived there all their life (i've been to India a good number of times) can get better at Hindi?
I already know it decently, but I don't know it as well as people who live in India, and I really want to.

Any suggestions?[/quote]

Hindi is not my mother tongue yet I can understand it and I am okay at speaking it. I have lived in the USA for my whole life too so I learned it by watching hindi movies.

You know it's spelled Telugu right? That's mine and I'm decent with it.

Hindi almost perfectly, some bengali, little kannada and sanskrit :D


probability
conditional probability
expected value
AMC

#30 from the 40th AHSME

Suppose that 7 boys and 13 girls line up in a row.  Let S be the number of places in the row where a boy and a girl are standing next to each other.  FOr example, for the row GBBGGGBGBGGGBGBGGBGG we have S=12.  THe average value of S is closest to...

i assume the answer is 2(7)(13)/(20)=9.1??  anybody verify??

Right.  There are 19 pairs of consecutive positions.  In each pair, the probability of BG is (7/20)(13/19).  In each pair, the probability of GB is the same, (13/20)(7/19).  Thus in each pair, the probability of being mixed is 2*7*13/(19*20).  Now multiply that probability by 19 to get the average number of mixed pairs.

just curious, which year is the 40th AHSME?

could someone explain this a little clearer, i am lost, thanks

Sorry, i was just blanking out last time i read it, now i get it.

Yes 9 is right... I don't know what the 40th is =D heh I think its around 1986 so you can try to find out.

In each pair, the probability of BG is (7/20)(13/19). In each pair, the probability of GB is the same, (13/20)(7/19).   But, the probabilities depend on the numbers of boys and girls sitting before them don't they?

No.  The probabilities that I gave are the unconditional probabilities, that is, without additional information.  For example, the (unconditional) probability that the 5th and 6th positions are B and G, respectively, is indeed (7/20)(13/19).  I wasn't trying to evaluate the conditional probability given extra information like the assignment of previous positions.  You are right that such extra information will affect the conditional probability, but that's not what I was evaluating.

im not sure, but i think because it is asking for the average, you can assume "perfect probability" so the number of each gender will be reduced by the exact amount (ex. 13/20 * 7/19 :approx: 11/17 * 6/16 :approx: 9/14 * 5/13) but its not exact because 13 and 7 are relatively prime... if you take infinite arrangements the average will be the same though.

No, you also are thinking about conditional probabilities, but those are irrelevant.  The average number of mixed pairs is exactly equal to the sum (over each pair) of the (unconditional) probabilities that the pair is mixed.  And the (unconditional) probability that a pair is mixed is exactly 2 * (7/20) * (13/19).

To add some details, let Xi be 1 if the ith pair is mixed, and 0 otherwise.  Then S, the number of mixed pairs, is exactly the sum of the Xi.  Therefore, E(S), the expected value (average) of S, is exactly the sum of E(Xi).  And E(Xi) is the same as the (unconditional) probability that the ith pair is mixed.  Which, as I've noted, is exactly 2*(7/20)*(13/19).


prove that if p and q are coprimes numbers and one of them its a even number then also 2*p*q and p^2+q^2 are coprimes

Assume that there exist a prime $r$ such that $r|2pq$ and $r|p^{2}+q^{2}$ now $r|2$ or $r|p$ or $r|q$ :
if $r|2$ $\longrightarrow$ $r=2$ $\longrightarrow$ $2|p^{2}+q^{2}$ but one of them is even an another is odd . Contradiction
if $r|p$ $\longrightarrow$ $r|p^{2}$ $\longrightarrow$ $r|q^{2}$ $\longrightarrow$ $r|q$ but $(q,p)=1$ , Contradiction .
$\longrightarrow$ $(p,q)=1$

let $\gcd(2pq,p^{2}+q^{2})=g$. then $g\mid (p+q)^{2}$ and $g\mid (p-q)^{2}$. but we have $(p,q)=1$ and exactly one of $p,q$ is even. so $(p+q,p-q)=(2p,p-q)=(p,p-q)=(p,q)=1$. this implies that $g=1$.


Let $a,b$ be integers which aren't the multiple of 3.Prove that $a^{4}+a^{2}b^{2}+b^{4}$ is the multiple of 3.

[hide="is this supposed to be in the marathon?"]We know that $x^{2}=1\mod3$ if $3\not| x$
we rewrite the expression like $(a^{2})^{2}+a^{2}b^{2}+(b^{2})^{2}$
which is equal to $1^{2}+1\cdot1+1^{2}=3=0\mod3$ AWD[/hide]


inequalities
function
inequalities proposed

[quote="rofler"]$ (a^2/b^3 \plus{} b^2/c^3 \plus{} c^2/a^3)^2(a^2 \plus{} b^2 \plus{} c^2)\ge(a^2/b^2 \plus{} b^2/c^2 \plus{} c^2/a^2)^3\ge27$

By Holders.[/quote]
your solutions nice but my solution than nice :)
By Cauchy-schwarz we have:
$ (\frac{a^2}{b^3}\plus{}\frac{b^2}{c^3}\plus{}\frac{c^2}{a^3})(a\plus{}b\plus{}c)\geq\ (\sqrt{\frac{a^3}{b^3}}\plus{}\sqrt{\frac{b^3}{c^3}}\plus{}\sqrt{\frac{c^3}{a^3}})^2
\geq\ 3$,and $ a\plus{}b\plus{}c\le\ 3$

This inequality than''stronger'' with condition:
$ \frac{a^2}{1\plus{}b^3}\plus{}\frac{b^2}{1\plus{}c^3}\plus{}\frac{c^2}{1\plus{}a^3}\geq\ \frac{3}{2}$:)  :)  :)

for the first, it's easy by using AM-GM:
$ \frac {a^2}{b^3} \plus{} \frac {b^2}{c^3} \plus{} \frac {c^2}{a^3}\geq\ 3\sqrt[3]{\frac{1}{abc}}$
put, $ \sqrt[3]{\frac{1}{abc}}\equal{}a$ then $ a^2\plus{}b^2\plus{}c^2\geq3\times\frac{1}{a^2}\Leftrightarrow a\geq1$ and finnaly,
$ \frac {a^2}{b^3} \plus{} \frac {b^2}{c^3} \plus{} \frac {c^2}{a^3}\geq 3$

[quote="quykhtn-qa1"]Let a,b,c be positive numbers such that:$ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$.Prove that:
$ \frac {a^2}{b^3} \plus{} \frac {b^2}{c^3} \plus{} \frac {c^2}{a^3}\geq\ 3$[/quote]
Let $ a > 0,$ $ b > 0,$ $ c > 0,$ $ n>0$ and $ a^n \plus{} b^n \plus{} c^n \equal{} 3.$ Prove that
\[ \frac {a^2}{b^3} \plus{} \frac {b^2}{c^3} \plus{} \frac {c^2}{a^3}\geq\ 3
\]

[quote="arqady"][quote="quykhtn-qa1"]Let a,b,c be positive numbers such that:$ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$.Prove that:
$ \frac {a^2}{b^3} \plus{} \frac {b^2}{c^3} \plus{} \frac {c^2}{a^3}\geq\ 3$[/quote]
Let $ a > 0,$ $ b > 0,$ $ c > 0,$ $ n > 0$ and $ a^n \plus{} b^n \plus{} c^n \equal{} 3.$ Prove that
\[ \frac {a^2}{b^3} \plus{} \frac {b^2}{c^3} \plus{} \frac {c^2}{a^3}\geq\ 3
\]
[/quote]
yes it's true. :)
you can solve second inequality?,arqady :)

[quote="quykhtn-qa1"]This inequality than''stronger'' with condition:
$ \frac {a^2}{1 \plus{} b^3} \plus{} \frac {b^2}{1 \plus{} c^3} \plus{} \frac {c^2}{1 \plus{} a^3}\geq\ \frac {3}{2}$:)  :)  :)[/quote]
It's obviously trues because: $ \sum\ a^2b^3 \leq\ 3$ with $ a^2\plus{}b^2\plus{}c^2 \equal{}3$
It's very well-known which can be solved by Cauchy Schwarts and Vasc's ineq.
The general problem: (with the same condition)  :)
$ \sum\ \frac{a^2}{k\plus{}b^3}  \geq\ \frac{3}{k\plus{}1}$

[quote="nguoivn"][quote="quykhtn-qa1"]This inequality than''stronger'' with condition:
$ \frac {a^2}{1 \plus{} b^3} \plus{} \frac {b^2}{1 \plus{} c^3} \plus{} \frac {c^2}{1 \plus{} a^3}\geq\ \frac {3}{2}$:)  :)  :)[/quote]
It's obviously trues because: $ \sum\ a^2b^3 \leq\ 3$ with $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$
It's very well-known which can be solved by Cauchy Schwarts and Vasc's ineq.
The general problem: (with the same condition)  :)
$ \sum\ \frac {a^2}{k \plus{} b^3} \geq\ \frac {3}{k \plus{} 1}$[/quote]
Nice proof,my friend :) .
But with this inequality,my proof use AM-GM and CAUCHY-SCHWARZ :)  :)

by Caushy-Schwarz:
$ \left(\sum\ \frac {a^2}{k \plus{} b^3}\right)(3k\plus{}\sum a^2b^3) \geq 9$
then $ \left(\sum\ \frac {a^2}{k \plus{} b^3}\right)(3k\plus{}3) \geq 9$ because $ \sum a^2b^3\leq3$
therefore: $ \sum\ \frac {a^2}{k \plus{} b^3}\geq \frac{3}{k\plus{}1}$

[quote="peine"]by Caushy-Schwarz:
$ \left(\sum\ \frac {a^2}{k \plus{} b^3}\right)(3k \plus{} \sum a^2b^3) \geq 9$
then $ \left(\sum\ \frac {a^2}{k \plus{} b^3}\right)(3k \plus{} 3) \geq 9$ because $ \sum a^2b^3\leq3$
therefore: $ \sum\ \frac {a^2}{k \plus{} b^3}\geq \frac {3}{k \plus{} 1}$[/quote]

I do not think this is true. :|
In fact, given $ \sum a^2b^3\le 3$, we DO NOT have :
$ (3k\plus{}\sum a^2b^3)\ge (3k\plus{}3)$, the inequality sign is reversed. :wink:

for polta,
we have $ \left(\sum\ \frac {a^2}{k \plus{} b^3}\right)(3k\plus{}3) \geq \left(\sum\ \frac {a^2}{k \plus{} b^3}\right)(3k\plus{}\sum a^2b^3) \geq 9$ now, I think you understand,

[quote="arqady"][quote="quykhtn-qa1"]Let a,b,c be positive numbers such that:$ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$.Prove that:
$ \frac {a^2}{b^3} \plus{} \frac {b^2}{c^3} \plus{} \frac {c^2}{a^3}\geq\ 3$[/quote]
Let $ a > 0,$ $ b > 0,$ $ c > 0,$ $ n > 0$ and $ a^n \plus{} b^n \plus{} c^n \equal{} 3.$ Prove that
\[ \frac {a^2}{b^3} \plus{} \frac {b^2}{c^3} \plus{} \frac {c^2}{a^3}\geq\ 3
\]
[/quote]
nice inequality,you can solve it only by AM-GM,
$ \frac {a^2}{b^3} \plus{} \frac {b^2}{c^3} \plus{} \frac {c^2}{a^3}\geq\ 3\sqrt[3]{\frac{1}{abc}}$
$ a^n\plus{}b^n\plus{}c^n \geq 3\sqrt[3]{a^nb^nc^n} \Leftrightarrow \frac{1}{abc} \geq 1$
then, $ \frac {a^2}{b^3} \plus{} \frac {b^2}{c^3} \plus{} \frac {c^2}{a^3}\geq\ 3$
more general, but easy,
let $ a_1,a_2,...,a_k,n>0$ and $ a_1^n\plus{}a_2^n\plus{}...\plus{}a_k^n\equal{}k$ prove that,
$ \sum_{i\equal{}1}^k \frac{a_i^t}{a_{i\plus{}1}^{t\plus{}1}}\geq k$

[quote="quykhtn-qa1"]Nice proof,my friend :) .
But with this inequality,my proof use AM-GM and CAUCHY-SCHWARZ :)  :)[/quote]
Of course, your ineq can be solved by AM-Gm and Cauchy Schwarts.
The similar result can be solved by the way above: : $ \frac{a^2}{1\plus{}b^4} \plus{} \frac{a^2}{1\plus{}b^4} \plus{} \frac{a^2}{1\plus{}b^4} \geq\ \frac{3}{2}$
How about the similar (with the same condition)
$ \sqrt {\frac {a^4}{k \plus{} b^3}} \plus{} \sqrt {\frac {b^4}{k \plus{} c^3}} \plus{} \sqrt {\frac {c^4}{k \plus{} a^3}} \geq\ \frac {3}{\sqrt {k \plus{} 1}}$
 :)

[quote="nguoivn"][quote="quykhtn-qa1"]Nice proof,my friend :) .
But with this inequality,my proof use AM-GM and CAUCHY-SCHWARZ :)  :)[/quote]
Of course, your ineq can be solved by AM-Gm and Cauchy Schwarts.
The similar result can be solved by the way above: : $ \frac {a^2}{1 \plus{} b^4} \plus{} \frac {a^2}{1 \plus{} b^4} \plus{} \frac {a^2}{1 \plus{} b^4} \geq\ \frac {3}{2}$ [/quote]

quykhtn-qa1:this inequality is than stronger(by billzhao) with the same condition:
$ \frac{a^2}{1\plus{}b^6}\plus{}\frac{b^2}{1\plus{}c^6}\plus{}\frac{c^2}{1\plus{}a^6}\geq\ \frac{3}{2}$ :)
[/quote=''nguoivn'']
nguoivn wrote:
How about the similar (with the same condition)
$ \sqrt {\frac {a^4}{k \plus{} b^3}} \plus{} \sqrt {\frac {b^4}{k \plus{} c^3}} \plus{} \sqrt {\frac {c^4}{k \plus{} a^3}} \geq\ \frac {3}{\sqrt {k \plus{} 1}}$ [/quote]
 :)

quykhtn-qa1-use HOLDER :)

[quote="Potla"][quote="peine"]by Caushy-Schwarz:
$ \left(\sum\ \frac {a^2}{k \plus{} b^3}\right)(3k \plus{} \sum a^2b^3) \geq 9$
then $ \left(\sum\ \frac {a^2}{k \plus{} b^3}\right)(3k \plus{} 3) \geq 9$ because $ \sum a^2b^3\leq3$
therefore: $ \sum\ \frac {a^2}{k \plus{} b^3}\geq \frac {3}{k \plus{} 1}$[/quote]

I do not think this is true. :|
In fact, given $ \sum a^2b^3\le 3$, we DO NOT have :
$ (3k \plus{} \sum a^2b^3)\ge (3k \plus{} 3)$, the inequality sign is reversed. :wink:[/quote]
peine'solution is true. IT IS SIMILAR nguoivn'solution :)

thank you quykhtn-qa1, for nguoivn's inéqualiy, I found another solution, I think it's beautiful
the function $ f(x)\equal{}\frac{1}{\sqrt{x}}$ is convex, then with Jensen,
$ \sum \frac{a^2}{3}\times\frac{1}{k\plus{}b^3} \geq \frac{1}{\sqrt{k\plus{}a^2b^3\plus{}b^2c^3\plus{}c^2a^3}}$
we have $ a^2b^3\plus{}b^2c^3\plus{}c^2a^3\leq 3$ then
$ \sum \frac{a^2}{3}\times\frac{1}{k\plus{}b^3} \geq \frac{1}{\sqrt{k\plus{}3}}$
$ \Leftrightarrow \sqrt {\frac {a^4}{k \plus{} b^3}} \plus{} \sqrt {\frac {b^4}{k \plus{} c^3}} \plus{} \sqrt {\frac {c^4}{k \plus{} a^3}} \geq\ \frac {3}{\sqrt {k \plus{} 1}}$

[quote="quykhtn-qa1"]
this inequality is than stronger(by billzhao) with the same condition:
$ \frac {a^2}{1 \plus{} b^6} \plus{} \frac {b^2}{1 \plus{} c^6} \plus{} \frac {c^2}{1 \plus{} a^6}\geq\ \frac {3}{2}$ :)
[/quote]

It's obviously trues by Am-Gm and $ a^2b^3 \plus{} b^2c^3 \plus{} c^2a^3 \leq\ 3$.

[quote="peine"]what about my inéquality? it's not very hard![/quote]
with inequality of nguoivn, use Cauchy-schwarz.not need jensen :wink:

why?Is that my solution with Jensen is wrong?
P.S: for my inéquality , see my thirs post.

[quote="peine"]what about my inéquality? it's not very hard![/quote]
i don't understand, :wink:

I wanted to say that no one proved my inequality. I proposed it in my third post. there is it,
let $ a_1,a_2,...,a_k,n>0$ a reel numbers, $ a_1^n\plus{}a_2^n\plus{}...\plus{}a_k^n\equal{}k$ prove that,
$ \sum_{i\equal{}1}^k \frac{a_i^t}{a_{i\plus{}1}^{t\plus{}1}}\geq k$

[quote="peine"]I wanted to say that no one proved my inequality. I proposed it in my third post. there is it,
let $ a_1,a_2,...,a_k,n > 0$ a reel numbers, $ a_1^n \plus{} a_2^n \plus{} ... \plus{} a_k^n \equal{} k$ prove that,[/quote]
$ \sum_{i \equal{} 1}^k \frac {a_i^t}{a_{i \plus{} 1}^{t \plus{} 1}}\geq k$[/quote]
By AM-GM inequality we have:
$ k\equal{}a_1^n\plus{}a_2^n\plus{}...\plus{}a_k^n\geq\ k\sqrt[k]{(a_1a_2...a_k)^n}$
=> $ a_1a_2...a_n\le\ 1$
THEREFORE $ \sum_{i\equal{}1}^k\frac{a_i^t}{a_{i\plus{}1}^{t\plus{}1}}\geq\ k \sqrt[k]{\frac{1}{(a_1a_2...a_k)^n}} \geq\ k$ :) :)

now,this inequality than stronger:$ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$.Prove that:
$ \frac {a^4}{b^3 \plus{} c^3} \plus{} \frac {b^4}{c^3 \plus{} a^3} \plus{} \frac {c^4}{a^3 \plus{} b^3}\geq \
\frac {a^4 \plus{} b^4 \plus{} c^4}{2}$ :)
see here:
http://www.mathlinks.ro/viewtopic.php?p=1417924#1417924  :)


what is ure weight, and height?

im $222$ pound,may be more, but im not fat. :wink:

[quote="ShoeFactory"]5'7 (CUZ IM ASIAN)

127 lbs[/quote]
um... 5'10 (CUZ I'M ASIAN) (approx. 177 cm)
something like 120 pounds.

excuse me, could u plz write ure height in centimeter,? :wink:

152 CM

5 feet, 0 inches,

43.5 KG

95 Pounds

98 kilograms or 216 pounds (yes that is a lot i know, but I am honest)

about 182 centimeters = maybe a little than 6 feet
[url]http://www.onlineconversion.com/length_common.htm[/url] is a nice converter:)

5'6.75" (or so, depending on who you ask) ('CAUSE I'M NOT ASIAN :D), 135 pounds.

50.7 kilos, which is about 111 pounds
5'6 1/2" - 5'8" depending on what time of day it is which is ~169-173 cm.

nice topic.and may be funny.I am 85 kilo and 184 cm height. :D

[quote="nutz_for2.718281828"]4 foot 10 inches, 83 pounds. I don't know my height in centimeters.[/quote]
Use Google.  :wink:

I'm 5 ft. 9.5 inches = 176.5 cm.
And 120 pounds = 54.5 kilos.

I'm about 5'8" or 5'9", I can't remember which.  I weigh about 100 pounds.

Using the metric system:
45.35 KG
172-175 CM

[quote="ragnarok23"]I hate you tall people. :bomb:  :P
5'1''
100 pounds.[/quote]

Supposing I count in the set of medium people, that's the first time anyone's ever called me more than short :D.

[quote="lizcheang"]Wow... tall people

I've heard such correlation :

The larger your shoe size, the smarter you are.
If you have a large shoe size, you are tall.

Therefore, you tall people are smart ;][/quote]

then what is ure shoe size? :wink:

[quote="lizcheang"][quote="akalra1"][quote="lizcheang"]Wow... tall people

I've heard such correlation :

The larger your shoe size, the smarter you are.
If you have a large shoe size, you are tall.

Therefore, you tall people are smart ;][/quote]

prejudice

i wear size 8s

my friend is a giant, but he does NOT have a giant brain[/quote]

Uh like 8 or 9... I don't know...

No, its not prejudice homie. It's just a myth that happens to be true in some cases. You never know, maybe your shoe size will ddrastically increase ;)[/quote]

*looks around*

does plastic surgery count

[quote="akalra1"][quote="lizcheang"][quote="akalra1"]yea

sounds stupid

*runs away*[/quote]

you are smart, you don't need plastic surgery. haha.[/quote]

i will walk in big shoes and wear big, thick socks so i will "fit" in big shoes[/quote]

 :roll:

probably some formula like this:

$s : iq = 1 : 10$ where $s$ is ur shoe size and $iq$ is ur iq
$s \approx f$ where $s$ is ur shoe size and$f$ is the lenght of ur face

my iq is about 80 then!

who knows!

[quote="mtms5467"][quote="lizcheang"]Wow... tall people

I've heard such correlation :

The larger your shoe size, the smarter you are.
If you have a large shoe size, you are tall.

Therefore, you tall people are smart ;][/quote]
I've heard that the length of your foot and the length of your face is the same. So are people with huge faces smarter than the rest?[/quote]

not necessarilly :wink:


Post here, those going to Iowa ARML 2005
(And include your team)

I'm going with Michigan. See you guys there!

Going with Indiana, alc are you Alex C. from lux middle school?

[quote="Rep123max"]I'm going from Texas.[/quote]
a question... wouldnt it be closer from Texas to go to SJSU then IOWA? (obviously i have no knowledge of geography whatsoever)

[quote="xxreddevilzxx"](obviously i have no knowledge of geography whatsoever)[/quote]

Indeed. :P

It's far closer for Texas to go to Iowa. The distance from Houston to El Paso (west Texas) would be like the distance to Iowa...Slight exaggeration...but w/e

If you're flying, it doesn't particularly matter- the western site sometimes had an Alabama (I think) team when it was in Las Vegas. San Jose isn't quite as easy to fly to as Las Vegas, but it's still probably easier than the other two sites.

Time zones might be an issue, though: the western site does the team and power rounds Friday night.

I'll be there, Minnesota Gold.  (At least, I'm assuming Gold team as I've been on it two years already; we don't officially find out our final team placements until the night before the meet!)

I'm going.  Minnesota.

[EDIT]  Gold team, it seems.  Although, as Silas said, we don't find out until 10:30 pm on June 3rd..  [/EDIT]

Wisconsin will pwn at Iowa ARML 2005 (Either that, or just have lots of fun and almost completely ignore the competition :lol:)

i thikn Texas has a pretty strong team... is Zach Abel going??

Should we have some form of AoPS meeting when we get there?

I'm going to Iowa... from Iowa.
I have to walk four blocks to get to the dorms.
Oh the agony!
 :P
Oh yeah, and shogia is on my relay.

[quote="Fountainhead"]Missouri A. I formally challenge y'all to a frisbee match when we get there.[/quote]

Too bad the weather didn't care much for us  :(
Chicago here, though I realize it's over.

Yeah, I also went w/ Chicago A.

Good job everyone!

JB

My Iowa team was doing so well.  We were tied for 3rd in the nation after the power and team rounds (perfect team round yay!).  Then it fell apart in the individual and relays.  We arent losing too many people so hopefully we will do better next year.

Yeah, it was interesting seeing the standings (after team and individual) during lunch.  Everything kind of got turned on its head--Exeter first, North Carolina second, Lehigh Valley fourth, TJ and NYC way down.  And I was proud that half the people who solved #8 at our site came from Minnesota (I was one of them).

Also, the new tiebreaker thing was great.  I loved going up there, even if I lost badly and the tiebreak turned out to be meaningless.

[quote="Rep123max"]Does anybody have a cool solution for #6?

Ellipse $E$ is defined by the equation $\frac{x^2}{25}+\frac{y^2}{16}=1$. Compute the radius of the largest circle that is internally tangent to $E$ at $(5, 0)$, and only intersects $E$ at $(5, 0)$.[/quote]

I think you can use the formula for curvature of a thingy if you happen to know it. There's also a solution using an affine transformation I think, and then of course there's just algebra, but that's not "cool", I would imagine.

[quote="jackb1115"][quote="Sly Si"]Also, the new tiebreaker thing was great.  I loved going up there, even if I lost badly and the tiebreak turned out to be meaningless.[/quote]

I don't think the tiebreaker thing was new.  In the ARML books they offer tiebreaker questions from past years.  They didn't do it last year because it was unnecessary.

JB[/quote]

actually in San Jose last yr I remember Jongmin Baek was up there doing it.

Yeah, I've been there in the past and seen tiebreakers a couple of times.  By "the new tiebreaker thingy" I meant where they broke the tie for the site championship; thus all the people that got 6 (myself included) got to be in it.

[quote="Rep123max"]Does anybody have a cool solution for #6?

Ellipse $E$ is defined by the equation $\frac{x^2}{25}+\frac{y^2}{16}=1$. Compute the radius of the largest circle that is internally tangent to $E$ at $(5, 0)$, and only intersects $E$ at $(5, 0)$.[/quote]

If "hunches" are cool...  Since the two foci are colinear with the point $(5,0)$, my "hunch" would be that the radius of curvature would be a mean of the distances from $E$ to the two foci.

The foci are found to be at $(\pm 3,0)$ by the pythagorean theorem.  So the distances from the point $E$ to the two foci are $2$ and $8$.

The arithmetic mean of $2$ and $8$ is $5$.  But that answer is ridiculous, since the circle clearly extends beyond the ellipse.

The geometric mean of $2$ and $8$ is $4$.  Again, the circle extends beyond the ellipse.

The harmonic mean of $2$ and $8$ is $\frac{2}{\frac{1}{2}+\frac{1}{8}}=3.2.$   If time is a factor, I trust my "hunch" and I move on.

If hunches aren't cool, we can use algebra.  Solve for $x$:

$x=5\Big({1-\frac{y^2}{16}\Big)^\frac{1}{2}}$

Then assume $y$ is very small, and take the first nontrivial term of the binomial expansion:

$x\approx 5\Big(1-\frac{1}{2}\times\frac{y^2}{16}\Big)$

$x\approx 5\Big(1-\frac{y^2}{32}\Big)$

The inscribed circle we seek takes the form:

$(x-5+r)^2+y^2=r^2$

Solve for x:

$x-5+r=\Big(r^2-y^2\Big)^\frac{1}{2}$

$x=\Big(r^2-y^2\Big)^\frac{1}{2}+5-r$

$x=r\Big(1-\frac{y^2}{r^2}\Big)^\frac{1}{2}+5-r$

Assume $y$ is very small, but $r$ is not very small.  Take the first nontrivial term of the binomial expansion:

$x\approx r\Big(1-\frac{1}{2}\times\frac{y^2}{r^2}\Big)+5-r$

Set the two equations equal:

$5\Big(1-\frac{y^2}{32}\Big) = r\Big(1-\frac{y^2}{2r^2}\Big)+5-r$

$5-\frac{5}{32}y^2=r-\frac{1}{2r}y^2+5-r$

$-\frac{5}{32}y^2=-\frac{1}{2r}y^2$

$\frac{5}{32}=\frac{1}{2r}$

$r=\frac{32}{5\times 2}=3.2$

If "cool" includes being a master of trivia, the radius of curvature of an ellipse is a minimum where the ellipse intersects the major axis, and is known to be $\frac{b^2}{a}.$  If you happen to know this little formula, the radius of curvature is simply $\frac{4^2}{5}=3.2$.

Is this the right answer?

[quote=rcv]Is this the right answer?[/quote]

Yes, the correct answer was 16/5.

What was the correct answer for T-7 ?  If it's 48, I'm going to kick myself...

man i suck at guessing, i guessed 15/4 for the ellipse question, darn.  :mad:

[quote="Danbert"]What was the correct answer for T-7 ?  If it's 48, I'm going to kick myself...[/quote]

you were off. it was 60.


[quote="ductrung"][quote="hungkhtn"]Prove that for all non-negative real numbers $ a,b,c$, we have
\[ \frac{a^{4}+b^{4}+c^{4}}{a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}}+\frac{2(ab+bc+ca)}{a^{2}+b^{2}+c^{2}}\ge 3. \]
[/quote]
Standard Vornicu-Schur will do: using $ x^{2}+y^{2}+z^{2}-xy-yz-zx=\sum_{cycl}(x-y)(x-z)$, the inequality is equivalent to
\[ x(a-b)(a-c)+y(b-c)(b-a)+z(c-a)(c-b)\ge 0 \]
where
\[ x=\frac{(a+b)(a+c)}{a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}}-\frac{2}{a^{2}+b^{2}+c^{2}}\]
(and similarly for $ y$ and $ z$). Note that
$ (a^{2}+ab+bc+ca)(a^{2}+b^{2}+c^{2})\ge (ab+bc+ca)(a^{2}+b^{2}+c^{2})$
$ \ge a^{3}(b+c)+a(b^{3}+c^{3})+bc(b^{2}+c^{2})\ge 2(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2})$
by the AM-GM inequality so that $ x,y,z\ge 0$. And clearly, $ x\ge y$ if we assume $ a\ge b\ge c$.[/quote]

Good Job! Ductrung! Thank you,

[quote="hungkhtn"]I remember that I did propose the following problem in Mathlinks, but there was not a solution. It is very very nice I think, and let us prove it.

Prove that for all non-negative real numbers $ a,b,c$, we have
\[ \frac {a^{4} \plus{} b^{4} \plus{} c^{4}}{a^{2}b^{2} \plus{} b^{2}c^{2} \plus{} c^{2}a^{2}} \plus{} \frac {2(ab \plus{} bc \plus{} ca)}{a^{2} \plus{} b^{2} \plus{} c^{2}}\ge 3.
\]
[/quote]
Suppose that $ c \equal{} min\{a;b;c\}$
We can rewrite this inequality :
$ (a \minus{} b)^2[\frac {(a \plus{} b)^2}{a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2} \minus{} \frac {2}{a^2 \plus{} b^2 \plus{} c^2}] \plus{} (a \minus{} c)(b \minus{} c)[\frac {(a \plus{} c)(b \plus{} c)}{a^2b^2 \plus{} b^2  c^2 \plus{} c^2a^2} \minus{} \frac {2}{a^2 \plus{} b^2 \plus{} c^2}] \geq 0$
From $ c \equal{} min\{a;b;c\}$,we have $ \frac {(a \plus{} b)^2}{a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2} \minus{} \frac {2}{a^2 \plus{} b^2 \plus{} c^2} \geq 0$ and $ \frac {(a \plus{} c)(b \plus{} c)}{a^2b^2 \plus{} b^2  c^2 \plus{} c^2a^2} \minus{} \frac {2}{a^2 \plus{} b^2 \plus{} c^2} \geq 0$
Done!

[quote=luofangxiang][quote=silouan]Here is another solution with AM-GM

We have that $ \frac {a^4 \plus{} b^4 \plus{} c^4}{a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2} \plus{} 2\frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2}\geq 3\sqrt [3]{\frac {(a^4 \plus{} b^4 \plus{} c^4)(ab \plus{} bc \plus{} ca)^2}{(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2)(a^2 \plus{} b^2 \plus{} c^2)^2}}$

So we have to show that $ (a^4 \plus{} b^4 \plus{} c^4)(ab \plus{} bc \plus{} ca)^2\geq (a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2)(a^2 \plus{} b^2 \plus{} c^2)^2$ which is

equivalent to
$ \sum (abc^2(a \minus{} b)(a^3 \minus{} b^3)) \plus{} abc(a^5 \plus{} b^5 \plus{} c^5 \minus{} abc(a^2 \plus{} b^2 \plus{} c^2))\geq 0$[/quote]

This is wrong[/quote]

I think silouan's solution is right!

[quote=hungkhtn]I remember that I did propose the following problem in Mathlinks, but there was not a solution. It is very very nice I think, and let us prove it.

Prove that for all non-negative real numbers $ a,b,c$, we have
\[ \frac{a^{4}+b^{4}+c^{4}}{a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}}+\frac{2(ab+bc+ca)}{a^{2}+b^{2}+c^{2}}\ge 3.\][/quote]
Write inequality as
\[\frac{a^4+b^4+c^4-a^2b^2-b^2c^2-c^2a^2}{a^2b^2+b^2c^2+c^2a^2} \geqslant \frac{2(a^2+b^2+c^2-ab-bc-ca)}{a^2+b^2+c^2}.\]
Accorduing to Schur Inequality then
\[a^4+b^4+c^4-a^2b^2-b^2c^2-c^2a^2 \geqslant (ab+bc+ca)(a^2+b^2+c^2-ab-bc-ca).\]
Therefore we need to show that
\[(ab+bc+ca)(a^2+b^2+c^2) \geqslant 2(a^2b^2+b^2c^2+c^2a^2),\]
equivalent to
\[\sum ab(a-b)^2+abc(a+b+c) \geqslant 0.\]
Which is true.

[quote=Nguyenhuyen_AG][quote=Akatsuki1010]I think silouan's solution is right![/quote]
This is wrong because
[quote=silouan]
So we have to show that $ (a^4 \plus{} b^4 \plus{} c^4)(ab \plus{} bc \plus{} ca)^2\geq (a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2)(a^2 \plus{} b^2 \plus{} c^2)^2$
[/quote]
wrong.[/quote]

True this is correct silouan's solution is wrong.


geometry
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Hello, and welcome to this game.

I will have 10 players in all. The first person to sign up will co-mod, unless they don't want to. You will all start out at level 0. You must get 100 xp. points to get to level 1.
To start, there is are several towns, cities, and kingdoms. The town you will all start in is called Pallatona. In Pallatona is an Inn, a shop, a bank, and a training area. The Inn is pretty much a your house, and restaurant. The shop is where you can buy things, and look at the news. The bank is where you can withdraw, deposit, or loan money. Finally, the training area, is where you can add to you're skills.

The skills all start out at 1, and they are as follows:Health, Attack(strength), Defense(how much the enemy's attack is reduced), Accuracy, and Charm.You all start out with 0 for your money. And you're current maximum Health is 20. Once you reach level 10, you can choose to face the Drakon, an evil beast. Read the following, it may contain vital information to your survival in the game.
[hide][hide]HA ha tricked you![/hide][hide][hide]If you are reading this, you get 10000 coins, p.m. me if you see this.[/hide][/hide][/hide]

CO MOD! I may not be that active monday to thursday

/in... i  :P hope you don't modkill ur game again

wow, this game doesn't sound familar.
legend of the black dragon and legend of the drakon. no, i don't notice anything :rotfl:
Im in

7 spots left. I agree with bluecarneal, that's why i have a co-mod.

OMG YOU HAVE CHARM AND FLIRTING TOO  :o
 :mad:  :mad:

Well this seems a little TOO similar to LOBD...I know it's not exactly the same, but could I make some suggestions to make it slightly different?

1. classes.  Like ninja and mage and warrior.
2. special skills.
3. fights against multiple enemies.

BTW, I'm not joining, just watching.

okay! You're classes you can choose are warrior, ninja, mage, chicken/coward, and jack-of-all-trades. You can train for different skills at the training area. Oh, There's also speed(chance to dogde attacks).

/in
Wants to be a mage, and a question - What happened to archery?

--
Hopes this does not die like ferret wars.

You all go to Sheno. Mathwiz85 finds a votex and he gets teleported to your all's location....

[color=orange][size=200]The Final Battle Of the World[/size][/color]

Both of the Drakons, Fire and Ice, have rampaged all over the world. And now, they have reached the town of Sheno. They prepare for battle. You all look up at them and shiver in your boots. Each drakon- 1000 hp. Each of you- 150 hp.

Since when do I find vortexes lying around... ok who cares
hah I start sniping the dragons eyes with my awsome skillz

Everyone except mathwiz85 is staring at the Drakons. He jumps to action and shoots the Drakons in the eyes. -100 health to each of them. They both retaliate and stare at mathwiz85, he freezes in fear. Then, the Fire Drakon lights mathwiz85 up....... -50 health to mathwiz85... -25 health per turn.....

aggg!! take off my shirt wrap it around my katana, jump on top of ice drakon while stabing it...

Good idea, you do that. The Ice Drakon swipes at you but misses. You stab it in the brain and cut a hole out of the top of its head. You stab the brain and pull it out. Thus killing the Ice drakon. You have 75 health left mathwiz85.

THIS IS AWSOME i am taking down the drakons single handedly...  :lol:

umm.  :idea:  =)
throw all the people standing around at the fire drakon after putting ice drakon remains on them(including the mod hover 10 off the ground telling me what happens)

[quote="Butler"]You do that. The Fire Drakon loses 850 health..... It stomps you and you have 1 leath left...[/quote] good means I have all my health...
Grab a taser from my fallen friends and tas that sucka

Uhhhh.....  You only have 1 health point left..... You taser it and it falls over dead. You gave it a heartattack....

THANKS FOR PLAYING!! HOPE YOU ENJOYED IT!!

AWSOME great game
how ever I think richard, sabein, and tennis going to be mad that they were offline...

kewl now you have no more games running
START THE NEW ONE!!!!

Uhh..... Sign-ups for ROTLF are open.....

Yeah I know, just sent my pm
I said start it as soon as we get enough pms, none of this waiting around stuff

I'm hoping to have at least 20 players..... If you want to discuss this further, go to my blog and use the shout box.

Mods please lock this thread. The game is over.


function
algebra
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calculus
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functional equation
algebra unsolved

Find [color=red]all  functions[/color]  $f$ , if they have domain the interval $(0 , +\infty)$  , have derivative for all $x$ in this interval  and
     $f(x) = mf(mx) , \forall x > 0$ for some positive real  $m$ which is different fom $1$ .

 [u]Babis[/u]

Let $g(x) = \frac{1}{f(x)}$.  We have

$m g(x) = g(mx)$

Infinitely differentiating gives $g(x) = c_1 x \implies f(x) = \frac{c_2}{x}$.

Or can I not assume it is infinitely differentiable?  Either way, I'm pretty sure the above functional equation has an obvious solution...

I was given the problem only with the existence of the first derivative.How did you prove that f is not zero (is different from 0 )for all $x \ in \mathbb{R}$. I'm also trying to prove that f is of the form you have found.

[u]  Babis[/u]

[i](please see also the posts in the calculus section. Perhaps some idea is usefull)[/i]

Well, from the first derivative we can ascertain $g'(x) = g'(mx)$.  So, for any $x_0$ such that $g'(x_0) = d$ is known we can find $g'(m^k x_0) = d$ for integers $k$.  After that I think it is necessary that we assume $g'(x)$ continuous.  Otherwise it can take an ugly piecewise form...

Yes!
   I agree , that probably something is not given in the problem.But the continuouity of  $g'$ seems not to be enough.I'll check it more deeply.

 babis


Let $ a,$ $ b$ and $ c$ are non-negative numbers, such that $ a^6\plus{}b^6\plus{}c^6\equal{}3.$ Prove that:
\[ a^3\plus{}b^3\plus{}c^3\geq ab\plus{}ac\plus{}bc\]
[hide="remark"] There is a nice proof and it's much easier than
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=175192[/hide]

Square both sides,
$ 3\plus{} 2\sum a^3b^3 \ge 3 \sum a^2b^2 \ge (ab\plus{}bc\plus{}ca)^2$ by AMGM.


If the equation  $ ax^4 \plus{} bx^3 \plus{} cx^2 \plus{} dx \plus{} e \equal{} 0$ ,with $ a , b , c ,d,e$  where $ a,b,c,d\in \{1,2,3,\ldots,49\}$(notice $ e$ not included)non zero digits, has only one integer root (49), find all  $ (a,b,c,d,e)$

it is  $ a,b,c,d\in \{1,2,3,\ldots,49\}$

the first one u said


Let $A_{n}$ be the number of parts of {1,...,n} that does not contain two consecutive integers.
1/ Show that :
$A_{n} = A_{n-1} + A_{n-2}$
2/ Calculate $A_{n}$

Good luck :)

This is quite easy, because any set with no two consecutive integers from $\{1, \cdots, n \}$ is either formed by $n$ and any 'good' set from $\{1, \cdots, n-2 \}$ (since it cannot canotains $n-1$) or it doesn't contain $n$ and is only formed by any good set from $\{1, \cdots, n-1 \}$. The converse is quite obvious.
Thus $A_n = A_{n-1} + A_{n-2}.$

Note that $A_1 = 2$ (the emptyset and $\{1\}$) and $A_2 = 3$ (the emptyset, $\{1 \}$, $\{2\}$).
Since the above relation is linear of order two the rest is straightforward....

Pierre.

Thanks Pierre, I was thinking it was very hard, but... you proved me wrong! ;)


Given $ a,b,c>0$.Prove that:
$ 2\left( {\frac{a}{{b \plus{} 2c}} \plus{} \frac{b}{{c \plus{} 2a}} \plus{} \frac{c}{{a \plus{} 2b}}} \right) \ge \frac{{a\left( {b \plus{} c} \right)}}{{{a^2} \plus{} ab \plus{} bc}} \plus{} \frac{{b\left( {c \plus{} a} \right)}}{{{b^2} \plus{} bc \plus{} ca}} \plus{} \frac{{c\left( {a \plus{} b} \right)}}{{{c^2} \plus{} ca \plus{} ab}}$

I will provide a proof of Can's last inequality he used, since I have used the $ \sum S_{a}(a \minus{} b)(a \minus{} c)$ method for the first time!  :lol:
We have to show
$ \frac {a \plus{} b \plus{} c}{ab \plus{} bc \plus{} ca} \ge \frac {a}{a^2 \plus{} 2bc} \plus{} \frac {b}{b^2 \plus{} 2ac} \plus{} \frac {c}{c^2 \plus{} 2ab}$
We can assume $ a \ge b \ge c$.
Note that $ \frac {a}{ab \plus{} bc \plus{} ca} \minus{} \frac {a}{a^2 \plus{} 2bc} \equal{} \frac {a(a \minus{} b)(a \minus{} c)}{(ab \plus{} bc \plus{} ca)(a^2 \plus{} 2bc)}$
So we have to show
$ \frac {a(a \minus{} b)(a \minus{} c)}{a^2 \plus{} 2bc} \plus{} \frac {b(b \minus{} a)(b \minus{} c)}{b^2 \plus{} 2ac} \plus{} \frac {c(c \minus{} a)(c \minus{} b)}{c^2 \plus{} 2ab} \ge 0$
$ (a \minus{} b)(\frac {a(a \minus{} c)}{a^2 \plus{} 2bc} \minus{} \frac {b(b \minus{} c)}{b^2 \plus{} 2ac}) \plus{} \frac {c(c \minus{} a)(c \minus{} b)}{c^2 \plus{} 2ab} \ge 0$
But $ \frac {c(c \minus{} a)(c \minus{} b)}{c^2 \plus{} 2ab}$ is obviously greater than $ 0$, so we have to show
$ (a \minus{} b)(\frac {a(a \minus{} c)}{a^2 \plus{} 2bc} \minus{} \frac {b(b \minus{} c)}{b^2 \plus{} 2ac}) \ge 0$
$ (a \minus{} b)\frac {(a \minus{} c)(abc \plus{} 2a^2c) \minus{} 2b^2c(b \minus{} c)}{(a^2 \plus{} 2bc)(b^2 \plus{} 2ac)} \ge (a \minus{} b)\frac {(a \minus{} c)(abc \plus{} 2a^2c \minus{} 2b^2c)}{(a^2 \plus{} 2bc)(b^2 \plus{} 2ac)} \ge 0$
Which is obviously true.

[quote="can_hang2007"][quote="leedt26"]Given $ a,b,c > 0$.Prove that:
$ 2\left( {\frac {a}{{b \plus{} 2c}} \plus{} \frac {b}{{c \plus{} 2a}} \plus{} \frac {c}{{a \plus{} 2b}}} \right) \ge \frac {{a\left( {b \plus{} c} \right)}}{{{a^2} \plus{} ab \plus{} bc}} \plus{} \frac {{b\left( {c \plus{} a} \right)}}{{{b^2} \plus{} bc \plus{} ca}} \plus{} \frac {{c\left( {a \plus{} b} \right)}}{{{c^2} \plus{} ca \plus{} ab}}$[/quote]
See here: http://canhang2007.wordpress.com/2009/11/10/inequality-64-unknown-author/[/quote]
Your proof is very nice Can_hang , thanks.  :lol:

[edit] Thanks, abdek :( . Maybe I should had gone to sleep by that time (12:53 am) :(
It's a shame that my solution was wrong; can the mod please delete this post? [/edit]

[quote="Potla"][quote="dgreenb801"]
$ \frac {a \plus{} b \plus{} c}{ab \plus{} bc \plus{} ca} \ge \frac {a}{a^2 \plus{} 2bc} \plus{} \frac {b}{b^2 \plus{} 2ac} \plus{} \frac {c}{c^2 \plus{} 2ab}$
[/quote]
I also proved it just using AM-GM and CS as follows:
[hide=My solution/Please verify]
Hence we are required to prove that
\[ \frac {9(a \plus{} b \plus{} c)}{ab \plus{} bc \plus{} ca}\le \frac {ab \plus{} bc \plus{} ca}{abc} \plus{} \frac {2(a^2 \plus{} b^2 \plus{} c^2)}{abc}\]
\[/quote]

 you have to prove it in the other way  :oops:


hi, could someone tell me where I can found the proof of these inequalities:

(1)

$(a+b)^{\alpha}\leq 2^{\alpha-1}(a^{\alpha}+b^{\alpha})\quad\text{for }a,b\geq 0 \quad\text{and}\alpha\geq1$

and

(2)

$|x|^{\alpha}\leq1+|x|\quad\text{for } x\in\mathbb{R}\quad\text{and } \alpha\in[0,1].$

please I just need the name of the reference as soon as..!

thanx,,

hello, do you mean
(1)

$ (a\plus{}b)^{\alpha}\leq 2^{\alpha\minus{}1}(a^{\alpha}\plus{}b^{\alpha})\quad\text{for }a,b\geq 0 \quad\text{and
}\alpha\geq1$
and

(2)

$ |x|^{\alpha}\leq1\plus{}|x|\quad\text{for } x\in\mathbb{R}\quad\text{and } \alpha\in[0,1]$.

Sonnhard.

[quote]
(1)

$ (a \plus{} b)^{\alpha}\leq 2^{\alpha \minus{} 1}(a^{\alpha} \plus{} b^{\alpha})\quad\text{for }a,b\geq 0 \quad\text{and }\alpha\geq1$

(2)

$ |x|^{\alpha}\leq1 \plus{} |x|\quad\text{for } x\in\mathbb{R}\quad\text{and } \alpha\in[0,1]$.
[/quote]

(1)just power mean
(2),because $ \alpha\in[0,1]$, so when $ |x|\le 1$, then $ |x|^{\alpha}\leq 1 \leq1 \plus{} |x|$, when $ |x| \ge 1$, then $ |x|^{\alpha}\leq |x| \leq1 \plus{} |x|$.


we just started doing something called organic compounds recently and i forgot to take notes while our teacher was lecturing. so basically, could someone tell me what the prefixes and suffixes on organic compounds are or provide a link? THANK YOU

well, for starters, you have numeric identifiers-
1- meth
2- eth
3- prop
4- but
5- pent
6- hex
7- hept
8- oct
9- non
10- dec
20- icos

Then you also have functional suffixes:

-ane: alkane
-ene: alkene
-yne: alkyne
-one: ketone
-al: aldehyde
-ol: alcohol
-oxy: ether (not common)

Hope that helps, but keep in mind it's just a rough list.

oh yea... i remember seeing those on the board. i have another question though that he might've talked about.
what are double bonds? i know u represent them w/ two lines.

double bonds represent unsaturation in an organic molecule. it is a type of covalent bond formed by the sharing of two pairs of electrons. eg. alkenes(a kind of hydrocarbons) have double bonds accounting for their unsaturation.

 u can refer any good book on organic chemistry for further explaination. organic chemistry by 'robert thornton morrison' and 'robert neilson boyd' is just one such book. u may find better ones in ur class lib.


Of the eggs produced by salmon, 80$ \%$ hatch, and of those, 25$ \%$
survive to migrate to the ocean. How many eggs are needed to
produce 100 salmon that migrate to the ocean?


limit
integration
trigonometry
calculus
calculus computations

Please solve this (if it's possible) without using power series:
\[ \displaystyle\lim_{x\to 0}\displaystyle\int^x_0 \frac{\sin t}{t}\,dt\]

Maybe using Fundamental Theorem of Calculus (first part)???

The Fundamental Theorem of Calculus (or the proof of it) is enough to tell you that
\[ \lim_{x\to0}\frac1x\int_0^x\frac{\sin t}{t}\,dt\equal{}\lim_{t\to0}\frac{\sin t}t\equal{}1.\]
Without dividing by $ x,$ we can say this:

If $ \int_0^a|f(t)|\,dt$ is a convergent integral for some $ a,$ then $ \lim_{x\to0}\int_0^xf(t)\,dt\equal{}0.$


algebra
polynomial
integration
real analysis
real analysis solved

$f:[0,1]\rightarrow \mathbb{R}$ is a polynomial such that:

$\displaystyle \int^1_0f(x) x^m = 0$ for all $m\in \mathbb{Z}^{*}$

Prove that $f$ is identically zero on $[0,1]$.

Moreover, if $f\in L_1(0,1)$ and $\int_0^1f(x)x^mdx=0$ for all $m\in\mathbb{Z}_+$ then $f\equiv 0$.
It is because the set $\{x^m\}$ is dense in $L_1(0,1)$. Though for your problem we don't need such strong fact. ;)

[quote="Myth"]Though for your problem we don't need such strong fact. ;)[/quote]

This problem is actually pretty trivial if done correctly.  Otherwise it gets ugly, but nevertheless, I thought it cute.

I saw a similar but more general problem in a Romanian contet: $f$ was continuous there, and that could be solved by Weierstrass's theorem, the one stating that we can find a sequence of polynomials uniformly converging to $f$. Then we could easily get $\int^1_0 [f(x)]^2 dx=0$. Of course, Myth's assertion generalizes this even further.

Yeah.  Using the Weierstrass approximation theorem is nice, because the case that I posted has that nice solution involving $f^2(x)$, but it doesn't generalize to non-polynomial functions.  However, the Weierstrass approximation lets us reduce everything to that case.


Here is a nice inequality (I think).It was posted by me in the proposed and own problems but nobody was interested.I hope it will be interesting for you.
Let a,b,c are real numbers s.t.$0\leq{a,b,c}\leq{1}$.Denote $S=\sqrt{a(1-b)(1-c)}+\sqrt{b(1-c)(1-a)}+\sqrt{c(1-a)(1-b)}$.Prove that $S\leq{1+\sqrt{abc}}$.When equality holds?

I don't really know what to make of this inequality.

Are you sure we don't have $0\leq a,b,c \leq 1$?

Oh I am veryvery sorry .Of course we have. :blush:

Common guys I am waiting for a nice solution (try with some substitutions) ;) .

Take into consideration that it is Pre-Olympiad Section... ;)

Just stating the obvious from AM-GM and seeing if I can go anywhere.

let a=cos^2(A), b=Cos^2(B), c=cos^2(C)

then S=sin(A)cos(B)Cos(C)+Sin(B)Cos(A)cos(C)+Sin(C)CosACosB

rearranging S=CosA(Sin(B+C))+cosBCosCSinA

need to prove s<=1+sinAsinBsinC

or cosAsin(B+C)+SinA(Cos(B+C)<=1

or sin(A+B+C)<=1

which is true..

Chakding meant $a=\sin^2 A$, $b=\sin^2 B$ and $c=\sin^2 C$.

Very nice!

At last you have solved the problem.And that's what I meant substitutions.
Perfect!!!  :lol:

I think, it is highly difficult to solve the problem without such substitution.
Maybe I am wrong, Armo? Do you know another solutions?

Yes, I think you are right Myth. As I remember where was a solution  with Causchy-Shwartz , but it was very long  and it was difficult to see when equality occurs.


If $n$ is an odd natural number prove that for every natural number $a$ it stands \[gcd(a+1,\frac{a^{n}+1}{a+1})|n.\] Bye!!

That's maybe because $gcd(a+1,\frac{a^{n}+1}{a+1})=gcd(a+1,n)$ :wink:

For all $x,y,m$ we have $\gcd(x,y) = \gcd(x,y+mx)$, in other words it only depends on the residue class $\mod x$. Especially $\gcd(x,y) = \gcd(x,y \mod x)$.

Note that $a \equiv-1 \mod (a+1)$. Now it's an easy task to show that $\frac{a^{n}+1}{a+1}= a^{n-1}-a^{n-2}+-...-a+1 \equiv n \mod (a+1)$.
This gives $\gcd(a+1,\frac{a^{n}+1}{a+1}) = \gcd(a+1,\frac{a^{n}+1}{a+1}\mod (a+1) ) = \gcd(a+1,n)$.
The latter is a divisor of $n$ be definition.


calculus
integration
algebra
function
domain
number theory
Ring Theory

algebraic number theory problem:

Let w=(-1 +(-3)^0.5))/2 and note that w^2=-1 -w.
Consider the integral domain Z +Zw, and let U(Z+Zw) be the set of units.

Prove that U(Z+Zw)=(+/-1, +/-w, +/-w^2).

[quote="LeoYard"]I'm a bit surprised that no one has replied or commented to my question.

I'll try rewording the last part of the question a little better this time.

Let w=(-1 +(-3)^0.5))/2 and note that w^2=-1 -w. Consider the integral domain Z +Zw, and let U(Z+Zw) be the set of units.

Prove that U(Z+Zw)=(+1,-1, +w, -w ,+w^2, -w^2).

I want to put up a bunch of questions taken from a text book I have titled 'Introductory Algebraic Number Theory'

Here's the next question:

Let m be an integer with m = 1 (mod 4) and m less than -3. Prove that the set of units U(Z +Z((1 + sqrt(m)/2) = (+1, -1)[/quote]

I need to add the following:

For a given integral domain A (say), a member 'i' (say) of this set is called a unit if there exists another element 'j' (say) in A such that i*j=1.

We denote the set of all the units of A as U(A)

In the first question, I'm saying that the only units for the integral domain (Z +Zw) are +1, -1, +w and -w.


USAMTS geometry AMC AIME USA(J)MO USAMO ARML

This is the [size=59]small[/size] area of the forum where you can post and brag about your scores on USAMTS, Year 15, Round 1

Since we can't have two polls in one, post to announce your commendations, etc.

P.S. Does anyone know what a prizeable total score is?  Is it only for achieving gold?

GOT A 21 BIOTCHES..........  Pretty good for a lazy ass korean

Hm. That's almost an anagram of "Biotech".

Anyway, I almost forgot --> 24, 1 comm.

25, 0 commended :/

But anyways, I didn't send in anything for round 2 since I don't really have time for this. I'd sorta prefer spending more time to study for the real competitions like AMC, AIME, USAMO, ARML, etc...

did you contact the guy i told you?  did you already have the tryout or whatever it's called?

25 with two commends on the twoI thought were least likely to get commended

25/25 with two commends, #1 and #4...

too bad i can get at most a 20/25 on round two...

gold is 96-100
silver is 76-95
bronze is 60-75
honorable mention is 40-59

[quote]too bad i can get at most a 20/25 on round two...[/quote]

You might get partial credit...I know I got #5 wrong for round 2, but it might be a "good try"...

I got 20/25 for skipping 4, but 1 comm on 1.

I proved the gaps, and the reason why the gaps cannot contain the answer, and then why the answer works.

i screwed up totally on number 3... misread it... thought it was the sum of the 2 numbers accross from each other... so i got a 19/25...

How many points do you lose for getting a question wrong on USAMTS? On number 3 - i think- i forgot to count some of hte cyclic quadrilaterals, and on number 5 i was 1 edge over the correct answer. How many points will I lose?


LaTeX
limit
calculus
derivative
integration
logarithms
calculus computations

Excuse my latex, i'm new to this.

1. how many different methods are there to do (show examples)
$ \lim_{x \to 3}\tfrac{x^3 \minus{} 27}{x \minus{} 3}$

2. prove
$ x^{179} \plus{} \tfrac{163}{1 \plus{} x^2 \plus{} sin^2 x} \equal{} 119$

3. take the derivative of $ f(x) \equal{} \int_{cos x}^{x^{2}} e^{sin t}\, dt$

4.
$ \lim_{x \to \plus{} \infty}x*(ln x)^{2}$
$ \lim_{x \to 0 \plus{} }x*(ln x)^{2}$

2 isn't worded properly. Should that maybe say "solve" instead of "prove"? That equation isn't true for all x. Just try $ x\equal{}0$.

As for the first one from 4, I don't see any reason that wouldn't tend to infinity as $ x$ gets large. Both $ x$ and $ \ln x$ tend to infinity, so their product will too (and the square won't change that).

I suspect that the wording for #2 is probably "prove that this equation has at least one real solution."

[hide="Hint:"]If you're trying to prove that an equation has a solution, your usual tool is the Intermediate Value Theorem.[/hide]


although it is solvable by ordinary rules of inequalities,
prove by INDUCTION that

$ n>\sqrt{n}\plus{}\sqrt[3]{n}\plus{}\sqrt[4]{n}$  $ \forall n\geq 9$


Ever look at a signature, or someone's tshirt, and you see a quote that makes you think "Where's that from?"
This is the place to post your favorite obscure quotes, and I guess if you want to make it a game, it could be like the others: first to guess the previous one correctly posts the next one...

We'll start with an easy one.

"42."

"42" comes from Hitchhiker's Guide to the Galaxy or a 2nd (1st?) grade math book.

Ok how about

"No he's not dead, He's pining!   ...for the fjords!"

Monty Python (googled it... sorta cheap i guess)!

Anyone else have any quotes that we can guess?

"We want cubes, and cubing will get us cubes, so we cube."

rrusczyk.

Next:
"What is it that leads you to credit that you can suppose that you believe that you think that you assume that you know that something is true?"    (same as my signature) (good luck!)

odds are nobodies ever even heard this one before, but:

"Hey,...hey,....hey,... don't be mean, you don't have to be mean, 'cause remember....... where ever you go,....... there you are."

Buckaroo Banzai.

"This proof is short and clever..."  :rotfl:
[hide]just like [hide]probability1.01[/hide][/hide]

I'm gonna have to go with a quote from the wongfuproductions movie yellow fever.

[img]http://www.post-jdm.com/bairen.jpg[/img]

I want that shirt!


[color=blue][b]
Determinati:
$I_n = \int {\frac{1}{{1 + x^{2n} }}dx}$
[/b][/color]

hello, for $ n=1$ we get
$ \int\frac{1}{1+x^2}\,dx=\arctan(x)+C$
and for $ n=2$ we have
$ {\int\frac{1}{1+x^4}\,dx=1/8\,\sqrt {2}\ln  \left( {\frac {{x}^{2}+x\sqrt {2}+1}{{x}^{2}-x
\sqrt {2}+1}} \right) +1/4\,\sqrt {2}\arctan \left( x\sqrt {2}+1
 \right) +1/4\,\sqrt {2}\arctan \left( x\sqrt {2}-1 \right) }$
Sonnhard.


prove
a/(pb+qc) + b(pc+qd) + c/(pd +qa) + d/(pa+qb) >= 4/(p+q)for positive numbers a,b,c,dp,q satifuing p>=q
     Does the inequality hold for p<q

this used to be a wrong solution :D
P.S. now its :spam:  :lol:

How to get that $\geq 4(a+b+c+d)$ ?  :?

I only get $\geq \left(\sum\sqrt{a}\right)^2$  :?

:oops_sign:  i was wrong but i don't know what i was thinking when i wrote it  :?
 I have no idea what i was talking about :oops:


How do i do these questions?
1)How can the word DECISIONS be arranged so that the "N" is somewhere to the RIGHT of "D" (my problem is the two I's in particular)
ANSWER: 45360

2)Bob is about to hang 8 shirts in the wardrobe. He has 4 different styles of shirt, 2 identical ones of each particular style. How many different arrangements are possible if no 2 identical shirts are next to one another?
ANSWER: 864

3) A motorist travels through 8 sets of traffic lights, each of which is red or green. He is forced to stop at 3 sets of lights
a) In how many ways could this happen? ANSWER: 56
b) What other number of red lights would give an identical answer to part a)? ANSWER: 5

4) Find how many 5 digit numbers are possible from the digits 9,8,7,6,5,4,3,2,1,0 if digits are to be in
a) ascending order? ANSWER: 126
b) descending order? ANSWER: 252

Any help on any of these questions really appreciated !!

1.
[hide]
$\frac{9!}{2!*2!*2!}$
because of the 2 i's and the 2's, and then you also want to count all orders of DN but not ND, so same idea as the repeated letters.  i think that's right
[/hide]

[hide="3 Solution"]a) If your total number of lights is 8 and your number of red lighs (stops) is 3, your number of green lights is (8-3)\[\frac{8!}{3!(8-3)!} = 56\]

b)Since the total number of lights stays the same, let r = # of red lights and g = number of green lights.  g = 8-r.
To get the same number of possibilities $3!5! = r!(8-r)!$ This equation is true only if $r = 3, 5$ Becuase you already had r being 3, your only answer is that $r=5$[/hide]

thanks alot for that guyz..just 2 more please if its possible :)

[hide="solution"]we can start off by using 9 at the 10000's digit.
there is one number if 3 is 1000's digit: 93210
there are three numbers if 4 is 1000's digit: 94321, 94310, 93210
...... and the pattern is that as every 1000's digit increases, the numbers increase to triangular numbers...

so, add them all together: 1+3+6+10+15+21 = 56

now, if 8 is at the 10000's digit, we still have the pattern, but we have one less number to use at 1000's digit, since we CANNOT use 8. so.... subtract 21 off the last one, and we get 35

keep going....
1+3+6+10+15+21 = 56 (9's in the 10000's digit)
1+3+6+10+15 = 35 (8's in the 10000's digit)
1+3+6+10 = 20 (7's in the 10000's digit)
1+3+6 = 10 (6's in the 10000's digit)
1+3 = 4 (5's in the 10000's digit)
1 = 1 (4's in the 10000's digit)
total = 126[/hide]

[hide="4."]

Any 5-element subset has 1 arrangement except ones starting with 0, which we need to subtract. This is the number of 4-element subsets from {1,2,3,4,5,6,7,8,9}. So $\binom{10}{5} - \binom{9}{4} = 252-126 = 126$.

Same thing for the second part except you don't need to subtract the ones starting with 0 because there are none. So its $\binom{10}{5} = 252$.[/hide]

[quote="paladin8"]
4.

Any 5-element subset has 1 arrangement except ones starting with 0, which we need to subtract. This is the number of 4-element subsets from {1,2,3,4,5,6,7,8,9}. So $\displaystyle \binom{10}{5} - \binom{9}{4} = 252-126 = 126$.

Same thing for the second part except you don't need to subtract the ones starting with 0 because there are none. So its $\displaystyle \binom{10}{5} = 252$.
[/quote]

i seem to work it all out. why didnt i think of it that way... maybe i need more problem solvin gpractice, especially the problems where i can identify combinatorics and probability connection :(


I have a problem .I think it is very nice.
1\Let A={1\k,1\k+1,.....,1\k+n}.Prove that we canot partition A In  to two subsets B,C s.t sum of element in each set is equal.

I think it's classical.
$v_2$ denote the $2$-adic valuation.
Assume that $\sum_{y \in B}y=\sum_{y \in C}y$ ($*$).
Let $x$ in $\{k,k+1,\cdots,k+n\}$ such that $v_2(x)$ is maximal.
In fact it's easy to show that $\forall y \in \{k,k+1,\cdots,k+n\}-\{x\}$, we have: $v_2(y)<v_2(x)$.
Now multiply both side ($*$) by $\frac{x}{2} \times \prod_{t \in \mathbb{N} , 2t+1 \leq k+n} (2t+1)$ then except for the term corresponding to  $x$ every other term in ($*$)' would be an integer so: contradiction.


MATHCOUNTS likes to write problems about the number that is the current year.  Can you find anything interesting about the number 2008?

I will start with one:

2008 is a happy number (A happy number is defined by the following process. Starting with any positive integer, replace the number by the sum of the squares of its digits, and repeat the process until the number equals 1 (where it will stay), or it loops endlessly in a cycle which does not include 1. Those numbers for which this process ends in 1 are happy numbers, while those that do not end in 1 are unhappy numbers. A computer search up to 1020 suggests that about 12% of numbers are happy, though no proof is known (Guy 2004:§E34). (wikipeda))

$ 2^2 \equal{} 4,  0^2\equal{}0,  0^2\equal{}0,  8^2\equal{}64$
$ 4\plus{}0\plus{}0\plus{}64\equal{}68$
$ 6^2\equal{}36,  8^2\equal{}64$
$ 36\plus{}64\equal{}100$
$ 1^2 \equal{} 1,  0^2\equal{}0,  0^2\equal{}0$
$ 1\plus{}0\plus{}0\equal{}1$
Now, can you come up with more?

The prime factorization of 2008: $ 2^3*251$


Circumcircle $ (N)$ of triangle $ E_aE_bF$ is just the 9-point circle of triangle $ ABC$. Draw the diameters $ E_aM_a$, $ E_bM_b$ of the circle $ (N)$. Continue with [url=http://www.mathlinks.ro/viewtopic.php?t=288219]MaMbF[/url]


The Bernoulli sequence $\{B_{n}\}_{n \ge 0}$ is defined by \[B_{0}=1, \; B_{n}=-\frac{1}{n+1}\sum^{n}_{k=0}{{n+1}\choose k}B_{k}\;\; (n \ge 1)\] Show that for all $n \in \mathbb{N}$, \[(-1)^{n}B_{n}-\sum \frac{1}{p},\] is an integer where the summation is done over all primes $p$ such that $p| 2k-1$.

It's not that hard and definitely posted before:
Express $\sum_{k=1}^m k^n$ as a polynomial in the variable $m$ in terms of the Bernoulli numbers, then apply that  $\sum_{k=1}^{p-1} k^n $ is $-1$ modulo $p$ if $p-1 | n$ and $0$ otherwise. Conclude.

On a note: that $(-1)^n$ in your term is pretty useless and the sign should be the other way around.

Found it: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=57&t=91510 .


Evaluate the limit of :

$ \sum_{k\equal{}1}^\infty \frac{k^2}{c^k}$

if $ |c| \geq 1$

Reffer to my solution [url=http://www.mathlinks.ro/viewtopic.php?p=1575817#1575817]here.[/url], although $ \sum_{n \equal{} 1}^{\infty} \frac {k^5}{c^k}\ (|c| > 1).$


vector
function
calculus
derivative
logarithms
trigonometry
calculus computations

Given the vector field $ F(x,y,z)\equal{}(\minus{}y,x,\minus{}z)$, find the equation of the field line through an arbitrary point $ (x_0,y_0,z_0)$. In particular, give a vector-valued function $ g(t)$ that satisfies $ g(0)\equal{}(x_0,y_0,z_0)$.

I don't know how to do this for field lines in 3-Space :(
Thanks for any help.

let $ \vec{g}(t)\equal{}(x(t),y(t),z(t))$. In order for this to be a field line, we require that:

$ \frac{d\vec{g}}{dt}\equal{}F(\vec{g}(t))$

Which basically means that the tangent of a field line at some point is the same as the vector from the vector field at that point. This gives us a system of three differential equations:

$ \dot{x}(t)\equal{}\minus{}y(t)$
$ \dot{y}(t)\equal{}x(t)$
$ \dot{z}(t)\equal{}\minus{}z(t)$

Which reduces the problem to differential equations. If you need more help from here, just ask.

Yea i got those three. Do I just solve the DEs and put them back into their corresponding spots?
Like, would I get $ g(t) \equal{} (\minus{}yt,xt,Ce^{\minus{}x})$ where the $ C$ can easily be determined.

And drawing field lines, I can easily do it for 2D stuff. Use the chain rule, get $ \frac{dy}{dx}\equal{}x(t)y(t)$ and solve the separable DE. But now how with the extra z-component? Since z depends on itself can I still use the 2D method? It would just be above or below the x-y plane -_-a.

It isn't quite that simple, as all three of the variables themselves depend on a fourth variable, $ t$. The derivatives are taken with respect to $ t$, so the solution process is a little bit more complicated. Take the last equation, $ \dot{z}(t) \equal{} \minus{} z(t)$. This would be solved like this:

$ \frac {dz}{dt} \equal{} \minus{} z\rightarrow \frac {dz}{z} \equal{} \minus{} dt\rightarrow \ln (z) \equal{} \minus{} t \plus{} A\rightarrow z(t) \equal{} Be^{ \minus{} t}\rightarrow z(0) \equal{} B \equal{} z_0 \Rightarrow z(t) \equal{} z_0e ^{ \minus{} t}$

Substituting the first and second equation into the second and first equation respectively, we find:

$ \ddot{y}(t) \equal{} \minus{} y(t)\Rightarrow y(t) \equal{} A\sin (t) \plus{} B\cos (t)$
$ \ddot{x}(t) \equal{} x(t)\Rightarrow z(t) \equal{} C\sin (t) \plus{} D\cos (t)$

To apply the initial values, we simply note that $ x(0) \equal{} x_0$, $ y(0) \equal{} y_0$, $ \dot{x}(0) \equal{} \minus{} y(0) \equal{} \minus{} y_0$, and $ \dot{y}(0) \equal{} x(0) \equal{} x_0$. Here are the results:
[hide="Answer"]
$ x(t) \equal{} x_0\cos (t) \minus{} y_0\sin (t)$
$ y(t) \equal{} y_0\cos (t) \plus{} x_0\sin(t)$

$ \vec{g}(t)\equal{}(x_0\cos (t) \minus{} y_0\sin (t), y_0\cos (t) \plus{} x_0\sin(t), z_0e ^{ \minus{} t})$
[/hide]


trigonometry
inequalities
inequalities unsolved

For $ a,b,c>0$ such that $ a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}abc\equal{}4$ prove that $ a\plus{}b\plus{}c<\equal{}3$.

Let $ x\equal{}\frac{a}{2}$ , $ y\equal{}\frac{b}{2}$ and $ z\equal{}\frac{c}{2}$ so $ x^2\plus{}y^2\plus{}z^2\plus{}2xyz\equal{}1$ therefore $ x,y,z$ can be expressed as
$ x\equal{}\cos(A)$ ,$ y\equal{}\cos(B)$ , $ z\equal{}\cos(C)$ where $ A,B,C$ are angles of an acute triangle. We have to show that
$ \sum \cos A \leq \frac{3}{2}$
Because $ cos$ is concave on $ (0,\frac{\pi}{2})$ from Jensen's inequality we have that
$ \frac{1}{3} \sum \cos(A)\leq \cos \left( \frac{1}{3} \sum A\right)\equal{}\cos\left( \frac{\pi}{3}\right)\equal{}\frac{1}{2}$ ,and the conclusion follows.

It is not that hard . Put $ a \equal{} 2x$ etc and the relation gives $ x^2 \plus{} y^2 \plus{} z^2 \plus{} 2xyz \equal{} 1$ and we have to prove that

$ x \plus{} y \plus{} z\leq\frac {3}{2}$ but there are are numbers  A,B,C angles of a triangle such that $ x \equal{} \sin \frac {A}{2}$ etc

so we have to prove that $ \sum\sin \frac {A}{2}\leq\frac {3}{2}$  which is true due to Jensen's inequality

EDIT : Svejk you beat me  :oops:  :P


superior algebra
superior algebra solved

Let $G$ be the group of invertible real upper triangular $2\times 2$ matrices. $H=\{kE:\ k\in\mathbb{R}\}$
Determine what $G/H$ is isomorphic to.

Let $(G,\cdot)$ be the group of invertible upper triangular matrices.

Let $S=(\mathbb{R},+)\ltimes (\mathbb{R^*},\cdot)$ be the semidirect product of $(\mathbb{R},+)$ and $(\mathbb{R^*},\cdot)$.
The group law on $G$ is $(a,b)(c,d)=(a+bc,bd)$, the neutral element is $(0,1)$ and $(a,b)^{-1}=(-\frac ab,\frac 1b)$.

Let $f: G\to S$, defined by $f(\left(\begin{array}{cc}a&b\\0&c\end{array}\right))=(\frac bc,\frac ac)$.
$f(I_2)=(0,1)$
$f(\left(\begin{array}{cc}a&b\\0&c\end{array}\right)\left(\begin{array}{cc}a'&b'\\0&c'\end{array}\right))= f(\left(\begin{array}{cc}aa'&ab'+bc'\\0&cc'\end{array}\right))=(\frac{ab'+bc'}{cc'},\frac{aa'}{cc'})=(\frac bc+\frac ac\cdot\frac{b'}{c'},\frac ac\cdot\frac{a'}{c'})=(\frac bc,\frac ac)(\frac{b'}{c'},\frac{a'}{c'})=f(\left(\begin{array}{cc}a&b\\0&c\end{array}\right))f(\left(\begin{array}{cc}a'&b'\\0&c'\end{array}\right))$.
So $f$ is a morphism.
$f(\left(\begin{array}{cc}b&a\\0&1\end{array}\right))=(a,b)$, so $f$ is onto.

$f(\left(\begin{array}{cc}a&b\\0&c\end{array}\right))=(0,1)\Leftrightarrow \frac bc=0$ and $\frac ac=1\Leftrightarrow (b=0,a=c)\Leftrightarrow \left(\begin{array}{cc}a&b\\0&c\end{array}\right)\in H$.
By the Fundamental Theorem of Isomorphism, $G/H\simeq S$.

I used intensively the fact that if $\left(\begin{array}{cc}a&b\\0&c\end{array}\right)\in G$ then $a,c$ are not zero.


linear algebra
matrix
algebra
polynomial
complex numbers

Um, do you know the Cayley-Hamilton theorem?

(since I'm too lazy to tex)

A=
[0.5 0.75]
[-1    0.5]

A^2=
[-0.5 0.75]
[-1    -0.5]

A-I_2=A^2

Thanks. ! Thanks. !  Thanks. !  Thanks. ! $blahblahblah$  :D  :D  :D
I don't know such theorem before. !

[quote="svrsk"]Ah. $blahblahblah$ for n = 3, Do you think is it possible to find mattrix A such that $A^2 = A - I$ ?  :blush:[/quote]

Yes, in fact it is possible  even for n=1,  (If you allow complex  numbers) .

No. Complex number are not allow. $gyan$, can you show me an example.?  ;)

Can I  ask why complex numbers are not allowed?  Most of the matrices I deal  with have complex numbers as their elements. Unless explicitly told, one can assume that complex numbers are fair game.

...But, in any case, what i was trying to do is to give a big HINT,  (Think in terms of eigen values)

the solution (in numbers) for $x$ when  $x^2=x-1$ is
$x = \frac {1(+-)\sqrt(3)}{2}$  or $\omega + 1$ and $\omega^2+1$. So any matrix whose eigen values are these will do. (Check the eigen values of the matrix given by blahblahblab  -  Hope this helps.

My knowledge about matrix is very simple. So I want to deal with real number.

$gyan$ I think that I understand what you say.(If I" wrong please tell me.)
for n = 2. It 's so easy by Cayley-Hamilton theorem

but for n = 3, Consider this matrix
[ -1 , 1, 0
-3, 0, 1
-6, 0, 2]

this matrix has eigen values = $-x^3 + x^2 - x$
or $-x(x^2 -x + 1)$
but it 's not has property : $A^2 = A - I$

Do you really find it.? If you can done it. Please tell me.

[quote="svrsk"]Is it possible to find matrix A so that $A^2 = A - \bf{I}$.?
A be any n x n matrix.
I try to solve for n = 1, 2 and found that it 's not possible. But I can't deduce for $n \geq 3$

Please help me to solve this problem.  :D[/quote]

HINT:  Suppose $ n=3 $. From $ A(I-A)=I $  it may be seen that  $ A $  is non-singular having the inverse $ A^{-1}=I-A $.  Therefore  (*)    [b]det[/b] $(A)\ne 0. $
 On the other hand $A^3-A^2+A = 0 $ which means that  $ p_A(\lambda):= \det(A-\lambda I) = -(\lambda^3-\lambda^2 +\lambda)= 0 . $ Now,  for $\lambda=0 $ we find
 $ p_A(0)=$[b]det [/b]$ (A)=0 $  which contradicts (*).    Consider further  $ n\ge 4 $.

How to construct an $n\times n$ matrix $M$ which satisfies the $n$-th degree polynomial $M^n+a_{n-1}M^{n-1}+a_{n-2}M^{n-2}+c\dots+a_1M+a_0I=0$

Put the negatives of the coefficients of the polynomial in the right column, a line of ones one step below the main diagonal, and zeros elsewhere:
$M=\begin{pmatrix}0&0&0&\cdots&0&-a_0\\1&0&0&\cdots&0&-a_1\\0&1&0&\cdots &0&-a_2\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\0&0&0&\cdots&0&-a_{n-2}\\ 0&0&0&\cdots&1&-a_{n-1}\end{pmatrix}$
This works in any field. In this case, the polynomial in question is $x^2-x+1$, which gives us $\begin{pmatrix}0&-1\\1&1\end{pmatrix}.$

For other sizes of matrix, it depends on how the polynomial factors. If the polynomial is irreducible ($x^2-x+1$ is irreducible over $\mathbb{R}$) the degree of the polynomial must divide the dimension of the matrix.

The construction above has its minimal polynomial equal to its characteristic polynomial, and is associated with rational canonical form:
Theorem: Any matrix over [b]any[/b] field is similar to one of the form
$\begin{pmatrix}B(p_0)&0&\cdots&0\\0&B(p_1)&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\0&0&\cdots&B(p_k)\end{pmatrix}$ where $B(p)$ represents the block as above associated with the polynomial $p$, $p_0$ is the minimal polynomial, and $p_i|p_j$ if $i>j.$

Thanks everyone to help me to solve this problem. !!!  :D