λοιπον με αφορμη αυτα που ειπωθηκαν στα χρονια πολλα περι οργανωσης ειναι σωστα.αν θελουμε μικροτερα sub-foroums και γενικα αδαδιαρθρωση του ελληνικου πρεπει να συμφωνησουμε.ή κανω λαθος ? ας ξεκινησω εγω¨.συμφωνω με τον Σιλουανο και θεωρω μιας και ειναι τοσο ενεργο (ισως και απο το ρουμανικο)εθνικο φορουμ να μπει ενας moderator.ακουω λοιπον προτασεις κι ιδεες  :roll:

Ναι τα έχουμε ξαναπεί αλλά ...  :ninja:

Το πιο σημαντικό για τώρα πιστεύω ότι είναι ένας moderator και IMHO
 moderator πρέπει να γίνει ο Silouan γιατί είναι το πιο δραστήριο μέλος του Forum.

Τα subforums θα ήταν ωραία αλλά δεν ξέρω αν εξαρτάται από εμάς ...  :agent:

Το forum χρειάζεται οπωσδήποτε  :cleaning: και πρέπει να γίνει σύντομα...

[quote="Ramanujan"]
Το forum χρειάζεται οπωσδήποτε  :cleaning: και πρέπει να γίνει σύντομα...[/quote]

Συμφωνώ.....Να πω όμως το εξής που μάλλον δεν το έχουμε καταλάβει .Για να  γίνει moderator κάποιος ή για να δημιουργηθούν sub-forums πρέπει ΟΛΟΙ να στείλουμε PM στον Vornicu ή στον McCroford (μάλλον στον δεύτερο γιατί ο πρώτος έχει αρκετή δουλειά αυτό τον καιρό...).
Δηλαδή δεν φτάνει να το ζητήσουμε τρία τέσσερα ατομα.....
Τέλος πιστεύω οτι ο κ.Στεργίου ή ο nickolas θα μπορούσα να το πετύχουν πιο εύκολα αυτό επειδή γνωρίζουν τον Vornicu .
Τους παρακαλώ να το κάνουν.......

[quote="silouan"]
Συμφωνώ.....Να πω όμως το εξής που μάλλον δεν το έχουμε καταλάβει .Για να  γίνει moderator κάποιος ή για να δημιουργηθούν sub-forums πρέπει ΟΛΟΙ να στείλουμε PM στον Vornicu ή στον McCroford (μάλλον στον δεύτερο γιατί ο πρώτος έχει αρκετή δουλειά αυτό τον καιρό...).
Δηλαδή δεν φτάνει να το ζητήσουμε τρία τέσσερα ατομα.....
Τέλος πιστεύω οτι ο κ.Στεργίου ή ο nickolas θα μπορούσα να το πετύχουν πιο εύκολα αυτό επειδή γνωρίζουν τον Vornicu .
Τους παρακαλώ να το κάνουν.......[/quote]

Μήπως θα ήταν καλύτερο να αποφασίσουμε όλοι έδώ και κάποιος από τους 2 που αναφέρεις να αναλάβει το πει ?

Γιατί είναι λίγο πιο όμορφο και πολιτισμένο από το να στείλουμε καμιά 20 αριά  pm's ....  :wink:

Εκτός και αν πρέπει να τον κουράσουμε με συνεχή μυνήματα για να γίνει δεκτό το αίτημα... :huh:

Δεν ξέρω βέβαια πόσοι συμφωνούν με δημιουργία των subforums και την ύπαρξη moderator...

Φαίνεται πως σε κάποιους αρέσει το σχετικό χάος στο ελληνικό forum  :alien:

Πάντως πιστεύω ότι εύκολα βάζουν κάποιον moderator...  :)

Edit : τουλάχιστον ας αποφασίσουμε εδώ και ας κανονίσουμε να στείλουμε τα  pm's....

[quote="Ramanujan"]
Το forum χρειάζεται οπωσδήποτε :cleaning: και πρέπει να γίνει σύντομα...

[quote="Ramanujan"]
Φαίνεται πως σε κάποιους αρέσει το σχετικό χάος στο ελληνικό forum  :alien:
[/quote]
[/quote]

 :oops:  :oops:  :blush:

Σοβαρά τώρα,συμφωνώ κι εγώ στην ύπαρξη ενός moderator,ο οποίος θα οργανώνει το forum και θα επιβάλει την τάξη εκεί που πρέπει. Συμφωνώ επίσης ότι θα ήταν καλύτερο απ'το να στέλνουμε όλοι pm's σε κάποιον,να αναλάβει κάποιος που ξέρει, να το ζητήσει από τους αρμόδιους, εκπροσωπόντας όλους μας.
Αν γίνει αυτό, στη συνέχεια θα είναι πολύ πιο εύκολο να δημιουργηθούν τα κατάλληλα subforums ώστε να επιτύχουμε την πλήρη τάξη.

Edit:Sorry πάντως αν κάποιοι ενοχλήθηκαν από αυτό που κάναμε. :|

λοιπον εχω μιλησει με αρκετα μελη και πιστευω οτι θα συμφωνησουν ολοι.σιγουρα καλυτερα   να μιλησει ενας απο τους Νικολας ή Στεργιου.οπως θα ειδατε υπαρχουν στιγμες που ο καθενας κανει οτι θελει.και πρωτα απο ολα να ψηφισουμε τον moderator δημοκρατικα, οσα μελη ειναι εγεγραμμενα στη σχετικη λιστα.προτεινω να γινει ανοιχτη ψηφοφορια αμμεσα για να τελειωνουμε.(ο Vornicu ειναι εξαφανισμενος αρκετο καιρο καλυτερα σε αλλον αν γινεται)

Το καλύτερο αν θα γίνει ψηφοφορία είναι να κάνουμε vote και όχι ανοιχτή.....
Ζητώ από τον κ. Στεργίου να ζητήσει από τον McCroford την δημιουργία subforum και την εκλογή κάποιου ως moderator .

Χαίρομαι που συμφωνούμε και φαίνεται να προχωράει η διαδικασία επιτέλους...  :lol:

μπραβο ρε παιδια . να σκεφτουμε κατι ο moderator θα χαιρει της κaθολικης εκτιμησης ολων .εχει τη δυνατοτητα να διαγραφει μηνυματα(post) που θεωρει προσβλητικα?

Σίγουρα ένας moderator έχει αυτήν την δυνατότητα, αλλά θα πρέπει να τη χρησιμοποιεί με μέτρο.

Λοιπόν , αγαπητοί φίλοι,

  τώρα που μαζευτήκαμε από τις διακοπές , θα συνενοηθώ με το Νικόλα για να δούμε τι μπορεί να γίνει.Επιχειρησαμε και στο παρελθόν να πείσουμε το Vornicu  για τη δημιουργία υποφόρουμ , αλλά δεν το δέχτηκε. Δεν ξέρω γιατί , αλλά να το ξέρετε και αυτό : Με αυτούς τους ανθρώπους είναι λίγο δύσκολο να συννενοηθείς , διότι έχουν τελείως διαφορετική νοοτροπία από εμάς.Τέλος πάντων. Θα ξαναδοκιμάσουμε. Αν δεν τα καταφέρουμε , θα ανοίξουμε μόνοι μας topic με ειδικό χαρακτήρα το καθένα και θα στέλνουμε τα προβλήματα στη θέση τους, ώστε ανοιγοντας ένα πεδίο να βλέπουμε μαζεμένα πολλά θέματα.
 Αυτά για τώρα!

 Καλές εμπνεύσεις.

[u]Μπάμπης[/u]

  [color=red]Α!!! Για moderator  τα πράγματα είναι απλά. Ο καθένας μας θα πρέπει να είναι πάνω από όλα ευγενικός με όλους, είτε συμφωνεί είτε διαφωνεί με τις απόψεις τους . Τα άλλα είναι πολύ εύκολα.
Αν κάποιος δε ''συμμορφώνεται''  , απλά οι άλλοι δεν απαντούν στα μηνύματά του. Αν κάποιος μας περιφρονεί , θα τον τιμήσουμε με το ίδιο νόμισμα.Σε ακραίες περιπτώσεις(υβριστικές) το μήνυμα θα διαγράφεται.[/color]

Λοιπον,θελω να πω κατι....
Ειναι αναγκη να υπαρχει το φοβητρο καποιου moderator,για να λειτουργει ευρυθμα το forum??
Eιναι αναγκη να υπαρχουν οι νομοι ,αν υπαρχη κατανοηση μεταξυ των ανθρωπων???Προσωπικα,πιστευω οχι....
Και ο λογος που τελικα υπαρχουν οι νομοι ειναι γιατι θελουμε ΟΛΟΥΣ να τους προσαρμοσουμε σε ενα κοινωνικο συνολο,παρα την θεληση τους.
Στη συγκεκριμενη περιπτωση,δεν προσπαθουμε να τους κανουμε ολους μελη του foroum.Οποιος ειναι εδω ,ειναι μετα απο προσωπικη του επιλογη,και ειναι αυτονοητο οτι αποδεχεται τους κανονισμους που το διεπουν.
Παρολα αυτα,επειδη κατατην γνωμη μου πιστευω οτι δεν θα πρεπει να λειπει και μια διασκεδαστικη χροια στο foroum :rotfl: ,γιατι διαφορετικα θα καταντησει ενα <μονοτονο> site,με μαθηματικακαι τιποτα αλλο,(παρολο που αυτος ειναι ο σκοπος για τον οποιο δημιουργηθηκε...),και κατα συνεπεια δεν θα μπορεσει να προσελκυσει νεα μελη,ΠΡΟΤΕΙΝΩ σ΄ολα τα μελη,(ανεξαρτητα απο το αν αναγραφονται στην λιστα ,ειτε οχι,ειτε ειναι τακτικα μελη,ειτε οχι,)τη δημιουργια ενος (και μονο ενα!)topic που θα μπορει να ειπωθει οτιδηποτε(ακομα και αν δεν εχει αμεση σχεση με τα μαθηματικα)αλλα ολα τα υπολοιπα θα περιεχουν μονο σηζητησεις με αμεσο και καθαρα μαθηματικο ενδιαφερον.Ετσι και ο σκοπος για τον οποιο δημιουργηθηκε,θα επιτελειται,και ταυτοχρονα δεν θα λειπουν τα αστεια και οι προσωπικες συχητησεις μεταξυ δυο η περισσοτερων ατομων.Πιστευω ειναι ανουσιο και δεν θα οδηγησει πουθενα το να παιζουμε με τις λεξεις,αν τα δυο ατομα ειναι ομαδα ατομων η οχι.Και ενας μονος του να ειναι,θα μπορει να γραφει οτι θελει σε εκεινο το topic.Ετσι οταν καποιος δεν θα θελει να ασχοληθει με τιποτα ανουσιο,πολυ απλα δεν θα επισκεπτεται εκεινο το topic.....Kai καθενας που θα το επισκεπτεται,θα το κανει υστερα απο δικη του επιλογη και οχι γιατι καποιος ηθελε να γραψει κατι ασχετο η να συνομιλησει με καποιον αλλον,στη μεση μιας μαθηματικης ακησης.
Δεν θελω να βαλω ψηφοφορια,για το αν συμφωνειτε με αυτο που προτεινω,εχουμε ηδη αρκετες  :lol: ,απλα σκεφτειτε το.Το να στελνουμε ολοι pmσε καποιον για να ορισει εναν αλλο που να μας επιτηρει, με παραπεμπει σε παιδια του δημοτικου που μαλωνουν για το αν ηταν γκολ η οχι και πανε στον διευθηντη του σχολειου για να λυσουν το ζητημα :D
Απλα σκεφτειτε το.....

ΡΑΛΛΗΣ :D

[quote]Ειναι αναγκη να υπαρχει το φοβητρο καποιου moderator,για να λειτουργει ευρυθμα το forum?? [/quote]

έτσι φαίνεται.... :( Γιατί νομίζεις ότι γίνεται αυτό που γίνεται στο ελληνικό forum..? Θα τολμούσε κάποιος το ίδιο στο κεντρικό forum ...?

[quote]Eιναι αναγκη να υπαρχουν οι νομοι ,αν υπαρχη κατανοηση μεταξυ των ανθρωπων???Προσωπικα,πιστευω οχι.... [/quote]

  :huh:  πολύ ρομαντικό σε βρίσκω...κοίτα όμως την πραγματικότητα... όπου υπάρχουν 2 η περισσότεροι χρειάζονται νόμοι....
ακόμα και σε μια παρέα φίλων ή σε μιά οικογένεια τηρούνται κάποιοι κανόνες για να συνυπάρξουν...
για να επιβιώσεις εσύ κι εγώ υπαρχουν οι νόμοι...

[quote]Και ο λογος που τελικα υπαρχουν οι νομοι ειναι γιατι θελουμε ΟΛΟΥΣ να τους προσαρμοσουμε σε ενα κοινωνικο συνολο,παρα την θεληση τους.[/quote]

Κανείς δεν αναγκάζει κανέναν...υπάρχουν νόμοι στο forum (δυστυχώς όχι στο ελληνικό )
γιατί έτσι ήθελε ο δημιουργός του. Όποιος διαφωνεί μπορεί να την κάνει... :P

[quote]Οποιος ειναι εδω ,ειναι μετα απο προσωπικη του επιλογη,και ειναι αυτονοητο οτι αποδεχεται τους κανονισμους που το διεπουν. [/quote]

Δεν βλέπω τους κανονισμούς που αναφέρεις... :roll:

[quote]δεν θα πρεπει να λειπει και μια διασκεδαστικη χροια στο foroum  :rotfl:  [/quote]

ακρίβώς αυτό έλεγα χθές σε ένα από τα μέλη... :)

[quote]διαφορετικα θα καταντησει ενα <μονοτονο> site,με μαθηματικα και τιποτα αλλο[/quote]

δεν το πιστεύεις αυτό όπως και κανένας εδώ μέσα...  :wink:

[quote]τη δημιουργια ενος (και μονο ενα!)topic που θα μπορει να ειπωθει οτιδηποτε[/quote]

πολύ καλή η ιδέα σου  :thumbup:

[quote]Το να στελνουμε ολοι pm σε καποιον για να ορισει εναν αλλο που να μας επιτηρει, με παραπεμπει σε παιδια του δημοτικου που μαλωνουν για το αν ηταν γκολ η οχι και πανε στον διευθηντη του σχολειου για να λυσουν το ζητημα  :D  [/quote]

ποιός βιάζεται να μεγαλώσει ???  :P

Δεν απευθύνομαι σε κάποιον συγκεκριμένο και δεν ενοχλούμαι από μικροπράγματα
αλλά είναι πραγματικά κρίμα να έχει το όμορφο αυτό forum τέτοια εικόνα
όταν υπάρχουν σ'αυτό άτομα που έχουν να δώσουν πολύ καλύτερα πράγματα ...  :!:

H καλύτερη λύση imho είναι να υπάρχει δημοκρατία στο forum δηλαδή
να γίνει αυτό που επιθυμούν οι περισσότεροι....

Χαίρομαι που συζητάμε...αν και δεν θα έπρεπε να
συζυτάμε για τέτοια θέματα...

Φιλικά  :)

Βασικά συμφωνώ με τον Ramanajuan .
Ta topic με αστείο χαρακτήρα αυτή τη στιγμή είναι πολύ περισσότερα από ένα και κάθε μέρα αυξάνονται .....Επίσης γίνονται <<αστείες παρεμβολές >> σε καθαρά μαθηματικές ασκήσεις .....Αυτό πρέπει να μετριάσουμε λίγο πιστευω .....
Και να φέρω ένα παράδειγμα δυσλειτουργίας....
Πες ότι εγώ έχω αχοληθεί με ένα πρόβλημα έχω κάποια συνέχεια και μετά από ένα μήνα απουσίας θέλω να δω τι παράπανω  συζητήθηκε .Δεν είναι λειτουργικό να ψάχνω σε 10 σελίδες τόπικ από τις οποίες οι 8 είναι αστειάκια.......
Συμφωνώ μαζί σου Ράλλη στο ότι πρέπει να γίνει ένας και μοναδικός χώρος για αυτά........

Σιλουανέ ,

 το συζήτησα με το Σκοιμπρή. Θα επικοινωνήσουμε με το Vornicu να του ζητήσουμε υποφόρουμ. Ίσως κάτι αλλάξει.

 Έδωσα και τις ανισότητες για φωτοτυπία. Σου στέλνω και πολλές λυμένες ασκήσεις στους μετ/σμούς.

 Θα επικοινωνήσουμε.

 Μπάμπης

Λοιπόν αγαπητοί φίλου μας να θέσω κι εγώ την δική μου άποψη στην διάθεση σας να την σχολιάσετε κι να την επεξεργαστείτε Λοιπόν όταν αποφασίζουμε να οργανωθούμε είναι μια κουβέντα που σηματοδοτεί την επεξεργασία 1000000 κι πλέον θεμάτων αλλά για να μην μακρηγορήσω συμφωνώ από όλες τις απόψεις μόνο σε αυτή της προσπάθειας να ξεχωρίσουμε τα πόστα από καθαρά μαθηματικά κι από εύθυμα αστεία ή άλλου τύπου πόστ οκ? Από κει κι πέρα είναι ανεπίτρεπτο να ζητάμε την μείωση των πόστ πού είναι εύθυμα ή άλλου τύπου καθότι θα είναι άδικο ακόμα κι έναν moderator να κάνει τέτοια δουλειά δηλαδή να διαγράφει μηνύματα ή να βάζει όρια σε αυτά τα πράγματα  Πρόταση μου για αυτό είναι να δημιουργηθεί ένα μεγάλο τόπικ πού θα το λέμε miscellaneous κι εκεί (συμφωνώ δηλαδή με τον Ράλλη) όπου εκεί θα βάζουμε όλα τα πόστ που αφορούν  άλλου τύπου μηνύματα εξωμαθηματικού περιεχομένου κι έτσι θα μπορεί το φόρουμ να γίνει πιο ελκυστικό Τέλος όσον αφορά το καθαρά μαθηματικό μέρος προτείνω την κατηγοριοποίηση του φόρουμ ανάλογα με τους τομείς των μαθηματικών (απλή λογική)
π.χ. (παραδείγματος μοσχάρι)
1.	Άλγεβρα
2.	Γεωμετρία  (αναλυτική Ευκλείδειος κ.α)
Κ.ο.κ.
Νομίζω ότι με αυτά τα δύο βασικά πράγματα θα συμμαζευτεί η κατάσταση κι μόλις δούμε ότι έχει συμμαζευτεί αρκετά θα προχωρήσουμε κι σε άλλες διορθωτικές κινήσεις είναι καλύτερα να κάνουμε κάποιες βασικές κινήσεις κι μετά να δούμε κι να ξαναδράσουμε αυτά για την ώρα
Χαιρετώ
Υ.Γ. Επειδή μπορεί η θέση μου να μην είναι ξεκάθαρη περί της ύπαρξης ή όχι moderator ξεκαθαρίζω συμφωνώ εν μέρει με τον  Ράλλη περί της προσπάθειας να είμαστε κόσμιοι κι λειτουργικοί για το φόρουμ χωρίς «Μπαμπούλες» 16 χρονών άντρες είμαστε κι μάλιστα καλοί κι συνειδητοποιημένοι πολίτες από όσο μας ξέρω κι νομίζω ότι μπορούμε να φέρουμε εις πέρας την αποστολή αυτή δεν νομίζετε? Αλλά θεωρώ ότι σε μερικές περιπτώσεις μπορεί να χρειαστεί η παρέμβαση κάποιου που όμως ο ρόλος του θα είναι κατευναστικός ως προς τα πνεύματα αλλά κι να συμβάλλει στην ομαλή λειτουργία αλλά όχι με πειθαρχικές μεθόδους (π.χ. διαγραφή μηνυμάτων) Φυσικά κι όταν θα υπάρχουν εριστικοί τότε θα πρέπει να υπάρχει κάποιος να τον σταματήσει φυσικά με σεβασμό πάντα στους υπολοίπους
Σε τέτοια περίπτωση θα συμφωνούσα κι εγώ στην επιλογή moderator όχι ότι τώρα διαφωνώ αλλά η θέση μου είναι ακριβώς « συμφωνώ με επιφυλάξεις» Τέλος οι
καλύτεροι κατά την ταπεινή γνώμη μου είναι με σειρά προτίμησης 1 Σιλουανός  2 ραμανατζάν 3 κώστας 4 Αρχιμήδης 6

Στην περίπτωση που δε γίνουν sub-forums μπορούμε να θεσπίσουμε ορισμένες συντομογραφίες που θα μπαίνουν πριν από το τίτλο του εκάστοτε topic. Τι σημαίνει IMHO? Έχει σχέση με το imo=in my opinion?


Find the minimum of $A=\frac{x}{\sqrt{y^{2}+z^{2}}}+\frac{y}{\sqrt{z^{2}+x^{2}}}+\frac{z}{\sqrt{x^{2}+y^{2}}}$ where $x,y,z$ are positive numbers

Sorry for my mistakes.
Find the minimum of A where $x,y,z\geq 0$
we have:
$\sqrt{\frac{y^{2}+z^{2}}{x^{2}}}\leq \frac{{1+\frac{y^{2}+z^{2}}{x^{2}}}}{2}=\frac{x^{2}+y^{2}+z^{2}}{2x^{2}}$
Hence $\frac{x}{\sqrt{y^{2}+z^{2}}}=\sqrt{\frac{x^{2}}{y^{2}+z^{2}}}\geq\frac{2x^{2}}{x^{2}+y^{2}+z^{2}}$
similarly for other remain elements.
We conclude the minimum of A is 2 where $a=0,b=c$


calculus
integration
logarithms
algebra
function
domain
complex numbers

Calculate the following indefinite integrals.

[1] $\int \frac{6x+1}{\sqrt{3x^2+x+4}}dx$

[2] $\int \frac{e^x}{e^x+e^{a-x}}dx$

[3] $\int \frac{(\sqrt{x}+1)^3}{\sqrt{x}}dx$

[4] $\int x\ln (x^2-1)dx$

[5] $\int \frac{2(x+2)}{x^2+4x+1}dx$

[quote="kunny"]You are wrong, Jimmy. :([/quote]

I know I read the problem wrong, I thought it said $e^{a+x}$, not $e^{a-x}$, I deleted the post after I noticed the mistake.

[quote="amirhtlusa"]2
$\frac{\ln({e^{2x}}+e^a)}{2}$
you simplify it  :lol:[/quote]

Yes.That's O.K. , amirhtlusa. :)

[quote="amirhtlusa"]3
$\frac{(\sqrt{x}+1)^4}{2}$[/quote]

 :)

Maybe we should post a solution instead of just the answer, I'll start with #2:

$\frac{e^x}{e^x+e^{a-x}} = \frac{e^{2x}}{e^{2x}+e^a}$

$u=e^{2x}+e^a$

$du = 2e^{2x} \ dx$

$\int \frac{e^x}{e^x+e^{a-x}} \ dx = \frac{1}{2} \int \frac{du}{u} = \frac{\ln(e^{2x} + e^a)}{2}$.

Thank you for your detail, Jimmy. :)

kunny

And for the sake of completness, here's #3

$\int \frac{(\sqrt{x}+1)^3}{\sqrt{x}} \ dx$

$u=\sqrt{x}+1$

$du = \frac{1}{2\sqrt{x}} \ dx$

$\int \frac{(\sqrt{x}+1)^3}{\sqrt{x}} \ dx = 2\int u^3 \ du = \frac{(\sqrt{x}+1)^4}{2}$

[quote="Jimmy"][quote="kunny"][quote="Jimmy"]5)

$\frac{2(x+2)}{x^2+4x+1} = \frac{2x+4}{x^2+4x+1}$

$u=x^2+4x+1$

$du=(2x+4) \ dx$

$\int \frac{2(x+2)}{x^2+4x+1} \ dx = \int \frac{du}{u} = \ln(x^2+4x+1)$[/quote]

What's the sign of $x^2+4x+1$?[/quote]

Yes, you are right, it should be $\ln|x^2+4x+1|$  (but you never said the domain did not include complex numbers :) )[/quote]

Yes,you are right, but as you see this Today's calculation of Integral contains elementary calculation, so the domian of $x$ is limited to real numbers.  :lol:


function
real analysis
real analysis unsolved

Show that there is a function $f:[0,1]\rightarrow[0,1]$ such that $A=\{(x,f(x))/x \in [0,1]\}$ is dense in $[0,1]\times[0,1]$

We can partition $[0,1]$ in classes according to the equivalence relation $x\equiv y\iff x-y\in\mathbb Q$. There must be $c$ such classes ($c$ is the continuum cardinality), so we can give them the entire $[0,1]$ as a set of indices. Our classes are $(A_t)_{t\in[0,1]}$. We now take $f(x)=t\iff x\in A_t$.

you can even suppose that $f(x+y)=f(x)+f(y)$  :)

Yes, I had written that in my original message, but then I told myself that there was no point, since the problem does not mention additivity, so I edited it :).


Three players take turns to colour the edges between 6n vertices. Each player uses a different colour. The first player to complete a triangle with each of its sides a different colour, wins. Show that the first player does not have a winning strategy.

What do you mean [b]exactly[/b] by "winning strategy"?

[hide]Well, if the second player's edge shares an endpoint with the first player's edge, the third player can make the triangle.  So the first player doesn't have a winning strategy.[/hide]

[hide]Even if everyone was playing optimally, the first player would actually just lose... he/she is a loser.

First the players will never have their segments connect. And then the first player will be the first to be forced to make a connection between two vertices that are already endpoints. When that happens, optimally, he will connect two of his own vertices - (1) otherwise, player 2 or player 3 will win.

(1) Consider player 1 connects between two vertices of different colored segments.

1.  Between one of his own and player 2.
Player 2 must block player 3 from forming the open triangle, and when player 2 blocks the necessary block, she opens up another triangle for player 3. Draw to see what I'm talking about.

2. Between one of his own and player 3.
Player 2 wins automatically.

3. Between one from each of the other players.
Player 2 wins again.

Of course, since everyone will have the same number of segments of their color, the first player will again, after optimizing his connections, be forced to, this time, connect two vertices that are not those of segments of the same color.

This is the same dilemma as (1).

Of course by the definition of no "winning" strategy, Matt is correct in his explanation. I just wanted to see the game played out.[/hide]

I suppose it is true, the fact that any strategy by the first player can immediately be defeated proves the fact. However, I believe that using white_horse's argument, you can show that the third player wins. I'm not quite sure if I'm right though. Anyone want to try to prove/disprove my conjecture?

Well, I think this game qualifies more under the game of psychology. Player 1 choses the fate of the game, although he/she cannot win it. Rather ironic.


calculus
derivative
integration
function
calculus computations

I have to help teach this class tomorrow for my Senior Seminar Math class.  The lesson is over Exponential and Logarithmic Functions and L'Hospital's Rule.  I have all the problems needed to review this lesson, but i can not solve any of them.  My teaching partner and I have had schedule conflicts and we are now on our own.  It is late starting but any help or advise would be greatly appreciated.  I have not had calculus in 5 years.

1.  y=√(2)*e^(√(2)x)      y` =?

2.  y=ln(sec^(2)x)         y`=?

3.  y=9^(2t)                 y`=?

4.  y=[2(x^2+1)]/[√(cos 2x)]   y`=?

EVALUATE INTEGRALS
5.  ∫ e^t * cos(3e^t - 2) dt

6.  ∫(top: π/6, bottom: -π/2)  (cos t)/(1-sin t)  dt

7.  ∫ [ln(x-5)]/(x-5)  dx

8.  ∫(top: e, bottom: 1) (1/x) * (1+ 7 ln x)^(-1/3)  dx

9.  ∫(top:3, bottom: √3)  dt/(3+t^2)

Solve for Y

10.  3^y = 3 ln x       y=?

11.  lim (x→0)  (3^x - 1) / x

12.  lim (x→0) (4 - 4e^x) / xe^x

13.  lim (y→0+)  e^(-1/y) ln y

My partner has the other 13 problems and his are basically the same.  I know this is a lot but any help with any of these would be greatly appreciated.

Keywords (roughly in order):  the chain rule, the product rule, the quotient rule, (optional: logarithmic differentiation), $ u$-substitution.  Also consult the list of common integrals for #9.  Do you have access to a textbook?

i do not have access.  It has been 5 years since i took calculus and no longer have any books.  This seminar class covers all the math a college senior majoring in math should have knowledge of.  It is too broad for 1 book

[quote="Kalle"]Huge what. You can't be serious[/quote]

i do not get what you are talking about


ARML probability geometry trigonometry ratio circumcircle inradius

1.
When the sum of the first $k$ terms of the series

\[ 1^2 + 2^2 + 3^2 + \cdots + n^2 + \cdots \]
is subtracted from the sum of the first $k$ terms of the series

\[ 1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \cdots + n(n+1) + \cdots, \]
the result is 210.  Find the numerical value of $k$.

2.
Three coplanar circles are concentric, and the lengths of their respective radii are 10, 20, and 30.  A square is circumscribed about the smallest circle, and squares are inscribed in the other two circles.  Find the length of the shortest line segment that could contain a point on each square.

3.
Find the numerical value of $k$ for which

\[ \frac{7}{x+y} = \frac{k}{x+z} = \frac{11}{z-y}. \]

4.
Spender Sam always carried only dollars and pennies, but he never carried more than 99 pennies.  When leaving home one day, Sam noted that the number of dollars he had was 150% of the number of pennies he had.  After spending 10 dollars and 78 cents, Sam discovered that the number of dollars he had remaining was 1 less than the number of pennies he then had.  With how much money did Sam leave home that day?

5.
Two cards are randomly selected from a standard 52 card deck of playing cards, without replacement.  Find the probability that at least on king and at least one heart are selected (though not necessarily on different cards).

6.
Find the ordered pair of positive integers $(a, b)$, with $a<b$, for which
\[ \sqrt{1+\sqrt{21+12\sqrt{3}}} = \sqrt{a} + \sqrt{b}. \]

7.
The length of the base of an isosceles triangle is 20.  In the plane of the triangle, a point 4 units from this base is 10 units from each leg.  Find the area of the triangle.

8.
If Arc denotes the principle value, find all values of $x$ which satisfy

\[ \sin{(4\text{Arc}\tan{x})} = \frac{24}{25}. \]

Here's 1.

[hide]We know the problem is asking for ${(1\cdot 2+2\cdot 3... k\cdot k+)} - {(1^2+2^2... k^2)}$.  Another way to write this is ${1\cdot 2-1^2+2\cdot 3-2^2... k\cdot k-k^2}$.  And since ${{k\cdot k+1} - {k^2} = k}$, we can just write this as $1+2...k$.  So basically we get $\frac{k(k+1)}{2} = 210$.  Solving for that, we get $k=20$[/hide]

Since several of us have the OHMIO coming up (and possibly the ARML if we do well enough, I've decided to revise this thread for use as practice.  These problems will probably also work well as practice for the AIME.

[hide="2."]First we note that the side length of a square circumscribed about a circle is equal to the diameter of the circle and that the square inscribed in a circle is equal to the diameter of the circle, and that the ratio of the side diagonal of a square to its side length is $\sqrt2: 1$.

The shortest line segment that can possibly contain points on all three squares is the difference between the circumradius of the smallest square and the inradius of the largest square, i.e., half the diagonal of the smallest square subtracted from half the side length of the largest square.  This answer occurs when the center of the circles, a vertex of the smallest square, and the center of a side of the largest square are colinear.

The smallest square has an inradius of $10$, therefore its circumradius is $10\sqrt2$.  The largest square has a circumradius of $30$, therefore its inradius must be $15\sqrt2$.  The difference is $(15-10)\sqrt2=5\sqrt2$.  Therefore, the smallest line segment that can possibly contain points on all three squares has length $\mbox{5\sqrt2}$.[/hide]

[hide="3"]Rearranging, we have $kx+ky=7x+7z$ and $kz-ky=11x+11z$. Adding these together, we have $kx+kz=18x+18z\Longrightarrow \boxed{k=18}$.[/hide]

[hide="5"]There are 3 non-heart kings and 12 non-king hearts. Thus, we can pick any combination first, yielding $2\cdot\frac{3\cdot 12}{52\cdot 51}$. Also, if we drew a king of hearts, we would automatically satisfy the criteria, thus, providing with $\frac{1}{26}$. Adding them up, we have $\frac{29}{442}$. Hope this is right... [/hide]

[hide="6"]$\sqrt{1+\sqrt{21+12\sqrt{3}}}= \sqrt{1+(3+2\sqrt{3})}= \sqrt{4+2\sqrt{3}}=\sqrt{1}+\sqrt{3}$. Therefore $(a, b)=\boxed{(1, 3)}$. [/hide]

[quote="tjhance"][quote="tjhance"][hide="8"]
using the double angle formula's twice, $\sin(4\tan^{-1}x)=4\sin (\tan^{-1}x)\cos^{3}(\tan^{-1}x)-4\sin^{3}(\tan^{-1}x)\cos (\tan^{-1}x)$

If we draw a triangle with angle $tan^{-1}x$, then the opposite side is $x$ and the adjacent side is $1$. The hypotnuse is then $\sqrt{x^{2}+1}$. That makes $\sin(\tan^{-1}x)=\frac{x}{\sqrt{x^{2}+1}}$ and $\cos(\tan^{-1}x)=\frac{1}{\sqrt{x^{2}+a}}$ Plugging these in:

$\left(\frac{x}{\sqrt{x^{2}+1}}\right)\left(\frac{1}{\sqrt{x^{2}+1}}\right)^{3}-\left(\frac{x}{\sqrt{x^{2}+1}}\right)^{3}\left(\frac{1}{\sqrt{x^{2}+1}}\right)=\frac{6}{25}$
$\frac{x}{(x^{2}+1)^{2}}-\frac{x^{3}}{(x^{2}+1)^{2}}=\frac{6}{25}$
$\frac{x-x^{3}}{x^{4}+2x^{2}+1}=\frac{6}{25}$
$25x-25x^{3}=6x^{4}+12x^{2}+6$
$6x^{4}+25x^{3}+12x^{2}-25x+6=0$

Now all that is left is to find the roots of that quartic. Does anybody have any clue how to do that?

[/hide][/quote]

Okay I found a more reasonable way to do this.

[hide]
It uses the identity $\cos 4\theta=8\cos^{4}\theta-8\cos^{2}\theta+1$

First, we have:
$\cos(4\text{Arc}\tan x)=\m\sqrt{1-\left(\frac{24}{25}\right)}=\pm\frac{7}{25}$
Use the indentity mentioned above and let $y=\cos(4\text{Arc}\tan x)$
$8y^{4}-8y^{2}+1=\frac{7}{25}$
$8y^{4}-8y^{2}+\frac{18}{25}=0$
I solved for $y^{2}$ with the quadratic formula to get
$y=\pm\sqrt{\frac{9}{10}}\text{ or }\pm\sqrt{\frac{1}{10}}$

Since $y=\sin(\text{Arc}\tan x)=\frac{1}{\sqrt{x^{2}+1}}$
So that gives us
$\frac{1}{10}=\frac{1}{x^{2}+1}\implies x=\pm 3$
$\frac{9}{10}=\frac{1}{x^{2}+1}\implies x=\pm \frac{1}{3}$

Testing for extraneous roots, we see that only $-3$ and $\frac{1}{3}$ work.

Likewise, when we cover the $\cos(\text{Arc}\tan x)=\frac{-7}{25}$ case, we see that $\frac{1}{2}$ and $-2$ work.
[/hide][/quote]
complex numbers should work too(eh too lazy to work it ouy)

[hide="Problem 4"]He has D dollars and P pennies. D=(3/2)P.

We know that
$(3/2)P-10, P-78$, or $(3/2)P-11, P+22$.

The first gives negative values, the second gives P=64, D=96.
$\boxed{\$ 96.64}$.[/hide]

[quote="mathgeniuse^ln(x)"]Does #7 make sense to anyofyou[/quote]

I don't think it's right either. I don't believe that's possible...

Yeah, whenever I try to draw #7, I get a rectangle, or in other words, the sides of the "triangle" doesn't intersect.

Ooh I think I got it... My problem was that I put the point inside the triangle istead of outside of it... Is that what you guys did?

[hide]
So extend down the altitude 4 units, draw a perpendicular of length 10 to the legs' extensions. You should see 2 right triangles. Label the altiude x and the leg of the triangle y, then by the pythagorean theorem:

$y^{2}+10^{2}=(x+4)^{2}$
$x^{2}+10^{2}=y^{2}$
And solving, you get:
$(x,y)=(23,\sqrt{629})$

So the area is $\frac{1}{2}(20)(23)=\boxed{230}$
[/hide]

A different approach than Xaero's for [hide="#5"] I went for finding the probability you didn't get them and subtracting it from 1.

Since there are 16 cards that are a heart, king, or both, there is a probability of $\left(\frac{36}{52}\right)\left(\frac{50}{51}\right)$ that you don't select a king or a heart on the first pick and don't select the king of hearts on the second pick.  Also, there is a probability of $\left(\frac{15}{52}\right)\left(\frac{37}{51}\right)$ that you select a heart or a king on the first pick (not including the king of hearts) but don't select one on the second pick.  This gives an overall probability of not choosing a king and a heart of $\frac{785}{884}$ and a probability of choosing a king and a heart of $\boxed{\frac{99}{884}}$.[/hide]

A note to tjhance's solution: the perpendiculars go to the legs, not their extensions.

I just realized that I don't think my solution wouldn't work anyway. I messed up when writing my equations.

Uh, I don't think 7 is possible either.

Draw two circles of radius 10 around the two points that are the endpoints of the base of the triangle.  They are tangent at one point.  The point in the problem is not on the base, but yet still 10 from each of endpoints of the base.  Uh...

here's my interpretation of the problem: the segments from the point to the legs and the segments through the base hit the leg at different angles, which is what makes it possible for them to be the same distance.

Whoops, I think I misread it to be 10 from each of the vertices of the base.  *Gets out notebook and pencil again*

Here you go:

[hide]Since I am a stupid person, I am going to put this thing unto the coordinate plane.

Let the point 4 units down from the base be the origin.  So we need to find a line crosses (10,4) and is 10 units away from the origin.  Now, I actually put something I learned in school to use on a competition problem (maybe there is another way) by using the normal form for a line.

$\sin(\theta)x+\cos(\theta)y-10=0$ ($\theta$ is the angle between the positive x axis and the normal to the line)
But since it crosses (10,4) we have:
$\sin(\theta)10+\cos(\theta)4-10=0$
$\sin(\theta)10-10=-\cos(\theta)4$
$\sin(\theta)10-10=-\sqrt{1-\sin^{2}(\theta)}4$
$100\sin^{2}(\theta)-200\sin(\theta)+100=16-16\sin^{2}(\theta)$
$116\sin^{2}(\theta)-200\sin(\theta)+84=0$

Plug into quadratic
$\sin(\theta)=\frac{200\pm32}{232}$
We see that 1 is not a solution because then we get that infinte triangle so $\sin(\theta)=\frac{21}{29}$.  And our equation is:
$\frac{21}{29}x+\frac{20}{29}y-10=0$
This line has a y intercept of 14.5 so the height of the triangle is 10.5 and the area is 105.[/hide]

Your method is better.  But as coordinates work and we are doing this in school, it is the first thing that comes to mind.  Are calculators allowed in ARML?

Calculators can be used during the team and power rounds, but not the individual and relay rounds.

which test are these questions from?

also, what kind of calculators are allowed?

Any calculator that is allowed on the SAT test date immediately prior to ARML is allowed on the ARML itself.


I'm having difficulty with some number theory questions, can someone please help me out? Thanks.

1) Prove that square root of 3 cannot be expressed as (a/b) for any two integers a,b.

2) Let p be an odd prime. Then it can be proved that
    [ (p-1)!/2 ]^2 is congruent to ((-1)^[(p+1)/2]) mod p
    Use the above to show that if p is congruent to 1 (mod 4), then (-1/p) = 1

[hide]ok for number 1 lets assume its true:

sqrt(3)= a/b
3 = a^2/b^2

we know that x^2 = 0, 1,-1 mod 5

but no combination of those divided by eachother can be 3 mod5 so thus there are no values that satify a and b[/hide]

[hide="Answer to 1"]
Assume the contrary that $\sqrt 3=\frac pq$ and $\frac pq$ is in lowest terms.
Then we have $p\sqrt 3=q$ or $3p^2=q^2$
Since $3p^2$ is divisible by 3, $q^2$ is also divisible by 3.
But since $q^2$ is a perfect square, it has no "unpaired" factors, so $q$ is also divisible by 3.
Thus we can write some number $r$, such that $3r=q$.
Substituting, $3p^2=9r^2$ or $p^2=3r^2$.
Thus there is another smaller pair of numbers that satisfies the condition, and that is a contradiction.
Q.E.D.
[/hide]

Doublepost, please merge them or whatever: http://www.mathlinks.ro/Forum/viewtopic.php?t=89226


The second North American Computational Linguistics Olympiad will be in early February, and online registration for it is open at http://www.naclo.cs.cmu.edu/index.php. No knowledge of linguistics is necessary or expected (very few high school students have studied it), so it's basically a series of logic puzzles--samples are available on the NAmCLO site or http://webscript.princeton.edu/~ahesterb/.

Looks interesting.  You did really well on this last year, didn't you?  :)

About time this forum exists but too late, I'm already over with high school. And this is my old post when I requested for a new forum :P [url]http://www.mathlinks.ro/viewtopic.php?t=82517[/url] Anyways, the problems and puzzles are still fun to do :)

http://webscript.princeton.edu/~ahesterb/ is having some odd problems, but until they're resolved all the puzzles are still accessible at http://webscript.princeton.edu/~ahesterb/puzzles.php .


geometry
trapezoid
ratio
similar triangles

Let $ ABC$ have a right angle at $ B$. Suppose $ AC\equal{}204$ and $ AB\equal{}96$. Let $ M$ be a point in the plane of the triangle $ ABC$ such that the three points $ A,B,M$ are collinear and $ B$ is between $ A$ and $ M$. Suppose that $ BM\equal{}48$. On $ M$, we draw the perpendicular at $ AM$. Let $ N$ be the point of intersection of $ AC$ with a perpendicular line. What is the number of square centimeters in the area of the trapezoid $ BCNM$?

Oh boy.  Similar triangles as a hint =D.
[hide]
Since the ratio of the areas between two similar triangles is simply the square of their side ratio.

$ \frac{\bigtriangleup AMN}{\bigtriangleup ABC}\equal{}\frac{9}{4}$.
To find the area of the trapezoid, we subtract 1, giving us $ \frac{5}{4}$ of the area of $ \bigtriangleup ABC$.

Since $ 96\equal{}12*8, 204\equal{}12*17$ we see that the other leg must be $ 12*15\equal{}180$

Thus our area is $ \frac{5}{4}*\frac{180*96}{2}\equal{}10,800$[/hide]


I was wondering is someone could help me with the following problems:

1. Two children are trying to shoot a marble of mass m into a small box using a spring-loaded gun that is fixed on a table and shoots horizontally from the edge of the table. View Figure The edge of the table is a height H above the top of the box (the height of which is negligibly small), and the center of the box is a distance d from the edge of the table. The spring has a spring constant k. The first child compresses the spring a distance x_1 and finds that the marble falls short of its target by a horizontal distance d_12.
By what distance, x_2, should the second child compress the spring so that the marble lands in the middle of the box? Express answer in terms of m, k, g, H and d.

2. In a ballistic pendulum an object of mass m is fired with an initial speed v_0 at a pendulum bob. The bob has a mass M, which is suspended by a rod of length L and negligible mass. After the collision, the pendulum and object stick together and swing to a maximum angular displacement theta. Find an expression for v_0, the initial speed of the fired object.

3. At time t=0 a grinding wheel has an angular velocity of 21.0 rad/s. It has a constant angular acceleration of 30.0 rad/s^2 until a circuit breaker trips at time t = 2.30 s. From then on, the wheel turns through an angle of 437 rad as it coasts to a stop at constant angular deceleration.
Through what total angle did the wheel turn between t=0 and the time it stopped?
At what time does the wheel stop?
What was the wheel's angular acceleration as it slowed down?

4. Consider a turntable to be a circular disk of moment of inertia I_t rotating at a constant angular velocity omega_i around an axis through the center and perpendicular to the plane of the disk (the disk's "primary axis of symmetry"). The axis of the disk is vertical and the disk is supported by frictionless bearings. The motor of the turntable is off, so there is no external torque being applied to the axis.
Another disk (a record) is dropped onto the first such that it lands coaxially (the axes coincide). The moment of inertia of the record is I_r. The initial angular velocity of the second disk is zero.
There is friction between the two disks.
After this "rotational collision," the disks will eventually rotate with the same angular velocity.
What is the final angular velocity of the two disks?
Because of friction, rotational kinetic energy is not conserved while the disks' surfaces slip over each other. What is the final rotational kinetic energy, K_f, of the two spinning disks?
Assume that the turntable deccelerated during time Deltat before reaching the final angular velocity ( Deltat is the time interval between the moment when the top disk is dropped and the time that the disks begin to spin at the same angular velocity). What was the average torque, \avg{\tau}, acting on the bottom disk due to friction with the record?

1)

i somehow feel that there are many unnecessary things have been given in the question --

time of flight = t = root(2H/g)

also

[img]http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/7/8/a/78abbb0e8b8e6a89dac8c1cb7f6fdd58a7546e89.gif[/img]

also in horizontal direction --

d = vt

putting in the values-

we get

[img]http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/6/7/c/67c4d4ebf845b8913edcce284a2a3a36fda71c9f.gif[/img]

hence

[img]http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/0/6/4/064246b8153aed5d9a2c79a455534e4f8b6deb17.gif[/img]

2)

conserving momentum before and after collision and assuming that the object has "horizontal" velocity v_0

mv_0 = (m+M)v  ---1

where v is the velocity of combined mass after collision

hence

[img]http://alt2.artofproblemsolving.com/Forum/latexrender/pictures/f/4/c/f4cf2c935576ff0a47b2b808a2052915e426721a.gif[/img]

now conserving energy after the collision and at maximum height --

[img]http://alt2.artofproblemsolving.com/Forum/latexrender/pictures/a/5/5/a55da5e53f08d7ecc0a4084cf4941d45e93bc413.gif[/img]

thus

[img]http://alt2.artofproblemsolving.com/Forum/latexrender/pictures/9/f/5/9f5228a444a873838a088d9bc4c5cb4cb1283515.gif[/img]

now just use eqn 1 to get the answer

Dear slogger,
When i was reading your post, instead of equations i only see the word 'image', so I am unable to understand your solution.
Anyway I have solved #1, #2, #3 now but i still need help with #4.


hello, for the product
$ A\equal{}\frac{k^3\minus{}1}{k^3\plus{}1}\frac{(k\plus{}1)^3\minus{}1}{(k\plus{}1)^3\plus{}1}\cdots\frac{(k\plus{}4)^3\minus{}1}{(k\plus{}4)^3\plus{}1}$
i have got
$ A\equal{}\frac{(k\minus{}1)k(k^2\plus{}11k\plus{}31)}{(k\plus{}5)(k\plus{}6)(k^2\minus{}k\plus{}1)}$
I have no idea how we can solve the general problem.
Sonnhard.


Is $ \arctan\left(\frac { \minus{} 16}{ \minus{} 10.2}\right)\approx237^{\circ}$?

Reasoning:

I know that $ \arctan\left(\frac {16}{10.2}\right)\approx57^{\circ}$ but since $ y$ and $ x$ are negative,you add $ 180^{\circ}$ since it should go into Quadrant IV.

Since when was $ \frac { \minus{} 16}{ \minus{} 10.2}$ not equal to $ \frac {16}{10.2}$?

The [i]principal[/i] arctan function $ \text{Arctan}(x)$ is defined so that its range is $ [ \minus{} \frac {\pi}{2}, \frac {\pi}{2} ]$.  This is so it is consistent as a function.  This is not the same thing as the angle the point $ (\minus{}10.2, \minus{}16)$ makes with the $ x$-axis.  But keep in mind that these are [i]two different questions.[/i]


algebra
polynomial
number theory proposed
number theory

Show that every non-zero polynomial can be multiplied by an appropriate non-zero polynomial, such that the exponent in each term of the product polynomial is divisible by $ 10^7\minus{}3$.

Replace $ 10^7 \minus{} 3$ with $ n$.  One answer is $ \prod_{k\equal{}1}^{n\minus{}1} P\left( e^{ \frac{2 \pi ik}{n} } x \right)$.  (This is easiest to see in the case $ n \equal{} 2$.)


If  $ x,y,z$ are positive numbers,  then

$ \frac {2}{x} \plus{} \frac {2}{y} \plus{} \frac {2}{z}\leq \frac {2}{y \plus{} z} \plus{} \frac {2}{z \plus{} x} \plus{} \frac {2}{x \plus{} y}\plus{}\frac {x}{yz} \plus{} \frac {y}{zx} \plus{} \frac {z}{xy},$

with equality if and only if $ x \equal{} y \equal{} z$.

i have that $ 2/(y\plus{}z) >\equal{} (y\plus{}z)/2yz$(prove easy)
so cyclic i get
 $ 2/(y\plus{}z) \plus{} 2/(z\plus{}x) \plus{} 2/(x \plus{}y)>\equal{} (y\plus{}z)/2yz \plus{}(z\plus{}x)/2xz \plus{} (x \plus{}y)/2xy \equal{} 2/2x \plus{} 2/2y \plus{} 2/2z \equal{} 1/x \plus{} 1/y \plus{} 1/z  (1)$
so now i have to prove that $ x/yz \plus{} y/xz \plus{} z/xy >\equal{}1/x \plus{} 1/y \plus{} 1/z$
because its symmetric i say$ x>\equal{}y>\equal{}z$and i have $ 1/yz>\equal{}1/zx>\equal{}1/xy$
so from the rearrangement inequality i get
$ x/yz \plus{} y/xz \plus{} z/xy >\equal{} z/yz \plus{} x/xz \plus{} y/xy \equal{} 1/x \plus{} 1/y \plus{} 1/z (2)$
by adding (1)+(2) i get$ 2/x \plus{} 2/y\plus{}2/z <\equal{} 2/(y\plus{}z) \plus{} 2/(z\plus{}x) \plus{} 2/(x \plus{}y) \plus{} x/yz \plus{} y/xz \plus{} z/xy$ qed :wink:

oops i forgot the equality.
about the cyclic: 2/(y+z) >= (y+z)/2y the equality only if x=y=z
and about the rearrangement the equality only if x=y=z or xy=yz=xz
but the equality for both is only the wanted x=y=z.i think now it is ok

[quote="Athinaios"]i have that $2/(y + z) > = (y + z)/2yz$(prove easy)
so cyclic i get
 $2/(y + z) + 2/(z + x) + 2/(x + y) > = (y + z)/2yz + (z + x)/2xz + (x + y)/2xy = 2/2x + 2/2y + 2/2z = 1/x + 1/y + 1/z (1)$
so now i have to prove that $x/yz + y/xz + z/xy > = 1/x + 1/y + 1/z$
because its symmetric i say$x > = y > = z$and i have $1/yz > = 1/zx > = 1/xy$
so from the rearrangement inequality i get
$x/yz + y/xz + z/xy > = z/yz + x/xz + y/xy = 1/x + 1/y + 1/z (2)$
by adding (1)+(2) i get$2/x + 2/y + 2/z < = 2/(y + z) + 2/(z + x) + 2/(x + y) + x/yz + y/xz + z/xy$ qed :wink:[/quote]

I think that '2/(y + z) > = (y + z)/2yz$(prove easy)' is wrong. 2/(y + z) [color=red]< =[/color] (y + z)/2yz$(prove easy)


trigonometry
trig identities
Law of Cosines

Hello!

My aunt (a math teacher) gave me this pretty simple problem:

From the top of an observation post that is 90 meters high, a ranger sights a campsite at an angle of depression of 10 degrees.  Turning in a different direction, the ranger sees another campsite at an angle of depression of 13 degrees.  The angle between these two lines of sight is 35 degrees.  How far apart are the campsites?

This should be pretty simple unless I'm doing something really stupid:

[hide]
We can use basic trig identities to find
the distance from the person to the first campsite, which we find to
be 90 sec(90-10) = 90sec(80).  Similarly, the distance from the
person to the second campsite is 90 sec(77).  So we can use the law
of cosines to get sqrt((90 sec(80))^2 + (90 sec(77))^2 - 2 * 90sec
(80) * 90sec(77) cos(35)), which is about 3659.05 (according to Mathematica).
[/hide]

However, she says that I did it wrong and that the answer is four hundred something, but refuses to tell me how she did it.  However, I don't think that I'm doing anything wrong... Any comments?[/hide]

Well maybe you have to account for the fact that the two lines of sight are in 3-d. But I doubt it would change the order of magnitude...  :?

I believe your expression is correct, but maybe you typed it into Mathematica incorrectly because I get 298 meters for it.


Determine the values of $a\in R$ sucha that the sequnce $(x_n)_{n\geq1}$ defined by $x_n=n\cdot\{an!\}$ converges.(where $\{x\}$ stands for the fractional part of $x$ , $\{x\}=x-[x]$).

$n\cdot\{an!\}=n\cdot(an!)$?  If so, $a=0$, I think.

Sorry for nopt being clear enough . I have edited now. $\{x\}$ stands for the fractional part of $x$


How many isomers (i'm not sure if this is the right word in english, but it means same no of H and C, but different connections between them) has $C_n H_{2n+2}$, $n \geq 4$?

That is not simple.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=52415

Not only is the problem not trivial, is also is not definite.

If one has one set of 4 different groups attached to a carbon atom, there are two different optically active molecules.  One needs to include the desire to count these two different optically active molecules as one or two in the counting.  If a complex molecule had two or more of these carbon atoms with 4 different groups then the situation becomes more difficult.

I believe that there is a general solution for a more limited question.  I can't remember what the solution is.


inequalities
function
integration
real analysis
real analysis unsolved

Prove that if $f$ is a continuously differentiable function with $f(1)=0$, then the following inequality holds:
\[ \int_0^1(f(x))^ndx \le\frac 1 n \int_0^1 (f'(x))^n dx \].

Well, he forgot to say that $n$ is even. :)

xirti you said it is a improvement of NMO

What was the initial problem ?
thank you

The initial problem was to prove the following inequality:

                    \[ \int_{0}^{1} (f(t))^{n} dt\leq n^{n}\int_{0}^{1} (f'(x))^{n} dx. \]


linear algebra
matrix
linear algebra unsolved

Let $A\in M_{2}(\mathbb{R})$ such that $tr(A)>2.$ Prove that for any $n\in\mathbb{N}$ we have $A^{n}\neq I_{2}.$ ;)

let $p$ and $q$ be the eigenvalues of A

suppose $A^n=I$ then $p^n$ and $q^n$ are eigenvalues of $A^n=I$ and thus $p^n=q^n=1$

or thus $p$ and $q$ are both of form
$cos(\theta)+sin(\theta)i$

this implies that $\Re(p+q) \leq  2$

but the trace of $A$ is exactly the sum of the eigenvalues of $A$
so that implies $\Re(p+q)>2$

contradiction


a,b,c are positive reals then prove that:

$ \boxed{(a \plus{} b \plus{} c)(\frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{a}) \ge 3 (a^2 \plus{} b^2 \plus{} c^2)}$

After Expandation it becomes

$ (\frac {a^3}{b} \plus{} \frac {b^3}{c} \plus{} \frac {c^3}{a}) \plus{} (\frac {a^2 c}{b} \plus{} \frac {b^2 a}{c} \plus{} \frac {c^2 b}{a}) \ge 2 (a^2 \plus{} b^2 \plus{} c^2)$

And:

$ (a^4c \plus{} b^4 a \plus{} c^4b) \plus{} (a^2 b^3 \plus{} b^2 c^3 \plus{} c^2 a^3) \ge 2(a^3 bc \plus{} c^3 ca \plus{} c^3 ab)$

But I don't see how.

[quote="FOURRIER"]a,b,c are positive reals then prove that:

$ {(a \plus{} b \plus{} c)(\frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{a}) \ge 3 (a^2 \plus{} b^2 \plus{} c^2)}$
[/quote]
Let $ c\equal{}\max\{a,b,c\}.$ Then
$ {(a \plus{} b \plus{} c)(\frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{a}) \ge 3 (a^2 \plus{} b^2 \plus{} c^2)}\Leftrightarrow$
$ \Leftrightarrow\frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{a}\minus{}a\minus{}b\minus{}c\geq\frac{3(a^2\plus{}b^2\plus{}c^2)\minus{}(a\plus{}b\plus{}c)^2}{a\plus{}b\plus{}c}\Leftrightarrow$
$ \Leftrightarrow\frac{a^2}{b}\plus{}\frac{b^2}{a}\minus{}a\minus{}b\plus{}\frac{c^2}{a}\minus{}\frac{b^2}{a}\plus{}\frac{b^2}{c}\minus{}c\geq\frac{2(a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}ac\minus{}bc)}{a\plus{}b\plus{}c}\Leftrightarrow$
$ \Leftrightarrow\frac{(a\minus{}b)^2(a\plus{}b)}{ab}\plus{}\frac{(c\minus{}a)(c\minus{}b)(b\plus{}c)}{ac}\geq\frac{2((a\minus{}b)^2\plus{}(c\minus{}a)(c\minus{}b))}{a\plus{}b\plus{}c}\Leftrightarrow$
$ \Leftrightarrow(a\minus{}b)^2\left(\frac{a\plus{}b}{ab}\minus{}\frac{2}{a\plus{}b\plus{}c}\right)\plus{}(c\minus{}a)(c\minus{}b)\left(\frac{b\plus{}c}{ac}\minus{}\frac{2}{a\plus{}b\plus{}c}\right)\geq0,$ which is obviously.

there's a simpler and more elementry solution

[hide="HINT"]$ \frac{a^2}{b}\equal{}\frac{a^4}{a^2b}$[/hide]

[quote="FOURRIER"]a,b,c are positive reals then prove that:

$ \boxed{(a \plus{} b \plus{} c)(\frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{a}) \ge 3 (a^2 \plus{} b^2 \plus{} c^2)}$[/quote]

My solution. By the Cauchy Schwarz we have

$ \frac{{a^2 }}{b} \plus{} \frac{{b^2 }}{c} \plus{} \frac{{c^2 }}{a} \ge \frac{{\left( {a^2  \plus{} b^2  \plus{} c^2 } \right)^2 }}{{a^2 b \plus{} b^2 c \plus{} c^2 a}}$

It suffices us to show that

$ \left( {a \plus{} b \plus{} c} \right)\left( {a^2  \plus{} b^2  \plus{} c^2 } \right) \ge 3\left( {a^2 b \plus{} b^2 c \plus{} c^2 a} \right)$

We assume $ a \ge b \ge c \equal{} 1$ then the inequality becomes

$ \left( {a \plus{} b \plus{} 1} \right)\left( {a^2  \plus{} b^2  \plus{} 1} \right) \ge 3\left( {a^2 b \plus{} b^2  \plus{} a} \right)$

$ f\left( a \right) \equal{} a^3  \plus{} \left( {1 \minus{} 2b} \right)a^2  \plus{} \left( {b^2  \minus{} 2} \right)a \plus{} b^3  \minus{} 2b^2  \plus{} b \plus{} 1 \ge 0$

We have

$ f'\left( a \right) \equal{} 3a^2  \plus{} 2\left( {1 \minus{} 2b} \right)a \plus{} b^2  \minus{} 2 \ge 0$

So $ f\left( a \right) \ge f\left( b \right) \equal{} b^3  \minus{} b^2  \minus{} b \plus{} 1 \equal{} \left( {b \minus{} 1} \right)^2 \left( {b \plus{} 1} \right) \ge 0$

Done.

[quote="Honey_S"][quote="FOURRIER"]a,b,c are positive reals then prove that:

$ \boxed{(a \plus{} b \plus{} c)(\frac {a^2}{b} \plus{} \frac {b^2}{c} \plus{} \frac {c^2}{a}) \ge 3 (a^2 \plus{} b^2 \plus{} c^2)}$[/quote]

My solution. By the Cauchy Schwarz we have

$ \frac {{a^2 }}{b} \plus{} \frac {{b^2 }}{c} \plus{} \frac {{c^2 }}{a} \ge \frac {{\left( {a^2 \plus{} b^2 \plus{} c^2 } \right)^2 }}{{a^2 b \plus{} b^2 c \plus{} c^2 a}}$

It suffices us to show that

$ \left( {a \plus{} b \plus{} c} \right)\left( {a^2 \plus{} b^2 \plus{} c^2 } \right) \ge 3\left( {a^2 b \plus{} b^2 c \plus{} c^2 a} \right)$

We assume $ a \ge b \ge c \equal{} 1$ then the inequality becomes

.[/quote]
I don't think we can assume $ a \ge b \ge c \equal{} 1$
The inequality $ (a\plus{}b\plus{}c)(a^2\plus{}b^2\plus{}c^2)\geq 3(a^2b\plus{}b^2c\plus{}c^2a)$ can solve by AM-GM easily
it's equivalent to $ a^3\plus{}b^3\plus{}c^3\plus{}\sum_{cyc} {ab^2}\geq 2\sum_{cyc} {a^2b}$
which obviously because: $ a^3\plus{}ab^2\geq 2a^2b$ :wink:

I think we can assume that, toanineq.

Fisrt we see that we just have to prove the inequality in the case $ a \ge b \ge c$

Second we also see that the inequality is homogenerous so we can assume that $ c \equal{} 1$

This nice problem is from Romania JBMO TST 2006 . See here

http://www.mathlinks.ro/viewtopic.php?p=515092#515092


trigonometry
integration
function
calculus
calculus computations

Let $\alpha$ de a real that is not an integer, and let $f(x)=e^{i\alpha x}$ for $-\pi<x<\pi$. Find Fourier series of $f$ and use it to show
\[\sum_{-\infty}^{\infty}\frac{1}{(n-\alpha)^{2}}=\left(\frac{\pi}{\sin{\alpha \pi}}\right)^{2}\]

The Fourier coefficients are

$\widehat{f}(k)=\frac1{2\pi}\int_{-\pi}^{\pi}f(x)e^{-ikx}\,dx$

$=\frac1{2\pi}\int_{-\pi}^{\pi}e^{i(\alpha-k)x}\,dx= \left.\frac1{2\pi i(\alpha-k)}e^{i(\alpha-k)x}\right|_{-\pi}^{\pi}$

$=\frac{(-1)^{k}(e^{i\pi\alpha}-e^{-i\pi\alpha})}{2\pi i(\alpha-k)}=\frac{(-1)^{k}\sin(\pi\alpha)}{\pi(\alpha-k)}$

(Note that in size, these coefficients are $O(k^{-1})$ as $|k|\to\infty.$ That's what we expect for a function which has a jump discontinuity.)

Now we can use the Plancherel identity:

$\sum_{k=-\infty}^{\infty}|\widehat{f}(k)|^{2}= \frac1{2\pi}\int_{-\pi}^{\pi}|f(x)|^{2}\,dx.$

For this function, $|f(x)|^{2}=1$ almost everywhere, which simplifies the integral greatly. This gives us

$\sum_{k=-\infty}^{\infty}\frac{\sin^{2}(\pi\alpha)}{\pi^{2}(\alpha-k)^{2}}=1.$

Rearrange a few things and we have our identity.


geometry
geometric transformation
reflection
conics
ellipse
hyperbola
analytic geometry

On the ray $ \overrightarrow{AG},$ construct point $ M_a,$ such that $ \overline {GM_a} \equal{} \frac {_1}{^2} \overline{AG}.$ This is midpoint of $ BC.$ Construct circle $ (Q)$ with diameter $ AM_a$ and let $ M_aE_b$ cut $ (Q)$ again at $ D.$ Since $ M_aE_b \perp AB$ $ \Longrightarrow$ $ c \equiv AD$ is the triangle sideline $ AB,$ with $ B$ unknown. Construct parallel $ e \parallel c$ through $ E_b.$ Reflect line $ c$ in line $ e$ into line $ h,$ so that $ e$ is midparallel of $ c \parallel h.$ Since $ B \in c$ and $ E_b$ is midpoint of $ BH$ $ \Longrightarrow$ the orthocenter $ H \in h.$ But $ H$ is also on a conic $ \mathcal K$ centered at the midpoint $ P$ of $ AE_b,$ passing through $ E_b$ and tangent to the circle $ (Q)$ at $ A, D.$ This conic is an ellipse, when $ E_b, M_a$ are on the same side of $ c$ and hyperbola otherwise (in your sketch). (See the lemma at the end of this post.) To find intersections the line $ h$ with the conic $ \mathcal K,$ it is necessary to find at first the conic foci $ F_1, F_2$ and its pedal circle $ (P).$

Let $ p$ be midparallel of $ c \parallel e$ through $ P,$  the conic major axis. Perpendicular $ m$ through $ P$ (the perpendicular bisector of $ AD$) is the conic minor axis. Perpendicular to $ QD$ at $ D$ is the conic tangent at $ D,$ cutting its major axis $ p$ at $ S.$ Perpenndicular $ DE_b$ to $ p$ cuts $ p$ at $ T.$ The conic cuts $ p$ at unknown $ U, V,$ such that $ PU^2 \equal{} PV^2 \equal{} PS \cdot PT.$ Reflect $ T$ in the minor axis $ m$ into $ T'$ and draw a circle with diameter $ ST'.$ This circle cuts $ m$ at $ W, W'.$ Circle $ (P)$ with center $ P$ and radius $ PW$ cuts the major axis $ p$ at $ U, V$; it is the conic pedal circle. The conic tangent $ DS$ cuts the pedal circle at $ X_1, X_2$ and perpendiculars to $ DS$ at $ X_1, X_2$ cut the major axis $ p$ at the foci $ F_1, F_2.$

To construct intersections of the line $ h$ with the conic $ \mathcal K,$ given by its foci $ F_1, F_2$ and its pedal circle $ (P),$ leads to a degenerate Apollonius problem of constructing 2 circles passing through 2 given points and tangent to a given circle. Indeed, let $ (F_1)$ be a circle with center at the focus $ F_1$ and with radius equal to the major axis length $ UV.$ Let $ F_3$ be reflection of the 2nd focus $ F_2$ in the line $ h.$ Let $ H, H'$ be centers of the circles passing through $ F_2, F_3$ and tangent to the circle $ (F_1)$ at $ K, K'.$ Obviously, the circles $ (H), (H)'$ are centered on the line $ h,$ which is the perpendicular bisector of $ F_2F_3.$ Assuming the foci $ F_1, F_2$ are inside of the pedal circle $ (P),$ (the ellipse case), the points $ F_2, F_3$ are inside of the circle $ (F_1)$ and the circles $ (H), (H')$ are internally tangent to $ (F_1).$ But then $ UV \equal{} F_1K \equal{} F_1H \plus{} HK \equal{} F_1H \plus{} HF_2,$ which means that $ H \in \mathcal K$ and similarly, $ H' \in \mathcal K.$ Consequently, $ \{H, H'\} \equiv h \cap \mathcal K$ are the 2 possible orthocenters of the unknown $ \triangle ABC.$

To construct $ H, H',$ draw an arbitrary circle $ (E)$ through $ F_2, F_3,$ centered on the line $ h,$ and make sure it intersect the circle $ (F_1)$ at $ L_1, L_2.$ $ L_1L_2$ is radical axis of $ (E), (F_1)$ and $ F_2F_3$ is radical axis of $ (E), (H).$ The 2 radical axes meet at the radical center $ R$ of $ (E), (F_1), (H).$ Radical axis of $ (F_1), (H)$ is their single common tangent through $ R.$ Therefore, construct tangents to $ (F_1)$ from $ R$ and let $ K, K'$ be the tangency points. Lines $ F_1K, F_1K'$ then cut the line $ h$ at $ H, H'.$ (It is not necessary to actually draw the circles $ (H), (H'),$ their centers are all we need.) Using just one orthocenter $ H,$ line $ HE_b$ cuts the line $ c$ at $ B.$ Perpendicular $ a$ to $ AH$ through $ B$ cuts the perpendicular $ b$ to $ BH$ through $ A$ at $ C,$ completing the $ \triangle ABC.$

[hide="Lemma about the conic"] Put origin at the center $ Q$ of the circle $ (Q)$ with diameter $ AM_a$ and x-axis through $ Q$ parallel to $ AD.$ Let $ Z$ be midpoint of $ AD.$ Let $ E_b \in DM_a \perp AD$ be fixed and and $ B \in AD$ arbitrary. $ P$ is midpoint of $ AE_b.$ Denote $ q \equal{} ZQ,\ a \equal{} AZ \equal{} ZD,\ b \equal{} ZB,\ p \equal{} DE_b.$ Coordinates of $ A, D, M_a, B, E_b, P$ are then $ A \equal{} [ \minus{} a, \minus{} q],$ $ D \equal{} [a, \minus{} q],$ $ M_a \equal{} [a, q],$ $ B \equal{} [b, \minus{} q],$ $ E_b \equal{} [a, p \minus{} q],$ $ P \equal{} [0, \frac {_p}{^2} \minus{} q]$ and equations of the lines $ BM_a, BE_b$ are

$ BM_a: \ \ y \minus{} q \equal{} \frac {2q}{a \minus{} b} (x \minus{} a),$

$ BE_b: \ \ y \minus{} (p \minus{} q) \equal{} \frac {p}{a \minus{} b} (x \minus{} a).$

$ BM_a$ cuts $ (Q)$ again at $ H_a$ and equation of $ AH_a \perp BM_a$ through $ A$ is

$ AH_a: \ \ y \plus{} q \equal{} \minus{} \frac {a \minus{} b}{2q} (x \plus{} a).$

To find equation of the locus of $ H \equiv BE_b \cap AH_a,$ we eliminate the arbitrary parameter $ b$ from equations of the lines $ BE_b, AH_a.$

$ \frac {p(x \minus{} a)}{y \minus{} (p \minus{} q)} \equal{} a \minus{} b \equal{} \minus{} \frac {2q(y \plus{} q)}{x \plus{} a},$

$ p(x^2 \minus{} a^2) \plus{} 2q(y \plus{} q)(y \plus{} q \minus{} p) \equal{} 0,$

$ \frac {2}{2a^2 \plus{} pq}\ x^2 \plus{} \frac {4q}{p(2a^2 \plus{} pq)}\left(y \plus{} q \minus{} \frac {_p}{^2}\right)^2 \equal{} 1,$

which is equation of a conic $ \mathcal K$ with center $ P \equal{} [0, \frac {_p}{^2} \minus{} q],$ with main axes parallel to the coordinate x-, y-axes. For $ p > 0$ (point $ E_b$ above the line $ AD$), we get an ellipse, for $ p < 0$ (point $ E_b$ below the line $ AD$), we get a hyperbola. Plugging in $ [x, y] \equal{} [\pm a, \minus{} q]$ satisfies the equation $ \Longrightarrow$ $ A, D \in \mathcal K.$ It is absolutely clear that the locus of $ H$ has to be either completely inside of $ (Q)$ or completely outside of $ (Q),$ therefore $ \mathcal K$ is tangent to $ (Q)$ at $ A, D.$ It is also easy to verify by direct calculation that the conic $ \mathcal K$ and the circle $ (Q)$ with equation $ x^2 \plus{} y^2 \equal{} a^2 \plus{} q^2$ have common tangents at $ A, D \equal{} [\mp a, \minus{} q]$ with equations $ \mp ax \minus{} qy \equal{} a^2 \plus{} q^2.$[/hide]

could you show your construction

thanks

Sure, here it is.
Can you tell me, how are problems generated in that handbook of triangle constructions ? Are all solutions known ? Any computer involvement ?

HEbB is not in line with my construction

 :(

Wonderful! Your method can solve many interesting problems with beautiful geometric properties  :lol:
In this problem, we may find a simpler solution.
Easy to construct the midpoint $ M_a$ of $ BC$. $ M_aE_b\parallel CH$, so the perpendicular line $ d$ from $ A$ to $ M_aE_b$ is just the sideline $ AB$. Draw the line $ d'$ symmetric to $ d$ wrt $ M_a$. Let $ K$ such that $ M_a$ is the midpoint of $ E_bK$. See that $ KC\parallel BE_b$, so $ KC\perp AC$. Hence $ C$ is just the intersection of $ d'$ and the circle of diameter $ AK$.

Very nice, thanks ! Could you also try to find something simpler for this one: [url=http://www.mathlinks.ro/viewtopic.php?t=286325]EaEbG[/url] ?


Let $ \Delta ABC$ be a triangle .points $ D,E$ are chosen on $ AB,AC$,respectively, such that $ \angle ADC\plus{} \angle AEB\equal{}90^\circ$. Let $ O$ be the circumcenter of $ \Delta AED$ and let $ Q$ be a point on $ BC$ such that $ \angle AQC\equal{} 2 \angle ADC$. prove that $ OQ \perp BC$.
(i don't know the source ,sorry if it has been posted)

This problem is equivalent to this problem:
[b]P1:[/b]Let $ \triangle ABC$ and $ O$ is its circumcenter . $ S$ be a point lies on the tagent of $ (O)$ at $ A$ (like the figure below) . $ SO$ meets [/tex](O)[/tex] at $ P,Q$ respectively .$ CQ$ meets $ AB$ at $ D$.$ SD$ meet $ AB$ at $ E$. Prove that : $ B,E,Q$ are collinear

And it could be generalized like this  (Figure 2) :
[b]P2:[/b]LEt $ (O;R)$ ,from a point $ S (SO > R)$ we draw three secants $ SAB,SCD,SEF$
$ P \equal{} CF \cap AE ;Q \equal{} DE\cap BF$  (
Prove that :$ A,P,Q$ are collinear

Proof:
Let $ I \equal{} (SAE)\cap SI$. $ Q' \equal{} SI \cap BF$
We have:
*$ \overline{PC}.\overline{PF} \equal{} \overline{PA}.\overline{PE} \equal{} \overline{PI}.\overline{SI}$
Hence :$ S,C,I,F$ are cyclic
$ \Rightarrow (IQ',IF) \equal{} (CD,CP)$(1)
*$ (IQ',IE) \equal{} (AB,AE) \equal{} (FB,FE) \equal{} (FQ',FE)$
$ \Rightarrow I,Q',F,E$ are cyclic
$ \Rightarrow (ED,EF) \equal{} (EQ',EF) \equal{} (IQ',IF)$ (2)
From (2) and (1) ,we get :$ E,Q',D$ are collinear
$ \Rightarrow Q' \equiv Q$
Q.E.D

P/s: My figure for P2 has lost point S .Sorry


I've got a Function Lab due in 3 hours, and I was wondering if someone could help with this one problem I'm stuck on.

5. Complete the square in the formula for the parabola \[y=3x^2+bx-2\]

I know the steps involved, however I keep making algrebraic mistakes.

Steps:

\[ax^2+bx+c=y\]
\[ax^2+bx=y-c\]
\[a(x^2+\frac{b}{a}x)=y-c\]
\[a(x^2+\frac{b}{a}x+\frac{b^2}{4x^2})=y-c+\frac{b^2}{4a}\]
\[a(x+\frac{b}{2a})^2=y+\frac{b^2-4ac}{4a}\]

Could someone please show me step by step how to correctly do this?

Hmm... you just have to replace the a, b and c of your formula with : 3, b and -2..

You should obtain : $3(x+\frac{b}{6})^2=y+\frac{b^2+24}{12}$

 :)

I need the solution in the final step which is... I guess "complete the square?" format... also steps are nice.. because I'm not sure I did the "moch" steps correctly.

$y=3x^2+bx-2$

Multiply both sides by 3:
$3y=9x^2+3bx-6$

Sum $\frac{b^2}4$ to both sides:
$3y+\frac{b^2}4=9x^2+3bx+\frac{b^2}4-6$

Rearrange some terms:
$3y+\frac{b^2}4=(3x)^2+2.3x.\frac b2+\left(\frac b2 \right)^2-6$

Perfect square:
$3y+\frac{b^2}4=\left(3x+\frac b2 \right)^2 - 6$


group theory
abstract algebra
superior algebra
superior algebra theorems

How can you prove this theorem using the isomorphism theorems?

Let K be a normal subgroup of G.  Let (pi) (a map from G to G/K) be the cannonical homomorphism.  THe map H to pi(H) defines a 1-1 correspondence between subgroups of G contained in K and subgroups of G.K.  Normal subgroups match up to normal subgroups.

The correct one is:

Let $G$ be group and $K \lhd G$ a normal subgroup. Then there exists a canonical bijection between the subgroups $\tilde{H}$ of $G/K$ and the subgroups $H$ of $G$ containing $K$, which is implied by the canonical homomorphism $\pi: G \to G/K$.
Also this bijection maps normal subgroups to normal subgroups.

what is the meaning of a canonical homomorphism?

Here the surjective homomorphism $\pi: G \to G/K$ is meant, that sends every $x \in G$ to it's 'residue' class $xK$ in $G/K$.

Theorem.Let $N$ be normal to group $G$.Then there exists a one-to-one correspondence between the subgroups of $G$ containing $N$ and the subgroups of $G/N$ by $H \iff H/N$ and this correspondence also gives a one-to-one correspondence between the normal subgroups of $G$ containing $N$ and normal subgroups of $G/N$
Proof.Define $f:G \rightarrow G/N$ by $f(g)=gN$.Then it is an epimorphism.Let $N\subset H\leq G\implies f(H)=H/N \leq G/N$.Now consider another subgroup of $G$ containing $N$ and show that they are equal.Hence $H\rightarrow H/N$ is well defined and injective.Now let $Y\leq G/N$ and then $N \subset f^{-1}(Y)$ and now show that $f^{-1}(Y)\leq G$ using homomorphism property of $f$.Thus $H\rightarrow H/N$ is bijective.Now let $N\subset H\lhd G$ and to show:$f(H)\lhd G/N$.Let $gN \in G/N$ and $hN\in H/N\implies...\implies ghg^{-1}\in N\subset H\implies f(H)\lhd G/N$.Now let $K\lhd G/N$ and let $L:=\left\{x\in G:f(x)\in K \right\}$.Show that $N\subset L$ and $L \lhd G$ and $f(L)=K$.Hence proved.


Prove that in every polygon, there exist two sides whose ratio is $\geq 1$ and < 2.

Yeah, it's simple... but I found it nice...

  Darij

[hide]

if it was false,

the smallest side is of length, the next smallest side is >=2n, the next smallest is >=4n so the largest side >=2^(k-1)*n, where k is the number of sides, and this is larger than the sum of the lengths other sides, and thats bad!

[/hide]


For every positive integer $n$, form the number $n/s(n)$, where $s(n)$ is the sum of digits of $n$ in base 10. Determine the minimum value of $n/s(n)$ in each of the following cases:

(i) $10 \leq n \leq 99$
(ii) $100 \leq n \leq 999$
(iii) $1000 \leq n \leq 9999$
(iv) $10000 \leq n \leq 99999$

[hide]I'm getting that the answer for i is 10,ii 100, iii 1000, iv 10000.
for i:
n=10a+b
n/s(n) = (10a+b)/(a+b) = 10-9b/(a+b)
Clearly, since a,b are non-negative, and a+b =/= 0, we can see that a nonzero value of 9b will lower the ratio.
Therefore the maximum is 10, when nothing is subtracted and b = 0.
Extending this concept to ii,iii,iv, we get 100-(90b+99c)/(a+b+c), 1000-(900b+990c+999d)/(a+b+c+d), and 10000-(9000b+9900c+9990d+9999e)/(a+b+c+d+e), for ii,iii,iv, respectively.
Applying the same logic yields 10,100,1000,10000 for i,ii,iii,iv, respectively.[/hide]

^^mchoi i don't think that's right

[hide]
$ \frac {n}{s(n)} \equal{} \frac {10a \plus{} b}{a \plus{} b} \equal{} 1 \plus{} \frac {9a}{a \plus{} b}$
Clearly, we need to minimize $ a$ and maximize $ b$ to minimize the value of $ \frac {n}{s(n)}$
So for (i), it'll be $ \frac {n}{s(n)} \equal{} 1 \plus{} \frac {9(1)}{(9 \plus{} 1)} \equal{} \boxed{19/10}$, with $ n \equal{} 19$
For ii, iii, and iv, its the same logic, for ii:

$ \frac {n}{s(n)} \equal{} \frac {100a \plus{} 10b \plus{} c}{a \plus{} b \plus{} c} \equal{} 1 \plus{} \frac {99a \plus{} 9b}{a \plus{} b \plus{} c}$, again here  we minimize $ a$ and $ b$ and maximize $ c$, getting $ n \equal{} 119$ and $ \frac {n}{s(n)} \equal{} \boxed{119/12}$, so yeah the rest the minimum is when $ \boxed{n \equal{} 1119}$, and $ \boxed{11119}$.

[/hide]

Well, I think you got (i) right, but for (ii) compare the value of $ \frac {n}{s(n)}$ for $ 199$ to the one you got, which should be $ \frac {119}{\boxed{11}}$ by the way.

At the risk of giving yet another incorrect answer to this problem, [hide]I would guess that $ 2 \cdot 10^n \minus{} 1$ gives the minimum in each interval $ [10^n,10^{n \plus{} 1} \minus{} 1]$. My reason for believing this is that the leading digit of each number overrides all the others in determining the numerator of the quantity we're trying to minimize. On the other hand, all digits play a role in increasing the denominator, so they should all be maximized. I should probably try to rigorize this later.[/hide]

-- Nevermind!

However, can we seek an inductive solution of some sort?

No risk, really?  :P  Unless my addition skills are off,
\[ \frac{10999}{28} \approx 392.82 < \frac{19999}{37} \approx 540.51\]

So, at the risk of throwing out another wrong answer, I'm predicting the answer to be
[hide]on the interval $ [10^n, 10^{n\plus{}1}\minus{}1]$:
\[ \text{max}\ (n/s(n)) \equal{} \begin{cases}11\cdot10^{n\minus{}1}\minus{}1&n\ge 3\\ 2\cdot10^n \minus{} 1&n \le 2\end{cases}\][/hide]
Apparently this problem is tougher than at first glance.

Alright, I think I've figured out how to do this. The idea in jimhu's post where he clears the lowest digit from the numerator is the key; he just didn't use it enough. The first step is to show that the leading digit must be $ 1$ by simple bounding. Then apply the mixed numbers method to get the lowest digit out of the numerator. Now its only location in the expression is the denominator. Now you know it must be $ 9$ to minimize the expression (or $ 0$, see below). Substituting this in allow you eliminate the next lowest digit. Repeat this.

The reason it turns out to be $ 11 \cdot 10^{n \minus{} 1} \minus{} 1$ and not $ 2 \cdot 10^n \minus{} 1$, at least for $ n \equal{} 3$, is the following. If you make your number $ (abcd)$ and use this method to get that $ c \equal{} d \equal{} 9$, you get a fraction in $ b$. But now when you use the mixed numbers method to get rid of the $ b$ from the numerator, the numerator becomes negative! Suddenly you want $ b$ to be a low value. So you make it $ 0$.

Right now I've only worked this method through for $ n \equal{} 4$, and I'm going to try to see how well it works for an arbitrary $ n$. It's possible that the number of $ 0$s and $ 9$s gets weird especially at high $ n$, but maybe not.

This problem is a lot trickier than it looks, indeed.

EDIT: Algebra is ridiculously messy if you try for arbitrary $ n$. Bashing this with excel has told me that having two $ 0$s after the leading $ 1$ is optimal when $ n \equal{} 14$. Sorry Andy, but your formula is wrong too. :(

In fact, my guess was wrong; there are larger counterexamples. We find that $ 1$, followed by a $ 0$, and then 13 $ 9$s, in fact yields a larger number than does a $ 1$, followed by a $ 00$, and then 12 $ 9$s. So four distinct, incorrect answers in a row.  :wink:

MellowMelon, my logic was basically the same as yours. But if we write the fraction to get rid of $ c$ and drive the numerator to be negative, some crude bounding gets me to $ n \equal{} 13$ or so.

Edit: And you beat me. (I used excel for verification also)  :roll:

Here is an answer I am confident of.

Let $ f(n) \equal{} \lfloor \log_{10}9(n \minus{} 2) \rfloor$ for $ n \geq 3$. Then the minimum of $ k/s(k)$ for $ n$ digit numbers ($ n \geq 3$) is achieved when $ k \equal{} 10^n \plus{} 10^{n \minus{} f(n)} \minus{} 1$.

Computed with the help of excel and the mixed numbers method mentioned in the above post.

hahaha, nice MellowMelon, but what did you use in excel to help you?  :o  I've never discovered it useful as a ti-89 lol

I needed a function taking one or two positive integers computed for lots of inputs, and it was preferable that the outputs be simultaneously visible. There are almost certainly better ways to get that than Excel, but it does the job. Main difficulty is getting it to display an entire number with lots of digits.

I employ Excel along with Mathematica in brute-forcing some problems. :P

[hide="Hint"]Fix $ s(n)$, then figure out what $ n$ must look like to minimize $ n/s(n)$.[/hide]


If $\sum_{k=1}^n {a_k} = n(2n+1)$, what is $a_n$ in terms of $n$?

If we calculate the sum of the squares of all even positive integers up to 1998 and subtract from it the sum of the squares of all the odd positive integers up to 1998 and then subtract the sum of all of the positive integers up to 1998 what will we get as a result?

Find the value of $\tan{(x)}$ if:

$\frac{\sin^2x}{3}+ \frac{\cos^2x}{7} = \frac{-\sin{(2x)}+1}{10}$

3)

Multiply through by 210 to attain:

\[70\sin^2 x+ 30\cos^2 x=-21\sin 2x +21.\]

Now use the identity $\sin 2x=2\sin x\cos x$ :

\[70\sin^2 x +42\sin x \cos x +30\cos^2x=21.\]

Now, this looks factorable.  But when we factor we want the product of two terms equal to zero.  Remember that $\sin^2 x+\cos^2 x=1$.  We can make it:

\[49\sin^2 x +42\sin x \cos x +9\cos^2 x=0.\]

Much better!  So:

\[(7\sin x +3\cos x)^2=0\Rightarrow 7\sin x +3\cos x =0\Rightarrow \tan x=-\frac 37.\]

To calculate $a_n$, calculate $\sum a_n$ and $\sum a_{n-1}$ and subtract to get $a_n$ by itself. Ans: $a_n=4n+1$

Errr, is that a typo or the actual question in the second question?

2)

First let's evaluate the sum of the odd squares to 1998 subtracted from the sum of the even squares to 1998:

\begin{eqnarray*}
(2^2+4^2+\cdots+1998^2)-(1^2+3^2+\cdots+1997^2) &=&(2^2-1^2)+(4^2-2^2)+\cdots+(1998^2-1997^2)\\
&=& (2+1)(2-1)+(4+3)(4-3)+\cdots+(1998+1997)(1998-1997)\\
&=& 1+2+3+4+\cdots+1997+1998
\end{eqnarray*}

Now, assuming that 19989 was a typo meant to be 1998 the final answer is just subtracting the sum of the first 1998 positive integers. Clearly this comes out to 0.

Yeah, its supposed to be 1998...sorry...


inequalities
algebra
function
domain
inequalities solved

Here's an inequality proposed but not used :(  for the 2000 Moscow olympiad :
Let x,y,z > 0 such that xyz = 1.
Prove that : x 2  + y 2  + z 2 + x + y + z  \geq  2(xy + yz + zx).

I know a solution, but only one so it will be nice to see others.

Pierre.

Really nice:
    Put x=a^2 and so on. Use Schur to find that sum a^4>=sum ab(a^2+b^2)-abc(a+b+c)>=2 sum a^2b^2-a-b-c. So, we have sum x^2>=2 sum xy-sum sqrt(x). But it's easy to see that sum sqrt(x)<=sum x, since xyz=1. And it's done.

Or you can prove it like this:
   Put x=a^2 and so on. Then 2 sum xy-sum x^2=(a+b+c)(a+b-c)(b+c-a)(c+a-b)<=abc(a+b+c)=a+b+c<=a^2+b^2+c^2.

4 minutes!!!!!
You are really incredible Harazi....

And this is a new kind of solution I was hoping.
Mine is based on a trick Namdung used to try to solve inequality (f(x,y,z)  \geq f(x, \sqrt (yz),  \sqrt (yz) ) where x = min {x,y,z}).

Pierre.

[quote="pbornsztein"]4 minutes!!!!!
You are really incredible Harazi....
[/quote]
Harazi is an ineq expert..

I really would like to be, but I'm far away from that objective.

ahem ... then I must be the all-time grand idiot in inequalites, if you are far away from an expert :) you're no cartoaje yet, but who knows ? :D

To be sincerely I wouldn't like to be as Cartoaje. He's the master of inequalities, but I want to be good in other domains, too.


topology
advanced fields
advanced fields unsolved

Which topological spaces can be embedded in compact topological spaces?

If we're talking about compact Hausdorff spaces, they're precisely the completely regular spaces (they have the Stone-Cech compactification, for instance, and the converse is obvious). I'm not sure what happens if we take into consideration pathological, non-Hausdorff spaces.

The intended question has no regularity conditions at all, on either the original space or the space it's embedded in.

Ok, and if as you said, the compact space is just compact with the usual definition without Hausdorff property, then any topological space $X$ can be identically mapped into itself, but supplied with indiscrete topology(which is compact).
So, if i'm not wrong the answer would be: all the topological spaces.

That's not an embedding- the embedded copy has to have the original topology as a subspace of the new space.

This isn't as hard as it looks. Try it again!

Apparently, every topological space can be embedded (as a dense subspace, even) in a compact space.

Let $X$ be an arbitrary topological space. The compact space $\widetilde X$ in which we'll embed $X$ will consist of the points of $X$, together with a point $p_{D}$ for each non-empty open subset $D$ of $X$. A basis of open sets of $\widetilde X$ consists of the open subsets of $X$ and the sets of the form $\{p_{D}\}\cup D$ for non-empty open subsets $D$ of $X$. Since the only neighborhood of $p_{X}$ is $\widetilde X$, the space is clearly compact.

Yes. You only really need one point ($p_{X}$).

:) Of course you do.. I actually remember finding the easier version some time ago, but today, when I tried to remember my proof, only this mess came out :).


The least value of the function $ ax^2\plus{}bx\plus{}c$  with $a>0$ is:

$\textbf{(A)}\ -\dfrac{b}{a} \qquad
\textbf{(B)}\ -\dfrac{b}{2a} \qquad
\textbf{(C)}\ b^2-4ac \qquad
\textbf{(D)}\ \dfrac{4ac-b^2}{4a}\qquad
\textbf{(E)}\ \text{None of these}$


I have no idea where to start with this, can anyone help?

Prove that the integer a has an inverse mod n if and only if gcd(a,n) = 1.

Help!

nice proof i totally understand it.. although how to prove if two integers multiply to 1 they must both be 1 or negative 1?

Only PROMYS or a rigorous number theory course would ask you to prove that. It gets down into the axioms.

Basic idea: Suppose $ ab \equal{} 1$ and $ |a| > 1$. WLOG $ a$ is positive or negate $ a$ and $ b$. Then $ a^k b^k \equal{} 1$, and $ b^k \equal{} 1/a^k$ can get arbitrarily close to $ 0$. This contradicts the well-ordering principle.

[quote="rorylando45"]OK here's what I wrote, does it all look watertight and OK?

[quote]For a to have an inverse mod n, there must exist an integer b such that a.b = 1 mod n; that is, a.b = k.n + 1 for some integer k.

For any common divisor of a and n we can call it m, and can write:

a = a’m
n = n’m

Therefore, a’mb = n’mk + 1
m(a’b – n’k) = 1

Now from inspection we can see that any two integers that multiply together to equal 1 must both equal 1, or both equal -1. We can disregard m = -1 because a = a’m so if m is negative then so must a’ because a is positive. So we can just accept that m = 1.

That has shown that if a has an inverse mod n then the greatest common divisor will be 1. I shall now who it backwards. If m = 1:

m(a’b – kn’) = (a’b – kn’) = 1
a’b = 1 + kn’
But a = a’m = a’ and n = n’m = n’ so:
a.b = 1 + k.n

Which is back to the start and the correct format for an inverse mod n of a.[/quote][/quote]

what does the dot mean? multiplication?
BTW, multiple quotes hurts my eyes


[quote="#H34N1"]Both the roots of the quadratic equation $x^{2}-63x+k=0$ are prime numbers. What is the number of possible values for $k$?[/quote]
[hide="hm"]The only possibility for k is 122[/hide]

[hide="full explanation"]
suppose the roots are $r$ and $s$
then you can get the equation:
$x^{2}-(s+r)x+sr$
that means $(s+r)=63$
since both $s$ and $r$ are prime, that means neither could be negative, else one number would have to be an even negative munber. neither can be $-2$ because then the other would have to be $65$, which is not prime.
that means the roots must both be positive. the only way to add up to an odd is with one even and one odd.
that means the even is 2 and the odd one is 61
so $k=112$
this explanation kind of sucks, someone else could probably word it better.

[/hide]


Hey,
I have been searching online for a few hours now and I can't seem to answer this problem.
I have been asked the question: "What is a Hamiltonian and What is it good for?"

I have searched everywhere for an outright definition but to no avail. Could anyone help me out with this I would be very greatful.

Sorry in advance if this is not the right forum for this question!

Maybe it can help you:

http://www.answers.com/topic/hamiltonian

there are many links that seem interesting... but I don't open them..
Bye, Uber


Find all positive integers $n$ such that $2^{n}\equiv 1$ (mod n).


[quote="Megus"]Reangarrement ?  :P  :P  :P[/quote]

Not quite. Or at least it's not obvious to me.

The inequality is a special case of the following inequality (probably discussed on the forum, but I can't find it now):

Suppose that $f: (a, b) \to \mathbb{R}$ is convex, $n>  2$ and $a< x_1 < x_2 <\cdots  < x_n<b$, then
\[
x_1 f(x_2) + x_2 f(x_3) + \cdots +  x_n f(x_1) \geq x_2 f(x_1) + x_3 f(x_2) + \cdots + x_1 f(x_n)
\]

It can be proved with induction.


AMC
AIME
greatest common divisor
algebra
polynomial
calculus
integration

You are now free to discuss the Alternate AIME.

Anyone have a list of answers they got? Especially 1-10. I want to check with other members of AoPS on mine.

These are the answers that I/my school got... I'd greatly appreciate some confirmation.

1. 013
2. 079
3. 802
4. 435
5. 054
6. 392
7. 125
8. 405
9. 250
10. 011
11. 889
12-15 ?

how did you guys get #5 to be 54? Everyone at my school agreed to 52.
you overcount the powers of 2^6 and 3^6 once.

we have A: {2^2, 3^2 ..... 44^2} 43 elements
and 2: {2^3, 3^2, .... 12^3}. 11 elements

both has 2^6 = (8)^2 and 4^3
and 3^6 = 27^2 as well as 9^3. So there is an overcount by 2. I'm confused. I hope I didn't make a dumb mistake.

I don't think you overcount; they're asking for ordered pairs, so (8,2) and (4,3) should be distinct, right?

My bad, perhaps I overthought the problem.

Can someone explain how number 4 is "435" or in one instance: "434"?

10^10 = 121 divisors (2^10 * 5^10) = (10+1) * (10+1) = 121
15^7 = 64 divisors
18^11 = 276 divisors

==> 121+64+276 = 461 possible divisors

However, you find that there are technically 27 similar divisors
5^7 ==> 8 divisors
3^7 ==> 8 divisors
2^10 => 11 divisors

==> 8+8+11 = 27

Still, you take away 2, (27-2) because 1 is 3^0, 5^0, and 2^0.. Thusmus tbe only counted once.

461 - 25 = 436...

Am I making a repeated stupid mistake? Sheessshhh. Thanks =D

Look at how many times you counted the divisor 1.
The first time through, you add it three times, so you've overcounted by two.
The next time through, you subtract it once for each of the three GCFs pairwise, so you subtract three, and now you've undercounted by one.
Then you add two back in, so now you've overcounted by one and your answer is indeed one larger than 435.

The people who got 434 probably didn't remember to include 1 back in the third time (that's what I did).

Completely off topic here, but dude Samson...you should be in class. -.- I thought getting online was not allowed...

[quote="DPopov"]dts, what did you end up with as your score?[/quote]

9 (made 3 stupid mistakes).  Alas, this puts my index at 222.  Hope that's enough... :?

#12: 307
Reflecting the triangle over OF; the image of E will then be on BC! Repeating this process around O, we obtain an octagon composed of 8 images of EOF. Let x be BF; then AE = 500-x. The area of the square is 810000; the area of the octagon is 4(400)(450) = 720000. The remaining area in the square is 2x(500-x) = 90000. Solving this, we get x = 250 plus or minus 50 root 7. BF is the larger of the two solutions, so we want 250 + 50 + 7 = 307.

#15: 036
I don't feel like trying to prove it using text only, but the basic idea is that the locus of the center of the third circle is a circle of radius 10 centered at (0, 12). We want a line y = ax (with positive a) to be have as small as possible a slope when intersect with this circle. That means y = ax is tangent to the third circle. The third circle has equation x^2 + (y - 12)^2 = 100; replacing, we get x^2 + a^2 x^2 - 24ax + 144 = 100 or (a^2 + 1)x^2 - 24ax + 44 = 0. This has exactly one solution (since the line is tangent), so we want the discriminant to be zero. This gives 576a^2 - 176(a^2 + 1) = 0, or 400a^2 = 176. This gives a^2 = m^2 = 176/400 = 44/100 = 22/50 = 11/25, so we want 11 + 25 = 36.

[quote]#15: 036
I don't feel like trying to prove it using text only, but the basic idea is that the locus of the center of the third circle is a circle of radius 10 centered at (0, 12). We want a line y = ax (with positive a) to be have as small as possible a slope when intersect with this circle. That means y = ax is tangent to the third circle. The third circle has equation x^2 + (y - 12)^2 = 100; replacing, we get x^2 + a^2 x^2 - 24ax + 144 = 100 or (a^2 + 1)x^2 - 24ax + 44 = 0. This has exactly one solution (since the line is tangent), so we want the discriminant to be zero. This gives 576a^2 - 176(a^2 + 1) = 0, or 400a^2 = 176. This gives a^2 = m^2 = 176/400 = 44/100 = 22/50 = 11/25, so we want 11 + 25 = 36.[/quote]

Hmm..., but the official answers say the answer is 169.

I think the locus is actually an ellipse. The two focii are the centers of the original circle. If you draw the picture, the distance from the circle to (-5,12) is 16-r and from (5,12) is 4+r. Adding them up gives the major axis to be 20.

About 2005 Alternate AIME #15:
[quote]I think the locus is actually an ellipse. The two focii are the centers of the original circle.[/quote]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=30627]This question[/url] addresses a similar issue.

Yakko

Interesting....

I think it depends on where the circles are. (see drawing)

[quote="filletwho"][quote="frodo"]Why have this year's AIME tests been relatively easy?  Is it like the people in charge of forming questions are running out of ideas or are they preparing for more difficult AIME problems to come?  :|[/quote]

Actually, i don't think these tests were necessarily easier. After looking at the problems, I think past AIMEs had similarly easy solutions, it's just that those problems were more disguised. It took a certain amount of insight to see those solutions, making the AIME problems unique and interesting. This year however, the problems were straightforward and either u knew the solution or u didn't.[/quote]

Actually on past AIME's there were easy questions too, but on this year's AIME 2 there were less computations than usual(problem like #7, 9, 11 were 15-seconds problems once you see the trick, but usually on AIME it would still takes about 5 minutes of computing after you see the trick).

I wonder if this year's AIME II was the shortest AIME ever. (exactly 2 pages)

[quote="mirebeka"]anyone know of a comprehensive solutions page like there was for the AIME I?[/quote] You can view the math jam transcript [url=http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=51]here[/url].  It has the answers and solutions to each question.

are the questions posted on the official AMC site?

[quote="jen7"]No, there were other years that had 2 pages of problems, but they had 1 page dedicated to their sponsors (I think).[/quote]Like 1998, right Jen--

And they'll never post the Questions, Sheepwarrior.  That way, they can make money.  (Not that anyone in their right mind really buys most of the problems they use now:  [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=15513]Mock-[/url]-These are so cool. --and Take a Problem Series to load up even more and with boku organization.)

Eddy Fish

[quote="Agent_00pi"]
#12: 307
Reflecting the triangle over OF; the image of E will then be on BC! Repeating this process around O, we obtain an octagon composed of 8 images of EOF. Let x be BF; then AE = 500-x. The area of the square is 810000; the area of the octagon is 4(400)(450) = 720000. The remaining area in the square is 2x(500-x) = 90000. Solving this, we get x = 250 plus or minus 50 root 7. BF is the larger of the two solutions, so we want 250 + 50 + 7 = 307.
[/quote]

Or, once you get the octagon, you can use any corner of the square: a right triangle with legs x and 500-x and hypotenuse 400.


factorial
algebra
polynomial
induction
combinatorics proposed
combinatorics

Prove the following equality. \[\frac{n!}{x(x+1)\cdots (x+n)}=\sum_{r=0}^{n}(-1)^{r}\frac{_{n}C_{r}}{x+r}\ (n=1,\ 2,\ \cdots).\]

Let $P_{r}(x) = \prod_{k = 0, k \neq r}^{n}(x+k)$.  Multiplying both sides by $\prod_{k = 0}^{n}(x+k)$ yields

$n! = \sum_{r = 0}^{n}(-1)^{r}\binom{n}{r}P_{r}(x)$.

Note that $P_{r}(-s) = 0$ for $s = 0, 1, \ldots, n, s \neq r$, and $P_{r}(-r) = (-1)^{r}r! (n-r) !$.  So for $s = 0, 1, \ldots, n,$

$\sum_{r = 0}^{n}(-1)^{r}\binom{n}{r}P_{r}(-s) = (-1)^{s}\binom{n}{s}(-1)^{s}s! (n-s)! = n!$.

But the right hand side is a polynomial of degree at most $n$, so it must be identically $n!$.

Induction also works, but it's less desireable.  Using similar notation as above let \[P_{r}^{n}(x) = \prod_{k = 0, k \neq r}^{n}(x+k)\] Note that \[P_{r}^{n}(x)=xP_{r-1}^{n-1}(x+1)\] Now using Pascals identity and the relation above we can find a recurrence for $F_{n}(x): =\sum_{r=0}^{n}(-1)^{r}\binom{n}{r}P_{r}^{n}(x)$.
\begin{eqnarray*}F_{n}(x)&=&(x+n)F_{n-1}(x)-xF_{n-1}(x+1)\\ \ &=&(x+n)(n-1)!-x(n-1)!\\ \ &=&n! \end{eqnarray*}


I learned about this website from school and here's an awesome game:
[url]http://www.notdoppler.com/theimpossiblequiz.php[/url]

It's called "The Impossible Quiz"  Post how far you get here!  I just can't get past level 42  :wallbash_red:

I rembeber that my friends emailed me the whole list of answers... but i don't have it nemore.. think i deleted it XD

Uh...I knew this for a long time...and I beat both quizzes a long time ago. Anyway, for level 42, the answer is 42, that's all I can say for a hint.  :)

Well, duh it's 42.  I don't know what I'm missing  I never understood that problem and they won't let me skip(even though I always have 1 left).  I've played it in school before and it's really fun though.

Hint: so how many 42s are there on the screen?

Everybody says that the BOGGY painting problem was difficult. Actually it clicked for me pretty quickly. It's the shooting star problem that ism stuck on

"Neigh"  Whinny and a picture of a horse.  That one was pretty easy if you ask me, as long as your in the right frame of mind.

I always got killed on problem 82, due to my bad mouse.  :P

there is impossible quiz 2, u know, but not as fun

Are you sure? I always thought the second was more fun, since you get a grade and there is more variety. The second one was also easier, though...I beat it in about one-fourth of the time it took me to beat the first.

On 42, I have like 3 lives and 3 saves... BUT THEN I DIE :(

cgyao15,
It's "blue, orange, green, green, yellow" take the first letter of each color and put it in order and get....

[quote="Berzerk"]ugh can't get past level 80 something.[/quote]

shooting star?

I hate the stupid "CHARGE UP THE LASER" level.  I've tried again and again and the time limit always kills me..............

59; just keep clicking your mouse, seriously

At the charge up lazer part, I usually finish charging the guy up when there is 2 seconds left.

You should check this out. http://i41.tinypic.com/1szlsz.jpg

[quote="meewhee009"]I finished charging that...
thing when there was 1 second left.
 :P[/quote]
I know i shouldn't be reviving....
but I finished the charging with 3 seconds left. I finally completed it !

[quote="Berzerk"][quote="andrewjjiang97"]Wow.. I died on question 109...the second to last[/quote]
The one where you make the spidermonkey poop out the arrow and click it?[/quote]

yep

BUT I JUST BEAT IT!!!!! W00T!!!


geometry
trapezoid
angle bisector
perpendicular bisector
geometry unsolved

$ \triangle ABC$ has $ \angle A \equal{} 60^\circ$ and $ \angle B \equal{} 70^\circ$.  We have $ D \in CB$ and $ E \in  CA$ such that $ AD \equal{} AB$ and $ EC \equal{} EB$.  Show that $ DE \equal{} DB$.

Let $ l$ be the angle bisector of $ \angle{AEB}$, then $ l \parallel BC$. So by $ S(l)$, suppose $ A \to A'$, then $ A'$ is on segment $ BE$ and $ AA' \perp l$, hence $ AA' \perp BC$.

Since $ AD=AB$, so $ AA'$ is the angle bisector of $ \angle{BAD}$. Since $ \angle{BAD}=40^o$, so $ \angle{A'AD}=20^o$. Because $ \angle{ABE}=20^o$, so $ AA'=BA'$. We also have that $ DA'=BA'$, so $ AA'=DA'$. So there is $ \angle{A'DA}=20^o$. Remember $ \angle{BAC}=60^o$, so $ \angle{DAC}=20^o$, which gives $ A'D \parallel AC$.

Construct $ DH \perp BC$ with $ H \in AC$. So $ HD \parallel AA'$. Since $ A'D \parallel AC$ and $ AA'=A'D$, so $ AA'DH$ is a rhombus. Next we prove that $ ED$ bisects $ \angle{ADH}$.

$ ED$ is the bisector if and only if $ \frac{HE}{AE}=\frac{DH}{DA} \Longleftrightarrow \frac{AH}{AE}-1=\frac{AA'}{AB}$. Or $ \frac{AA'}{AE}-1=\frac{AA'}{AB}$. This is equivalent to $ \frac{1}{AE}=\frac{1}{AB}+\frac{1}{AA'}$.

Take point $ A''$ on the extention of $ BE$ (not shown in the figure) such that $ \angle{AA''B}=40^o$. So we have $ AA''=AA'$. Since $ \angle{ABA''}=20^o$, so $ \angle{A''AB}=120^o$. Because $ \angle{EAB}=60^o$, so $ AE$ is the angle bisector of $ \angle{BAA''}$. Hence $ \frac{1}{AE}=\frac{1}{AB}+\frac{1}{AA''}=\frac{1}{AB}+\frac{1}{AA'}$. Now we conclude that $ ED$ is the angle bisector of $ \angle{ADH}$.

Since $ \angle{ADH}=20^o$, so $ \angle{EDA}=10^o$. So $ \angle{CED}=30^o$. Since $ \angle{C}=\angle{CBD}=50^o$, so $ \angle{CEB}=80^o \Longrightarrow \angle{DEB}=50^o = \angle{DBE}$. Hence $ BD=ED$.

Let M be the middle of the arc BC not containing A in the circle (ABC) and N the intersection of the parallel to BE through A with BM. Since angle ABM is 100 deg, it follows that triangle ABN is 20-80-80, AEBN isosceles trapezoid, with AE = BN and ADN an equilateral triangle. Hence the triangles AED and NBD are congruent, due to simmetry w.r.t. the perpendicular bisector of AN.

[i]Note:[/i] [b]If O is the centre of (ABC) - on AD, obviously - one can prove that AE = OD.[/b]

Best regards,
sunken rock

Nice solution, sunken rock! To clarify (I didn't get this for a while), N is the intersection of the parallel to A through BE and [u]BM[/u]. I don't think the circle is needed, you can just construct triangle ABN.

Thanks for your remark, I've edited it.
Actually we can give up triangle ABN, see that M (the middle of arc BC) is the symmetrical of O w.r.t. BC, therefore ODM is an isosceles triangle; due to the fact that AEO is isosceles as well ( angle CEO being 40 deg, it follows that angle AOE is 20 deg),
angle AOE = angle OAE. Since OA = OM we get the 2 triangles AEO and ODM congruent, hence OE = OD and angle DEO = 10 deg, consequently angle BED = 50 deg and DE = BD.
This way we proved my remark, AE = OD, too.

Best regards,
sunken rock


trigonometry
symmetry
geometry
trig identities
Law of Sines
angle bisector

Consider the square $ ABCD$  and a point $ E$ on the side $ AB$ , The diagonal $ AC$ intersects The segment $ DE$ at point $ P$. The perpendicular from $ P$ to $ DE$ intersects the side $ BC$ at $ F$ . show that $ EF\equal{}AE\plus{}FC$

[quote="Abdek"]Consider the square $ ABCD$  and a point $ E$ on the side $ AB$ , The diagonal $ AC$ intersects The segment $ DE$ at point $ P$. The perpendicular from $ P$ to $ DE$ intersects the side $ BC$ at $ F$ . show that $ EF \equal{} AE \plus{} FC$[/quote]
Wlog $ AB\equal{}BC\equal{}CD\equal{}DA\equal{}1$.
Let $ u \equal{} \angle ADE$, $ x \equal{} AE \equal{} \tan u$, $ y \equal{} FC$.

We need to prove $ AE\plus{}FC \equal{} EF \equal{} \sqrt{EB^2 \plus{} FB^2} \equal{} \sqrt{(1\minus{}x)^2\plus{}(1\minus{}y)^2} \iff$
$ (x\plus{}y)^2 \equal{} (1\minus{}x)^2\plus{}(1\minus{}y)^2 \iff$
$ xy\plus{}x\plus{}y \equal{} 1$

Law of sines: $ \frac{AP}{\sin 90^\circ\minus{}u} \equal{} \frac{AE}{\sin 45^\circ\plus{}u} \iff$
$ AP \equal{} \frac{\sin u}{\sin 45^\circ\plus{}u} \equal{} \frac{\sqrt{2}x}{x\plus{}1}$.
$ AP\plus{}PC\equal{}AC\equal{}\sqrt{2} \iff PC \equal{} \sqrt{2} \left ( 1 \minus{} \frac{x}{x\plus{}1} \right ) \equal{} \frac{\sqrt{2}}{x\plus{}1}$

Law of sines: $ \frac{FC}{\sin 45^\circ \minus{} u} \equal{} \frac{PC}{\sin 90^\circ \plus{} u} \iff$
$ y \equal{} FC \equal{} \frac{\sin 45^\circ \minus{} u}{\cos u} \frac{\sqrt{2}}{x\plus{}1} \equal{} \frac{x\minus{}1}{x\plus{}1} \iff$

$ xy\plus{}x\plus{}y\equal{}1$, and we are done :)

See that $ DPFC$ is cyclic; from $ CP$ angle bisector ($ AC$ diagonal), $ PF \equal{} PD \ ( \ 1 \ )$.
See that $ PB \equal{} PD\ ( \ 2 \ )$ by symmetry, so $ PB \equal{} PF$, hence $ \angle PBF \equal{} \angle PFB\ ( \ 3 \ )$. From $ PEBF$ cyclic we get $ \angle PEA \equal{} \angle PFB \ ( \ 4 \ )$ and $ \angle PEF \equal{} \angle PBF \ ( \ 5 \ )$, from (3), (4) and (5) we get $ \angle PEA \equal{} \angle PEF$, i.e. $ DE$ is the angle bisector of $ \angle AEF$, consequently $ D$ is the excenter of $ \triangle BEF$ and $ Q$, the foot of the perpendicular from $ D$ to $ EF$ makes $ QE\equal{}AE$ and $ QF \equal{} FC$, as tangents from $ E$ and $ F$ to that circle, done.

Best regards,
sunken rock

Dear Mathlinkers,
an article concerning "Geometric miniatures on a square" with original proofs has been put on my website
http://perso.orange.fr/jl.ayme    vol. 7 Miniatures géométriques sur un carré  p. 30
Sincerely
Jean-Louis


function calculus quadratics algebra polynomial

[quote="iamhe"]How can you find how many negative or positive roots a function has?
Example:
How many negative real roots does the function y=2x^4-x^3+4x^2-5x+3 have?
4
3
2
1
0[/quote]

Descartes' rule of signs  :wink:

For negative x:

x^(2n)>0
x^(2n+1)<0 thus -x^(2n+1)>0

So, going termwise, each term is positive for negative x, hence their sum is positive, and thus non-zero. So, no roots.

But doesn't Descartes rule say the number of sign changes or minus 2, or 4, or 6...?
IDK, maybe I'm misinformed... :?:

Yes it does. But $ y(\minus{}x) \equal{} 2(\minus{}x)^4\minus{}(\minus{}x)^3\plus{}4(\minus{}x)^2\minus{}5(\minus{}x)\plus{}3 \equal{} 2x^4 \plus{} x^3 \plus{} 4x^2 \plus{} 5x \plus{} 3$ has no sign changes anyway.

I didn't notice that you wanted negative [b]and[/b] positive roots. Descartes's rule of signs works for the negative numbers, but not for the positive ones. Namely, it doesn't give you a definite answer, but only a bunch of possible answers. Here is a complete solution for this case.

[hide="With very basic calculus"]

y' is a cubic, so it either has 1, 2, or 3 roots.

Checking, y'' is always positive (this is a quadratic, we can check the discriminant), so y' is always increasing and there is one root. Since the leading coefficient in x of y is positive, the single extremum represented by the single root of y' is a minimum.

Since there is a single local minimum, we can have three scenarios:
- the minimum is above the x axis, so no roots
- the minimum is on the x axis, one root
- the minimum is below the x axis, two roots

Note that where y'=0, we can use this for simplification:
y has the same sign as 32y = 32y+(1-8x)y' = 61x^2 - 112 x + 91
*The above only applies at the relative minimum!

The discriminant of that quadratic is negative, so no roots, and since the leading coefficient is 61, that quadratic is always positive. Thus, the minimum value of y is positive, and no real roots.
[/hide]

[quote="iamhe"]How many real roots does the function $ y \equal{} 2x^4 \minus{} x^3 \plus{} 4x^2 \minus{} 5x \plus{} 3$ have?[/quote]$ y \equal{} \frac {x^2(4x\minus{} 1)^2}{8} \plus{} \frac {(31x \minus{} 20)^2}{248} \plus{} \frac {43}{31} > 0.$

Descartes' Rule of Signs state that:

 :arrow: The number of positive  REAL roots of $ P(x)$ is equal to the number of variations of signs of $ P(x)$ or is less than by a positive even integer.

 :arrow: The number of negative REAL roots of $ P(x)$ is equal to the number of variations of signs of $ P(\minus{}x)$ or is less than by a positive even integer.

but, take for instance, if the polynomial is of degree $ 5$ and the number of positive and negative is $ 2$ and $ 1$, respectively, then you must have $ 2$ imaginary roots by the Fundamental Theorem of Algebra.

I have a question:  What is the REAL Fundamental Theorem of Algebra? The one which states that "Every polynomial function of degree $ n\geq 1$ has exactly n complex zeros" or the one which states that "Every polynomial of degree $ n \geq 1$, with complex coefficients,  has [u]at least one complex zero[\u]?"

Can you see this Graph:
[img]http://www.fileden.com/files/2008/8/1/2029422/adad.JPG[/img]

Can you prove that $ y\equal{}2x^4\minus{}x^3\plus{}4x^2\minus{}5x\plus{}3>0$ ? :oops:

Derivate that and see ween the derivate is equals to zero, and there is the minimus of the funtion gaven... simple :P

And you could just post once :mad:


The product of two consecutive page numbers is 18,360. What is the sum of the two page numbers?

Also, could someone possibly explain Kouichi Nakagawa's method?  I don't really get it...  Someone taught me something similar to this (I think), but I'm not sure how it relates to this.

This calculation is a calculation method of the extraction of square root by the calculation on a paper.

Japanese explanation is possible, but explanation in English is difficult because ability for language study is not high.
So Japanese and English introduce the URL that they seem to be related to for the time being.

[url=http://ja.wikipedia.org/wiki/%E7%AD%86%E7%AE%97]筆算[/url] - Wikipedia (jp)

[url=http://ja.wikipedia.org/wiki/%E9%96%8B%E5%B9%B3%E6%B3%95]開平法[/url] - Wikipedia (jp)
[url=http://en.wikipedia.org/wiki/Methods_of_computing_square_roots]Methods of computing square roots[/url] - Wikipedia (en)

[b]Translation[/b]
筆算:calculation on a paper
開平法:extraction of square root / Methods of computing square roots

Although I like Kouichi Nakagawa's method much better, we can also write an equation where $x$ = the smaller integer.

\[x(x+1)=18,360\]
\[x^2+x=18,360\]
\[x^2+x-18,360=0\]
\[(x+136)(x-135)=0\]

So the smaller number is $135$ and the larger number is $136$, therefore, the sum of the two page numbers is $135+136=\boxed{271}$.

what you just showed is basically finding it by hand, unless you use the quadradic formula which would be very long. I like Kouichi Nakagawa's method just I need some more explanation.

You can sort of estimate the answer, because $140^2 = 19600$, and $130^2 = 16900$, and you have to have a multiple of 5 times a multiple of 2.


inequalities
logarithms
inequalities unsolved

Let $a=3x$, $b=\frac{8y}{x}$, $c=\frac{9z}{y}$ and $d=\frac{6}{z}$.
Then $\sqrt[4]{abcd}=6$ and we get \[\frac{1}{(1+a)(1+b)(1+c)(1+d)}\le\left(\frac{1}{1+\sqrt[4]{abcd}}\right)^{4}=\frac{1}{7^{4}}.\] The inequality comes from applying Jensen's Inequality, with $f(x)=\log\left(\frac{1}{1+e^{x}}\right)$, which is everywhere concave.

I could not understand the jensen part. I know the jensen, but how did that inequality came from jensen? If we add logs, then it will be in multiplication form. But what about e?

Sorry, I probably should have spelled that out a bit more.

By Jensen, $f(\log(a))+f(\log(b))+f(\log(c))+f(\log(d))\le 4f\left((\frac{\log(a)+\log(b)+\log(c)+\log(d)}{4}\right)$, so $\log\left(\frac{1}{1+a}\right)+\log\left(\frac{1}{1+b}\right)+\log\left(\frac{1}{1+c}\right)+\log\left(\frac{1}{1+d}\right)=\log\left(\frac{1}{(1+a)(1+b)(1+c)(1+d)}\right)\le4\log\left(\frac{1}{1+\sqrt[4]{abcd}}\right)$.

Raising $e$ to the power of each side (since $e^{x}$ is increasing) gives the desired inequality.