Γειά σας παιδιά! Χαίρομαι που έμαθα για το Forum και σας συγχαίρω για την προσπάθεια που κάνετε. Ελπίζω κάποια στιγμή να γνωρίσω κάποιους   απο εσάς απο κοντά.  :lol:

Καλώς όρισες galois01!  :)

PS (σε όλους):
[b]Παρακαλώ[/b] [u]μην[/u] ανοίγετε ανούσια νέα threads, γιατί ...θα χάσουμε τη μπάλα! Μπορείτε να χρησιμοποιείτε το thread "Μέλη του forum".

Κάτι  μου  λέει  φίλε  Mostel  ότι  η  μπάλα  έχει  χαθεί  εδώ  και  καιρό.......... Είχα  να  μπώ  καιρό  στο  ελληνικό  φόρουμ και  παρατηρώ  μία  νέκρα .......... γιατί  ρε  παιδιά  :?: ............Και  όλα  έδειχναν  οκ  ...  μήπως  μετά  την  Βαλκανιάδα  δεν  λύνει  κανένας  ασκήσεις . :?: ......Οκ   καταλαβαίνω  είχατε  εξετάσεις  (Πανελλήνιες  ή  και  ενδοσχολικές )  ..........είναι  κρίμα  να  έχουμε  τα μέσα  και  να  μην  τα  εκμεταλλευόμαστε  :wink: ..... θα  μου  πείς  ποιος  μιλάει κάποιος  που  έχει  να  γράψει  χρόνια  στο  φόρουμ  :blush: ........ Και  κάτι  άλλο  ρε  παιδιά  δεν  το  λέω   κακοπροαίρετα ( Τουλάχιστον  οι  περισσότεροι εδώ  μέσα  ξέρετε  ότι  δεν  γράφω  για  να  προκαλέσω )....  το  φόρουμ  αυτό  είναι  φόρουμ  ολυμπιάδων ( η  γενικά  δύσκολων  μαθηματικών  προβλημάτων )   και  όχι  Α,Β,Γ Λυκείου ....  ( έλεος )  δεν  είναι  δα  και  τόσο  δύσκολα  τα   θέματα  των  πανελληνιών .

[quote="gdeskos"] Και  κάτι  άλλο  ρε  παιδιά  δεν  το  λέω   κακοπροαίρετα ( Τουλάχιστον  οι  περισσότεροι εδώ  μέσα  ξέρετε  ότι  δεν  γράφω  για  να  προκαλέσω )....  το  φόρουμ  αυτό  είναι  φόρουμ  ολυμπιάδων ( η  γενικά  δύσκολων  μαθηματικών  προβλημάτων )   και  όχι  Α,Β,Γ Λυκείου ....  ( έλεος )  δεν  είναι  δα  και  τόσο  δύσκολα  τα   θέματα  των  πανελληνιών .[/quote]

Φίλε gdesko, ίσως έχεις δίκιο για αυτό που λες...
Αφενός όμως επειδή είναι (μάλλον) το μόνο ελληνικό forum που ασχολείται με τα μαθηματικά θεωρώ ότι είναι ωφέλιμο να τίθενται και θέματα λίγο χαμηλότερου επιπέδου απ' ότι των Ολυμπιάδων, κι αφετέρου έτσι το φόρουμ προσελκύει μαθητές που σίγουρα αγαπάνε τα μαθηματικά κι απλώς δεν έχουν ασχοληθεί με τα "δύσκολα" (κάποια στιγμή θα γίνει κι αυτό :wink: ...αρκεί να τους δοθεί η ευκαιρία)

Ρε παιδιά ότι και να πούμε ο Γιώργης έχει δίκαιο .
Σκεφτείτε και κάτι άλλο .
Οτιδήποτε έχει ξεκινήσει με ενθουσιασμό και ότι όλοι θα βοηθήσουν και ότι όλοι θα είναι μέσα έχει πάει στράφι . (βλ πχ το βιβλίο  :o )

Σοβαρή άσκηση ολυμπιάδων έχει πάρα πολύ καιρό να συζητηθεί εδώ .

Φίλε  Bilot4  σίγουρα  το  φόρουμ  είναι  και  για  μαθητές  που  απλά  ασχολούνται  με  τα μαθηματικά  και  όχι  με  τις ολυμπιάδες
        Αλλά  υπάρχει  και  αλλά .............. εδώ  τα  πράγματα  έχουν  ώς  εξής  .Το  βιβλίο  από  οτι  βλέπω  (  ιδέα  όχι  απλά  τρομερή αλλά  και  μοναδική  στα χρονικά )  δεν  προχωράει.......... Ασκήσεις  δύσκολες  δεν  μπαίνουν  ......... και  γενικά  δεν  υπάρχουν  δημοσιεύσεις.... δηλαδή  όχι  απλά  μπαίνουν  εύκολες  ασκήσεις  δεν  μπαίνουν  και  δύσκολες .........  Δέν  ήθελα  να  μπώ  σε  συγκρίσεις  αλλά  θα  τις  κάνω  και  αυτές  μονάχα  με  καλή  διάθεση  (  το  τονίζω )  .........Πέρσι  για  παράδειγμα  δεν  προλαβαίναμε  να  λύσουμε  μια  άσκηση  και  πεταγώντουσαν  10  ....  Τώρα  τίποτα .Δεν  νομίζω  να  τελείωσαν  οι  ασκήσεις ... :D   .........   Λοιπόν  εγώ  νομίζω  οτι  απλά  είμαστε  σε  περίοδο  ξυρασίας
Το  βιβλίο  είναι  μεγάλη  δουλεία  και  θέλει  ενασχόληση ...Εγώ   προτείνω    να  ασχοληθεί  το  φόρουμ  με  ασκήσεις  και  τεχνικές  ολυμπιάδων  (  Για  τους  μαθητές ) και  με  δύσκολες  το  τονίζω  ασκήσεις  Ανάλυσης  άλγεβρας ( για τους  φοιτητές ...  με  ασκήσεις  όπως  IMC  και  Putnam ) δεν  νομίζω  ότι  τα  μέλη  του  δεν  μπορούν  να    λύσουν  τέτοιες  ασκήσεις  απλά  κάτι  τους  απομάκρυνε .......Μαζί  και  εμένα ....

         Σιλουανέ  ............ εσύ  τώρα  που  είσαι  ομάδα  πρέπει  να  βοηθήσεις  λίγο  ποιο  πολύ  το  φόρουμ  :wink:

[quote="gdeskos"]Σιλουανέ  ............ εσύ  τώρα  που  είσαι  ομάδα  πρέπει  να  βοηθήσεις  λίγο  ποιο  πολύ  το  φόρουμ  :wink:[/quote]

Ξέρεις ότι πάντα βοηθούσα . Αλλά δεν πρόκειται να είμαι μονίμως η τροφή άλλων που δεν μπαίνουν στο φόρουμ , ούτε ο βλάκας που θα προσπαθεί μόνος του για να τα παίρνουν έτοιμα και να τα διαβάζουν άλλοι . :ninja:

Παιδιά η γνώμη μου είναι η εξής .  Σαφώς και δεν πρέπει να πέσει το επίπεδο των ασκήσεων και σίγουρα πρέπει συχνά να παρουσιάζονται ασκήσεις υψηλού επιπέδου. Μονο και μόνο με τι διακίνηση των ιδεών που θα προκύψει θα βγαίνουμε όλοι κερδισμένοι. Από την άλλη όμως δεν είναι άσχημο να υπάρχουν και πιό απλές ασκήσεις οι οποίες θα δώσουν ώθηση σε μαθητές ( και γιατί όχι και σε εμάς )... Πιστεύω ότι δεν πρέπει να γυρίζουμε την πλάτη σε κανένα...

Φιλικά  Βαγγέλης .Μ.

[quote="silouan"]

Ξέρεις ότι πάντα βοηθούσα . Αλλά δεν πρόκειται να είμαι μονίμως η τροφή άλλων που δεν μπαίνουν στο φόρουμ , ούτε ο βλάκας που θα προσπαθεί μόνος του για να τα παίρνουν έτοιμα και να τα διαβάζουν άλλοι . :ninja:[/quote]

  ΑΠΛΑ   ΤΟ  ΜΟΝΟ  ΠΟΥ  ΕΧΩ  ΝΑ  ΠΩ   ΕΧΕΙΣ  Δ Ι Κ Ι Ο   :(

Θα συμφωνήσω με τον vanglelismak .Στο forum καλό είναι να βάζουμε ασκήσεις για τις τάξεις του λυκείου ,πιστεύω ωστόσο ότι το πιο σημαντικό είναι να μπαίνουν ασκήσεις από ολυμπιάδες ώστε να μπορούν να βοηθηθούν όσοι επιθυμούν να συμμετέχουν με αξιώσεις στους διαγωνισμούς της Ε.Μ.Ε ,αλλά και όσοι θέλουν να κάνουν κάτι πέρα από τις ασκήσεις του σχολείου.Για το λόγο αυτό θα πρέπει να συμμετέχουμε ολοι,όσο μπορεί ο καθένας βάζοντας κάποια θέματα ολυμπιάδων και να μην προσπαθεί μόνος του ο silouan  :!:

φιλοι μου όλα αυτά καλά είναι αλλα ένα ξεχωριστό μέρος για τια ασκήσεις Α,Β,Γ λυκείου κι πανελλαδικές δεν έιναι κακό επίσης θα ήθελα να προσθέσω την συγνώμη μου για την απραξία μου κι να πώ  ότι αξίζει να βλέπουμε κι άλλα θέματα πανελάδικών όχι γιατι είναι τόσο δύσκολα αλλα γιατί είναι "δύσκολα" λόγω της κατάστασης ξέρετε τι εννοώ  :wink:
χαιρετώ

Θεωρώ ότι ο κ gdeskos καλά θα κάνει, αν θέλει να ανανεώσει το forum, να αρχίσει να προτείνει ασκήσεις, παρά να κάνει διαπιστώσεις του τύπου "που καταντήσαμε" και "έτσι έχουν τα πράγματα", γιατί με τις τελευταίες δεν καταφέρνει τίποτα (όπως και κανένας άλλος). Αν δεν μπουν δυο άτομα να αρχίσουν να προτείνουν  θέματα, ασκήσεις, προβλήματα κλπ, όσο και να λέμε ότι δεν είναι καλά τα πράγματα δεν θα αλλάξει κάτι. Επιπρόσθετα, οι ασκήσεις που απευθύνονται στους μαθητές Α, Β, Γ Λυκείου σε καμία περίπτωση δεν είναι αρνητικό στοχείο, αφού κάνει το forum περισσότερο προσβάσιμο σε μεγαλύτερο αριθμό ατόμων. Στην ουσία αυτά τα threads ήταν που εμφάνισαν και την μοναδική κίνηση του Forum τον τελευταίο καιρό.

Σε ότι αφορά το βιβλίο, ωραία ιδέα, δεν διαφωνώ, αλλά μάλλον την υποδέχθηκαν οι περισσότεροι με πιο πολύ εμθουσιασμό, παρά συνειδητοποίησαν τι απαιτήσεις θα έχει αυτό το πράγμα. Η συγγραφή του βιβλίου δεν είναι ούτε εύκολη υπόθεση, ούτε "άντε θα προτείνουμε δυο ασκήσεις και τελείωσε". Πρέπει να βρεθούν κάποια άτομα τα οποία θα ασχοληθούν σ ο β α ρ ά και υ π ε ύ θ υ ν α, γνωρίζοντας ότι έχουν να κάνουν πολλή δουλειά (αφού πρώτα ξεπεραστεί ο αρχικός ενθουσιασμός. Ειδάλλως βιβλίο δεν γίνεται

ΥΓ. Σιλουανέ, εσύ τώρα που είσαι ομάδα (κατά τη γνώμη μου) κοίταξε να ασχοληθείς με αυτό το θέμα (την ολυμπιάδα) και άσε το forum προς το παρόν. Είναι θέμα προτεραιότητας...

[quote="alexandros_plus"]Θεωρώ ότι ο κ gdeskos καλά θα κάνει, αν θέλει να ανανεώσει το forum, να αρχίσει να προτείνει ασκήσεις, παρά να κάνει διαπιστώσεις του τύπου "που καταντήσαμε" και "έτσι έχουν τα πράγματα", γιατί με τις τελευταίες δεν καταφέρνει τίποτα (όπως και κανένας άλλος).  quote]

  Είπα  πώς  ότι  λέω....  το  λέω  καλοπροαίρετα  δεν  κάνω  κρητική στο  φόρουμ  και  στα  μέλη  του !  Ακόμη  μόνος  μου  είπα  ότι  φταιω  και  εγώ  για  την  κατάσταση ( δές  τα  παραπάνω  ποστ  μου ).......... Όταν  θα  γίνεις  φοιτητής  ούτε  εσύ  θα  έχεις  πολ΄υ  χρόνο
Δέν  μίλησα  υποτιμητικά , όπως  έκανες  εσύ  , (ο  κ gdeskos καλά  θα  κάνει  ...) και  γι  αυτό  θα  σου  πώ  κάτι κακοπροαίρετο αλλά  αληθινό  ........ Βασικά  από  οτι  ξέρω  έχεις  γύρω  στα  πενήντα post  τα  οποία  τα  μισά  είναι  σε  τόπικ  του  στύλ  αστειάκια , μέλη  του  φόρουμ  ..........  Και  οι  ασκήσεις  είναι  της  τάξης ( θέμα  τεστ  στο  σχολείο ) και  γι  αυτό μάλλον πίστεψες  ότι σε  έθιξα... . Λοιπόν  τέλος   η  ειρωνία  από  την  μεριά  μου  (  καλά  θα  κάνεις  να  κάνεις  και  εσύ  το  ίδιο ) . Τώρα  θα  σου  πώ  κάτι  σαν  συμβουλή  σαν  παλαιότερος   στο  φόρουμ .

1) Το  να  προτείνεις  δεν  έχει  ουσία  αν  δεν  λύνεις  να  το  ξέρεις  .....
2) Αν  θές  προτείνω  ασκήσεις  δύσκολες  και  βάζω  και  τις  λύσεις  αλλά  το  να  λύνεις  δικιές  σου  ασκήσεις  είναι  λίγο  συκέ :wink:
3) Όταν  πρωτοφτιάχτηκε  αυτό  το  φόρουμ  το  ξέρανε  5,6  άτομα  όλοι  σχεδόν  παιδία  από  μαθηματικές  ολυμπιαδες ... Οι  ασκήσεις  που  πρότειναν  αυτοί  οι  5  ήταν  ασκήσεις  που  μία  διάβαζες  10  έλυνες  μετά ...  Το  φόρουμ  τράβηξε  τον  κόσμο  επειδή  θέματα  όπως  οι  Ολυμπιάδες   είναι  αγνωστα  στο  "Ελλάντα"   .....  Σε  έναν  μέσο  μαθηματικό  αν  πας  και μιλήσεις  για  ολυμπιάδες  θα  σου  πεί :  τι  είναι  αυτά  μην  ασχολείσαι  και  είναι  δύσκολα  και  θα  απαγοητευτείς ........Αν  τον  βάλεις  δε  να  λύσει .... δύσκολα  θα  σου  λύσει  κανα  θέμα  του  Ευκλείδη ..........
(  Ευτυχώς  όμως  υπάρχουν  και  άξιοι  μαθηματικοί  που  δίνουν  και  την  ζωή  τους  για  τα  πραγματικά  μαθηματικά ....για  να  μην  τα  ισοπεδώνουμε
όλα ) Ασκησούλες  Λυκείου   βρίσκεις  από  την  άλλη  εκοτομύρια  με  το  τσουβάλι  ......Αν  θές  μπορω  να  σου  στείλω  καμία  20αρια  βιβλία δεν  είναι  ανάγκη  να  στραφεί  το  φόρουμ  εκεί .  Καταλήγω  λοιπόν  ότι  το  φόρουμ  αυτο  είναι  για  τις  ολυμπιάδες και  όχι   για  απλές  ασκησούλες  .Αν  ωστόσο  θές  να  βάλεις  τέτοιες  ασκήσεις  βάλε  αλλά   από  ότι  βλέπω  μονάχα  τέτοιες  μπαίνουν  (  εξαιρούνται  οι  γεωμετρίες  Mostel  και  οι  λύσεις  r_boris )

  Υ.Σ.    Ίσως  έχεις  κάπου  κάποιο  δίκιο  για  την  αποχή  μου  γι  αυτό  θα  βάλω  μία  άσκηση   και  ότι  γίνει  .................

Αξίζουν μπράβο μερικοί για τον χρόνο που αφιερώνουν στο να βρουν και να βάλουν άσκηση αλλά δεν είναι κανένας υποχρεωμένος να βάζει ασκήσεις .
Θα ήταν καλό να μας δώσετε και έναν ορισμό της δύσκολης άσκησης γιατί κάποιοι έχουν ασχοληθεί περισσότερο με θέματα ολυμπιάδων και κάποιοι λιγότερο ( εγώ είμαι στην δεύτερη κατηγορία) .

Υποστηρίζω απόλυτα αυτή την άποψη :

[quote]Αφενός όμως επειδή είναι (μάλλον) το μόνο ελληνικό forum που ασχολείται με τα μαθηματικά θεωρώ ότι είναι ωφέλιμο να τίθενται και θέματα λίγο χαμηλότερου επιπέδου απ' ότι των Ολυμπιάδων, κι αφετέρου έτσι το φόρουμ προσελκύει μαθητές που σίγουρα αγαπάνε τα μαθηματικά κι απλώς δεν έχουν ασχοληθεί με τα "δύσκολα" (κάποια στιγμή θα γίνει κι αυτό  ...αρκεί να τους δοθεί η ευκαιρία)[/quote]

Ένα βασικό πρόβλημα που ήταν η ακαταστασία στο forum πιστεύω λύθηκε μιας κι έχουμε moderator που τον εμπιστεύομαι απόλυτα .

Αγαπητοί φίλοι,

έτσι δε βγαίνει τίποτα!

Ας αρχίσουμε όλοι από την αρχή, όσο μπορεί ο καθένας, για να επαναφέρουμε το φόρουμ στην παλιά του αίγλη.

Καλό καλοκαίρι, καλά μπανάκια και ...καλές εμπνεύσεις !

 :D

PS: Το "Ελληνικά" ποτέ δε βγήκε εκτός παιχνιδιού, απλώς ...είναι σε time out!  :)

Μονο δυο λόγια γι' αυτά που γράφει ο gdeskos (δεν θα αναλύσω το τι σήμαινε το "κύριος" και γιατί το έβαλα). Αρχίζω να απαντάω ανάποδα σε αυτά που λες. Το forum (όπως λ.χ. και μια κοινωνία) αναπότρεπτα αλλάζει καθώς αλλάζουν οι άνθρωποι από τους οποίους αποτελείται. Θετικό ή αρνητικό το φαινόμενο δεν θα προσπαθήσω να ερευνήσω αυτό. Πάντως νομίζω ότι τα stickies με τις ασκήσεις Α, Β, Γ Λυκείου οι περισσότεροι τις υποδέχθηκαν με καλό μάτι. Αν θες μπορείς να προτείνεις τα 20 βιβλία σου στον moderator του ελληνικού forum, αλλά δεν θα έλεγα ότι θα κέρδιζες κάτι.

Όντως έχω 54 Post από τα οποία ζήτημα αν, όχι τα μισά, αλλά τα 10 έχουν καθαρά μαθηματικό περιεχόμενο, το οποίο μπορεί να το κρίνει όποιος θέλει. Δες όμως και τα δικά σου post, όχι μόνο των άλλων και προαπάθησε να τα  κρίνεις κι αυτά. Επίσης είναι δικαίωμα του καθενός να διαβάζει ότι θέλει, να δημοσιευει ή να μη δημοσιευει, να λύνει ή να μή λύνει, να προτείνει ή να μην προτείνει.

Απλά λέω ότι αν θες να δεις αλλάγή στο forum μπορείς κάλλιστα να την κάνεις, αρχίζοντας να δημοσιευεις το περιεχόμενο που θες. Μόνο μην κολλάς σε πράγματα του τύπου "εμείς οι παλαιότεροι, οι έμπειροι, οι έξυπνοι που ξέρουμε κι εσείς οι νέοι, οι άσχετοι που δεν ξέρετε πώς λειτουργεί το forum μας", γιατί το forum δεν ανήκει σε κανέναν.

ΥΓ Προσωπικά δεν με έθιξες καθόλου. Αυτό το περιεχόμενο ήθελα να δημοσιευσω και αυτό δημοσίευσα. Ωστόσο αυτό που έκανες δεν μπορείς να αρνηθείς ότι όντως ήταν "κρητική" (sic) (και το αν ήταν καλοπροαίρετη/κακοπροαίρετη, εποικοδομητική/μη εποικοδομητική δεν με αφορά και δεν το κρίνω).

Αυτό ειχα να πω και ας σχολιασθεί όπως ο καθένας νομίζει ότι πρέπει.

παιδια γεια σας , εχω καιρο να μιλησω  αλλα βλεποντας την συζητηση ειπα να παρεμβω. απλα θελω να να σας πω οτι πρεπει να συνεχισετε να ασχοληστε με αυτο το μοναδικο φορουμ .δεν θα κανετε πουθενα τετοιες συζητησεις ουτε τετοιο περιβαλλον για μαθηματικα θα βρειτε.
το οτι δεν μπαινουν  ή οτι δεν μιλουν καποιοι δεν σημαινει αδιαφορια αλλα ισως κατι αλλο πιο συναισθηματικο και βαθυ,. προσωπικα πιστευω οτι καποιοι ενοιωσαν οτι δεν μπορουν να προσφερουν ποια στην κοινοτητα μας,ή απογοητευτηκαν  απο καποια περιστατικα
γι αυτο και αποχωρισα και εγω ,
μπορει να μην μπαινω 8 ωρες καθε μερα πια, αλλα παρακολουθω και ειμαι περιφανως για ολους εδω μεσα ...
καλη συνεχεια ...

ειδικους χαιρετισμους σε mostel και Ramanujan

Παιδιά όντως καλά τα μέλη του φόρουμ και τα αστειάκια αλλά δεν είναι αυτοσκοπός. Ας βάζουμε πού και πού κανένα, ας μαθαίνουμε και τα νέα μέλη αλλά κυρίως ας δημοσιεύσουμε ασκήσεις.

Αυτές οι συζητήσεις για το stfu άσκηση του μήνα και τις γεωμετρίες του Στέλιου ήταν πολύ καλές. Ας συνεχίσουμε έτσι. Όλοι όσοι γράφουν εδώ έχουν υψηλό επίπεδο και μερικοί μάλιστα είναι άπαικτοι οπότε δε λείπει τίποτα από το φόρουμ. Λίγη όρεξη μόνο. Θα προσπαθήσω κι εγώ να συμβάλλω περισσότερο.

Φιλικά,
Κώστας

Να και μια άσκηση και απο ΄μένα που είμαι νέος...

Πιθανώς κάποιοι να την έχουν ξανα δεί. Αν όχι αξίζει να την προσπαθήσετε γιατί δεν είναι τόσο εύκολη όσο φαίνεται στην αρχή.

ΑΣΚΗΣΗ ΓΕΩΜΕΤΡΙΑΣ.

Δίνεται ισοσκελές τρίγωνο ΑΒΓ   ( ΑΒ=ΑΓ ) με  Α = 20 μοίρες. Στην προέκταση της ΑΓ ( προς το μέρος του Γ )παίρνουμε σημείο Ε τέτοιο ώστε η γωνία ΓΒΕ = 30 μοίρες. Η κάθετη από το Γ στη ΒΕ τέμνει την προέκταση της ΑΒ στο Δ. Αν η ΔΕ τέμνει την προέκταση της ΒΓ στο Ζ , να βρεθεί ( σε μοίρες ) η γωνία ΒΖΔ.

Με ενδιαφέρει κάθε λύση με οποιαδήποτε στοιχεία θεωρίας...

Φιλικά Βαγγέλης Μακ.

Φίλε Βαγγέλη,

η άσκηση έχει ξαναπροταθεί, από εμένα. Τη λύση την έχει δώσει και ο κύριος Στεργίου.

Δες εδώ:

http://www.artofproblemsolving.com/Forum/viewtopic.php?mode=attach&id=7684

Και γενικά το topic:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=137279

Φίλε  mostel  δεν το γνώριζα... μιας και είμαι νεο μέλος και δεν έχω παρακολουθήσει τι έχει γίνει πιό πρίν.

Πραγματικά με ενδιαφέρει όμως να δω μια συλλογή από λύσεις...
Με την πρώτη ευκαιρία θα σας προτείνω και εγώ μια λύση  ( βλέπετε δεν είμαι εξοικειωμένος ακόμα με το να κάνω  posts  από τη μια και από την άλλη δεν έχω πολύ χρόνο )

Αγαπητοί φίλοι,
 είχα πρόβλημα με το connex στη ΧΑλκίδα και δεν μπόρεσα  να μπω νωρίτερα.Διάβασα όμως τώρα την αλληλογραφία και τους προβληματισμούς σας με προσοχή .
   [color=blue]α) [/color]Είναι ήδη  τέλη Ιουλίου και όλα τα μέλη είναι κουρασμένα.Δεν χρειάζεται υπερένταση αυτόν τον καιρό. Καλοκαίρι είναι και αξίζει να ξεκουραστείτε.
   [color=blue]β)[/color] Οι φοιτητές και μέλη του φόρουμ είναι σε περίοδο φόρτου μελέτης. Πέφτει πολύ ύλη μαζεμένη και πρέπει να δώσουν προτεραιότητα στις επικείμενες εξετάσεις.
 [color=blue]  γ)  [/color]Η ολυμπιακή χρονιά έχει τελειώσει για τους περισσότερους. Τα μέλη των ομάδων μας αυτόν τον καιρό έχουν πολύ διάβασμα.Πρέπει να λύσουν δύσκολες ασκήσεις , να διαβάσουν τα θέματα πολλών και δύσκολων διαγωνισμών που προηγήθηκαν , να διευρύνουν τις θεωρητικές τους γνώσεις κλπ. Ο χρόνος γι αυτούς είναι πολύτιμος και καλό  είναι σε αυτή τη φάση να ασχοληθούν κατά το δοκούν. Έτσι και αλλιώς αυτή τη στιγμή στο ελληνικό φόρουμ δεν έχουμε κάποιον που να μπορεί να απαντήσει μέσα σε 5 ώρες σε οποιαδήποτε απορία ή πρόβλημα κάποιου από τα παιδιά που είναι στην ομάδα και χρειάζεται βοήθεια.Το να γράφουν λοιπόν ασκήσεις και να περιμένουν λύση δεν τους βοηθάει σε αυτή τη φάση.Το να βάζουν ασκήσεις που ξέρουν τη λύση , πάλι δε βοηθάει κανέναν , διότι ο καθέναςν αχολείται με εκατοντάδες προβλήματα που έχει ήδη συναντήσει και περιμένουν λύση.
  [color=blue] δ) [/color]Για το βιβλίο δεν υπάρχει βιασύνη. Θα προχωρήσουμε από το φθινόπωρο πιο δυναμικά. Ας μαζέψει ο καθένας ότι καλό βρει και από το Σεπτέμβρη το βάζει στο αντίστοιχο κεφάλαιο. Μη νομίζετε ότι το υλικό που έχουμε είναι λίγο. Απλά θέλουμε και άλλα προβλήματα για να επιλέξουμε τα πιο γενικά από αυτά.Η ύλη για το βιβλίο θα μαζευτεί γρήγορα. Αλλού είναι τα δύσκολα, αλλά για τις δυσκολίες θα σας μιλήσω όταν ήδη θα τις ... έχουμε ξεπεράσει , για να μην απογοητευτούν όσοι δε γνωρίζουν τις διαδικασίες.
  Λοιπόν , ας κρατήσουμε την αισιοδοξία μας και το μεράκι μας άσβεστα και ας μην οδηγηθούμε σε αρνητική στάση επειδή ο ρυθμός έχει πέσει. ΑΥτό ήταν αναμενόμενο .
 Εκτός αυτού είχα και γω εμπλοκή με το βαθμολογικό που για ένα μήνα με είχε καθηλωμένο. Ας είμαστε λίγο πιο χαλαροί τώρα. Μόλις γυρίσουν και τα παιδιά των ομάδων , θα ανατροφοδοτήσουν το φόρουμ με νέους προβληματισμούς και ασκήσεις. Ειδικά αν  αποδώσουν σύμφωνα με τις δυνατότητές τους και είναι ευχαριστημένοι , τότε όρεξη να έχετε να διαβάζετε ή να λύνετε ασκήσεις!Με άλλα λόγια αν πιάσουν και κανένα .. τιμημένο μετάλλιο , τότε με τρία άτομα το βιβλίο θα πάει για δεύτερο τόμο !
 ΑΥτά τα λίγα. Τους καλοκαιρινούς μήνες ας ρίχνουμε εδώ στο φόρουμ καμιά ματιά , να βάζετε κανένα θεματάκι για να περνάει η ώρα στην παραλία ή πριν βγούμε στην παραλία , να ανταλλάσσετε απόψεις και ιδέες και όλα θα κυλήσουν ήρεμα. Με 40 βαθμούς είναι λίγο δύσκολο να κάνεις αφαρμογές στις αριστερά κυρτές και δεξιά κοίλες ανισότητες :) .Θα τα λέμε λοιπόν σε πιο χαμηλό ρυθμό και σε όλα τα επίπεδα. Και όσοι έχουν πιο πολύ διάθεση , ας βάλουν και  καμιά πενηνταριά γερά ολυμπιακά θέματα για το βιβλίο , στις αντίστοιχες ενότητες!

 [b] Καλό Καλοκαίρι σε όλους  - Καλές διακοπές  και    Καλή επιτυχία στα παιδιά των ομάδων μας ![/b]

Όπως είχα υποσχεθεί...να μια διαφορετική προσέγγιση της άσκησης που είχε προτείνει ο mostel...
Η λύση είναι αρχείο word  και λείπει το σχήμα.  Λεπτομερείς αιτιολογήσεις λείπουν επίσης...

Όποιος θέλει ας τη δεί και ας κάνει τια απαραίτητες παρατηρήσεις για τυχόν λάθη.


analytic geometry
graphing lines
slope
geometry
trapezoid

Source: Massachusetts Mathematics Leagues States 2005

First quadrant Circle $O$ is tangent to the x- and y-axes at points $Q$ and $P$ respectively. $\overline{BC}$ and $\overline{AD}$ are parallel tangents to the circle with slopes of -1. If the area of $ABCD$ equals 48, and $r$ is the radius of circle $O$, determine $r^2$.

Edit: Points $A$ and $B$ are on the y-axis, while $C$ and $D$ are on the x-axis.

You need to know something more about the tangents.  Otherwise, they could continue indefinetely.

Ahh, right... I was being foolish and lazy by not putting up a diagram, and certainly this problem doesn't need a diagram, except for this part. A and B are on the y-axis, while c and d are on the x-axis

Say the center of the circle is (r,r).  Then the median of the trapezoid will be on the line -x + 2r = y.  The length of the median is then 2rt2 * r.  The height of the trapezoid is 2r.  Now we have 4*rt2*r^2 = 48.  r^2 = 6*rt2.

Nice solution, WarpedKlown.  I got the same answer, but I didn't see the elegant way to express the area of the trapezoid.


In how many ways can 15 different recruits be divided into three equal groups? In how many ways can they be drafted into three different regiments?

[quote="buzzer11"][hide]
To pick the first five, we have: $ \binom{15}{5}$.

For the next five, we have: $ \binom{10}{5}$.

Thus we get $ 108108$.

However, we have overcounted, because the order in which we choose the 3 groups does not matter.

Thus, $ \frac{108108}{6}= \boxed{18018}$

I think I made an error somewhere...  :) [/hide]

EDIT: why are none of the tags working? and LATEX?[/quote]
You must have disabled some things because I quoted you and it worked...

Buzzer - you almost certainly had 'Disable BBCode in this post' checked off when you posted.


the problem is
Consider the following algorithm to generate a sequence of numbers. Start with an
integer n. If n is even, divide by 2. If n is odd, multiply by 3 and add 1. Repeat this
process with the new value of n, terminating when n = 1. For example, the following
sequence of numbers will be generated for n = 22:
22 11 34 17 52 26 13 40 20 10 5 16 8 4 2 1
It is conjectured (but not yet proven) that this algorithm will terminate at n = 1 for
every integer n. Still, the conjecture holds for all integers up to at least 1, 000, 000.
For an input n, the cycle-length of n is the number of numbers generated up to and
including the 1. In the example above, the cycle length of 22 is 16. Given any two
numbers i and j, you are to determine the maximum cycle length over all numbers
between i and j, including both endpoints.

Input

The input will consist of a series of pairs of integers i and j, one pair of integers per
line. All integers will be less than 1,000,000 and greater than 0.

Output

For each pair of input integers i and j, output i, j in the same order in which they
appeared in the input and then the maximum cycle length for integers between and
including i and j. These three numbers should be separated by one space, with all three
numbers on one line and with one line of output for each line of input.
Sample Input

1 10
100 200
201 210
900 1000

Sample Output

1 10 20
100 200 125
201 210 89
900 1000 174

I have a solution but it's not efficient.
here is mine:

Your code has some minor issues, for example:
you probably shouldn't put main as a "long int"
you shouldn't use '\0' for every 0
" if(a) " is the same as " if(a!='\0') ".

[color=blue]/*
Think about precomputation, instead of brute force.
Simply run your function on every number from 0 to 1 million (storing them in an array), and then find the max in between every i and j.
You can make this even faster by avoiding recomputation
(i think this should work, i haven't actually tried it)
*/[/color]

EDIT: -_- '' apparently brute force will do?? ok this is what you should do (if you're lazy). just shorten down your code, get rid of the array that stores the inputs, and it should probably work.

Note: this is UVa's 3n+1 problem

[quote="madadpoor"]I have a solution but it's not efficient.[/quote]

I hear dynamic programming works well.  :)

thanks miller4math.
I just converted '\0' to 0 and did'nt change anything other.
then I compile it with dev-c++ and it works efficiently without any time spending.
formerly I compile it with visual-c++ and it output incorrectly.
I think I should emigrate for visual-c++ to dev-c++. :(  :(
I thought visual-c++ is the best compiler, but it isn't good in arrays.

I have solved the problem before. The key is to use memoization (Dynamic Programming). That is, after computing the cycle-length for n store it in an array; next time your program needs to know the cycle-length of n, don't recompute it, use the value in the array. If the cycle-length of n has not been computed before the value of A[n] (A is your array) will contain "-1" or whatever your variable to signal uncomputed cycle-length is.

Robert Otome

[code]
#include <cstdio>
struct mysrtuct
{
	bool isTest;
	int count;
}a[1000001];
long long stack[100000],start,add = 0;

int main()
{
	a[0].isTest = a[1].isTest = true;
	for (int i=2;i<=1000000;i++,start = add = 0)
	{
		long long j = i;;
		stack[start++] = i;
		while ( j != 1 )
		{
			if ( j <= 1000000 ) a[j].isTest = true;
			(j%2 == 0)?j=j/2:j=3*j+1;
			stack[start++] = j;
			if ( j <= 1000000 && a[j].isTest == true ) break;
		}

		for (int i = start-1,add = a[stack[i]].count;i;add++)
			if ( stack[--i] <= 1000000 )
				a[stack[i]].count = (add + 1);
	}
}

[/code]


articles
induction
probability
MATHCOUNTS
inequalities
AoPS Books

Hi, so, as many of you may know (from previous references on the board), I am creating a mathematics journal at my school this year with a few classmate (who is joining this forum very soon, hopefully). Anyways, I was wondering whether any of you have any ideas for articles, fillers, etc... that I could put in the journal. You must remember though, that my school, though, a good one, is not a very mathy school like Exeter or TJ..more like a liberal arts school. Thus, any articles and such would have to be interesting for non-math-crazy people as well. Thanks.

What sort of mathematical levels are the readers? What topics do they want to see? How advanced are they?

Like I said, they are typical high school math students...not as mathematically inclined as the people on this forum.

Give them some proofs with contradiction and induction probably. They are very useful techniques that aren't used much in high school curricula.

But as I said, they are not the most mathematically-inclined folk. I need to include some "fun" and "interesting" math.

The game of Nim ? :)
Or is that beyond their level ?

* Cryptarithms (or alphametic) are fun  :D and do not require high maths ! They are also a good opportunity to see the use of computers to solve maths problems.

An easy one :

.AT
+.A
---
TEE

(. = space)

Tougher :

FORTY
+.TEN
+.TEN
-----
SIXTY

Many more at http://www.cut-the-knot.org/cryptarithms/st_crypto.shtml (found with google)

* Euclid's proof of Pythagora theorem is wonderful and can't leave people indifferent :mrgreen: :

http://www.sunsite.ubc.ca/LivingMathematics/V001N01/UBCExamples/Pythagoras/pythagoras.html

HTH

Perhaps you could pick up a few real-life questions, like "The probability of people doing the English essay" or "the equation of how fast the bus moves and how we can improve upon it"
I know they're bad examples, but like my MATHCOUNTS coach problem (which has been solved; we got a coach!), math get a lot more interesting when it relates to us.

Perhaps you could pick up a few real-life questions, like "The probability of people doing the English essay" or "the equation of how fast the bus moves and how we can improve upon it"
I know they're bad examples, but like my MATHCOUNTS coach problem (which has been solved; we got a coach!), math get a lot more interesting when it relates to our everyday life.

I agree, belenos. I just [b]love[/b] those types of problems.

Teach the kids how to simplify radicals and rationalize the denominators of things like 1/sqrt(2).  I know that you said they aren't mathematically motivated, but if even a few people learned it, then it would be an improvement.  It really appears that I'm the only one in my AP Calc BC class that can simplify radicals and such without being phased.

Oh, and it can be fun too ;)

[quote]Teach the kids how to simplify radicals and rationalize the denominators of things like 1/sqrt(2). [/quote]

They're not that dumb. There are many good students, just not like the people here.

Mmm, we should really advertise the book along with the homepage.
If I had known AoPS books before, I would've stayed with my math class and read AoPS during the class time :)

The reason I'm saying this is because rationalizing denominator is like the very first thing discussed in AoPS 1.

I agree with you. I'm really smart compared to normal people :) , but I look real stupid here since everyone's a super-genius...

[quote="Tare"]"The probability of people doing the English essay.[/quote]

A funny and not too difficult problem in probability is :

given a group of n people, what is the probability P(n) that two are born the same day (day = day + month). For which n is P(n) >= 1/2 ? The answer is surprisingly low !

i have also founded a small math publication at my school. i think its probably smaller than what fierytycoon has in mind, but its been fun. it presents some information about contests, contests results. there is an article section for problem solving techniques + solved problems (mostly simpler probs, amc style. ocassionally an olympiad.) Unfortunately, a very low number of issues have been created yet. as soon as my school math club is in session (hopefully ill be an officer) ill up my efforts to get readers.

[quote="belenos"]A funny and not too difficult problem in probability is :

given a group of n people, what is the probability P(n) that two are born the same day (day = day + month). For which n is P(n) >= 1/2 ? The answer is surprisingly low ![/quote]

23. That's really neat.

[color=#FF00E0]An interesting topic might be basic inequalities -- Trivial inequality, AM-GM-HM, that kind of thing.[/color]

[quote="ComplexZeta"]For the sake of convenience, let's assume that there is no such thing as February 29th. The number of ways you have no people with the same birthday out of 23 people is 365P23, and the total number of ways is 23^365. Divide the two, and you get .492702765567..., which is less than 1/2. The conjugate is more than 1/2. You can check that with 22 the reverse will be true.[/quote]
Correct me if I'm wrong, but isn't the total number of ways 365^23 rather than 23^365?

there are 23 people. each one can have his birtday one of 365 days. thus there are 365*365*.... = 365^23 ways. i think complexzeta just wrote down the wrong thing, because 365p23/365^23 is almost 1/2. 365p23/23^365 is very small.

Err yeah. I think that's what I did when I solved it on paper otherwise the answer wouldn't have been close. Sorry about that.


You may rewrite is as $(a+1)(b+1)(c+1) \geq 64abc$
which is $(2a+b+c)(a+2b+c)(a+b+2c) \geq 64abc$.
Then use AM/GM for each of the three factors in the LHS, noticing that $2a+b+c = a+a+b+c$ and same for the two others.

Equality occurs iff $a=b=c$.

Pierre.


counting
distinguishability
number theory
least common multiple
LaTeX
Purple Comet

The alphabet in its natural order ABCDEFGHIJKLMNOPQRSTUVWXYZ is T0. We apply a permutation to T0 to get T1 which is JQOWIPANTZRCVMYEGSHUFDKBLX. If we apply the same permutation to T1, we get T2 which is ZGYKTEJMUXSODVLIAHNFPWRQCB. We continually apply this permutation to each Tm to get Tm+1. Find the smallest positive integer n so that Tn = T0 .

Thanks in advance!

And these:

Evaluate the sum $ 1^2 \plus{} 2 ^2 \minus{} 3^2 \minus{} 4 ^2 \plus{} 5^2 \plus{} 6 ^2 \minus{} 7 ^2 \minus{} 8^ 2 \plus{} \dots \minus{} 1000 ^2 \plus{} 1001^2$.

A rectangular storage bin measures 10 feet by 12 feet, is 3 feet tall, and sits on a flat plane. A pile of dirt is pushed up against the outside of the storage bin so that it slants down from the top of the storage bin to points on the ground 4 feet away from the base of the storage bin as shown. The number of cubic feet of dirt needed to form the pile can be written as m + n π where m and n are positive integers. Find m + n.

[Diagram can be found under #14 of the Purple Comet Middle School 2007 meet.]

We have some identical paper squares which are black on one side of the sheet and white on the other side. We can join nine squares together to make a 3 by 3 sheet of squares by placing each of the nine squares either white side up or black side up. Two of these 3 by 3 sheets are distinguishable if neither can be made to look like the other by rotating the sheet or by turning it over. How many distinguishable 3 by 3 squares can we form?

[hide="Hint for the First Problem"]
There are cycles of letters; i.e., in spot 1, the letters that appear in it in T0, T1, T2, ..., cycle.  Find the LCM of all the lengths of these cycles.
[/hide]

Thanks Bubbles. I got the answer now. :)

@Fanfan - The answer key says something different.

Nope, it's not. The answer is apparently 1001.

izzy....

using logic, if the series goes + + - - and the last term is - and is the largest, the answer should be negative

quickster....

the inside of your parenthesis should be

$ (4 \plus{} 6 \plus{} 12 \plus{} 14 \plus{} .... \plus{} 190 \plus{} 196 \plus{} 198) \equal{} 2(2 \plus{} 3 \plus{} 6 \plus{} 7 \plus{} ... \plus{} 95 \plus{} 98 \plus{} 99)$ which is $ 2(4950 \minus{} (1 \plus{} 4 \plus{} 5 \plus{} 8 \plus{} 9 \plus{} ... \plus{} 96 \plus{} 97)) \equal{} 2(4950 \minus{} (1 \plus{} 101(24))) \equal{} 5050; \minus{} 2\cdot5050 \equal{} \minus{} 10100$

so fanfan is right

Based on the pattern, 1001^2 should be positive, since it is 1 mod 4.

/me kills myself

i didnt see the $ \color{red} 1001^2$ at the end

[quote="isabella2296"]@Fanfan - The answer key says something different.[/quote]Izzy you changed the problem...

You said the problem was

'What is the value of $ 1^2\plus{}2^2\minus{}3^2\plus{}4^2\plus{}5^2\plus{}6^2\minus{}7^2\minus{}8^2\plus{}\cdots\minus{}99^2\minus{}100^2$'...

What? I didn't change it. I copied, pasted, and LaTeXed.


I was reading in Engel about Sophie Germain's Identity: a^4+4b^4 = (a^2+2b^2+2ab)(a^2+2b^2-2ab). However, I was wondering, why is this identity so special as to deserve a special name like "Sophie Germain's Identity"? a^4 + 4b^4 seems pretty random, and I am sure that there are many other similar-looking expressions that can be proven to never be prime. Any insights?

I made a special note of pointing this identity out to my Putnam seminar since I think it is important. A former USAMO honorable mention winner who was a TA at SUMaC 2001 (which was the year I attended) told me to pay special attention to this identity. I guess the point is that if a problem that uses this identity comes up in a competition and you know this identity, you will get the problem right. If this problem comes up and you don't know the identity, you basically have no chance to get it (although I derived it on my own without much effort after I was told about it).

Yeah, but don't you think that the expression a^4+4b^4 looks sorta...random? I am sure there are other similar expressions in the form of ax^n + bx^n that are always prime or not prime. Why is this one used in olympiads so often (and its even pretty simple to prove)? Or is it just my weirdness in thinking that it seems very random?

According to Prof. Ryavec (the professor who runs our Putnam practices), it goes back to a problem on a university contest: Find all positive integers n for which n^4+4^n is prime. He said no university student solved it, so they gave it to the high school students, and some of the high school students solved it. As a result of this problem (which is a simple application of the Germain Identity), everyone knows this identity.

So one problem and one instance where high school students figured it out faster than college students caused everyone, including Engel, to consider the identity important?  Methinks there ought to be more support than that...or is it just an extremely [i]fun[/i] fact?

[quote]Find all positive integers n for which n^4+4^n is prime.[/quote]
Are you sure it's 4^n? Or is it to the fouth power? How would you apply the formula if it's n^4+4^n?

Clearly this can only work when n is odd. So we'll replace n by 2k+1. Then we get n^4+4^n=(2k+1)^4+4^(2k+1)=(2k+1)^4+4*(2^k)^4. The only prime of this form is 5.


let 0<x,y<1 ,show x^y+y^x>1,
 :wink: ,this may have been posted earelier,please let me know,
 thanks.

It maybe posted many times as this problem is somewhat popular.. (I remember that this was from France.)

:?: ,will anybody  please post details?
i need itimmensly,
 :|

it is an old problem.
$(1+\frac{x}{y})^{\frac{1}{x}}\ge 1+\frac{x}{y}\cdot \frac{1}{x}= 1+\frac{1}{y}> \frac{1}{y}$
so $\frac{x+y}{y}> y^{-x}$, $y^{x}> \frac{y}{x+y}$. From the similar method, $x^{y}> \frac{x}{x+y}$. Thus, $x^{y}+y^{x}> 1$


Find the closed-forms of

$ \sum_{k\equal{}1}^n (\minus{}1)^{k\plus{}1} \frac{2k}{2k\plus{}1}$

and

$ \sum_{k\equal{}1}^n (\minus{}1)^{k\plus{}1} \frac{2k\minus{}1}{2k}$

$ n \in \mathbb{N}$

This is calculus. It doesn't belong in this forum.


How big of a difference is there between Division A and Division B? My team has a few really good students and some who are the best and we're thinking about doing a half team in one division and a full team in another. Could you compare them to AMC 10/12/AIME?
How are the individual tests graded? I see people say that the questions are worth 3,3,3,4,4,4,4,5,5,5 while the wiki site says that they are each worth one. Also, what is the cutoff for individual finals? Is it based on number of people participating or score?
Thanks in advance!

there seems to be a good deal of confusion on this topic

the questions are graded 2,2,3,3,4,4,5,5,7,7 ( so a perfect would be 42 :D)

the cutoff for individual finals varies quite a bit cuz i think the level of difficulty changes each year ( its only been 4 years). the easier questions would be about AIME 1-2 level and im not sure about the harder ones.

getting a 20 is good enough to make top 4 in your individual test.

and a 10 is good enough to ALMOST guarantee  top 10

i think you can make the finals by placing in at least one event. (and performing decently in the other)

Last year, the top ten in each subject got to take the individual finals (so you could get 10th place in one subject, score a 0 on the other subject, and make individual finals).
I heard that the difficulty has varied a lot (like the first year, it was ridiculously hard, and the second year, it was ridiculously easy).
I'm not sure what the point distribution is this year, the website doesn't seem to say anything about it.

This is what the website says about divisions:
[quote]Teams may register for the A or the B Division. The A Division is designed for the more experienced teams, and the B division is recommended for teams newer to competitions (Divison B will be slightly less challenging than division A) . Divisions will be given different tests on the day of the competition. The Power Test will be identical for both divisions, and teams in different divisions will be held to different grading standards. Awards for the two divisions will be separately calculated.

Teams that do not come from a single high school must compete in the A Division.[/quote]


inequalities
calculus
integration
inequalities proposed

I'm not sure I post this in correct place. And I'm not sure I use English correctly,because my country doesn't speak English.
I've searched for my question in this webboard,and..there not have or may be I don't see them. :roll:

My question is ..
Where the inequality come from?.some of beautiful problem must expand all term and use Muirhead's theorem to prove.
I don't understand why person who make the problem can simplify it to wonderful inequality.

[hide="example"]Let x,y,z be a positive real numbers.Prove that
$(xy+xz+yz)(\frac{1}{(x+y)^{2}}+\frac{1}{(y+z)^{2}}+\frac{1}{(x+z)^{2}})\geq\frac94$[/hide]

I guess they use computer check and then make conjecture.
[b][color=red]If i guess true.Could you introduce me about that program?. :) [/color][/b]

Thank you
Tummy

Someone tell me that ...It come from deep heart   :huh:

I really like you question, and I know it is very interesting. In fact, I ask it by myself many times and find out that every people have their own ways. Some people like to create inequality according to other inequalities, such as "change variables, change some constants...", but I dont like this way. Remember that a good creation should not be relied on Computer, it must be your inspiration. However, the following things are helpful.

(1) The formation of another inequality.

You can use it to create your own inequality.

Example: when VASC and Zhaobin find out the inequality

Let $x,y,z$ be positive real numbers such that $x+y+z=1$. Prove that
\[\left(\frac{1}{\sqrt{x}}-1\right)\left(\frac{1}{\sqrt{y}}-1\right)\left(\frac{1}{\sqrt{z}}-1\right) \ge \lt(\sqrt{3}-1\rt)^{3}. \]
I tried my own (similar in form, but not same in solution, it is much harder)

Let $x,y,z$ be positive real numbers such that $x^{2}+y^{2}+z^{2}=3$. Prove that
\[\left(\frac{2}{\sqrt{x+y}}-1\right)\left(\frac{2}{\sqrt{y+z}}-1\right)\left(\frac{2}{\sqrt{z+x}}-1\right) \ge \left(\sqrt{2}-1\right)^{3}. \]
According to the famous IRan's 96 inequality

If $a,b,c$ are non-negative then
\[\frac{1}{(a+b)^{2}}+\frac{1}{(b+c)^{2}}+\frac{1}{(c+a)^{2}}\ge \frac{9}{4(ab+bc+ca)}\]
I find out (a more difficult and even more beautiful because the fraction $\frac{9}{4}$ in the above from is not good). The following has not any number fraction.

[u]Problem (Hungkhtn's, SII book, chapter II)[/u]
If $a,b,c$ are non-negative then
\[\frac{1}{(a+2b)^{2}}+\frac{1}{(b+2c)^{2}}+\frac{1}{(c+2a)^{2}}\ge \frac{1}{ab+bc+ca}\]
Remember that the form of the inequality is the most important feature of the progress of creating inequality. Basing on a certain form of a certain inequality, we can create a lot and a lot of other inequalities.

(2) Generalization.

Ok. I think this way is natural. Generalization can make up a general inequality of a particular case. I don't think I need to explain more this way.

Example:

When VASC create the following statement for Romanian IMO Test

If $a,b,c>0$ and $a+b+c=3$ then
\[\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\ge a^{2}+b^{2}+c^{2}. \]
I generalize it to "4 variables"

If $a,b,c,d>0$ and $a+b+c+d=4$ then
\[\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}+\frac{1}{d^{2}}\ge a^{2}+b^{2}+c^{2}+d^{2}. \]
This problem has a very simple solution without mixing variable or any advanced method. I wait it. Let's make a try.

Furthermore, the following stronger inequality hold

[u]Problem 5.3.4 (Hungkhtn's, SII Book) [/u]
Let $a_{1},a_{2},...,a_{n}$ be non-negative real numbers such that $a_{1}+a_{2}+...+a_{n}=n$. Prove that
\[n^{2}\left(\frac{1}{a_{1}^{2}}+\frac{1}{a_{2}^{2}}+...+\frac{1}{a_{n}^{2}}-n\right) \ge 10(n-1)\left(a_{1}^{2}+a_{2}^{2}+...+a_{n}^{2}-n\right). \]
Remember that the above problem is very very hard. I wait a solution with out EV or SMV, too.

Let's discuss more. This  way (generalization) is good but it is not always good. Sometimes it makes the inequality becomes terrible. Sometimes the particular case is nice enough not to try any more generalization. The solution to the generalization is always complicated, too. Yes I remember some of my friend who send problems to my Vietnamese book proposed some generalization but I really not like. For example, we have already the following very nice inequality of HoJooLy

Let $a,b,c$ be positive real numbers such that. Prove that
\[\sqrt{a^{4}+a^{2}b^{2}+b^{4}}+\sqrt{b^{4}+b^{2}c^{2}+c^{4}}+\sqrt{c^{4}+c^{2}a^{2}+a^{4}}\ge a\sqrt{2a^{2}+bc}+b\sqrt{2b^{2}+ca}+c\sqrt{2c^{2}+ab}. \]
One of my friend proposed (however, I think his solution is incomplete)

Let $a,b,c$ be positive real numbers such that. For what $k$ when
\[\sqrt{a^{4}+ka^{2}b^{2}+b^{4}}+\sqrt{b^{4}+kb^{2}c^{2}+c^{4}}+\sqrt{c^{4}+kc^{2}a^{2}+a^{4}}\ge a\sqrt{(k+1)a^{2}+bc}+b\sqrt{(k+1)b^{2}+ca}+c\sqrt{(k+1)c^{2}+ab}. \]
Generalizing an old inequality is easy work, however, you are only [b]polite[/b] and [b]gentle[/b] if and only if you both generalize and are able to prove your generalization, not generalize only.

(3). Making up inequalities form definitive way. I think you may have some suggestion: using contradiction or integral or expanding a square. By this way, you can create true inequalities immediately but hardly have a nice inequality. Let's see an example

Starting inequality:

If $a,b,c$ are non-negative then
\[\left(\frac{a}{b+c}\right)^{k}+\left(\frac{b}{c+a}\right)^{k}+\left(\frac{c}{a+b}\right)^{k}\ge \min \left(2,\frac{3}{2^{k}}\right). \]
We get the following one

Problem 4.2.2 (Hungkhtn's, SII book) Find the best positive real constant $k$ for which the inequality
\[\frac{1}{2+{a^{k}}}+\frac{1}{2+{b^{k}}}+\frac{1}{2+{c^{k}}}\ge 1, \]
is true for all non-negative real numbers $a,b,c$ such that $a+b+c=3$.

(I am sure that you firstly never realize two above inequalities are the same in solution).

(4). Just by inspiration.

Let imagine that in a nice day, you stand in a beautiful garden with scene of many flowers, let your imagination hover in the sky... you may can create a nice inequality just by your inspiration. I know this work is harder, but every time doing that, I feel happiest (because my inequality isn't the same as anyone else's).

I intend to write this matter (how to create an inequality) as the final chapter in SII book (although it is discussed in Vietnamese book, it is not deep as I want). Just I am afraid I haven't  time to do that because IMO is approaching. I also want to discuss more with all of you about this matter, because $n$ people are better than $1$ only. Feel free to send me message Tummykung if you like to explain or discuss about your other personal matters.

For example

$Problem$ $1$. (source http://www.mathlinks.ro/Forum/viewtopic.php?t=143943)
Let $a,b,c$ are three side length of a triangle such that $a+b+c=1$. Prove that:
$\sum{\frac{a^{3}}{b(a+b-c)}}\geq 1$.

I'll try to answer the question "How that inequality was made ?" and prove this inequality.
I don't know hwo made this inequality, but I think it was something like this:  :lol:
_____________________________________________________________

Let x,y,z be positive real numbers

$thougts$
At first I will try to write some inequalities, that can be proved with one step and then I'll try to mix them.

By $A(m)\geq G(m)$, we have: $\sum{xy}+2\sum{x^{2}}\geq (\sum{x})^{2}$ $\Rightarrow$ $(\sum{xy})^{2}+2(\sum{x^{2}})(\sum{xy})\geq (\sum{x})^{2}(\sum{xy})$ (i1).
By $A(m)\geq G(m)$, we have: $(\sum{x^{2}})^{3}\geq (\sum{x^{2}})(3\sum{x^{2}y^{2}})\geq (\sum{x})^{2}(\sum{x^{2}y^{2}})$ (i2).
By $Cauchy$ $Shwarts$ inequality, we have: $(\sum{x^{2}})(\sum{x^{2}z^{2}})\geq (\sum{zx^{2}})$ (i3).

$thougts$
This inequalities are trivial and I think they aren't new, so what we gonna do ?
OK, just lets mix them.

By (i2) and (i3) $(\sum{x^{2}})^{4}\geq (\sum{x})^{2}(\sum{x^{2}y^{2}})(\sum{x^{2}})\geq (\sum{x})^{2}(\sum{zx^{2}})^{2}$ $\Rightarrow$
$(\sum{x^{2}})^{2}\geq (\sum{x})(\sum{zx^{2}})$ (i4).

$thouts$
OH, that's interesting (i4) is not symmetric and it's not so trivial "one step inequality", that's good! So, let's continue.

Summing (i4) and (i1), we have: $(\sum{xy})^{2}+2(\sum{x^{2}})(\sum{xy})+(\sum{x^{2}})^{2}\geq (\sum{x})(\sum{zx^{2}}+\sum{x}\sum{xy})$

$thougts$
Let's give to this inequality "nicer face".

<=> $4(\sum{x^{2}}+\sum{xy})^{2}\geq 4\sum{x}(\sum{z}x^{2}+\sum{x}\sum{yz})$
<=> $(\sum{(x+y)^{2}})^{2}\geq 4\sum{x}(\sum{z(x^{2}+\sum{xy})})$
<=> $(\sum{(x+y)^{2}})^{2}\geq 4\sum{x}(\sum{z(x+y)(x+z)})$
<=> $\frac{(\sum{(x+y)^{2}})^{2}}{\sum{z(x+y)(x+z)}}\geq 4\sum{x}$

$thougts$
Uh, it's ugly!!!!! How can I reduce it ?  :maybe:
What we have here: x+y, y+z, z+x ..... OH, got it !!!

Let $x+y=a$, $z+x=b$, $y+z=c$ $\Rightarrow$ $a,b,c$ are sides of triangle. So,
<=>  $\frac{(\sum{a^{2}})^{2}}{\sum{ab(b+c-a)}}\geq \sum{a}$, but by $Cauchy$ $Shwart$
$\sum{\frac{a^{4}}{ab(b+c-a)}}\geq \frac{(\sum{a^{2}})^{2}}{\sum{ab(b+c-a)}}$ $\Rightarrow$

$\sum{\frac{a^{3}}{b(b+c-a)}}\geq \sum{a}$, its homogenous so lets $\sum{a}=1$
$\Rightarrow$ $\sum{\frac{a^{3}}{b(a+b-c)}}\geq 1$.

$thougts$
Oh, I'm exhausted. But look, I've found new inequality!!!  :lol:
_____________________________________________________________

 :rotfl:

thank you so much.
I think the know of idea to make intequality would make idea to solve it ,too!
 :lol:


geometry perimeter logarithms AMC AIME probability algebra

I made two marathons in other forums and they were very successful so I'm hoping it works pretty well in here, too.

Rules:

1. You post the question in AMC level
2. Answer
3. New question

First:
Simplify

(2 :^2:) :^3:
----------------
(3 :^3:) :^2:

I suck at math

but tell me if this is right

 4
----
 27

[quote="dinoman"]64
---------
729[/quote]

Give the new problem

Try to write solutions and hide them, like this:

[code]
[hide=Solution to Problem 1]I win.[/hide]

[hide=Problem 2]You lose.[/hide][/code]

So...

[hide="Problem 2"]
What is the largest possible area of a triangle with perimeter $24$?[/hide]

[hide]$\dfrac{8^2*\sqrt{3}}{4}=\boxed{16\sqrt{3}}$[/hide]

[hide="New Problem"]What is the area of a triangle with sides $26$, $28$and $30$? Try not to use Heron's.[/hide]

[hide="Solution"]
This area will be $4$ times the area of a triangle with sides $13$-$14$-$15$.  Such a triangle is just a $5$-$12$-$13$ triangle placed together with a $9$-$12$-$15$ triangle, so the area is $5 \cdot 6 + 6 \cdot 9 = 84$.  Thus $84 \cdot 4 = \boxed{336}$.
[/hide]

[hide="NP"]
Find the product of all solutions to $x^{\log_{10}{x}} = 10$.

Note: the above problem is from an ACTUAL AMC 12.
[/hide]

[hide]Taking log base $x$ of both sides, we get the equation: $log_{10}x = log_{x}10$. A fundamental law of logarithms states that $log_{a}b = \frac{1}{log_{b}a}$. Setting $y = log_{10}x$, we get the equation $y = \frac{1}{y}$. The solutions to these equations are $y = 1, -1$. Plugging in these two values for $log_{10}x$, we get that $x = 10, \frac{1}{10}$. The product of these solutions gives the answer: $\boxed{1}$.[/hide]

[hide]The first $2007$ positive integers are each written in base $3$. How many of these base-$3$ representations are palindromes? (A palindrome is a number that reads the same forward and backward.) [/hide]

[quote="mhy123"][hide]Solution

We take the log base 10 of both sides. So we have:

$\log x^{\log x} = 1$. Applying a well-known theorem with logs, we see that $(\log x)^2 = 1$, or $(\log x)^2 - 1 = 0$. Factoring, $(\log x - 1)(\log x + 1) = 0$, so we have solutions of $x = 10, \frac {1}{10}$. Thus, the product of these is just $\boxed {1}$. [/hide

[hide]New Problemo

(Does this have to be strictly AMC? AMC season is over, and everybody's looking forward to the AIME. The following problem is from an AIME, but it is in the range of #1-#5, so it should be relatively okay with AMC 12 level.)

The product $N$ of three positive integers is $6$  times their sum, and one of the integers is the sum of the other two. Find the sum of all possible values of $N$. [/hide][/hide][/quote]

i think i beat you there, buddy :P

Sorry man. Solve VIPMaster's problem before you attempt mine, everyone.

(Two different solutions to the same problem.. nice)

[hide="Solution"]$2007$ base $3$ is $2202012$. The first half of a palindrome will determine the rest, so all we need to do is find the first few digits. The largest palindrome smaller than $2202012$  is $2201022$, so we must find base $3$ numbers less than or equal to $2201$. Each string of digits of length $n$ will determine a palindrome of length $2n$ and $2n-1$(for example, $2201$ can start both $2201022$ and $22011022$). However, this is only true for strings of length $1$, $2$, and $3$ because strings of length $4$ can only create $7$-digit palindromes($8$ digits would be too long.) So the total number of palindromes is the number of $4$ digit numbers in base $3$ less than $2202$ plus twice the number of $3$ or less digit numbers. The greatest $3$ digit number in base $3$ is $222$, which is $26$, so we can have any of the numbers $1-26$. Also, $2201$ base $3$ is $73$, so we can also have the numbers $27-73$. Therefore, our total is $46+2(26)=98$ [/hide]
[hide="New Problem"]On a standard die one of the dots is removed at random with each dot equally likely to be chosen. The die is then rolled. What is the probability that the top face has an odd number of dots?[/hide]
EDIT:Sorry mhy123, I didn't see your new problem.

Solution[hide]There are 21 dots in total. 12 are on even faces and 9 are on odd faces. If a dot on an even face is removed, the die will be 4 faces odd and 2 even. If a dot on an odd face is removed, the die will be 4 faces even and 2 faces odd. Thus, the probability of an odd number is (12/21)(4/6)+(9/21)(2/6), which is 11/21[/hide]

[quote="lg5293"][hide="solution #6?"]
Let $a, b, c$ and let $c = a + b$. Then we have $ab(a+b) = 12(a+b)$. Since $a, b > 0$, we can divide both sides by $a + b$ to get $ab = 12$. WLOG, let $b > a$, so we have the solutions $(a, b) = (1, 12), (2, 6), (3, 4)$, which gives us the values of N as $169, 96, 84$ respectively. Their sum is $\boxed{349}$
[/hide]

[hide="Problem 7 from pi37"]
On a standard die one of the dots is removed at random with each dot equally likely to be chosen. The die is then rolled. What is the probability that the top face has an odd number of dots?
[/hide][/quote]

It says what the probability is.

Actual answer
[hide]there is a 4/7 chance of picking a dot on an even face and 3/7 of picking an odd face. Picking an even face creates 4 odd sides and 2 even and odd gives 4 even and 2 odd. 4/7 x 2/3 + 3/7 x 1/3= 11/21, so 11/21 is the answer[/hide]

[hide]New Problem: How many distinct Complex zeroes exist for the following polynomial: p(x) = x^7 + x^2 + 1[/hide]


Last year, it was a not so good year for me. I had a overall gpa of a 2.9 last year unweighted(like a 3.2 weighted).

It was my sophmore year, and I know U of M doesn't take in freshman year. I decided to work harder I got a 4.21 my first semster, and a 4.41 my second. My ACT is a 30, and so far I have been in Students Against Drunk Driving, Theatre(about 120 hours in it), tutoring, and in Track. I'm in NHS but just got in my senior year, and applied before I got accepted.

So far in my senior year, my grades have been a 4.3. Is there a good chance I can get into U of M.

Your GPA's after sophomore year seem good, but then again, that could be relatively low compared to others in your class.  Your ACT score seems okay, not particularly great but not bad at all.  Your EC's seem solid.  I have seen people with similar numbers get in so I think you have as good a shot as anybody else.


Recently at the market the people who grow flowers, let us call them the V's, raised the price on their flowers. When they increase the price 1 dollar, the amount sold is reduced by 20%. When the V's decrease the price 1 dollar, they sell 10% more. How much should the V's charge to make the most money?

I say there, there needs to be more information....

Yeah, we need to know the original price of the flowers.

Oh, oops, they originally sold them for $\$$6 a bunch. :blush:

Well, if the price they sold has to be an integer, just try some situations (raise the prices and drop them and see what happens) and the answer is pretty obvious..


[quote="mr. math"]how do you do this problem?

a car travels 35 mph going from point A to B, then at 45 mph from point B to C. it takes 1 hr and 40 min to complete the trip from A to C, and point B is exactly midway between A and C. What is the distance between A and C. Give your answer as a decimal number of miles.

please include explanation[/quote]
[hide]
$\boxed{65.625}$
$r=\frac{d}{t}$
$35=\frac{d}{t}$
Since distances are the same and minutes=100,
$45=\frac{d}{100-t}$
Solving, $t=\frac{4500}{80}=56.25$ in minutes, so in hours,
$t=\frac{56.25}{60}$ and substituting back in, $d=32.8125$ and the total distance is $32.8125\cdot2=65.625$[/hide]

the key here is that rate*time=distance. If you go 50 mph for 2 hours you go 100 miles: easy to see. there. now you know exactly how to do every rate-time-distance problem thrown your way!!!


find the points where the given function is not defined and is therefore not continuous. For each point tell whether or not this discontinuity is removable.

G(u)= u+1/(u^2-u-6)

The only way in which $G(u)$ can fail to be defined is by having the denominator equal to zero. Such discontinuities are not removable, since the limits on both sides of discontinuity are infinite.

So the function is not defined at U=3, -2 and removable at -1?

[quote="Kent Merryfield"]...unless the numerator and denominator share common factors in the appropriate way. This particular function (considered on $\mathbb{R}$) has a pole at $1$ but  removable discontinuity at $-1.$[/quote]

Err, the denominator factors as $(u-3)(u+2)$... so there shouldn't be a discontinuity at $-1$ in the first place... right? I think it is discontinuous at $u = 3, u =-2$ neither of which are removable.

I interpreted $G(u)$ as $u+\frac{1}{u^{2}-u-6}$... Properly placed parentheses would be helpful.

Ah - and I was reading the denominator as $u^{2}-u^{6},$ hence my confusion.


limit
inequalities
calculus
calculus computations

Prove if $ \lim_{x\to c} f(x)\equal{}L$, then there are $ \delta, B>0$ such that $ 0<|x\minus{}c|<\delta$ implies $ |f(x)|<B$.

use the definition of continuity at a certain point . then an
application of the triangle inequality gives the result. :lol:


function
inequalities
number theory unsolved
number theory

I have a nice problem.
With a Calculator .Calculate $\{3^{\sqrt{2}}\}$  till 6 digits in base 10.

Guys this is so new,It is for this week .Work on it.

Forget about this post .just go to post 8

Hmm.a really stupid idea:

$\sqrt{2}=1.414213562373...$
So by a good chance we will have that the first $6$ digits of these are equal

$\{3^{\sqrt2}\}=\{3^{1.414213562373...}\}$

We should calculate the

$\{3^{1.414213562373...}\}$

and better to say ,we should calculate this:

$3^{\frac{141}{100}}=\sqrt[100]{3^{141}}$

But for god sake how shall i solve this ?
In the exam , you just have a simple calculator,just with functions
$+,-,/,*\sqrt[2]{x}$

 :wallbash_red:  :wallbash_red:

Yes ,you are allowe  to use calculator,
I think Arne idea is to somehow use the$\sqrt{(10-1)}$:
$3^{\sqrt2}=(10-1)^{\frac{\sqrt2}{2}}$

Yes, you are right, but I don't think it will work.

Well guys , this is the [b]Myth[/b] idea:

If you are allowed to calculate square roots only then my suggestion is to use base 2 representation for $\sqrt{2}$, i.e. you can calculate \[3^{\sqrt{2}}\approx 3^{1+1/2^2+1/2^3+1/2^5+1/2^7+1/2^{13}+1/2^{16}+1/2^{17}+1/2^{18}+1/2^{19}+1/2^{22}+1/2^{23}}.\] It will give you 6 correct digits. To calculate this number you should use Horner's scheme to increase accuracy (for example, to calculate $3^{1+1/2^2+1/2^3}$ you calculate $\sqrt{3}$, then $3\sqrt{3}$, then $\sqrt{3\sqrt{3}}$, then $\sqrt{\sqrt{3\sqrt{3}}}$ and finally $3\sqrt{\sqrt{3\sqrt{3}}}$).
It is a strange problem. Maybe there is another excellent trick for accurate numeric calculation.

 :fool:  :fool:  :read:

For others who like me ,dont know what is [b]H[/b]orner's scheme,I found this but it is not clear at all.
Would some one explain what is this scheme?

[url]http://hscool.netclub.ru/online/base/b1s1/e1.html[/url]

Guys ,It was not Gorner But Horner :

Here is a usefull link:
PS.Cooperated by grobber
[url]http://en.wikipedia.org/wiki/Horner's_rule[/url]

And how did you do that simple calculation with just a simple calculator ;) ?

then show your complete solution please
i bet its wrong .With out any offence
As far as i know ,just 2 ones managed to solve this from the 40 people participating in the exam.

To:lomos_lupin

I'm sorry , I was crazy.

I use simple calculator.    :blush:

Take it easy,friend.
This problem is a hard and so ugly(indeed ugly).
I apologise for my last post,but i was a bit angry cause i could not see the solution at all :)


:)

$ x \in \mathbb{R}$

$ (x\plus{}1)^2cos(x)\equal{}2$


[quote="chien than"]Find $x;y;z;t \in Z^{+}$ such that:$x^{x}+y^{y}+z^{z}= t^{t}$[/quote]
Let $x\le y\le z$, then or $x=y=0,z=t$ or $t\ge z+1.$
For second case $t^{t}\ge (z+1)(z+1)^{z}\longrightarrow t\le 2$ - no solutions.
Therefore solutions are $x=y=0,z=t$, $x=z=0,y=t$, $y=z=0,x=t$.

But $x,y,z,t>0$!
Let $1\leq x\leq y\leq z<t$
If $z=1$ then $x=y=1$. We have $3=t^{t}$, weed
If $z=2$ then $t\geq 3$. Therefore $t^{t}\geq 3^{3}>3.2^{2}\geq x^{x}+y^{y}+z^{z}$, weed
If $z\geq 3$ then:
$t^{t}\geq (z+1)^{z+1}>z^{z+1}=z.z^{z}\geq 3.z^{z}\geq x^{x}+y^{y}+z^{z}$
So this equation don't have natural solutions $\neq 0$


Hello. :)

I am looking for a control toolbox (programs) for CASIO 9860 calculator, to help me in control systems, namely:

[list][*][url=http://en.wikipedia.org/wiki/Nyquist_plot]Nyquist plot[/url]
[*][url=http://en.wikipedia.org/wiki/Routh_stability_criterion]Routh stability cretirion[/url]
[*][url=http://en.wikipedia.org/wiki/Root_locus]Root locus[/url][/list]

I haven't found anything useful so far in the internet, so any help would be greatly appreciated. :) :) :)

Does anyone happen to know anything?  :)  :)  :)


p is prime only if p! mod(p^2) = -p mod (p^2)

Can someone prove it?

Is there any link with Wilson's theorem?

Thank you for any help.


calculus
derivative
inequalities
function
algebra unsolved
algebra

Determine the lower and upper bounds for S_r = cos(x)^r + cos(y)^r + cos(z)^r where x,y,z are the angles of a non-obtuse triangle, and r is a positive real number.

The easy case : if 0 < r =< 1.
Then, using the concavity of cos^r on [0; \pi /2] (the second derivative is f''(x) = r*cos r-2 (x)(r*sin 2  - 1)  \leq  0), Jensen's inequality leads to :

S_r  \leq  3*cos r ( (A+B+C)/3) = 3/2  , and equality occurs when ABC is equilateral.

In another hand, for given 0 <x,y,z \leq 1 , the function f(r) = x r + y r  + z r  is decreasing. Thus, is ABC is acute, that leads to
S_r  \geq S_1 = cosA + cosB + cosC = 1 + 4sin(A/2)sin(B/2)sin(C/2) > 1.
Moreover, for A =  \pi /2, we easily have
S_r = cos r B + cos 2 ( \pi /2 - B)
= cos r B + sin r B
 \geq cos 2 B + sin 2 B
= 1, and equality cannot occurs since 0 < B < \pi /2.
Thus : 1 < S_r  \leq  3/2 r .

Orlando, are you sure that the result is known for r > 1? Because, according to Mitrinovic's 'Recent advances in geometric inequalities', it seems to be still unknown...

Pierre.

Dear Pierre,

I feel the cases r = 1 and r = 2 are widely known. The problem appeared in Crux as problem 1352 and I think Murray Klamkin submitted a solution for it.

Ok, I see that in Klamkin's proposal, it is asked for lower and upper bounds, but it is not known that these bounds are best possible.

Klamkin, R.I. Hess and Crux Editors jointly proved that for r \geq 2 :
3/2 r   \leq  S_r  \leq  1, and these bounds are best possible, which easily commes from the well-known
3/4  \leq  S_2  \leq  1
and from the power-mean inequality.

Hess claimed, but with a deficient argument, that the bounds given above for r  \leq  1 are still valid for r  \leq  ln(9/4)/ln(2),
that 1  \leq  S_r  \leq  2 1-r/2 for ln(9/4)/ln(2)  \leq  r  \leq  ln(3)/ln(2)
and that 3/2 r  \leq  S_r  \leq  2 1-r/2  for ln(3)/ln(2)  \leq  r  \leq  2.

As far as I know it is still open to prove Hess' claims.

Pierre.


a triangle ABC , I is the incenter of the triangle, r is the radius,p = (a+b+c)/2

proof:
AI×BI×CI×p = abcr


Show that
\[\frac{d(x,y)}{1+d(x,y)}\]
(where $d(x,y)$ is a metric) defines a new metric.

http://en.wikipedia.org/wiki/Metric_%28mathematics%29

see Equivalence of metrics


I know that there is a formula for number of divisors of a natural number. Is there a formula for the sum of divisors of a natural number?

Yes. Look at http://mathworld.wolfram.com/DivisorFunction.html formulas (10)...(18).


geometry
trigonometry
MATHCOUNTS
calculus
analytic geometry
graphing lines
slope

I take Geometry after Algebra II, so I'm not sure if we have exact curriculums. But I'm pretty sure that you'll cover trig in Alg II (we didn't) if not all of it in pre-calculus. I would not skip, since you might not know what you'll be getting into.

I think it depends on the quality of the Geometry class they'll offer. I was priveliged to have a really good teacher in 8th grade, who taught all middle schoolers taking Geometry. That class was really good, and if I were able to skip it, I would've missed out on lots of things. Now in my high school, the geometry class isn't so good, and anyone with enough knowledge of geometry to do mathcounts problems can get a good score.

I've decided I'm not going to skip any more classes (I'm pretty sure I can skip Algebra 2 and most likely Precalc) but it would just be a waste of time, since that would just screw up my stuff at home.

[quote="hm29168"]I just want your perspective on this.

I am currently in 7th grade taking 8th grade math, and at the end of 8th grade math, I have an option of taking the exam for 9th grade Geometry. If I pass the exam, I am allowed to skip to the next course which is Algebra II. If not, I just take 9th grade Geometry. If I pass, should I skip the course?

9th grade Geometry is basicly geometry and proofs, but they add some trig (Law of Sines, Law of Cosines) in there, and I don't want to miss out on any trig, as there is no formal trigonometry class later in high school. Also, I'm a bit concerned of taking math with people 2 grades older than me.

Any input is appreciated.[/quote]

Skip. Take a pre-calculus class to learn trig.  I'm skipping all the way to calc BC next year so. ya.

There NEVER is a class devoted to trig.  Unless your school doesn't even offer pre-calc, which means it's stupid. =P

I asked my math teacher if I could take the exam and he flat out refused me, as I have already skipped a grade in math.

Well, that's the end of that. Thanks for everyone's input.

The Math Department head of my high school refused to let me take the calc BC exam at the end of this (freshman year). She even went to the point of indirectly insulting my mathematical abilities. I guess they actually want you to take the course for the grade.

There are times when you should consider taking this sort of things to a higher level. I have had times where the polite route has failed and I have spoken to my parents and convinced them to write letters to the school to "request" skipping etc. Most of the time they are responsive to it.

I relearned everything I learned in geometry this year in precalc.  My Alg II class in RI wasn't as rigorous as the one in my new place of residence.  And yet nothing that I didn't learn was really that important, we 'reviewed' it this year and I actually enjoyed it more because of the faster pace.  As long as you take precalc before calc, you should be fine.

Edit:  Didn't notice you made an update.  I'd keep at it.  If you can't take it, so what?  Just keep trying and you've got the cance.

well. it depends on the type of teacher. Like my calc teacher was going over point-slope form while we were doing derivatives. I skipped pre-calc, and I am fine with calc. (b/c he goes over every little detail).

Also, if you are fine with geometry, then you should skip. That way, in high school, you can get over the math courses easily and add in another elective or two.

Mmm. I guess it just depends on your level of Geometry and the course/teacher itself.

It is, however, "widely believed that American students are weak in Geometry", so, unless you're strong in it, I don't think it could hurt. I'm in Geometry right now--and plan to take "Honors Advanced Algebra (Algebra II)" with everyone else. Parts of it should be challening, but parts will be review.

After reading your posts and talking to my parents, I think I shouldn't give up when I haven't even taken the exam. I mean, I don't even know if I'm going to pass the exam, so I might as well try. Either way I have nothing to lose. I'll try to talk to my teacher again.

I really enjoy this website! The people here are so supportive! Thanks again for your input!

totally skip i'm in 8th grade and my school is stupid so i am bored out of my mind in algebra 1. i'm taking geometry final too to skip into algebra 2, so i think that that's a great idea to skip cuz i mean being bored in math is NOT FUN! trust me i've been bored in math class for like a bajillion years. :lol:  :icecream:  :juggle:  :jump:

my mom normally doesn't like me accelerating, but the only reason i'm taking calc right now is because i'm bored out of my mind in geometry.
Seriously, if you bore yourself in class, then you will lose your love for math.

that NEARLY happened to me. but there's this paradise called AoPS I go to. It's very interesting..

Your situation sounds exactly like mine a couple of years ago, except that my teacher was much more cooperative. I've never regretted skipping geometry. Everything you need to know about geometry you can learn from AoPS books and classes (or other math competition resources) in much more detail and in a much, much more interesting way. As far as I can tell from talking to people who have taken geometry, its main function is to waste your time writing out pointless 2(or 3)-column proofs which completely warp the idea of "proof" and destroy the appreciation people might otherwise have for rigor. Unless you absolutely can't learn geometry outside of school, you should make every effort to skip it.

Hmmm...Lynelle if this guy were in Palo Alto though, that would be pretty bad because Ms. Slack is the most awesome Geometry teacher ever...Unfortunately the Gunn and Paly Geometry classes like I said above suck nowadays, just pure mathcounts type questions but easier.

In my experience, the worst thing you can do for yourself mathematically is sit through a math class where you know all the material. The test they would give you would be for them to see if you know the material, but also for you to see if you would be bored by the material. This is not to mention the social aspects of being the smartest one in the class, everyone asking you for help, etc.

It also won't help if you land yourself into a math class with no knowledge at all. While this may prove challenging for you, I don't believe AoPSers are used to a challenging school math class.  :wink:


Jo was at a pizza parlor w/ 16 of her friends. If she orders a massive, round deep-dish pizza, what is the minimal number of cuts needed to ensure that everyone gets at least one slice?

(is there a formula for this, or at least one for the kids getting equal slices?)

[hide="Hint (:"]Well, whenever you split a region, the maximum number of regions for n cuts possible to split is expressed as $ 2 \plus{} 2 \plus{} 3 \plus{} 4 \plus{} 5 \plus{} 6 \cdots n.$[/hide]

right. i mean i know how to do the problem normally. i was just wondering what the formula was to find it, and i know there is one b/c i used to know it..

That is the formula.  (Unless I'm misunderstanding your question.   :huh: )

[hide="hint"] What does $ 1\plus{}(1\plus{}2\plus{}3\plus{}4\plus{}\dots\plus{}n)$ look like?[/hide]

first term, plus last term, time number of terms, divided by 2, plus one?

That, or more precisely, $ \frac {n(n \plus{} 1)}{2} \plus{} 1$. (Yours is essentially the same).

soo...i got 137, but the answer is 6... maybe the AT LEAST has something to do w/it?

also, is that formula for EQUALLY sized pieces?

You reversed the formula. You plugged in $ 16$ for $ n$, but you were supposed to solve for $ n$. As a brief check, 137 cuts is a bit outrageous for only 17 pieces. In fact, anything above 17 doesn't make sense.

Also, there are $ 17$ people, Jo and her 16 friends. We have that $ \frac {n^2 \plus{} n}{2} \equal{} 17$, so $ n^2 \plus{} n \minus{} 34 \equal{} 0$. Note than $ n \equal{} 5$ yields $ \minus{} 4$, but $ n \equal{} 6$ yields $ 8$, so $ \boxed{6}$ is the minimum valid number of cuts.

[quote="tzhang1"]Would it be 5 cuts? $1 + (1 + 2 + 3 + 4 + 5)$[/quote]

No that only gives you enough slices for 16 people. You forgot to include Jo.

If everyone gets an equal slice, the minimum number of cuts is $ \lceil \frac{x}{2} \rceil $

Since it's deep dish, you could cut it into eighths, and then cut it in half horizontally to get 16. 5 cuts.

Actually it can be done with 5 cuts. First cut makes two halves. Then put those two halves on top of each other and make another cut down the center. Second cut makes four pieces. Stack all those pieces on top of each other. After the fifth cut, we have 32 pieces.


Let $ A > 0$ and $ B$ Hermitian.Prove that  If  $ AB \plus{} BA \geq  0$ then   $ B \geq 0$ ($ A,B \in M_{n}(R)$)

Let $ \lambda\in Spec(B)\subseteq\mathbb{R}$ with corresponding eigenvector $ \bar{v}\not\equal{}\bar{0}$, then $ 0\leq \bar{v}^t(AB\plus{}BA)\bar{v}\equal{}2\lambda \underbrace{\bar{v}^tA\bar{v}}_{>0}\Longrightarrow\lambda\geq 0\Longrightarrow B\geq 0$


David Jackson's job pays him 20,000 dollars this year. Suppose that in every future year he works, his salary is 1.1 times what it was the year before. After how many years will his salary be over 32,000 dollars[color=darkred]?[/color]

This is a very easy function question.

that's quite likely to appear on some middle school tests :D
[hide]
$20000\times1.1^{x}=32000
1.1^{x}=1.6
x=\log_{1.1}1.6=\frac{\lg1.6}{\lg1.1}
\upshape and run your calculator
x\approx4.9
\upshape therefore after 5 years.$
[/hide]

I don't think average middle schoolers learn natural logarithrm. ;)

Plus, try to let this post answered by middle-schoolers.

Thanks

that's definitely a high school problem
that's why I answered it

This is from my school practice paper...

I thought it was easy..

and we didn't have to use natural logarithrm

There are many easy ways...

draw a graph ..

or

find the pattern - which can be part of natural logarithrm

Were you allowed to use calculators?

20000 x 1.1 x 1.1 x 1.1 x 1.1 etc. etc. etc.

[quote="Treething"]Were you allowed to use calculators?[/quote]
Without a calculator.  Without logarithms.  This is a fair middle school problem.  Do you know how to multiply by eleven?  Do you know how to multiply by two?  Do you know how to handle shifting the decimal place?

[hide="solution"]
\begin{eqnarray*}1.1^2&=&1.21 \\
1.1^3&=&1.331 \\
1.1^4&=&1.4641 \\
1.1^5&=&1.61051 \\
&&1.61051\times 20000=32210.20 \\
\end{eqnarray*}[/hide]

Yup!

it could've done by middleschool abilities

Let the number of years be x
20000(1.1)  x =32000
A form 3 student can do that...with calculator

Calculators are more than welcome to be used!!

so use it if you need to...


Please help me prove the identity in the attached picture.

$\cos(x-c)=\cos(x)\cos(c)+\sin(x)\sin(c)$
$\tan(x)=\frac{\sin(x)}{\cos(x)}$
$\sin(x+b)=sin(x)cos(b)+sin(b)cos(x)$
$\cot(x)=\frac{\cos(x)}{\sin(x)}$
Now can you do it?

I'm sorry, the "identity" I gave is not true.

But I have another problem... (this time without a mistake, I hope)

Find the exact value of the attached expression without a calculator.


linear algebra
matrix
linear algebra unsolved

Find all real values of $ \alpha$ such the matrix $ A\equal{}(a_{i j}) _{n x n}$ defined by $ a_{i j} \equal{} cos((i\minus{}1).j\alpha)$, for $ 1\le i,j \le n$, has a null determinant.


I have noticed that some people (especially the French members) use ]0,1[ instead of (0,1). is this the natural notation in France for open intervals?

We use

]a;b[ = { x \in R; a<x<b} instead of (a;b)
]a;b]= { x\in R; a<x \leq b} instead of (a;b]

In Belgium we have the same notation as in France :)


Using a delta epsilon argument, prove that $ \lim_{x\to 2} x^2 \minus{} 3 \equal{} 1$.

[hide]
For e>0 we want to find d>0 such that $ |f(x)\minus{}1|<\epsilon$ whenever $ |x\minus{}2|<\delta.$

$ |f(x)\minus{}1|<e\implies$
$ |(x^2\minus{}3)\minus{}1|\equal{}|x^2\minus{}4|\equal{}|x\minus{}2|\cdot|x\plus{}2|<\epsilon$

Assume $ \delta \leq 1.$
Then $ |x\minus{}2|< \delta \rightarrow 1 < x< 3 \rightarrow 3 <|x\plus{}2|<5.$

$ |x\minus{}2|\cdot|x\plus{}2|< |x\minus{}2|\cdot 5< \epsilon$
Whenever $ |x\minus{}2|<\frac{\epsilon}5.$

Let $ \delta \equal{} min(1, \frac{\epsilon}5)$
Then if $ 0<|x\minus{}2|<\delta, |f(x)\minus{}1|<\epsilon.$[/hide]


HELLO GUYS. I HOPE THE AMCS WILL BE EASY THIS YEAR

My only hope is that I don't fail them.

ZOMG THEY'RE COMING UP SOON

that was some pretty fail spam there
[color=red][size=59]Mod notes: No spamming.This topic has been locked for spamming reasons. Please do not spam. (ie. read mr.rruscyks post in the missouri forum [/size][/color]


Posted.
http://www.mathlinks.ro/Forum/viewtopic.php?t=30952
My solution is same as the one Valiowk posted.
I would like to see another solution..


nice solution! one question: Was there an easier way to see that plugging in $ \frac{1}{10}$ into ${ \prod_{n=1}^{9} (x - \frac 1n})$ gives $ \frac{1}{10^9}$ other than writing out all nine terms and then multiplying?

wait....also shouldnt it be $ \frac{-1}{10^9}$?

You are right about the sign. The correct solution is $ P\left(\frac{1}{10}\right)\equal{}\frac{10^6\minus{}2009}{10^9}$ and with regard to the ease of calculation, observe that generally we have:

$ \prod_{n\equal{}1}^{N\minus{}1}\left(\frac{1}{N}\minus{}\frac{1}{n}\right)\equal{}\prod_{n\equal{}1}^{N\minus{}1}\frac{n\minus{}N}{nN}\equal{}\left(\frac{\minus{}1}{N}\right)^{N\minus{}1}\prod_{n\equal{}1}^{N\minus{}1}\frac{N\minus{}n}{n}\equal{}\left(\frac{\minus{}1}{N}\right)^{N\minus{}1}$

[quote="Zhero"][quote="hjbrasch"]$ P(x) \minus{} x^3 \equal{} 2009\prod_{n \equal{} 1}^9\left(x \minus{} \frac {1}{n}\right)\Longrightarrow P(\frac {1}{10}) \equal{} \frac {10^6 \plus{} 2009}{10^9}$[/quote]

I don't believe this solution is correct. You are forgetting that the quotient polynomial; the LHS of this equation has a degree of 2009, whereas the RHS only has a degree of 9. The correct equation is
$ P(x) \minus{} x^3 \equal{} 2009 Q(x) \prod_{n \equal{} 1}^9 \left(x \minus{} \frac {1}{n}\right)$, where $ Q$ is a polynomial, and we must find$ Q(10)$.

Hopefully I'm not missing anything horribly important. :/[/quote]

Reread the problem; the original problem has been changed.

I probably just fail, but I can't see anything. Could you please be more precise?


Prove that there exists infinitely many (m , n) that satisfy the given conditions.
$ a.$ m is not equal to n
$ b.$ m, n are positive integers
$ c.$ m and n have the same prime factors
$ d.$ m + 1 and n + 1 have the same prime factors.

About c and d - Does "have the same prime factors" mean not being relatively prime (like 12 and 14), or meaning something like having $ 12\equal{}2^2\cdot 3$ and $ 18\equal{}2\cdot3^2$?

[quote="Brut3Forc3"]About c and d - Does "have the same prime factors" mean not being relatively prime (like 12 and 14), or meaning something like having $ 12 \equal{} 2^2\cdot 3$ and $ 18 \equal{} 2\cdot3^2$?[/quote]I guess it's the latter. Otherwise, it seems quite trivial.

[quote="Brut3Forc3"]About c and d - Does "have the same prime factors" mean not being relatively prime (like 12 and 14), or meaning something like having $ 12 \equal{} 2^2\cdot 3$ and $ 18 \equal{} 2\cdot3^2$?[/quote]
They are not relativley prime.

[quote="rofler"][hide="Hidden by mod"]m=3 n=12n+3[/hide][/quote]

But then $ m$ only has the prime factor 3, and $ n$ is not necessarily a power of 3.

[hide="Solution"]Listing out some small examples, we see that $ (m,n) \equal{} (2,8)$ works since 2 and 8 have the same prime factors and 3 and 9 have the same prime factors.  Also, $ (m,n) \equal{} (6,48)$ works since 6 and 48 have the same prime factors and 7 and 49 have the same prime factors.  We notice that in both cases that $ n \plus{} 1$ is the square of $ m \plus{} 1$, which ensures that they have the same prime factors.  This means that $ n \equal{} (m \plus{} 1)^{2} \minus{} 1 \equal{} m(m \plus{} 2)$, so $ n$ and $ m$ have the same prime factors as long as the prime factors of $ m \plus{} 2$ are a subset of the prime factors of $ m$, which we can ensure by making $ m \plus{} 2$ a power of 2.  This gives us the infinite family of solutions

$ (m,n) \equal{} (2^{k} \minus{} 2,(2^{k} \minus{} 1)^{2} \minus{} 1)$

for $ k\ge 2$.[/hide]


[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14193[/img]

 :(

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14195[/img]

 :clap:


AMC
AIME
probability
USA(J)MO
USAMO
AMC 8
AIME I

I know that I got 9 correct answers on the AIME but I still did not qualify for the USAMO. Did this happen to anyone else? I was very careful about bubbling and filling out the sheet. I guess there is a small chance that I messed up on the sheet.... doubt it. Did this happen to anyone else? Any ideas?

PS I am a freshmen. hehe

They lopped off a few because there were too many qualifiers from the floor?

I am not so sure about that though. Why would they only include 37 then?

They probably couldn't read one of your bubbles. That happened to be on the AMC 8 last year. Wait until your AIME results come back; they usually tell you if they couldn't read one of the answers.

More likely it is the sad event of a misbubbling or processing error (or you didn't remember your answers right, but you probably recorded them on a slip of paper).  Or, (very unlikely), somebody cheated in the same room you took the test.

I am like the most careful bubbler on tests ever!! alright well i'll see what happens and post it later

The bubbles might be part of the problem.  Usually, to avoid that problem, I don't do any bubbling until the last 20 minutes of the test, after I have checked my answers.

JB

[quote="SpaceTerr808"]I am like the most careful bubbler on tests ever!![/quote]Other possibilities:
1. The score sheets were not sent
2. The score sheets were not sent on time
3. The score sheets were lost in the mail
4. AIME I was administered on the date of AIME II (this actually happened a few years ago and disqualified a 7th grader)

Regardless, call AMC to clear the suspense!  :)

[quote="JP"]another probablity is that your school did not choose to host USAMO.  because last year on USAMO qualifiers' letters it said in the beginning that you are gonna to take USAMO because you and your school chose to take it.[/quote]

I doubt it.

We were paranoid that one of our kids who had opted not to take the USAMO on the form would make it and not be invited.  He had a near perfect score on the AMC 10.  Our school principal wrote to the AMC Director and asked if in such an event, we could change our mind about taking the USAMO.

We got a reply from AMC that their policy is to invite everyone who qualifies even if their answer to that question was a 'no'.

It is clear that the AMC is more interested in promoting math talent than being a stickler about such insignificant things.

Possible bubbling error.


Λυστε τη διοφαντικη
$ 1\plus{}x\plus{}x^2\plus{}x^3\equal{}2^y$

Αν χ άρτιος, τότε $ 1 \plus{} x \plus{} x^2 \plus{} x^3$ περιττός, άρα $ y\equal{}0$ και άρα $ x\equal{}0$. Αν $ x \equal{} 2k\plus{}1$ περιττός, τότε $ 2^y \equal{} 4(k\plus{}1)(2k^2 \plus{} 2k \plus{} 1)$ και επειδή ο $ 2k^2 \plus{} 2k \plus{} 1$ είναι περιττός, πρέπει $ k\equal{}0$ ή $ k\equal{}\minus{}1$. Για $ k\equal{}0$ έχουμε την λύση  $ x\equal{}1,y\equal{}2$ ενώ για $ k\equal{}\minus{}1$ πρέπει $ 2^y \equal{} 0$ που είναι αδύνατον.

Άρα οι μόνες λύσεις είναι οι $ (0,0)$ και $ (1,2)$.

ειναι $ (x^2\plus{}1)(x\plus{}1)\equal{}2^y$Προφανως $ x\geq 0$
αν $ x \geq 3$
τοτε $ x\equiv \minus{}1 (mod4)$ τοτε $ x^2\plus{}1 \equiv 2(mod4)$ και αφου ειναι δυναμη του 2 ατοπο
αρα μονες λυσεις
$ (x,y)\equal{}(0,0)$ ή $ (1,2)$


Die A has 4 red faces and 2 green faces. whereas die B has 2 red faces and 4 green faces. A fair coin is flipped once. If it lands on a head, die A is tossed; otherwise, die B is tossed. Subsequently, the same die is tossed more than 2 times.

a. Find the probability of obtaining a red face in the first throw.

b. Find the probability that red faces turn up in the first two throws.

c. What is the probability that no two consecutive green faces occur in the three throws.

d. If the first two throws are both red, find the probability that
    i. a red face is obtained at the third throw.
   ii. it is die A that is being tossed.

[hide="a."]
$ \left(\frac{1}{2}\right) \cdot \left(\frac{4}{6}\right) \plus{} \left(\frac{1}{2}\right) \cdot \left(\frac{2}{6}\right)\equal{}\boxed{\frac{1}{2}}$

(note that this is probability that red will be obtained on any throw; then probability of obtaining green is also $ \frac{1}{2}$)[/hide]

[hide="b."]
$ \frac{1}{2} \cdot \frac{1}{2}\equal{} \boxed{\frac{1}{4}}$[/hide]

[hide="c."]
two consecutive green faces occur in these cases: $ ggr, rgg, ggg$ and the probability that it happens is $ 3 \cdot \left(\frac{1}{2}\right)^3\equal{}\frac{3}{8}$ so the probability that it doesn't happen is $ \boxed{\frac{5}{8}}$[/hide]

[hide="d."]
i. since every throw is independent, the probability is $ \boxed{\frac{1}{2}}$

ii. still independent, thus $ \boxed{\frac{1}{2}}$[/hide]

wont b be 5/18
c be 11/18
d ii be 4/5

as the toss is only done once?

You are right, I misinterpreted question.  :blush:


In a group of $ n$ people, any two persons have exactly one common friend. Prove that there exists a person in the group who is a friend of every other person in the group.

[i]Remark.[/i] If $ A$ is a friend of $ B$, then $ B$ is a friend of $ A$.

It is known as the friendship theorem...

http://matholymp.com/TUTORIALS/Friendship.pdf

Pierre.

[Moderator edit: See also http://www.mathlinks.ro/viewtopic.php?t=55513 ; http://www.mathlinks.ro/Forum/viewtopic.php?t=48500 .]

Thanks
my solution is quite different, so could you tell me is it correct??

Assume for contradiction that there isn't anybody who knows all of them.
Let $A$ be the vertex, which degree $n-1 \geq k \geq 4$ is the greatest. We know that $k$ is even, it's in this pdf so we won't prove this here.
Let $B_1,B_2...B_k$ be the vertices, which have a common edge with the $A$; $\psi=\{B_1,B_2...B_k\}$ and $\phi$ are the sets of vertices which don't have a common egde with $A$. We know that graph with vertices $A,B_1,B_2,...B_k$ is a "windmill" just like in this pdf. We know that $B_i$ has an even number $p_i$ of common edges with the vertices from $\phi$, the same prove like for $A$. Let $C_{i1},C_{i2},...C_{ip_i}$ be the vertices from $\phi$ which have a common edge with $B_i$. The graph with vertices $B_i,C_{i1},C_{i2},...C_{ip_i}$ must be a windmill too.
Let the $d$ be the number of elements in $\phi$. Consider $d-p_i$ pairs $(C_{ij}),x$ where for $x$ we will put vertex which doesn't have a common edge with $B_i$. Every new pair makes that in graph $W$ with vertices from $\phi$ we have two new edges in $W$, but this new edge cannot be between vertices which are adjacent to the same $B$ vertex. Because there are at least $(d-p_1)+(d-p_2)+...+(d-p_k)$ pairs so we have at least $\frac{1}{2}(2(d-p_1)+2(d-p_2)+...+2(d-p_k))=kd-(p_1+p_2+..+p_k)=kd-d=d(k-1)$ edges in graph $W$.There are exactly $d$ vertices in $W$ so one vertex has at least $k-1$ common edges with vertices from $W$. But this vertex has two more common edges: with $B_y$ and a vertex from $W$ which is with it in windmill's "wing", so it has at least $k+1$ edges. We have a contradiction because $k$ was the degree of the whole graph

it has been proven that only for l=1,2 it is possible to construct a graph
that every two nods are connected with one and excactly one path of length l containing
different nods .
As far as I remember , it was proved at first for l<9 , then 8<l<12
and finally by a tricky counting method for l>11.

The problem from the topic is a result for l=2 .


A carpenter is hanging molding that is $\frac{1}{8}$" thick around the ceiling of a rectangular room.  If the ceiling measures 18 ft 7 in by 14 ft 4 in, and molding is sold in 8-foot lengths, what is the minimum number of lengths the carpenter will need?

[hide]The perimeter of the ceiling is $2(18\cdot12+7)+2(14\cdot12+4)=790$.  In 8 feet, there are 96 inches.  $\frac{790}{96}\approx8.22$.  Round up because he has to buy the entire roll.  $\boxed{9}$[/hide]


induction
function
algebra unsolved
algebra

Let be given a sequence of real numbers $a_1,a_2,a_3,...$ satisfying:
$1< a_1< 2$ and$a_{n+1}=1+a_n-\frac{a_n^2}{2} \forall n\ge 1$.
Prove that:
$|a_n-\sqrt{2}| \le \frac{1}{2^n} \forall n\ge 3$

Notice that $a_{n+1}$ is never zero for $a_n\in(1,2)$, so it is strictly increasing or decreasing in this interval. Applying this, it is easy to see that,

$a_2\in(1,\frac{3}{2})$

$a_3\in(\frac{11}{8},\frac{3}{2})$

This gives $2^3(a_3-\sqrt{2})\in(11-8\sqrt{2},12-8\sqrt{2})\subset (-1,1)$, as required.

I will try to proceed by induction, as you might have guessed. :)

Now assume that $2^n(a_n-\sqrt{2})\in[ -1, 1 ]$.

Then $2^{n+1}(a_{n+1}-\sqrt{2})=2^{n+1}(1+a_n-\frac{1}{2}a_n^2-\sqrt{2})=[2^n(a_n-\sqrt{2})]( 2(1-\sqrt{2}) -\frac{1}{2^n}[2^n(a_n-\sqrt{2})] )$

The first factor is in $[-1,1]$ by assumption. The larger parantheses require some more work. $\frac{1}{2^n}[2^n(a_n-\sqrt{2})]$ is in $[-\frac{1}{2^n},\frac{1}{2^n}]$. $2(1-\sqrt{2})$ is about $-0.828$. So clearly the largest possible value does not exceed $1$. The smallest is at least $-0.828-1/8$ (since $n\geq 3$) which is still larger than $-1$. So,

$2^{n+1}(a_{n+1}-\sqrt{2})\in[-1,1]$.

And we're done.

consider the function $f(x)=1+x-x^2/2$for all$1<x<2$.$f'(x)=1-x$.By induction we have:$1<a_n<3/2$for all n>1.Using Lagrange theorem,we have:exists 1<c<3/2 such that:$(f(a_n)-f(sqrt{2}))/(a_n-sqrt{2})=f'(c)\Rightarrow |a_{n+1}-sqrt{2}|<1/2|a_n-sqrt{2}|$.by induction we have:$|a_n-sqrt{2}|<1/2^n$


Let $ X_1, X_2$ be any two r.v.'s (not necessarily independent but can be identically distributed) with corresponding CDFs $ F_1$ and $ F_2$. If $ F_1 \ge F_2$ on all of $ \mathbb{R}$, then is it possible for $ P(X_1 > X_2) > \frac{1}{2}$?

Yes, it is possible.  Let $ X_2$ have a uniform distribution on the set $ \{ 0, 1, 2 \}$.  Let $ X_1 \equal{} (X_2 \plus{} 1) \bmod 3$.  Then $ X_1$ and $ X_2$ have the same distribution, but $ \Pr(X_1 > X_2) \equal{} \frac{2}{3}$.

Hey, we've got a brand-new forum around here! We might as well use it. Let's post questions like this in the probability/statistics forum, OK?

Nice example, Ravi. Replace $ 3$ by $ n$ in the example, and it's clear that for any $ \epsilon>0,$ we can create identically distributed $ X_1$ and $ X_2$ such that $ P(X_1>X_2)>1\minus{}\epsilon.$


For first few values of $ n$, the relation is verified directly.
Then we use induction. Suppose up till $ n$ the relation is true.
For $ n+1$, there is $ 1$ set containing just $ n+1$. This set contributes with $ (n+1)^{2}$. Then $ n+1$ can be grouped with all subsets not containing $ n$, and these sets generate $ (n+1)^{2}(n!-1)$. And sets not containing $ n+1$ contribute with $ (n+1)!-1$.
We have to show that $ (n+1)^{2}+(n+1)^{2}(n!-1)+(n+1)!-1=(n+2)!-1$, which is quite obvious.

[quote="Erken"]Let $ S=\{1,2,\dots n\}$.Denote all possible subsets of $ S$,that don't contain consequtive natural numbers: $ A_{1},A_{2}\dots A_{k}$ for some $ k\in N$.If $ f(A_{i})$ is products of elements in $ A_{i}$ then prove that $ \sum_{i=1}^{k}(f(A_{i}))^{2}=(n+1)!-1$.[/quote]

Rather that $ S$, let $ S_{n}=\{1,2,\dots n\}$ and let $ a_{n}=\sum_{AppropriateSubsetsOfS_{n}}(f(A_{i}))^{2}$.
The subsets of $ S_{n}$ that don't contain consecutive natural numbers are :
$ 1).$ Those which don't contain the number $ n$ itself : the appropriate subsets of $ S_{n-1}$
$ 2).$ $ \{n\}$
$ 3).$ Those which contain the number $ n$ itself but not alone : $ \{n\}\cup A$ where $ A$ is any appropriate subset of $ S_{n-2}$

And so : $ a_{1}=1$, $ a_{2}=1^{2}+2^{2}=5$ and $ a_{n}=a_{n-1}+n^{2}+n^{2}a_{n-2}$, and hence the result.

Just a comment:
I had essentially the same solution as freemind.  This doesn't change the problem, but it might look nicer if you also included the empty set (with product $ 1$) and let the sum in question be $ (n+1)!$.


function
calculus
derivative
analytic geometry
graphing lines
slope
calculus computations

I have uploaded a picture of a graph and I need help figureing out which is the position, velocity, and acceleration functions.  Thanks for your help.

I'm a bit rusty, but:

A - Position
B - Velocity
C - Acceleration

This can be seen from the fact that at the peak of a function, the derivative will be 0.

Thanks for your help, but could someone please explain this in a little bit more detail.  I would really like to understand why this is that.

I can't make it work without adopting a "working hypothesis" and then checking through the properties to see whether it's possibly true or necessarily false.

Let's try the working hypothesis that [b]b[/b] is the position. Now look at the graph of [b]b[/b] and look for the slope. Starting from the left: the slope of [b]b[/b] starts out near zero, then tips over into sloping downwards. So it's derivative would have to start out near zero and then go negative. The graph of [b]b[/b] has a minimum at the axis, so its derivative will have to be zero there, and will have to be coming up towards zero as it approaches the axis. To the right of the axis [b]b[/b] is increasing, so its derivative has to be positive. Then it flattens out, so the derivative gets close to zero. In fact, it's slightly decreasing , so the derivative has to go slightly negative again.

So is the any curve that starts out near zero, goes negative, comes back to zero at the axis, goes positive, then comes back down to near zero and goes slightly negative. Does that describe any of these curves? It almost describes [b]c[/b]; it's missing the part about being slightly negative to the far right.

Maybe it's just a bad drawing? I don't like that, but let's run with it: assume it is a bad drawing and that [b]c[/b] is the derivative of [b]b[/b].

Now, what is the derivative of [b]c[/b]? It would have to be negative to the left, positive in a patch in the middle around the axis, then negative again to the right. That graph is clearly not here; in particular, that is not a description of [b]a[/b]. So the working hypothesis of [b]b[/b] as position, [b]c[/b] as velocity is false because [b]a[/b] isn't the acceleration.

Try again: could [b]a[/b] be the position? It starts out slightly increasing, so it's derivative should be slightly positive. Then it tips over to steeply decreasing, so its derivative must go substantially negative around and near the axis. Then it starts slowly increasing, so the derivative becomes slightly positive again. Slightly positive to the left, substantially negative in the middle around the axis, slightly positive to the right - that's a perfect description of [b]b[/b]. And we previously had [b]c[/b] as (almost) the derivative of [b]b[/b]. And the derivative of [b]c[/b] is nowhere to be found, so [b]c[/b] has to be the acceleration. So [b]a[/b] is the position, [b]b[/b] is the velocity, and [b]c[/b] is almost the acceleration, but the picture is wrong because at the far right, either [b]b[/b] shouldn't be decreasing or [b]c[/b] should be negative.

(dozerl- did you re-draw the picture from some original? What exactly do [b]b[/b] and [b]c[/b] do at the far right in the original?)

Yes I did redraw the picture.  [b]b[/b] and [b]c[/b] approach y = 0 but never actually do.

But if a postive function $f(t)$ tends "nicely" to zero as $t\to\infty,$ then $f'(t)$ must be negative for large $t.$ Do you understand what's wrong with that picture?


A and B are whole numbers and ${A/11}+{B/3}={31/33}$

[quote="xpmath"][hide]That means that $3A+11B=31$
And you can immediately see that $A=3,B=2$[/hide][/quote]

Another approach would be to take cases of what $B$ can be equal to, because $A$ and $B$ are whole numbers. Since there are very few cases, this is also a possibility.

yah i got that too
[hide] you know that 3*11=33. multiplying the equation by the common denominator 33, and finding what values of A and B satisfy the resulting equation, you get A=3 and B=2[/hide]

woOt!

Also, A and B are integers when $B \equiv 2 (\mbox{mod }3)$.

I couldn't solve this because I was thinking

What in the world does $/$ mean?
Use $\div$ or something next time.

Um, I think $/$ is definitely the preferred symbol over $\div$.  Look all over the forum and through the AoPS books and articles.

im gonna have to agree with archimedes on that one..
/ is kinda the universal symbol.. :oops:

I personally go with the fraction symbol, since it also is used for ratios and such.

Do I hear a little voice in my head saying *lock*?  Back on topic please...after all, use \frac{*numerator here*}{*denominator here*} .

Ex, for your problem:  \frac{A}{11}+\frac{B}{3}=\frac{31}{33}  is  $\frac{A}{11}+\frac{B}{3}=\frac{31}{33}$


Chocolate costing $ \$10$ per pound is mixed with nuts costing $ \$4$ per pound to make a mixture costing $ \$6$ per pound. What fraction of the mixture's weight is chocolate? Express your answer as a common fraction.

6 is 1/3 of the way from 4 to 10, so the answer is 1/3.

That explanation may work for many people, but not necessarily for all.

If you take a certain amount of Chocolate (say, x pounds) and add to it a certain amount of nuts, (y), then you have that

10X+4Y=6(x+y); expanding gives
10X+4Y=6X+6Y; moving around gives
4X=2Y,
X=1/2Y, and that the ratio of X (chocolate) to all (X+Y) is
1/2 Y to (1/2+1)Y is
1/3.


geometry
parallelogram
geometric transformation
homothety
ratio
geometry proposed

A proposed Geometry problem (I thing significant)

Is there possible to exist parallelogram ABCD similar to a given one such that there is one internal point P such that PA=a,PD=b and PC=c where a, b, c are given segments?

Sotiris E. Louridas

(*) I have given one solution plane to the Greek forum ‘www.mathematica.gr’

Welcome dear [b][size=100]Sotiris[/size][/b] in this Forum with interesting problems and some mathlinkers, very strong in Geometry.

$ \bullet$ This problem is familiar with the one posted [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=161987][b]Here[/b][/url], two years ago.

So, based on the remark of the generalization which has been mentioned there, we can construct a triangle $ \bigtriangleup AC'D'$, similar to the half of the given parallelogram, with the point $ P'$ inwardly to it, such that $ \frac {P'A}{P'D} \equal{} \frac {a}{b}$ and $ \frac {P'A}{P'C} \equal{} \frac {a}{c}$.

It is easy after by a homothety, to take the triangle $ \bigtriangleup ACD,$ with the point $ P$ inwardly to it, such that $ PA \equal{} a$ and $ PD \equal{} b$ and $ PC \equal{} c,$ as the half of the desired parallelogram and the solution of the proposed problem, will be completed.

Best regards, Kostas Vittas.

PS. I am sorry dear all my friends because I deleted my previous post with a wrong solurtion ( Evidently, I was asleep  :(  ).

The plane of the solution
We consider one parallelogram EZHQ similar to the ABCD
(E-A, Z-B, C-H and D-Q) and want to be exist one interpoint
$ P_1$(internal of the parallelogram EZHQ) such that
$ \frac{{P_1 Q}}{{P_1 E}} \equal{} \frac{b}{a} \wedge \frac{{P_1 H}}{{P_1 Q}} \equal{} \frac{c}{b}.$  So the point $ P_1$  must be the section of two circles of Apollonius with basics ,EQ and ratio $ \frac{b}{a},$QH and ratio$ \frac{c}
{b}$ respectively .
The investigation is essential because mast to obtain what the  section is one interpoint.

S.E.Louridas