Yes, and, more generally :
$i^n = \left\{\begin{array}{ }{1\text{ if } n = 4k\\i \text { if } n = 4k+1\\-1 \text{ if }n = 4k+2\\-i \text{ if }n = 4k+3}\end{array}\right$

:)

I basically remove multiples of $i^4=1$.

Yes, $k \text{ and } n \in \mathbb{N}$. :)

To help u remeber when dealing with complex rnumbers just remember the saying: (I won, I won!)

$\displaystyle \\ \text {I} \Rightarrow i \\ \text {won,} \Rightarrow 1 \\ \text {I} \Rightarrow i \\ \text {won!} \Rightarrow 1\\ \text {Also the negatives go in the middle so you have: } \\ i^1=i \\ i^2=-1 \\ i^3=-i \\ i^4=1$
 :P

:rotfl: That is a nice way!
Thanks for the share. :)


Let a,b,c,d be non-negative real numbers such that :a^2+b^2+c^2+d^2=1
Find the maximum value of the expression:
Q=(1-a)(1-b)(1-c)(1-d)

hello, by the method of Lagrange Multipliers i have got
$ Q\le\frac{1}{2}(3\minus{}2\sqrt{2})$
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, by the method of Lagrange Multipliers i have got
$ Q\le\frac {1}{2}(3 \minus{} 2\sqrt {2})$
Sonnhard.[/quote]
You can post other solutions?


Βάζω αυτό που είχα προαναφέρει . Δε λέω πολλά , θα μου πείτε εσείς τη γνώμη σας .
[hide]Νάτη για σένα [/hide]

[color=red]1 .Ανισότητα Αριθμητικού-Γεωμετρικού Αρμονικού Μέσου (ΑM-GM-HM)[/color]

Αν $a_{1},...,a_{n}>0$  τότε
$\frac{a_{1}+a_{2}+...+a_{n}}{n}\geq\sqrt[n]{a_{1}...a_{n}}\geq\frac{n}{\frac{1}{a_{1}}+...+\frac{1}{a_{n}}}$

[color=blue]ΧΡΗΣΗ[/color]
Γενικά όταν μας δίνουν το γινόμενο αριθμών και μας ζητούν το ελάχιστο κυκλικού ή συμμετρικού αθροίσματος .
Επίσης όταν μας δίνεται το άθροισμα συμμετρικών ή κυκλικών παραστάσεων και θέλουμε να βρούμε το μέγιστο του γινομένου ή του αρμονικού αθροίσαμτος .
Επιπρόσθετα όταν τα μονόνυμα των αθροισμάτων  δεν έχουν κοινό συντελεστή βολεύει να τα ανάγουμε στο ίδιο συντελεστή
(δηλαδή αν θέλουμε να <<μειώσουμε>> την παράσταση $3x^{2}y+2y^{x}$ [b]συνήθως[/b] βολεύει να τη γράψουμε στη μορφή $x^{2}y+x^{2}y+x^{2}y+y^{x}+y^{x}$)
Τέλος αν έχουμε κλάσματα προσπαθούμε να φτιάξουμε παρονομαστές που να απαλοίφονται ώστε μέσα τη ρίζα να έχουμε αριθμό απλά.
[color=brown]Για εξάσκηση [/color]
Αν $a,b,c$ είναι πραγματικοί αριθμοί να αποδειχθεί ότι
$\frac{a^{2}-b^{2}}{2a^{2}+1}+\frac{b^{2}-c^{2}}{2b^{2}+1}+\frac{c^{2}-a^{2}}{2c^{2}+1}\leq 0$

2.Ανισότητα Buniakowski -Cauchy-Swartz (BCS)
Αν $a_{1},..a_{n},b_{1},....,b_{n}$  είναι πραγματικοί αριθμοί τότε
$(a_{1}^{2}+...+a_{n}^{2})(b_{1}^{2}+...+b_{n}^{2})\geq (a_{1}b_{1}+...a_{n}b_{n})^{2}$

Άμμεση απόρια είναι η ανισότητα Andreescu ή λήμμα Cauchy ή Cauchy in Engel form (πάντως αν τα διορθώνει ο Κοντογιάννης μην γράψετε Andreescu γιατί θα σας μηδενίσει  :P )
Αν $x_{1},...x_{n}\in R$ και $a_{1},...a_{n}\in R^{+}$ τότε
$\frac{x_{1}^{2}}{a_{1}}+...+\frac{x_{n}^{2}}{a_{n}}\geq\frac{(x_{1}+...+x_{n})^{2}}{a_{1}+...+a_{n}}$

και η γενίκευσή της
Αν $0\leq r\leq m-1$ και $a_{1},...a_{n},b_{1},...,b_{n}>0$  τότε
$\frac{a_{1}^{m}}{b_{1}^{r}}+...+\frac{a_{n}^{m}}{b_{n}^{r}}\geq\frac{(a_{1}+...+a_{n})^{m}}{n^{m-r-1}(b_{1}+...+b_{n})^{r}}$

[color=green]Χρήση[/color]
Όταν έχουμε γινόμενο αθροίσματος άθροισμα τετραγώνων ,
και όταν έχουμε κλασματικές μορφές χρησιμοποιούμε Andreescu
Με κατάλληλες διασπάσεις προσπαθούμε να εφαρμόσουμε την παρπάνω .
(π.χ αν θέλουμε να δείξουμε ότι $(a^{2}+1)(b^{2}+1)\geq (ab+1)(a+b)$

Tότε βλέπουμε ότι το δεύτερο μέλος έχει δύο <<μορφές Cauchy >> . Για να την δείξουμε πρέπει να τις καλύψουμε και τις δύο .
Συνεπώς πρώτα $(a^{2}+1)(b^{2}+1)\geq (ab+1)^{2}$
και μετά $(a^{2}+1)(1+b^{2})\geq (a+b)^{2}$
Οπότε με πολλαπλασιασμό κατα μέλη προκύπτει η ζητούμενη.
Επίσης στον Andreescu ένα ισχυρό κόλπο είναι ο πολλαπλασίασμός αριθμιτή και παρονομαστή για την εξεύρεση τετραγώνων .
(π.χ $\frac{a}{x}=\frac{a^{2}}{ax}$ και $\frac{a^{2}}{xm}=\frac{a^{4}}{a^{2}xm}$ )
Και τα δύο μας βγάζουν από δύσκολες περιπτώσεις και επιπλέον
μπορεί μια ανισότητα να μη βγαίνει (να γυρνάει ανάποδα με Andreescu κατευθείαν ενώ με με πολλαπλασιασμό αριθμητή και παρονομαστή να βγαίνει .Αυτό έγγυται στη διαφορά δύναμης βαθμού  των μεταβλητών που δημιουγούμε )

[color=orange]Εφαρμογή [/color]
Αν $a,b,c\geq 0$ να δειχθεί ότι
$\frac{a^{2}}{2b^{2}-bc+2c^{2}}+\frac{b^{2}}{2c^{2}-ca+2a^{2}}+\frac{c^{2}}{2a^{2}-ab+2b^{2}}\geq 1$

3. Ανισότητα Αναδιάταξης
Αν οι ακολυθίες $(a_{1},...a_{n})$ kai $(b_{1},...,b_{n})$ έχουν την ίδια διάταξη και $c_{1},..,c_{n}$ είναι οι αριθμοί $(b_{1},...,b_{n})$ με διαφορετική ίσως σειρά τότε
$a_{1}b_{1}+...+a_{n}b_{n}\geq a_{1}c_{1}+..+a_{n}c_{n}$
Αν  οι ακολυθίες $(a_{1},...a_{n})$ kai $(b_{1},...,b_{n})$ έχουν αντίθετη  διάταξη και $c_{1},..,c_{n}$ είναι οι αριθμοί $(b_{1},...,b_{n})$ με διαφορετική ίσως σειρά τότε
$a_{1}b_{1}+...+a_{n}b_{n}\leq a_{1}c_{1}+..+a_{n}c_{n}$

[color=red]Χρηση [/color]
Μεγάλο όπλο για την απόδειξη ανισοτήτων που είναι ομοιογενείς
αλλά κυκλικές όχι συμμετρικές .Αρκεί κάθε φορά να βρούμε τις κατάλληλες διατεταγμένες ν-αδες
Όταν έχουμε κλασματικές μορφές κοιτάζουμε μήπως ισχύει αυτή οι διάταξη μεταξύ αριθμητών και η ίδια ή αντίστοιχη μεταξύ παρονομαστών

ΕΦΑΡΜΟΓEΣ
1 Αν $a,b,c\geq 0$ νδο $a^{3}+b^{3}+c^{3}\geq a^{2}b+b^{2}c+c^{2}a$

2. Να δειχθεί ότι an $a,b,c>0$
$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{3}{2}$  Aνισότητα Nesbitt

[color=red]4. Ανισότητα Tschebychev [/color]
Αν οι ακολυθίες $(a_{1},...a_{n})$ kai $(b_{1},...,b_{n})$ έχουν την ίδια διάταξη  τότε
$\frac{a_{1}+...+a_{n}}{n}\cdot\frac{b_{1}+...+b_{n}}{n}\leq\frac{a_{1}b_{1}+...+a_{n}b_{n}}{n}$

Αν οι ακολυθίες $(a_{1},...a_{n})$ kai $(b_{1},...,b_{n})$ έχουν αντίθετη διάταξη   τότε
$\frac{a_{1}+...+a_{n}}{n}\cdot\frac{b_{1}+...+b_{n}}{n}\geq\frac{a_{1}b_{1}+...+a_{n}b_{n}}{n}$

ΧΡΗΣΗ
Tην χρησιμοποιούμε όταν έχουμε γινόμενο κυκλικών ή συμμετρικων παραστάσεων και οι δυνάμεις αυτών δεν είναι βολικές για Cauchy

ΕΦΑΡΜΟΓES
1 .Αν $a,b,c>0$ να δείξετε ότι
$(a^{5}+b^{5}+c^{5})(a^{2}+b^{2}+c^{2})\leq 3(a^{7}+b^{7}+c^{7})$

2.Αν $a_{1},...a_{n}>0$ και $a_{1}+...+a_{n}=k$  να δειχθεί ότι
$\sum_{i=1}^{n}\frac{a_{i}}{2k-a_{i}}\geq\frac{n}{2n-1}$

[color=blue]5 Ανισότητα Holder (Γενίκευση της Cauchy )[/color]

Aν $m_{1},..,m_{k}$ θετικοί me $m_{1}+...+m_{k}=1$ και $b_{i}, x_{i},a_{i}>0$  τότε
$(a_{1}+...+a_{n})^{m_{1}}(b_{1}+...+b_{n})^{m_{2}}...(x_{1}+...+x_{n})^{m_{k}}\geq a_{1}^{m_{1}}b_{1}^{m_{2}}...x_{1}^{m_{k}}+....+a_{n}^{m_{1}}b_{n}^{m_{2}}...x_{n}^{m_{k}}$

ΧΡΗΣΗ
Ίδια χρήση με αυτή της Cauchy απλά είναι πιο ισχυρή αφού δουλεύει για κάθε δύναμη και όχι μόνο για τετράγωνα .

ΕΦΑΡΜΟΓΕΣ
1. Αν  $a,b,c>0$  νδο $9(a^{3}+b^{3}+c^{3})\geq (a+b+c)^{3}$

2.An $a^{4}+b^{4}+c^{4}=3$  να δειχθεί ότι
$\frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a}\geq\frac{3}{2}$

6.Ανισότητα των βαρών (θα την βαλλω με δύο τρόπους)
 α) Γνωστή απλή μορφή
Αν $a_{i},x_{i}>0$ με $x_{1}+....+x_{n}=1$  τότε
$x_{1}a_{1}+...+x_{n}a_{n}\geq a_{1}^{x_{1}}...a_{n}^{x_{n}}$

 b)Γενικευμένη μορφή
Αν $x_{1},...x_{n}\geq 0$ και  $w_{1},...w_{n}\geq 0$ με θετικό άθροισμα τότε έστω η συνάρτηση
$f(r)=(\frac{w_{1}x_{1}^{r}+...+w_{n}x_{n}^{r}}{w_{1}+...+w_{1}})^{\frac{1}{r}}$

η οποία είναι μια μη φθίνουσα συνάρτηση και με τη συμφωνία ότι
για $r=0$ προκύπτει η ανισότητα του α)
Γενικά η συνάρτηση είναι γνωσίως αύξουσα εκτός αν όλα τα $x_{i}$ είναι ίσα και πιθανόν όταν $r\leq 0$ και κάποια $x_{i}$ είναι μηδέν τότε η $f$ είναι η ταυτοτική συνάρτηση  .

7. Ανισότητα των δυνάμεων
Για $w_{i}=1$ στην παραπάνω παίρνουμε την ανισότητα των δυνάμεων

ΧΡΗΣΗ
Συνήθως χρησιμοποιύνται σε εκθετικές ανισότητες οι οποίες
εκφυλίζονται ορισμένες φορές αμέσως .
Επίσης παίζοντας με τα μονόνυμα και λύνοντας συστήματα για να προκύψουν οι κατάλληλες δυνάμεις μπορούμε με την ανισότητα των βαρών να λύσουμε πολύ κομψά δύσκολες ασκήσεις

8. Ανισότητα Schur
An $x,y,z$ είναι μη αρνητικοί αριθμοί kai $m>0$ τότε
$a^{m}(a-b)(a-c)+b^{,}(b-c)(b-a)+c^{m}(c-a)(c-b)\geq 0$

Γενικά όμως ισχύει η ανισότητα Vornicu -Schur που υποσστηρίζει ότι αν $f$ είναι μια συνάρτηση που παίρνει θετικές τιμές και η $f$ είναι αύξουσα ή κυρτή τότε
$f(a)(a-b)(a-c)+f(b)(b-c)(b-a)+f(c)(c-a)(c-b)\geq 0$

(είχα παρουσιάσει μια απόδειξή της στο παρελθόν )
Επίσης η ανισότητα για $m=1$ και $a+b+c=1$ παίρνει τη μορφή
$\frac{1+9abc}{4}\geq ab+bc+ca$ την οποία την χρησιμοποίησε μια φορά ο Ράλλης και μια φορά εγώ στο ανισότητες 2 .

ΓΕΝΙΚΕΥΣΗ
Ανισότητα Suranyi
Αν $a_{1},...a_{n}$ θετικοί πραγματικοί τότε
$(n-1)(a_{1}^{n}+..+a_{n}^{n})+na_{1}..a_{n}\geq (a_{1}+..+a_{n})(a_{1}^{n-1}+...+a_{n}^{n-1}$

(Αποδείξτε την !!!)
****Σημείωση
Υπάρχει και περαιτέρω γενίκευση η επονομαζόμενη Zaimi-Marku
αλλά δεν θεωρώ απαραίτητο να τη βάλλω .

ΧΡΗΣΗ
Είναι πολύ ισχυρή ανισότητα βοηθάει πάρα πολύ σε απόδειξη με απαλοιφή όπου χρησιμοποιούνται συμμετρικά αθροίσματα .
Επίσης μια ένδειξη για τη χρησιμοποίησή της είναι η περίεργη τιμή ισότητας .Δηλαδη αν μας δίνουν μια ανισότητα να την αποδείξπυμε και δούμε ότι η ισότηα ισχύει για $a=b$ και $c=0$ ας πούμε τότε αυτό είναι μια πολύ καλή ένδειξη για τη χρησιμοποίησή της .

[color=red]9 .Ανισότητα Maclaurin [/color]
Αν $a_{1},...,a_{n}\geq 0$ τότε
$S_{1}\geq S_{2}\geq...\geq S_{n}$ όπου
$S_{k}=\sqrt[k]{\frac{\sum a_{1}...a_{k}}{\binom{n}{k}}}$

XΡΗΣΗ
Βοηθάει πολύ όταν πάμε να αποδείξουμε μια ανισότητα με τη μέθοδο των συμμετρικών πολυωνύμων δηλαδή θέτοντας
$x+y+z=s$ ,   $xy+yz+zx=q$  , $xyz=w$  κτλ

10.Ανισότητα Muirhead
Θα λέμε ότι η ν-άδα $a_{1},...,a_{n}$ μεγιστοποιεί τη ν-άδα  $b_{1},...,b_{k}$ εάν
$a_{1}\geq...\geq a_{n}$ και $b_{1}\geq b_{2}...\geq b_{n}$ και για κάθε $k=0,...n-1$ ισχύει
$a_{1}+...+a_{k}\geq b_{1}+...+b_{k}$  και $a_{1}+..a_{n}=b_{1}+...+b_{n}$

kai θα συμβολίζουμε
$\left(a_{1},...,a_{n}\right)\succ \left(b_{1},...,b_{n}\right)$

Έτσι για παράδειγμα
$\left(4,2\right)\succ \left(3,3\right)$  και
$\left(4,1,1,0\right)\succ \left(3,2,1,0\right)$

Αν λοιπόν τώρα $a_{1},a_{2},a_{3},b_{1},b_{2},b_{3}$ πραγματικοί αριθμοί
και $x,y,z>0$ τότε
$\sum_{sym}x^{a_{1}}y^{a_{2}}z^{a_{3}}\geq \sum_{sym}x^{b_{1}}y^{b_{2}}z^{b_{3}}$

ΠΡΟΣΟΧΗ !!! Τα αθροίσματα είναι συμμετρικά
[color=blue]ΧΡΗΣΗ [/color]
Όταν τίποτε άλλο δεν πιάνει κάνουμε απαλοιφή παρονομαστών ή παρενθέσεων και χρησιμοποίούμε Muirhead για τα συμμετρικά αθροίσματα που βγαίνουν στην πορεία .Σε συνδυασμό με την Schur είναι ένας πολύ καλός <<δολοφόνος >> ανισοτήτων .
ΕΦΑΡΜΟΓΕΣ
1.Αν $a,b,c>0$ τότε $(1+a^{3})(1+b^{3})(1+c^{3})\geq (1+abc)^{3}$
2. An $x,y>0$ τότε $x^{5}y+xy^{5}\geq x^{4}y^{2}+y^{4}x^{2}$

Καλή ανάγνωση .... :)

Σιλουανός...

Τέλεια Σιλουανέ! Με μια ματιά που τα έριξα βλέπω πως έχεις κάνει καταπληκτική δουλειά! Σε ευχαριστούμε για τον κόπο που έκανες να γράψεις όλα τα παραπάνω. Θα μας βοηθήσουν πολύ! Είσαι άξιος συγχαρητηρίων !!

 :thumbup:  :coolspeak:  :10:  :clap:  :clap2:

Απλά θέλω να πω κάτι .Οι Εφαρμογές που έχω βάλλει παραπάνω είναι πολύ απλές για δύο λόγους (τουλάχιστον οι περισσότερες) .
α)Δεν ήθελα να ανοίξω κανένα βιβλίο.
β) Τα έγραψα όλα όπως τα έχω στο μυαλό μου για να θυμάμαι τις διάφορες ανισότητες .Θυμόμουνα και δυο-τρεις δύσκολες και τις έβαλα έτσι για κερασάκι στην τούρτα  :D

Σιλουανέ συγχαρητήρια :clap2:
τα βρίσκω όλα υπέροχα και νομίζω θα με βοηθήσουν πολύ στο διάβασμά μου για τον αρχιμήδη :lol:
θα μπορούσες plz να βάλεις τις λύσεις στις ασκήσεις της ανισότητας Holder?  :!:
δυστυχώς εγώ δεν μπόρεσα να τις βρω :oops:

Κατάλαβα :lol:
thnx πολύ mostel :coolspeak:

Σου άρεσε τελικά Nick ?Το έγραψα ως μαθητής  :wink: ?

Συγχαρητήρια, Σιλουανέ!
Εξαιρετική δουλειά
:coolspeak: :thumbup:  :clap2:

oxi!!!

[hide]to egrapses os metr.....[/hide]

Ετσι ετσι  ... :D  :D

Καλά υπερβάλλεις..... :)   .
Το διάβασες αναλυτικά ?
Αλήθεια ασχολήθηκε κανεις με όλες τις εφαρμογές ?
Η δεύτερη εφαρμογή στην ανισότητα Holder είναι αρκετά ζόρηκη αλλά πολύ όμορφη.....
Αξίζει επίσης να πω ότι η ανισότητα που προτείνω για πρώτη εφαρμογή της AM-GM είναι αυτή ουσιαστικά που με έκανε να αρχίσω να ασχολούμαι ένθερμα με τις ανισότητες αφού στον Προκριματικό Διαγωνισμό Νέων το 2005 που έλαβα μερος ήταν το μόνο θέμα που δεν είχα λύσει .....
O Nick τότε είχε βρει μια πολύ ωραία και σύντομη λύση ενώ και κάποιοι άλλοι έλεγαν ότι την έλυσαν με BCS αλλά είχαν βάλλει κάτω από τη ρίζα αρνητικούς αριθμούς .  :P
Θυμάσαι Nick ? :D

@silouan:μπορεις να βαλεις και τις αποδειξεις τους?μην παμε και τις μπαζουμε μεσα στις ακισεις αβερτα..νακσερουμε τι μας γινεται...

Πολύ καλά έκανες και ρώτησες Χρήστο! Άλλωστε, όπως έχω πει και παλαιότερα, δεν έχει νόημα να χρησιμοποιείς κάτι, αν δεν ξέρεις από που προήλθε αυτό!

[b]Όμως[/b], μερικές από τις παραπάνω ανισότητες έχουν αρκετά δυσνόητες αποδείξεις. Π.χ. εγώ αυτές που ξέρω για ΑΜ-ΓΜ βασίζονται είτε στο λήμμα Ehlers, είτε σε επαγωγή. Δεν ξέρω όμως κατά πόσο σου είναι γνωστά αυτά, αφού η επαγωγή ουσιαστικά διδάσκεται στη Β' Λυκείου (και αν). Πάντως, δυστυχώς, πιστεύω πως καλύτερα είναι να μάθεις αρχικά να τις χειρίζεσαι σε εφαρμογές. Επίσης, αν ενδιαφέρεσαι για αποδείξεις, υπάρχουν όλες λεπτομερώς είτε στο wikipedia, είτε στο mathworld.

Κοίτα Χρήστο . Για ΑΜ-GM και Cauchy -Swartz έχω τις απδείξεις στο ειδικά αρθράκια που είχα κάνει στα οποία είχα δώσει λινκ .
Η ανισότητα της αναδιάταξης αποδεικνύεται με στοιχειώδη τρόπο αλλά δεν λέει κάτι (τουλάχιστον για εμένα),το ίδιο και η Tscebychev .Για τη Holder έχω αποδείξεις στο μυαλό μου που χρησιμοποιούν ανάλυση άρα δεν υπάρχει λόγος  , για τη Schur και τη Vornicu Schur υπάρχουν αποδείξεις που είχα βάλλει στο τόπικ ΑΝΙΣΟΤΗΤΕΣ 1 και για τις Muirhead και Maclaurin έχω δει τις αποδείξεις τους αλλά δεν τις ξέρω απέξω .

gia tin proti tis am-gm tha valo ti lysi pou ekana tora kai oxi auti pou eixa kanei tote.....
pollaplasiaste to proto melos epi 4 kai meta spaste ta kasmata  kai telo am-gm...

Η εφαρμογή 1 της ανισότητας 10 δεν λύνεται πολύ πιο απλά με ανισότητα Holder :?:
 :oops: Γιατί για να πω την αμαρτία μου δεν την κατάλαβα την ανισότητα 10 (πολύ μπερδεμένη μου φαίνεται...) :blush:
*τι να κάνω?...ο καθένας με όσα ξέρει

[b]Silouane[/b] Poli kali douleia! (eksipakouete alla aisthanome kai anagkasmenos na se eyxaristhsw gia to post giati se kapoio allo meros tou forum hsoun apogohteumenos gia tetoiou eidous tutorial-post logw tis xamilis adapokrisis  :wink: )

[quote="bilot4"]Η εφαρμογή 1 της ανισότητας 10 δεν λύνεται πολύ πιο απλά με ανισότητα Holder :?:
 [/quote]

Ναι πράγματι . Αλλά αυτές απλά είπα είναι εφαρμογές της στιγμής που είχα εκείνη την ώρα στο μυαλό μου .Όσο για την Muirhead διάβασε ξανά και ξανά τη θεωρία που γράφω και μετά προσπάθησε να λύνεις τις εφαρμογές.

ok...thnx για τις συμβουλές :lol:

Σε παρακαλώ μην τις λες συμβουλές , γιατί αισθάνομαι άσχημα , εξάλλου ποιος είμαι εγώ να συμβουλεύω τους άλλους . :oops:
Πάρτο καθαρά ως συμπαράσταση στο δυσκολο αγώνα σου  :wink:

Ok silouane...oti nomizeis :)
aplos dn 8eoro oti einai kako na dineis symvoules...
os pio empeiros, exeis olo to dikaioma :wink:
 :D
Eyxaristo opos kai na' xei...

Μπάβο Σιλουανέ  :wink:
Μπορεί κάποιος να μου ποiα είναι η ανισότητα Minkowski?

Στην ουσία είναι η:

$|\vec a+\vec b|\leq|\vec a|+|\vec b|$

Ή

$\sqrt[n]{(a_{1}+b_{1})^{n}+\ldots+(a_{k}+b_{k})^{n}}\leq\sqrt[n]{a_{1}^{n}+\ldots+a_{k}^{n}}+\sqrt[n]{b_{1}^{n}+\ldots+b_{k}^{n}}$

Εννοείται ότι είναι θετικά αυτά  :)

 :lol:

Υπάρχουν και πολλά άμεσα πορίσματα της Minkowski τα οποία βρίσκονται στο βιβλίο του κύριου Στεργίου. Δυστυχώς δεν μπορώ να τα γράψω γιατί δεν έχω το πρόγραμμα. Συγχαρητήρια Σιλουανέ.

Επίσης η ανισότητα Nesbitt λύνεται πιο απλά με Tschebychev και αν θυμάμαι καλά με ένα ωραίο τρικ που χρησιμοποιεί αρμονικό μέσο. Στη wikipedia πρέπει να το είχα βρει.

Πως μπορουμε να αποδειξουμε οτι

$ \sqrt{(x_1\plus{}y_2)^2\plus{}(x_2\plus{}y_2)^2\plus{}...\plus{}(x_n\plus{}y_n)^2}\leq$ $ \sqrt{x_1^2\plus{}x_2^2\plus{}...\plus{}x_n^2}\plus{}\sqrt{y_1^2\plus{}y_2^2\plus{}...\plus{}y_n^2}$

Νομίζω βγαίνει έτσι:
όπως είπε και ο mostel $ |\vec x \plus{} \vec y|\leq|\vec x| \plus{} |\vec y|$
Αν θεωρήσουμε τα διανύσματα $ \vec x$ και $ \vec y$ στο n-διάστατο χώρο,  :o
το μέτρο του αθροίσματός τους είναι το $ \sqrt {(x_1 \plus{} y_2)^2 \plus{} (x_2 \plus{} y_2)^2 \plus{} \ldots \plus{} (x_n \plus{} y_n)^2}$
ενώ το άθροισμα των μέτρων ξεχωριστά είναι το $ \sqrt {x_1^2 \plus{} x_2^2 \plus{} \ldots \plus{} x_n^2} \plus{} \sqrt {y_1^2 \plus{} y_2^2 \plus{} \ldots \plus{} y_n^2}$.
Δεν είμαι σίγουρος για την απόδειξη, ούτε την έχω δεί πουθενά, οπότε αν είναι χαζομάρα πείτε μου!

mia proxeiri apodeiksi:
tetragwnise k B-C-S

Athinaios ευχαριστω..

Orfikoss, βασικα αυτο που ηθελα να αποδειξω ειναι οτι $ |\vec x \plus{} \vec y|\leq|\vec x| \plus{} |\vec y|$ για δυο διανυσματα του $ \mathbb{R}^n$

[quote="giannis18"]EDIT: ΛΑΘΟΣ[/quote]

:P , εκανα delete την απορια μου γιατι μου λυθηκε αμεσος μολις κοιταξα το sticky με τις ανισοτητες οπου ο Silouan χρησημοποιει τη συγκεκριμενη ανισοτητα η οποια τελικα προκειπτει απο Holder

Και έψαχνα σα χαζός να δω πού πήγε η απορία σου και έλεγα "παραισθήσεις έχω"; :rotfl:  :rotfl:


geometry
3D geometry
calculus
calculus computations

We have a circle with radius $ 1$. Then we cut off a sector. We make a cone from the rest. What angle should the cut sector have, so that the volume of the cone will be the greatest.

[hide="My thoughts:"]Volume of the cone, $ V \equal{} \frac {1}{3}\pi r^2 h \equal{} \frac {\pi}{3}(1 \minus{} h^2)h$.

Differentiating with respect to $ h$ and equating with $ 0$ for the maxima (since minima will be observed when $ \theta \equal{} 0$), we get $ h \equal{} \sqrt {\frac {1}{3}}$, hence, $ x \equal{} \sqrt {\frac {2}{3}}$.

The curved surface = $ 2\pi x r \equal{}$ Area of Sector Used for the cone.
Further calculations should be simple.
[/hide]

Thank you, Rajiv.

Yes, now it should be simple. My problem was, that I started from the other side (first expressed $ r$ with respect to $ \phi$ and then $ h$ w.r.t. $ r$. The expression for $ V$ came out to be a "little" complicated.

Let's generalize a bit. If the circle has initial radius $ r_1$, then it also has an initial circumference $ C_1$. This relation is given by $ C_1 \equal{} 2\pi r_1$. Now picture the circle w/ radius $ r_1$ from above. We shear out a pizza slice shaped portion that makes up $ \frac {\varphi}{2\pi}$ of the disk. The beauty here is that because circles are truly uniform, the amount of area/perimeter removed with respect to the whole is directly proportional to $ \varphi$. We now apply this to our equation for $ C_1$ to obtain a closed form for $ C_2$. Here,
\[ \displaystyle C_2 \equal{} C_1\left(1 \minus{} \frac {\varphi}{2\pi}\right)
\]
After subbing the formula for $ C_1$ back in we may express $ C_2$ in the following format:
\[ \displaystyle C_2 \equal{} 2\pi\left(r_1 \minus{} \frac {r_1 \varphi}{2\pi}\right) \equal{} 2\pi r_2\Rightarrow \boxed{r_2 \equal{} r_1\left(1 \minus{} \frac {\varphi}{2\pi}\right)}
\]
I insisted on the overhead visualization because $ r_2$ is the radius of the cone's base and $ r_1$ its slant height. The altitude of the cone is found by solving for $ h$ in
\[ \displaystyle h^2 \plus{} (r_2)^2 \equal{} (r_1)^2\Rightarrow \boxed{h \equal{} \frac {r_1}{2\pi}\sqrt {4\pi\varphi \minus{} \varphi^2}}
\]
From there it's fairly straightforward using the volume of a cone.
 :)
[b]Edit:[/b] The final expression ($ \text{d}V/\text{d}\varphi$) may seem complicated at first, but it factors easily and finding the roots is practically trivial.


modular arithmetic
number theory
number theory proposed

Find all pairs $ (p, q)$ of primes such that $ {p}^{p}\plus{}{q}^{q}\plus{}1$ is divisible by $ pq$.

i won't consider trivial cases
we must just use the fact that if $ p^p \equiv \minus{} 1 \pmod {q}$ then either $ ord_q(p) \equal{} 2 \implies q | p \plus{} 1$ or $ ord_q(p) \equal{} 2p \implies 2p |q \minus{} 1$
and for the first problem that u posted
i'm sure i saw it in this forum but i can't find it :(

Here is my approach.

If $ p\equal{}q$ or one of $ p, q$ is $ 2$, we can get $ (p, q)\equal{}(2, 5), (5, 2)$.

So let $ p, q$ are odd, and WLOG $ p<q$.

Since $ {q}^{q}\plus{}1$ is divisible by $ p$, we can get a interger $ k$ that satisfies $ {q}^{q}\plus{}1\equal{}kp$.

So, we can get $ ({q}^{q}\plus{}1)^{p}\plus{}k^p\equal{}k^p(p^p\plus{}1)$.

But, since $ p^p\plus{}1$ is divisible by $ q$, $ k^p\plus{}1\equiv0(mod q)$.

So, $ k^{2p}\equiv1(mod q)$. So, $ ord_qp$ is a divisor of 2p.

So, it can be $ 1, 2, p, 2p$.

Since $ k^p\plus{}1\equiv0(mod q)$, it can be $ 2, 2p$.

If it is $ 2$, $ k^2\equiv1(mod q)$. So $ k\equiv1(mod q)$ or $ k\equiv\minus{}1(mod q)$.

But $ k^p\plus{}1\equiv0(mod q)$. So $ k\equiv\minus{}1(mod q)$.

Let $ k\equal{}bq\minus{}1$. Then $ q^q\plus{}1\equal{}(bq\minus{}1)p$. So $ p\equiv\minus{}1(mod q)$.

So, $ q$ is the divisor of $ p\plus{}1$. But since $ p<q$, contradiction.

So, we know that $ ord_qp \equal{} 2p$.

But, because or Fermat's Little Theorem, $ p^{q\minus{}1}\equiv1(mod q)$.

So, $ 2p$ is a divisor of $ q\minus{}1$.

If we switch $ p, q$ and do this process again, we can get $ 2p$ is a divisor of $ q\plus{}1$ or $ 2q$ is a divisor of $ p\minus{}1$.

But, since $ 2q>2p>p\minus{}1$, $ 2q$ can't be a divisor of $ p\minus{}1$. So, $ 2p$ is a divisor of $ q\plus{}1$ and $ q\minus{}1$.

So, $ 2p$ is a divisor of $ 2$. Contradiction!

So the answer is $ (p, q)\equal{}(2, 5), (5, 2)$.

[quote="Stephen"]Here is my approach.

If $ p \equal{} q$ or one of $ p, q$ is $ 2$, we can get $ (p, q) \equal{} (2, 5), (5, 2)$.

So let $ p, q$ are odd, and WLOG $ p < q$.

Since $ {q}^{q} \plus{} 1$ is divisible by $ p$, we can get a interger $ k$ that satisfies $ {q}^{q} \plus{} 1 \equal{} kp$.

So, we can get $ ({q}^{q} \plus{} 1)^{p} \plus{} k^p \equal{} k^p(p^p \plus{} 1)$.

But, since $ p^p \plus{} 1$ is divisible by $ q$, $ k^p \plus{} 1\equiv0(mod q)$.

So, $ k^{2p}\equiv1(mod q)$. So, $ ord_qp$ is a divisor of 2p.

So, it can be $ 1, 2, p, 2p$.

Since $ k^p \plus{} 1\equiv0(mod q)$, it can be $ 2, 2p$.

If it is $ 2$, $ k^2\equiv1(mod q)$. So $ k\equiv1(mod q)$ or $ k\equiv \minus{} 1(mod q)$.

But $ k^p \plus{} 1\equiv0(mod q)$. So $ k\equiv \minus{} 1(mod q)$.

Let $ k \equal{} bq \minus{} 1$. Then $ q^q \plus{} 1 \equal{} (bq \minus{} 1)p$. So $ p\equiv \minus{} 1(mod q)$.

So, $ q$ is the divisor of $ p \plus{} 1$. But since $ p < q$, contradiction.

So, we know that $ ord_qp \equal{} 2p$.

But, because or Fermat's Little Theorem, $ p^{q \minus{} 1}\equiv1(mod q)$.

So, $ 2p$ is a divisor of $ q \minus{} 1$.

If we switch $ p, q$ and do this process again, we can get $ 2p$ is a divisor of $ q \plus{} 1$ or $ 2q$ is a divisor of $ p \minus{} 1$.

But, since $ 2q > 2p > p \minus{} 1$, $ 2q$ can't be a divisor of $ p \minus{} 1$. So, $ 2p$ is a divisor of $ q \plus{} 1$ and $ q \minus{} 1$.

So, $ 2p$ is a divisor of $ 2$. Contradiction!

So the answer is $ (p, q) \equal{} (2, 5), (5, 2)$.[/quote]

I guess when I was busily writing this, he had already finished his writing and posted before I post my work.

[quote="srinath.r"][quote="Stephen"]
So, $ k^{2p}\equiv1(mod q)$. So, $ ord_qp$ is a divisor of 2p. [/quote]
Sorry stephen ,but shouldnt $ o_{q}(k) \mid 2p$ instead of $ o_{q}(p) \mid 2p$

[quote="Stephen"]
So, we know that $ ord_qp \equal{} 2p$.

But, because or Fermat's Little Theorem, $ p^{q \minus{} 1}\equiv1(mod q)$.

So, $ 2p$ is a divisor of $ q \minus{} 1$.
[/quote]

Doesnt this coincide with the previous "quote" reply ? Sorry if I am wrong .[/quote]

I agree that my solution is wrong. I will edit my solution...

If $ p \equal{} q$ or one of $ p, q$ is $ 2$, we can get $ (p, q) \equal{} (2, 5), (5, 2)$.

So let $ p, q$ are odd, and WLOG $ p < q$.

Since $ {q}^{q} \plus{} 1$ is divisible by $ p$, we can get a interger $ k$ that satisfies $ {q}^{q} \plus{} 1 \equal{} kp$.

So, we can get $ ({q}^{q} \plus{} 1)^{p} \plus{} k^p \equal{} k^p(p^p \plus{} 1)$.

But, since $ p^p \plus{} 1$ is divisible by $ q$, $ k^p \plus{} 1\equiv0(mod q)$.

So, $ k^{2p}\equiv1(mod q)$. So, $ ord_qk$ is a divisor of $ 2p$.

So, it can be $ 1, 2, p, 2p$.

Since $ k^p \plus{} 1\equiv0(mod q)$, it can be $ 2, 2p$.

If it is $ 2$, $ k^2\equiv1(mod q)$. So $ k\equiv1(mod q)$ or $ k\equiv \minus{} 1(mod q)$.

But $ k^p \plus{} 1\equiv0(mod q)$. So $ k\equiv \minus{} 1(mod q)$.

Let $ k \equal{} bq \minus{} 1$. Then $ q^q \plus{} 1 \equal{} (bq \minus{} 1)p$. So $ p\equiv \minus{} 1(mod q)$.

So, $ q$ is the divisor of $ p \plus{} 1$. But since $ p < q$, contradiction.

So, we know that $ ord_qk \equal{} 2p$.

But, because or Fermat's Little Theorem, $ k^{q \minus{} 1}\equiv1(mod q)$.

So, $ 2p$ is a divisor of $ q \minus{} 1$.

If we switch $ p, q$ and do this process again, we can get $ 2p$ is a divisor of $ q \plus{} 1$ or $ 2q$ is a divisor of $ p \minus{} 1$.

But, since $ 2q > 2p > p \minus{} 1$, $ 2q$ can't be a divisor of $ p \minus{} 1$. So, $ 2p$ is a divisor of $ q \plus{} 1$ and $ q \minus{} 1$.

So, $ 2p$ is a divisor of $ 2$. Contradiction!

So the answer is $ (p, q) \equal{} (2, 5), (5, 2)$.

I hope it is right. :(

$pq|p^p+q^q+1$
clearly $gcd(p,q)=1$
w.l.o.g $p>q$
$q=2 \implies p=5$
Now let $p>q>2$
$p|q^q+1 \implies p|q^{2q}-1$
Let $Ord_p(q)=a$
$a|2q$
$a=1 \implies p|q-1$ which is impossible.
$a=2 \implies p|q^2-1 \implies p|q+1 \implies p=q+1$ which is impossible.
$a=q \implies p|q^q-1 \implies p|2$ which is impossible.
So $a=2q \implies 2q|p-1 \implies q|p-1 \implies q|p^p-1 \implies q|2$ which is impossible.
So our only solutions are $(p,q)=\boxed {(2,5)} ,\boxed {(5,2)}$

[quote=stephencheng][quote="Stephen"]Find all pairs $ (p, q)$ of primes such that $ {p}^{p} \plus{} {q}^{q} \plus{} 1$ is divisible by $ pq$.[/quote]

WLOG we let at first $ q>p$.
Now $ p|p^p\plus{}q^q\plus{}1 \Rightarrow q^{2q} \equiv 1 \pmod{p}$. As $ q>p$ we have $ \gcd(2q,p\minus{}1)\equal{}1$ or $ 2$.
So $ q^2 \equiv 1 \pmod{p} \Rightarrow q\equal{}kp \pm 1$ for some $ k \in N$

If $ q\equal{}kp\plus{}1$ we have $ p^p\plus{}q^q\plus{}1 \equiv 2 \equiv 0 \pmod{p} \Rightarrow p\equal{}2 \Rightarrow (p,q)\equal{}(2,5)$
If $ q\equal{}kp\minus{}1$ we have $ p^{2p} \equiv 1 \pmod{q}$ . As $ \gcd(2p, kp\minus{}2)\equal{}1$ or $ 2$ or $ 4$ , we have $ p^4 \equiv 1 \pmod{q} \Rightarrow q|(p\plus{}1)(p^2\plus{}1) \Rightarrow p\equal{}2 or kp\minus{}1|p\plus{}k \Rightarrow p\equal{}2$ or $ 3 \Rightarrow (p,q)\equal{}(2,5)$

So $ (p,q)\equal{}(2,5)$ is the only solution.[/quote]

but (2,3) is solution


[quote="Jane_Bennet"]I think if you use the substitution $ x \equal{} \tan{\frac {\alpha}{2}},\ y \equal{} \tan{\frac {\beta}{2}},\ z \equal{} \tan{\frac {\gamma}{2}}$, where $ \alpha \plus{} \beta \plus{} \gamma \equal{} \pi$, you can reduce the equations to

$ \frac {3}{\sin{\alpha}}\ \equal{} \ \frac {4}{\sin{\beta}}\ \equal{} \ \frac {5}{\sin{\gamma}}$

Hence $ \alpha,\ \beta,\ \gamma$ are the interior angles of a 3–4–5 triangle.[/quote]
  with the same way  we see that $ 3^2\plus{}4^2\equal{}5^2$,so  $ \gamma\equal{}\frac{\pi}{2}$ so $ z\equal{}\tan{\frac {\gamma}{2}}\equal{}1.  y\equal{}\frac{1}{2}$,and $ x\equal{}\frac{1}{3}$


[img]http://img209.imageshack.us/img209/7929/adszwi4.png[/img]
Thanks and best regards!

Can you show me your solution way ?
Thanks and best regards!

Which line is the number $ 15$ labeling?

[hide="Solution"]Using simple trigonometry, we see that $ \displaystyle \sin 15^\circ = \frac {3\sqrt {2}}{CE} \implies CE = 6\sqrt {3} + 6$.
Similarly, we can determine that $ \displaystyle DC = 3\sqrt {6} + 6\sqrt {2}$.
Now, by the angle-angle similarity theorem, we can see that $ \triangle ABC \sim \triangle DEC$.
Setting up a ratio, we have:
\begin{align*} & \frac {EC}{BC} = \frac {DC}{AC} \\
& \implies \frac {6\sqrt {3} + 6}{x + \left(3\sqrt {6} + 6\sqrt {2}\right)} = \frac {3\sqrt {6} + 6\sqrt {2}}{3 + \left(6\sqrt {3} + 6\right)} \\
& \implies \frac {6\sqrt {3} + 6}{x + 3\sqrt {6} + 6\sqrt {2}} = \frac {\sqrt {6} + 2\sqrt {2}}{2\sqrt {3} + 3} \\
& \implies 30\sqrt {3} + 54 = 24\sqrt {3} + 42 + x\left(\sqrt {6} + 2\sqrt {2}\right) \\
& \implies 6\sqrt {3} + 12 = x\left(\sqrt {6} + 2\sqrt {2}\right) \\
& \implies \boxed{x = 3\sqrt{2}} \end{align*}
[/hide]

[hide]
123456789's method works, but the solution is actually $ 3\sqrt{2}$, which I verified on a calculator
[/hide]

Yea, sorry. Solved it wrong. It's fixed now.

[quote="fishythefish"]Is something wrong on my computer, because where the picture is supposed to be it just says "Image".[/quote]Works fine for me...is the word "Image" a hyperlink?

Thank [b]123456789[/b]. Your solution is very nice.

Nope. This is what it looks like:

Image

It's exactly the same as the other text.


WoW!!!! New stuff, it's always great. Thanks Mr. rusczyk!

[quote="Mindspa"]Don't post nonrelated topics.[/quote]

Sorry... this was posted about right after it was created... but anyways, my apologies.


let $ a_1 \plus{} a_2 \plus{} ... \plus{} a_k \equal{} 2.n!$ such that :$ a_i \in$ {$ 1,2,...,n$}
 is there a subset of $ a_i$ like $ {b_1,...,b_l}$ that  $ n!|b_1 \plus{} b_2 \plus{} ... \plus{} b_l$


Consider the line determined by the points $A=(-4,7)$ and $B=(78,19)$. If a fly walks along this line $\frac{1}{5}$ of the way to $B$ from $A$, at what point will he be standing?


calculus
integration
trigonometry
function
LaTeX
algebra
rational function

Compute

$ \int_0^{2\pi}\frac{1\plus{}2cos\theta}{5\plus{}4cos\theta} d\theta$

 :)

Is there a general method?  :huh:

Consider, in the complex plane, the unit circle as our contour, and denote it by $ C$. On this contour, $ z \equal{} e^{i \theta}$, and so $ d \theta \equal{} \frac{dz}{iz}$. Also, $ \cos \theta \equal{} \frac12 \left(z \plus{} \frac{1}{z}\right)$. Hence,

$ \int_0^{2 \pi}\frac{1 \plus{} 2 \cos \theta}{5 \plus{} 4 \cos \theta}\,d \theta \equal{} \oint_C \frac{1 \plus{} z \plus{} 1/z}{5 \plus{} 2z \plus{} 2/z} \cdot \frac{dz}{iz} \equal{} \minus{}i \oint_C \frac{1 \plus{} z \plus{} 1/z}{2z^2 \plus{} 5z \plus{} 2}\,dz$

$ \equal{} \minus{}i \oint_C \underbrace{\frac{z^2 \plus{} z \plus{} 1}{2z(z \plus{} \frac12)(z \plus{} 2)}}_{\equal{} f(z)}\,dz \equal{} \minus{}i \cdot 2 \pi i \left(Res[f(z), 0] \plus{} Res[f(z), \minus{} \frac12]\right)$

$ \equal{} 2 \pi \left(\frac12 \minus{} \frac12\right) \equal{} 0$.

A minor visual note: Use \cos for the function in $ \text{\LaTeX}$. Most standard functions have a pretty version like this.

A method for finding antiderivatives of any rational function of $ \sin x$ and $ \cos x$: substitute $ t=\tan\frac{x}{2}$. We get $ dx=\frac{2}{1+t^2}\,dt$, $ \sin x=\frac{2t}{1+t^2}$, and $ \cos x=\frac{1-t^2}{1+t^2}$.
This transforms the integral into a rational function of $ t$, which can then be handled by the usual partial fractions method.

This particular integral is over a full period, which would be $ \int_{-\infty}^{\infty}\,dt$. That makes other methods that don't rely on finding antiderivatives reasonable.


Hello all,
[quote]In how many ways can 5 seniors, 3 juniors, 4 sophomores, and 3 freshmen be seated in a row if a senior must be seated at each end? Assume that the members of each class are distinct.[/quote]

In my attempt Ii found the answer to be 1 x 3! x 3! x 3! x 4! x 1= 5184 possible arrangements. Is this correct?

Please do not post solutions.

thanks in advance.

I got a different answer.

[hide="Hint"]
First seat the seniors on the end. How many ways are there to place on a senior of the left-most end? And then the right-most end?
[/hide]

There are 5 arrangements on the left most end and 4 arrangements on the right most end.

Would it then be 5 x 3! x 3! x 3! x 4! x 4 = 103680?

Thanks for your quick response.

[quote="AndrewTom"]I agree that theree are $ 20$ ways of seating a senior at each end. But the question doesn't state that the juniors (for example) must sit togetgher (the seniors can't). So isn't it $ 20\times 13!$?[/quote]
That makes sense. Putting a senior at each end leaves us with 3 seniors, 3 juniors, 4 sophomores and 3 freshmen, for a total of 13 students.

No order has been specified whether it should've been S S S J J J So So So F F F. I think this brings up another question which I will post in another thread.

Can someone confirm this?

Thanks both of you for your responses.

edited message:
[quote="chrischris"]I agree with AndrewTom.

Since the members of each class are distinct, we only need to divide them into seniors and non-seniors.[/quote]
20 x 13! = 124,540,416,000 arrangements

Thanks for your response!


trigonometry
limit
search
real analysis
real analysis unsolved

Let $ (a_n)_{n\in N }$ where $ a_n \equal{} n^2|sin(n)|$ ,$ n\in \mathbb{N}$.Prove that $ a_n$  doesn't have limit

If $ \{a_n\}$ was convergent, then it would be Cauchy. Therefore for each $ \epsilon>0$, there should exist $ n_{0}$, such that for any $ n, m\geq n_0$, we should have $ | a_n-a_m |<\epsilon$.

Choosing $ m=2n$, we get: $ | 4n^{2}|\sin(2n)|- n^{2}|\sin(n)| |<\epsilon$, which is equivalent to  $ n^{2}|\sin(n)(8|\cos{(n)}|-1) |<\epsilon$ and the latter to $ |\sin(n)(8|\cos{(n)}|-1) |<\frac{\epsilon}{n^2}$.

Now we send $ n \to  \infty$ and conclude $ \lim_{n \to \infty}(|\sin(n)(8|\cos{(n)}|-1) |)=0$.  (1)

Probably it is obvious, that the last statement if false, but we can argue in this way:

$ \sin{n}>\frac{\sqrt{2}}{2}$ infinitely many times, since moving through the integers, we can not jump over $ (\frac{\pi}{{4}}+2k\pi, 3\frac{\pi}{4}+2k\pi)$, $ k\in \mathbb{N}$ since its lenght is $ \frac{\pi}{2}>1$.

Therefore we can construct an infinite subsequence $ n_{1}, n_{2},\ldots$, (we take all for which that holds) for which $ \sin{n}>\frac{\sqrt{2}}{2}$. For (1) to hold, $ 8|\cos{(n)}|-1 |$ must converge to $ 0$ for this subsequence, which means, that $ |\cos{(n)}|$ should converge to $ \frac{1}{8}$. But then also $ \sin{n}$ should converge on this subsequence. Therefore $ n_{i} (\mathrm{mod} 2\pi)$ should be $ \frac{\pi}{2}-k$ or $ \frac{\pi}{2}+k$, for some $ k<\frac{\pi}{4}$, when $ i \to \infty$. But since the difference between $ n_i$ and $ n_{i+1}$ is at most $ 6$ and $ 2\pi$ is not $ 6$, consecutive integers in the sequence ($ \mathrm{mod} 2\pi$) must jump between $ \frac{\pi}{2}-k$ and $ \frac{\pi}{2}+k$. But then $ 2k=1$ on one hand and $ 2(\pi-k)=6$ or $ 5$ od the other. Contradiction.

there is an open question:
 Does there exist subsequence $ n_{k}$ such that :
$ n^{2}_{k}|\sin n_{k}| \to 0$ as $ k \to \infty$ ?
 (this is about irrationality measure of $ \pi$

suppose $ a(n)$ converge to  a real number $ L$ , then for any subsequence $ n(k)$ of positive integer

with n(k) tend to +oo when k tend to +oo

$ a((k))$ converge also to $ L$

sin(n) is dense in interval -1 ; 1

consider a subsequence $ n(k)$ such that sin(n(k)) tend to 1 when k tend to +oo

then $ n^2(k)$ is equivalent to $ L$ wich is a contradiction

I agree that it's obvius question if $ L < \infty$.
Let's change the question:
Prove or disprove :
$ \lim n^{2} |\sin n | \equal{} \infty$
 Now Arguments about  fundamental sequence or density of $ \sin n$ is out of question...
you need to think how well $ 0$ can be approximated by $ \sin n$..

Supose $ i_n$ is a sequence of natural numbers such that $ sin(i_n)\rightarrow 0$
$ \lim_{n\to \infty }i_n^2|sin(i_n|)$ $ \equal{} \lim_{n \to\infty }i_n\frac {|sin(i_n)|}{\frac {1}{i_n} }$ $ \equal{} \lim_{n \to \infty }i_n \equal{} \infty$
Is that ok? :maybe:

[quote="Extremal"][quote="c_adryan"]
$ \lim_{n \to\infty }i_n\frac {|sin(i_n)|}{\frac {1}{i_n} }$ $ \equal{} \lim_{n \to \infty }i_n$
Is that ok? :maybe:[/quote]
this is not true . why ?[/quote]
yes..i was wrong :blush:

so..does $ n^2|sin(n)|$ goes to $ \infty$ or not?

Let $ \mu$ be the irrationality measure of $ \pi$, then for any $ \epsilon>0$ there are infinitely many pairs $ m,n\in \mathbb{N}$ such that $ \left\vert n \pi \minus{} m\right \vert < \frac {1}{n^{\mu \minus{} 1 \minus{} \epsilon }}$ (see, for example, [url=http://mathworld.wolfram.com/IrrationalityMeasure.html]here[/url]).  This implies that there are infinitely many $ m\in \mathbb{N}$ such that $ \left \vert \sin m\right \vert < \frac {1}{\left(3m\right)^{\mu \minus{} 1 \minus{} \epsilon}}$, and hence $ \liminf_{m\to \infty}\left \vert m^k \sin m\right \vert \equal{} 0$ if $ k < \mu \minus{} 1$.  So, $ \lim _{m\to \infty}\left\vert m^2 \sin m\right \vert \neq \infty$ if one can show that $ \mu > 3$.  Conversely, if $ \mu < 3$ then it follows similarly $ \lim_{m\to \infty}\left\vert m^2 \sin m\right \vert \equal{} \infty$.

A quick search suggests that the best known bounds are something like $ 2\leq \mu \leq 8.0161$, so it seems the answer to your question is as yet unknown.


inequalities
quadratics
trigonometry
function
calculus
derivative
inequalities proposed

If x,y,z are the angles of a triangle prove that

1) (secx/2)^2  + (secy/2)^2  + (secz/2)^2  \geq 4

2) secx  + secy + secz  \geq 6

3) (secx)^2 + (secy)^2 + (secz)^2  \geq 12 in acute triangles

4) (secx)^2 + (secy)^2 + (sec)^2  >3        in obtuse triangles

5) secxsecy + secysecz +seczsecx  \geq 12

2) Use the well-known relation
cos(x) + cos(y) + cos(z) = 1 + r/R and Euler's inequality R  \geq  2r to deduce that cos(x) + cos(y) + cos(z)  \leq  3/2.
Then use that for a,b,c > 0 we have (a+b+c)(1/a + 1/b + 1/c)  \geq 9, to deduce (2).

(3) Follows directly from (2) and from the inequality between quadratic and arithmetic means.

(4) is trivial since for every real a =/= 0 mod[\pi /2], we have 1/(sec(a)) 2 . Summing gives (3).

Pierre.

(5) may be proved with the equality (*):
cos(x)cos(y) + cos(y)cos(z) + cos(z)cos(x) = (p 2  + r 2  - 4R 2 )(4R 2 ),
where p is the semi-perimeter.

Using that (a+b+c) 2/3  \leq  a 2  + b 2  + c 2  = 8R 2 (1 + cos(x)cos(y)cos(z) )  \leq 9R 2   (from the well-known inequality cos(x)cos(y)cos(z)  \leq  1/8),
we deduce that p 2  \leq  27R 2 /4.
Using Euler's inequality, it follwsfrom (*) that
cos(x)cos(y) + cos(y)cos(z) + cos(z)cos(x)  \leq  3/4.
Now, from (u +v + w)(1/u + 1/v + 1/w)  \geq 9 for every u,v,w > 0, we get (5).

Pierre.

(1) Note that 1/(sec(x/2)) 2 = 1 + (tg(x/2)) 2 .
Summing analoguous relations, we get :
1/(sec(x/2)) 2 + 1/(sec(y/2)) 2 + 1/(sec(z/2)) 2 = 3 + (tg(x/2)) 2 + (tg(y/2)) 2 + (tg(z/2)) 2. (**)

But is is easy to prove that :
tg(x/2)tg(y/2) + tg(y/2)tg(z/2) + tg(z/2)tg(x/2) = 1.
Since each of these numbers is nonnegative, it follows from Cauchy's inequality that
(tg(x/2)) 2 + (tg(y/2)) 2 + (tg(z/2)) 2   \geq  tg(x/2)tg(y/2) + tg(y/2)tg(z/2) + tg(z/2)tg(x/2) = 1.
Using (**), we get the conclusion.

Pierre.

Note that these inequalities are best possible in the sense that each equality case occurs in (1), (2), (3), (5) iff the triangle is equilateral.
And that, in (4) we may approach from 3 as near as we want with triangles close from the degenerated one with angles 180, 0, 0.

Pierre.

[quote="manlio"]I don't think this is a solved problem as the very interesting case of obtuse triangles is not considered here.

For example it is very interesting to prove

2) secx + secy + secz  \geq  6

when x,y,z are angles of an obtuse triangle.

Am I right ?[/quote]

But it's not true for obtuse triangles.

I will rename the angles of the triangle into A, B, C. Then your inequalities are:

[b]-[/b] The following inequalities hold for all triangles ABC:

[b]1.[/b] $\sec^{2}\frac{A}{2}+\sec^{2}\frac{B}{2}+\sec^{2}\frac{C}{2}\geq 4$.

[b]4.[/b] $\sec^{2}A+\sec^{2}B+\sec^{2}C \geq 3$.

[b]-[/b] The following inequalities hold not for all, but for all acute triangles ABC:

[b]2.[/b] $\sec A+\sec B+\sec C \geq 6$. (For an obtuse triangle, this is clearly wrong: if one of the angles is > 90�, but very near to 90�, then its cosine is very negative and hence the sum can never get $\geq 6$).

[b]3.[/b] $\sec^{2}A+\sec^{2}B+\sec^{2}C \geq 12$.

[b]5.[/b] $\sec B \sec C+\sec C \sec A+\sec B \sec C \geq 12$. (Again, by the same argument as for [b]2.[/b], this is wrong for an obtuse triangle ABC).

Concerning the proofs, for number [b]1.[/b] I would use Jensen. In fact, the function $f( x) = \sec^{2}x$ is convex in the interval ${[ 0;\;\frac{\pi}{2}}]$. Before you start proving this by differentiation, let me tell you why this is trivial: The function $f( x) = \sec^{2}x$ is the composition of the function sec x, which is convex in ${[ 0;\;\frac{\pi}{2}}]$, and the function $x^{2}$, which is monotonically increasing and convex on the positive halfaxis. But one can easily see that if $R$ is the set of the real numbers and $I$ and $J$ are two intervals on the real number axis (so we have $I\subseteq R$ and $J\subseteq R$), and if we have two functions $f: I\mapsto J$ and $g: J\mapsto R$ which are convex on the respective intervals where they are defined (i. e. the function f is convex on $I$, and the function g is convex on $J$), and if the function g is monotonically increasing on the interval $J$, then the function g(f) is convex on $I$, too. Hence, our function $f( x) = \sec^{2}x$ is convex in the interval ${[ 0;\;\frac{\pi}{2}}]$.

Thus we can use Jensen (the angles $\frac{A}{2}$, $\frac{B}{2}$, $\frac{C}{2}$ all lie in the interval ${[ 0;\;\frac{\pi}{2}}]$) and get

  $ \sec^{2}\frac{A}{2}+\sec^{2}\frac{B}{2}+\sec^{2}\frac{C}{2}\geq 3 \sec^{2}\frac{( \frac{A}{2}+\frac{B}{2}+\frac{C}{2})}{3}= 3 \sec^{2}\frac{A+B+C}{6}= 3 \sec^{2}\frac{\pi}{6}= 4$,

proving the inequality [b]1.[/b].

Exactly the same reasoning shows that [b]3.[/b] holds for all acute triangles ABC.

The inequality [b]2.[/b] for an acute triangle ABC is a consequence of Jensen for the convex function sec x.

As Pierre showed, the inequality [b]4.[/b] is trivial, but it cannot be sharpened.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=4972]Finally I would add another inequality for acute triangles ABC:

[b]6.[/b] ${{{( \sec A-1) }\cdot ( \sec B-1) }\cdot ( \sec C-1) }\geq 1$.

But this is not so easy.[/url]

  Darij


Which base 8 positive 3-digit integer when multiplied by 2 has its digits switched? (i.e $[abc]_{8}*2 = [cba]_{8}$).

[hide]Wow, i was being dumb. :/

A is even and has to be less than 4 at the same time, so a = 2.
Then c would have to be either 1 or 5 to satisfy this, so obviously c is 5.

Then we can write an equation to find b...

2(5+8b+2*64) = 2 + 8b + 5*64

10 + 16b + 4*64 = 2 + 8b + 5*64.

8b = 64 - 8

b = 8-1 = 7.

[b]275 [/b][/hide]


function
inequalities theorems
inequalities

what's VASC's Right-Convex Function Theorem

Find in vasc's book  :wink: Have you got that book?

it is a little similar to Half concave and half convex theorem


A quadrilateral with perpendicular diagonals is inscribed into a circle with radius 13cm. One of the diagonals has length of 18cm and the distance between the center of the circle and the intersection of the diagonals equals 4 square roots of 6 cm. Find the area of the Quadrilateral.

[b]1.[/b] Let the two diagonals be $ AB,CD$ and let them intersect at $ E$. Let $ O$ be the centre of the circle. Have wolog $ AB\equal{}18$. Let $ M$ be midpoint of $ AB$, wolog $ E$ lies on $ AM$. Then, as $ O$ is the centre and $ AB$ is a chord, $ OM\perp AB$. Also, $ AM\equal{}\frac{AB}{2}\equal{}\frac{18}{2}\equal{}9$.
[b]2.[/b]Now, $ OA\equal{}13$ so $ OM\equal{}\sqrt{AO^2\minus{}AM^2}$ as $ OM\perp AB$ at $ M$ so $ OM\equal{}2\sqrt{22}$. Then, as $ OE\equal{}4\sqrt{6}$ and $ OM\equal{}2\sqrt{22}$ and $ \angle{OME}\equal{}90^o$ then $ ME\equal{}\sqrt{OE^2\minus{}ME^2}\equal{}\sqrt{96\minus{}88}\equal{}\sqrt{8}\equal{}2\sqrt{2}$
[b]3.[/b] Let $ N$ be the midpoint of $ CD$. Then, $ ON\perp CD$ at $ N$ and as $ AB\perp CD$ then $ NEMO$ is a rectangle. Then, $ ON\equal{}ME\equal{}2\sqrt{2}$. Then, as $ \angle{CNO}\equal{}90^o$ then $ CN\equal{}\sqrt{CO^2\minus{}ON^2}\equal{}\sqrt{169\minus{}8}\equal{}\sqrt{161}$.
[b]4.[/b] $ CD\equal{}2CN\equal{}2\sqrt{161}$. So, area of $ ABCD$ is $ \frac{1}{2}AB*CD*sin{AEC}\equal{}\frac{1}{2}18*2\sqrt{161}\equal{}18\sqrt{161}$.

The area of the quadrilateral is $ 18\sqrt{161}$


geometry
rectangle
combinatorics open
combinatorics

A devil chases an angel on an infinite chessboard. At each move, the devil can eliminate one of the squares, and the angel can make a leap in any direction, covering a distance of at most n squares. Here, n is a positive integer previously fixed, and is called the "power" of the angel. The devil's aim is to trap the angel on an island surrounded by a hole of width at least n.

Can the angel indefinitely escape the devil, if his power is sufficiently high? Can the devil defeat an angel of any finite power?

Your thoughts?

Hmm the proof below is not very rigorous but still:

for closing her in a pxq field (or a field with this as circum-rectangle), the devil needs at least $4n^2 + 4pn + 4qn$ squares.

so that means the angel did 4n+4p+4q moves by then. So the rectangle (square in fact) must have its borders at least 4n+4p+4q moves away from the starting position.

$p \geq q \geq 1 + 8p + 8q + 8n$

So I'd say no. Even if the angel has smarter moves than what I have incalculated, it seems not possible to me. Although I'm aware that this is certainly NOT a good proof.

besides, I just discovered this tiny little link:
http://mathworld.wolfram.com/AngelProblem.html

hmm seems the problem has not been solved yet? incredible how easy-looking problems can be so hard :P

Yes, it has not been solved, but many simpler cases have been.

For example, an angel of power 1 can be trapped.

But even more amazing to me is that an angel of any power that promises to always increase (or never decrease by more than a finite number) its distance from the origin can be caught.

I've tried to modify the strategy used to catch this type of angel to catch the general angel. But it seems hard considering how many mathematicians have failed. Oh well, trying can only make me better. :)

hmm... do you have a proof of the angel with increasing power? seems awkward :)

I don't think you understand what i mean. Its power doesnt increase, but its distance from the origin does. Say it is at (10,10), it cant go to (9,9) because that is closer to the origin.

We define the angels laxity as the distance it can go toward the origin after it has moved a certain distance away from it. For example, an angel of laxity 1 on (10,0) can go to (9,0), but never any closer than that.

I am saying that any angel of finite power and laxity can be caught.

[quote="jeffrey"]Say it is at (10,10), it cant go to (9,9) because that is closer to the origin. We define the angels laxity as the distance it can go toward the origin after it has moved a certain distance away from it. For example, an angel of laxity 1 on (10,0) can go to (9,0), but never any closer than that.[/quote]

Eum.. I'm not getting this: why can't you go from 10,10 to 9,9 but can go from 10,0 to 9,0??

a couple questions:

can the angle move in a full 360-degree array of directions, or must it go left, up, right, or down?

(i.e.) from (0,0) can an angel of power 2 move to (1,1) or is it bounded to go to (+/-2,0) or (0,+/-2)

are diagonals considered 1 or 2 steps if you can move diagonally?

if an angel has a power of 2 does it have to move 2 every time or can it choose to move only 1?

[quote="tom11784"]if an angel has a power of 2 does it have to move 2 every time or can it choose to move only 1?[/quote]

yes: it says at most n :D

An angel of power n can move anywhere in the square with sidelength 2n+1 surrounding it.

Peter-I was unclear in my previous post. My first example referred to an angel with laxity 0. I then defined laxity and gave a similar example which made it confusing.

my question remains:

why CAN you go from 10,0 to 9,0 with a 1-angel? If you would've said from 12,10 to 12,9 I'd have understood, but 10,0 seems strictly further than 9,0 in this system...

Laxity is defined as the maximum distance the angel can go back to  the origin. Think of the angel's position in polar coordinates. If its laxity is x then an angel on ( :theta: , r) may never be inside the circle around the origin with radius r-x (he may be on the edge though).

The angel described in the problem has infinite laxity, but it has been proven that an angel with finite power and laxity can be caught.


How many inches are in the positive difference between 40 feet and 13 yards?

There are 3 feet in a yard. Thus, 13 yards is $ 13\times3\equal{}39$ feet. 40 feet minus 39 feet is 1 foot, which is $ \boxed{12}$ inches.


The figure below is composed of the rectangles $ A$, $ B$, $ C$, and $ D$ with whole number values for length and width. The areas of regions $ A$, $ B$, and $ C$ in square meters, are shown in the diagram. What is the area in square meters of rectangle $ D$?
[asy]draw((0,0)--(0,11)--(13,11)--(13,0)--cycle);
draw((0,6)--(13,6)); draw((8,11)--(8,0));
label("$A$",(0,11),SE); label("$B$",(8,11),SE); label("$C$",(8,6),SE); label("$D$",(0,6),SE);
label("40",(6,9)); label("25",(11,9)); label("30",(11,2)); label("?", (6,2));[/asy]

The side lengths of D are 8 and 6, so the area is $ \boxed{48}$.


There are $N$ cities in the country. Any two of them are connected either by a road or by an airway. A tourist wants to visit every city exactly once and return to the city at which he started the trip. Prove that he can choose a starting city and make a path, changing means of transportation at most once.

:maybe:  It seems not difficult by induction. Note that the statement is equivalent to the fact that if you colour the vertices of $K_{N}$ with $2$ colours (say $A$ and $B$, seen as sets of vertices) there always exists a cycle which passes exactly once for each vertex and in which there are at most $2$ changes of colour.  The base case $N=3$ is trivial.
Suppose it is true for $K_{N-1}$, consider the cyclical path in $K_{N-1}$ as the sequence of its vertices $a_{1},a_{2}\ldots a_{N-1}$  and choose a new vertex $V$. If the path is monochromatic by making $V=a_{N}$ we are done, since we add at most $2$ changes of colour. If not suppose  $a_{1}a_{2}\in A$ and $a_{2}a_{3}\in B$ and take without loss of generality $Va_{2}\in A$. Then if  $a_{1}V\in A$ (or $a_{3}V \in A$), the new path where $a_{1},V,a_{2}$  (or $a_{2},V,a_{3}$) are consecutive and the orther between the rest of vertices is mantained has also $2$ changes of colour. Now, if $a_{2}V\in B$, then the path in which $a_{1},V,a_{2}$ are consecutive has also exactly $2$ changes of colour (one of them at $V$) and we are done. Or maybe not.

You appear to have confused yourself.  If we assume $Va_{2}\in A$, then the cycle with $V$ between $a_{2}$ and $a_{3}$ has two changes of color.

You mean $4$ changes?
But I chose the path with $V$ between $A_{2}$ and $A_{3}$ only if $VA_{3}\in A$ and in that case $A_{1}A_{2}\in A,\;A_{2}V\in A,\;VA_{3}\in A,\;$
$A_{3}A_{4}\in B \; \ldots \; A_{m-1}A_{m}\in B,\;A_{m}A_{m+1}\in A; \! \ldots \!$; so we have $2$ changes in total and we can choose either $A_{3}$ or $A_{m}$ (the two points in which we change colour) as starting point. Sorry, maybe I misunderstood totally the problem.

No, I meant most of what you said was unnecessary.  Just take $V$ between $a_{2}$ and $a_{3}$ unconditionally, there will always be just two changes.

:lol: Ahh, all right, thank you; it is true.


group theory
abstract algebra
induction
vector
function
linear algebra
linear algebra unsolved

Here is a nice exercise; excuse me if it has already been posted.

Let's say that an (infinite) group $G$ has the [i]Burnside property[/i] if every subgroup of $G$ of finite exponent is finite. [In other words, if $H < G$ is a subgroup satisfying $h^{N}= 1$ for some $N \in \mathbb{N}$ and all $h \in H$, then $H$ is finite. ] Show that, for every $n$, the group $\mathrm{GL}_{n}(\mathbb{C})$ has the Burnside property.

yep, already posted and solved  :lol:  What is your solution ?

Most concisely put: view the inclusion $G \subset \mathrm{GL}_{n}(\mathbb{C})$ as a faithfull representation of $G$. Consider the $\mathbb{C}$-algebra $\mathbb{C}[G] \subseteq M_{n}(\mathbb{C})$ of all finite linear combinations $\sum_{g \in G}a_{g}g$, $a_{g}= 0$ for all but finitely many $g$; then, similarly, we may view the inclusion $\mathbb{C}[G] \subseteq M_{n}(\mathbb{C})$ as an $n$-dimensional faithful representation of the algebra $\mathbb{C}[G]$. Suppose this representation is reducible. Then, upon conjugation, we may assume that all elements of $G$ are simultaneously of the form $\left( \begin{array}{cc}A(g) & C(g) \\ 0 & B(g) \end{array}\right)$ with all $A(g) \in M_{r}(\mathbb{C}), B(g) \in M_{n-r}(\mathbb{C})$, for some $0 < r < n$. But then each such element necessarily has $C(g) = 0$, or else it would not be of finite order. Furthermore, $\{ A(g) \mid g \in G \}\subset \mathrm{GL}_{r}(\mathbb{C})$ and $\{ B(g) \mid g \in G \}\subset \mathrm{GL}_{n-r}(\mathbb{C})$ are subgroups of finite exponents, and hence, by induction, may be assumed finite. Thus $G$ is also finite.

Thus everything is reduced to the irreducible case  :roll: , and we may assume that the tautological representation $\mathbb{C}[G] \subseteq M_{n}(\mathbb{C})$ of the algebra $\mathbb{C}[G]$ is irreducible. But this already implies $\mathbb{C}[G] = M_{n}(\mathbb{C})$, as it is well-known and easy to show that $M_{n}(\mathbb{C})$ is irreducible and has no proper irreducible subalgebras. Now, since by assumption every eigenvalue of every element of $G$ is an $N$-th root of unity, it is clear that the traces of the elements of $G$ take no more than $N^{n}$ values. Since the trace pairing $\langle A, B\rangle : = \mathrm{tr}(AB)$ on $M_{n}(\mathbb{C})$ is nondegenerate, the assertion follows. Indeed, it shows that the number of elements of $G$ does not exceed $n^{2}N^{n}$.

Were there even more concise solutions given? :)

In the same style, but shorter: take $g_{1},...,g_{k}$ a basis of the vector space spanned by $G$. It is pretty immediate that the function which assigns to $g$ the $k$-tuple $(tr(gg_{1}),tr(gg_{2}),...,tr(gg_{k}))$ is one-to-one and since it has finite image, $G$ is finite.


[quote="gozpat"]no. let a=b=c=1[/quote]
Are you sure , gozpat?

$ a \equal{} b \equal{} c \equal{} 1$ is obviously no counterexample. As the inequality is homogeneous, we may ignore the condition $ a$, $ b$, $ c \ge1$. Then $ a \equal{} b \equal{} 1$, $ c\to 0$ is a counterexample.


[b]This is the Higher Arithmetic question ,,,
,,, let me know how to solve the question below ,,, I have no idea ,,,
[/b]

[u]Question :[/u]

Find the smallest number n, which has the following properties:
- Its decimal representation has 4 as the last digit;
- It is divisible by 6 and 8.
- It is a square number

$ n$ is divisible by $ 6$ and $ 8$ $ \Leftrightarrow 24|n$

As $ 24\equal{}2^3 \times 3$, $ 2^4 \times 3^2\equal{}144|n$ and it's easy to see that $ 144$ satisfies the condition.

[b]Thanks alot ,,, !

Such simple ,,, I should have got it myself ,,, ![/b]

To actually get it rigorously:

Since it is divisible by 6 and 8, it must be a multiple of 24. Since it ends in 4, the number 24 is multiplied by must be 1 mod 5. 24 is 2^3*3. So the number it is multiplied by must end in 1 or 6, have an odd number of factors of 2 and 3, and have an even number of factors of every other prime number. The smallest such number is 6, so 24*6=144.


geometry
3D geometry
tetrahedron
summer program
MathPath

There is a regular tetrahedron TRON with side length 7. L is the midpoint on side TN. I is the midpoint on side RO. Find LI.

Specify.  Which corners are we talking about?

Obviously two skew edges, since in a tetrahedron each vertex is adjacent to each - no ambiguity is possible.

This was a MathPath qualifying quiz problem, except the edges were 1 unit long. The answer is $\frac{\sqrt{2}}{2}\times 7$ I believe.

For this problem I strongly suggest building one out of paper or something. I couldn't visualize it so I asked my math teacher for four small construction paper equilateral triangle and made it with tape. After that it only took me a minute to solve :D

[quote="ckck"]Can you explain how you got the answer?[/quote]
Hint: Right triangles are the key. Draw some lines.

Yes, you can use Pythagorean a few times, but there's a very easy alternative solution. Think about a tetrahedron inscribed in a cube, as shown on the diagram.

Obviously, if $d$ is the edge of the cube, then $LI=d$ and the edge of the tetrahedron is $d\sqrt{2}$. Hence, if we're given the edge of the tetrahedron $t$, then $d={t\over\sqrt{2}}={t\sqrt{2}\over 2}$, which in this case makes $d={7\sqrt{2}\over 2}$

One easy way :  Imagine a cube. Take  the two diagonal points on one face and the     bottom two points on the diagonal of the opposit face. What you have now is  noting but a tetrahedron.

IOW     points of cube  $A(0,0,0),C(a,a,0),(0,a,a),(a,0,a)$  form a tetrahedron. (If side of cube is $a$, each side of the tetrahedron  is $a \sqrt 2$ rest is easy.  (Two points are now  middle points of opposite faces of  such a cube)

Added later: Just saw, some one just posted this  just right now ..:)

[quote="JavaMan"]Farenheit your diagrams are always lovely  :lol: do you use geometers sketchpad?[/quote]

Yes :)


MATHCOUNTS
summer program
Mathcamp
ratio

this was the first problem in the mathcamp thing a year ago, so its not cheating, lolo, this isnt mathcounts level, but since i wanted to boost this forum a bit, oh well

Every day at Mathcamp, Frank and Lilian competed to see who could solve more problems in problem-solving class. Each day at least one of them was able to do more than half the problems. Of the days when Frank solved more than half the problems, he won twice as often as he lost. Of the days when Lilian solved more than half the problems, she won five times as often as she lost. There were 25 class days total. On how many days did Lilian win? (You can assume there were no ties.)

[hide]this isn't the greatest way to do it, but its simple enough:

first start off with the equation 2x+5y = 25 where x is Frank's losses when he had more than half and y is Lilian's losses when she had more than half. So, since x and y are pos. integers, only two pairs satisfy it, (5,3) and (10,1). try them.

let F be frank and L be lillian

F(win) = 10 and L(win) = 15

F(lose but >1/2) = 5 and L(lose but >1/2) = 3

F(lose and <1/2) = 10 and L(lose and <1/2) = 7

we get the last two by subraction. since L(total losses) = F(win) and F(total losses) = L(win), we must be right.[/hide]

[hide]I didn't look at Paladin's solution, so here goes:

Since Frank wins in a 2:1 ratio when he gets most of the problems right, he must have gotten most of them right a multiple of 3 times. He also must have won a certain number of times, too:

G |W |L

3 |2  |1

6 |4  |2

9 |6  |3

12|8 |4

15|10|5

18|12|6

21|14|7

24|16|8

G= gets most of problems right

W= wins

L=loses

Lilian is in a similar position, except that her ratio is 5:1:

G  |W  |L

6  |5   |1

12|10|2

18|15|3

24|20|4

So, the question is, how many times does Lilian win? The answer is the number with a counterpart in Frank's win column that adds to 25 when added to the number in her column. This number is 15.[/hide]

This method is similar to Paladin's, but a bit simpler:

[hide]

The number of times that Frank won is a multiple of 2, say 2x.

The number of times that Lisa won is a multiple of 5, say 5y.

Then 2x+5y=25.

The possible solutions are (x,y)=(5,3) and (x,y)=(10,1).

However, if Lisa won 2x=20 times, she lost at least 10 times, which is impossible since 20 + 10 > 25, so we have (x,y)=(5,3).

[/hide]

--Dan


calculus
derivative
function
trigonometry
complex analysis
calculus computations

From my calculation using either the chain rule or the definition of an inverse function, I obtained the result pi as its derivative. However, there's a little funky about this function. As you might find in your calculator when you insert the original function and its derivative, the derivative actually is composed of 2 parts, one that is pi and the other negative pi. So the derivative is not just pi, but also includes -pi. Does anyone know why this is the case?[/img]

Forget the derivative and try graphing the function. It's a sawtooth wave, turning around at odd multiples of $ \frac {\pi}{2}$ when $ \sin$ hits $ \pm 1$. You need a piecewise definition to make sense of it, and any formula you get will only apply on one piece.

Alternately, there's the complex analysis explanation: $ \arcsin$ has many branch points. This composition does as well, and can be $ 2n\pi \plus{}\pi x$ or $ (2n\plus{}1)\pi \minus{}\pi x$ for any integer $ n$. You're detecting multiple versions and thus both possible derivatives.


function
calculus
integration
real analysis
real analysis theorems

$ \mathbf{Problem}$ Let $ f: [0,1]\rightarrow\mathbb{R}$ with
\[ f(x)\equal{}\left\{
\begin{array}{rl}
1 & \mathrm{if}\ x\equal{}1/n, n\in\mathbb{N}\\
0 & \mathrm{otherwise}
\end{array}
\right.
\]
Prove that $ f$ is integrable.

I know that a function is integrable if it has finitely many discontinuities, but this particular example has a countably infinite number of discontinuities.  I know that on $ [1/n,1]$, $ n\in\mathbb{N}$ and $ n\geq 2$ I can show that $ f$ is integrable, but does that necessarily mean that the union of all such intervals yields an integrable function?

This is the Riemann integral, of course.

The basic idea here- we can make a partition with a small interval at zero containing many peaks, and small intervals around each of the finitely many other peaks. The sum of the lengths of these bad intervals can be made as small as we want.

There is a theorem that says that a function is Riemann integrable if and only if its set of points of discontinuity has Lebesgue measure zero. Any countable set has Lebesgue measure zero.
Of course, you don't have the tools to prove or use this theorem.

Thanks jmerry, and I think I figured out how to do it.  It's a problem in $ \S$4.8 of Marsden's book Elementary Classical Analysis.

We know a function is integrable if
\[ \sup L(f,P) \equal{} \inf U(f,P),
\]
that is, the supremum of the set denoted by
\[ L(f,P) \equal{} \sum_{i \equal{} 1}^{n}\inf_{[x_{i \minus{} 1},x_i]}(f)(x_{i} \minus{} x_{i \minus{} 1})
\]
and
\[ U(f,P) \equal{} \sum_{i \equal{} 1}^{n}\sup_{[x_{i \minus{} 1},x_i]}(f)(x_{i} \minus{} x_{i \minus{} 1})
\]
over a uniformly continuous $ f$ and any partition $ P$.  This is also commonly denoted as
\[ \underline{\int_a^b}f(x)\ dx \equal{} \overline{\int_a^b}f(x)\ dx.
\]
Consider the partition
\[ P_n \equal{} [0,\sqrt {2}/n]\cup \bigcup\{\mathrm{the\ rest\ of\ the\ interval\ subdivided\ into\ intervals\ of\ width\ no\ larger\ than}\ 1/n^2\}
\]
then $ L(f,P_n) \equal{} 0$ and $ U(f,P_n) \leq \left(\frac {\sqrt {2}}{n}\right) \plus{} \frac {n \minus{} 1}{n^2}$.  And since
\[ L(f,P_n)\leq \sup L(f,P) \leq \inf U(f,P) \leq U(f,P_n)
\]
and both sides converge to 0 as $ n\rightarrow \infty$, then $ \sup L(f,P) \equal{} \inf U(f,P) \equal{} 0$ and $ f$ is integrable.  Thus $ \int_0^1f(x)dx \equal{} 0$ since if $ \underline{\int_a^b}f(x)\ dx \equal{} \overline{\int_a^b}f(x)\ dx$ then $ \int_a^bf(x)dx \equal{} \underline{\int_a^b}f(x)\ dx \equal{} \overline{\int_a^b}f(x)\ dx$ and we see that these are all 0.

Edit: How's that look jmerry?

Details- that's not the exact upper sum. You've overestimated both the width of the intervals containing bad points and the number of bad points.

I think the key is that $ 0$ is a limit point of $ \left\{\dfrac{1}{n}\right\}$ and by definition of limit point any $ \epsilon$ ball around $ 0$ must contain infinitely many points of $ \left\{\dfrac{1}{n}\right\}$.

The point is not that that ball contains infinitely many members of that sequence but that it contains all but finitely many members. There's a difference in meaning between those two phrases.

I'll remind you that the function which is $ 1$ on the rationals and $ 0$ on the irrationals is not Riemann integrable. Take $ 0$ (or any other point), and that is a limit point of the rationals - but that doesn't help us integrate it.

Here is a general theorem that is worth noting.

[b]Definition:[/b] A non-trivial subset $ S$ of $ \mathbb{R}$ has [u]null Lebesque measure[/u] when for any $ \epsilon > 0$ there exists finitely many compact intervals $ I_1,...,I_n$ such that $ S\subseteq \cup_{k\equal{}1}^n I_k$ and $ \text{L}(I_1,...,I_n)<\epsilon$. Where $ \text{L}(I_1,...,I_n)$ is the total combined length of the intervals.

The useful theorem is that suppose $ f : [a,b]\mapsto \mathbb{R}$ is a bounded function, let $ S$ be the set of discontinuities of $ f$ on $ [a,b]$ if $ S$ has null Lebesque measure than $ f$ is an integrable function on $ [a,b]$. Of course if $ S$ is empty then this is immediately true because continous functions are integrable, so we need to prove it when $ S$ is non-empty. Let us expound on what [b]jmerry[/b] was saying and prove this. Let $ m \equal{} \inf\{ f(x): x\in [a,b]\}$ and $ M \equal{} \sup\{ f(x): x\in [a,b]\}$.
We know we can cover $ S$ with $ I_1,...,I_n$ by assumption, also the total length can be made negligibly small. Let $ A \equal{} \cup_{k\equal{}1}^n I_k$ and $ B \equal{} [a,b] \minus{} \text{int}(A)$. Now $ A$ contains the discontinuities while $ B$ contains all the continuities. Let $ P$ be a partition of $ [a,b]$ which contains the endpoints of $ I_k$. Then $ U(f,P) \equal{} U_A(f,P)\plus{}U_B(f,P)$ where by $ U_A(f,P)$ we mean we group all the terms having the interval in $ A$ and likewise $ U_B(f,P)$ we group all the terms having the interval in $ B$. Similarly, $ L(f,P) \equal{} L_A(f,P) \plus{} L_B(f,P)$. Let $ \epsilon > 0$. So we can refine the partition $ P$ on $ B$ so that $ U_B(f,P) \minus{} L_B(f,P) < \epsilon$ because $ f$ is continous on $ B$ sub-intervals. While $ U_A(f,P) \minus{} L_A(f,P) \equal{} \sum_i (M_i \minus{} m_i)(x_i \minus{} x_{i\minus{}1}) \leq \sum_i (M \minus{} m)(x_i \minus{} x_{i\minus{}1} \equal{} (M\minus{}m)\text{L}(I_1,...,I_k)$. So $ U(f,P)\minus{}L(f,P) \equal{} \epsilon \plus{} (M\minus{}m)\text{L}(I_1,...,I_k)$ and we can make this as small as we please. So $ f$ is an integrable function.

Finally, a useful way to determine if a set has null Lebesque measure is to note if $ \{x_k\}$ is a convergent sequence then $ \{x_1,x_2,...\}$ has null Lebesquare measure. In your case the function is discontinous at $ x_k \equal{} \frac{1}{k}$ the rest follows.


AMC
AIME
geometry
MATHCOUNTS
number theory
AoPS Books
\/closed

The AoPS Summer class schedule is now posted on the [url=http://www.artofproblemsolving.com/Classes/AoPS_C_Enroll.php]enrollment page[/url].

This schedule may or may not be complete.

Additionally, the AIME Problem Series may soon change to being the AIME Problem Series B which will follow the same format as the original AIME Problem Series, but will include different problems.  Students should be aware that some of the problems in the AIME Problem Series B will be problems used in last month's AIME Seminar and a very small set of the problems covered in class will have been Message Board problems from the original AIME Problem Series.

However, from this point moving forward the AIME Problem Series A and B will be nearly identical curriculum with a couple of minor differences of focus within topic areas, but will include an entirely disjoint set of problems.

Students who would only take the AIME B should hold off on enrolling for the next few weeks while I see if I can have the entire class finished by then.

[quote="Rep123max"]So there is still a chance for Intermediate Number Theory? I really wanted to take lots of classes this summer, but the Olympiad classes are still out of reach, leaving only Intermediate Algebra which I have already taken  :([/quote]

We may still offer an Intermediate Number Theory class.  If we do, it might start later than our other classes and extend into September or even October.  That class is behind my current priority of finishing an Introduction to Number Theory textbook, so it will likely be May or June before I know whether or not we will be offering that class.  Many students have requested that class, so our hope is that we can accomplish that task soon.

[quote="MCrawford"]textbook[/quote]

Textbook?? O.O

We have begun writing textbooks in the areas of the subject matter than we teach.  We are beginning at what we call Introductory level -- MATHCOUNTS and most AMC material.  These books will be much more highly focused than the current AoPS books and accessible to a wider body of students.  They will cover everything our classes cover and more.  Nearly all of the text will be devoted to discussing problems and solutions unlike most textbooks.  They will be more in depth as they will each be devoted to one subject area at a time.

Hopefully this will give any student interested in accelerating in mathematics in or out of their school a greater opportunity.


A sequence $a_{n}$ is defined as follows: $a_{1}=p$ is a prime number with exactly $300$ nonzero digits, and for each $n\geq 1$, $a_{n+1}$ is the decimal period of $\frac{1}{a_{n}}$ multiplies by $2$. Determine $a_{2003}$

$n_{l}$ odd for any $l$, therefore $\phi(n_{l})$ is even when $n_{l}>1$. It give $n_{l}\le (n_{l-1}-1)/2$ if $n_{l-1}>1$.

Come on, just tell me how $\phi$, order, etc. is related here. I just don't get the reasoning at all.

[quote]Therefore $n_{2003}\le 2^{-2002}p \Longrightarrow a_{2003}=2$.[/quote]

This contains two falsehoods.  First, $n_{2003}$ is not necessarily less than or equal to $2^{-2002}p$.  Secondly, neither $n_{2003}= 1$ or $n_{2003}= 0$ imply that $a_{2003}= 2$.

It looks like you are assuming that $p$ has 300 digits, rather than 300 nonzero digits.  I'm not sure if ARMO really meant 300 nonzero digits or not.


1 is the positive root, that's why these kinds of manipulations are [b]forbidden[/b].  :D

God knows how many times I've seen this thread.

Please do a search function before posting.

Thank you.


Given n non-negative numbers a1, a2, ...an(n>=3) satisfying a1^2+a2^2+..+an^2=1. Prove that:
(a1+a2+...+an)^2>=3.(a1a2+a2a3+..+ana1)^2

[hide="Solution"]We want to prove that $ (a_1\plus{}a_2\plus{}\cdots\plus{}a_n)^2\ge 3(a_1a_2\plus{}\cdots\plus{}a_na_1)^2\implies 1\plus{}2\sum_{\text{cyc}}a_1a_2\ge 3\left(\sum_{\text{cyc}}a_1a_2\right)^2$.

Rearranging, we have $ 2\sum_{\text{cyc}}a_1a_2\minus{}3\left(\sum_{\text{cyc}}a_1a_2\right)^2\ge \minus{}1$.

Adding $ \sum_{\text{cyc}}a_1a_2$ to both sides, we want to prove that $ 3\sum_{\text{cyc}}a_1a_2\minus{}3\left(\sum_{\text{cyc}}a_1a_2\right)^2\ge \sum_{\text{cyc}}a_1a_2\minus{}1$.

However, $ 1\equal{}a_1^2\plus{}a_2^2\plus{}\cdots\plus{}a_n^2\ge \sum_{\text{cyc}}a_1a_2$ by Rearrangement Inequality.

Hence, it suffices to prove that $ \sum_{\text{cyc}}a_1a_2\minus{}\left(\sum_{\text{cyc}}a_1a_2\right)^2\ge 0$, which is true since $ 1\equal{}a_1^2\plus{}a_2^2\plus{}\cdots\plus{}a_n^2\ge \sum_{\text{cyc}}a_1a_2$. $ \blacksquare$[/hide]

This reminds me of Cauchy's.

P.S. I don't think is High School Basics.

May be this problem isn't high school basic :( (i am sorry) because it's  a problem towards mathematical olympad!!!

[quote="smiley4t"]I think the way of  FantasyLover is wrong. May be you are mistaken: a1a2+a1a3+..a1an+a2a3+...an-1an with a1a2+a2a3+a3a4+..+ana1.I hope you will have a better solution. :)  :)[/quote]

EDIT: Meh, never mind.

Apparently I failed expanding :wink:

we can also prove it using cauchy shwarz ineq.

[hide="Hint"]Homogenize to $ (a_1^2\plus{}a_2^2\plus{}...\plus{}a_n^2)(a_1\plus{}a_2\plus{}...\plus{}a_n)^2\ge 3(a_1a_2\plus{}a_2a_3\plus{}...\plus{}a_na_1)^2$ , and then break it in two inequalities .[/hide]


I have drawn up the angles (attached image) as best I can (which is not very good, so don't count on drawing being to scale).  The following is given:
- AOB is obtuse
- OC is perpendicular to OA
- OD bisects AOB
- OE bisects BOC

Thus, I know that AOC is 90, that AOD is equal to DOB, and that COE is equal to EOB.  None of this helps me with the problem, however, which asks for the measure of DOE...

...what am I missing?

PS - I'd rather learn how to solve this without having to guess-and-test backwards, but... the answer choices given are:

a.  60
b.  52.5
c.  45
d.  37.5
e.  30

The answer is (c), $\angle DOE = 45^\circ$.

$\angle DOE=\angle DOB-\angle EOB=\frac{1}{2}\angle AOB-\frac{1}{2}\angle COB=\frac{1}{2}\angle AOC$


Apartment rentals in Fairview run approximately $ \$$0.90 per square foot. Jillian has determined that she can afford $ \$$630 per month for rent. What is the largest apartment, in square feet, she should consider at the given rate?

Since Jillian is dividing her money among the square feet, we have to divide:

$ \frac {630}{.9} \equal{} \boxed{700}$


Find all functions $f: \mathbb R \to \mathbb R$ which satisfy $ f(x+y)=f(x)\cdot f(y) \cdot f(xy)$ for all $x,y\in\mathbb{R}$.

[quote="mathfunny"]Find all functions satify $ f(x \plus{} y) \equal{} f(x).f(y).f(xy), \forall x,y\in\mathbb{R}$[/quote]

Let $ P(x,y)$ be the property $ f(x \plus{} y) \equal{} f(x).f(y).f(xy)$

$ 1)$ If $ \exists x_0$ such that $ f(x_0) \equal{} 0$, then $ P(x \minus{} x_0,x_0)$ implies $ f(x) \equal{} 0$ $ \forall x$

$ 2)$ suppose now $ f(x)\neq 0$ $ \forall x$. Let then $ x\neq 0$ and $ y\neq 0$ :
$ P(x,\frac {1}{xy} \plus{} y)$ implies $ f(x \plus{} y \plus{} \frac {1}{xy}) \equal{} f(x)^2f(\frac {1}{xy})f(y)f(\frac {1}{x})f(\frac {1}{y})f(xy)$

$ P(x \plus{} \frac {1}{xy},y)$ implies $ f(x \plus{} y \plus{} \frac {1}{xy}) \equal{} f(x)f(y)^2f(\frac {1}{xy})f(\frac {1}{y})f(xy)f(\frac {1}{x})$

And so $ f(x) \equal{} f(y)\equal{}c$ $ \forall x,y\neq 0$

Then $ P(x,x)$, $ x\neq 0$  gives $ c\equal{}c^3$ and so $ c\equal{}\minus{}1$ or $ c\equal{}\plus{}1$
And $ P(x,\minus{}x)$, $ x\neq 0$ gives $ f(0)\equal{}c^3\equal{}c$

And so we have only three solutions : (EDITED : first post said five solution)
$ f(x) \equal{} 0$ $ \forall x$
$ f(x) \equal{} \minus{} 1$ $ \forall x$
$ f(x) \equal{} \plus{} 1$ $ \forall x$


Given a triangle ABC with $\angle{C}=90^o$. $D \in AB$ and $CD \perp AB$. Let $I_1$ and $I_2$ be the incentres of $\triangle{ACD}$ and $\triangle{BCD}$. Prove that $\angle{AI_2C}=\angle{BI_1C}$.

where        does     it come f    rom? maybe could be
useful the fact
that A,B,I:1 and I:2 are co.cyclic

Lemma: $A, B, I_{2}, I_{1}$ are concyclic points.
By angle chasing $\Delta BDI_{2} \sim \Delta BCI$ and $\Delta ADI_{1} \sim \Delta ACI$. Hence $\frac{DI_{1}}{AD} = \frac{CI}{CA}$ and $\frac{DI_{2}}{BD} = \frac{CI}{BC}$ $\Longrightarrow$ $\frac{DI_{1}}{DI_{2}} = \frac{BC}{CA}\cdot\frac{AD}{BD}$. On the other hand, $\frac{AD}{BD} = \frac{CA^2}{AB}\cdot\frac{AB}{BC^2}$. Hence $\frac{DI_{1}}{DI_{2}} = \frac{CA}{BC}$ $\Rightarrow$ $\Delta DI_{1}I_{2} \sim \Delta ABC$. And by angle chasing the thesis follows.
So $\measuredangle I_{2}AI_{1} = \measuredangle I_{2}BI_{1}$ and again with angle chasing we obtain $\measuredangle AI_{2}C = \measuredangle BI_{1}C$.


ratio
calculus
topology
linear algebra
graph theory

My personal suggestion is that you choose a different topic.  Fractals are one of two topics (the other is "the Fibonacci numbers and Golden Ratio in nature") that a lot of people stumble into as attractive projects for this kind of project.  Largely, I think this is because they come associated with a lot of pretty pictures.  Unfortunately, most of the actual mathematics of fractals is not really accessible at the high school science fair level, and high school science fair projects on fractals run a serious risk of being nothing more than those pretty pictures.

Now, obviously, this need not necessarily be true: one could start with something like a definition of fractals and fractal dimensions and then go on to compute the dimension of various easy-to-construct fractals, or the like.  But, honestly, I think there are much, much better subjects for a math science fair project.  A couple of suggestions (that arise in my field of study, combinatorics):
1) Linear recursive sequences.  (These are sequences that generalize the Fibonacci recurrence.  There's a well-developped theory of how to solve these that one can discuss.  Also, if you already know calculus, there's a parallel theory of linear differential equations and you could discuss the two in parallel and note similarities and differences between them.)
2) Catalan numbers.  (This is a sequence that answers remarkably many enumerative questions.  There are a lot of ways one could go -- choose a single object counted by Catalan numbers and use it to derive the recursion and/or the formula for them; choose several objects counted by Catalan numbers and show that they enumerate the same; etc.)
3) Graph coloring.  (Different options.  One could state the 4 color theorem but prove the 5 color theorem.  Etc.)

Thanks for the suggestions! I assume these are related to Combinatorics, right?
Also, is graph coloring related to topology?

[quote="JBL"]Fractals are one of two topics (the other is "the Fibonacci numbers and Golden Ratio in nature") that a lot of people stumble into as attractive projects for this kind of project.  Largely, I think this is because they come associated with a lot of pretty pictures.[/quote]
Hehe.  In 7th grade, my daughter Meena did a project on "Fibonacci Numbers in Nature" for her school's science fair.  In 8th grade, she did a project on Chaos/Fractals.  She ended up doing more than pretty pictures, but I can see where you are coming from.

remy -- yes, I would describe all three of these as falling into combinatorics.  The first also has elements of linear algebra (and possibly differential equations), and the third is soundly in the sub-field of combinatorics called graph theory.  Graph theory is in some sense connected to topology (in spirit if not always in practice), but I don't know of a particular link between graph coloring and topology (which isn't to say that there isn't one, just that I can't think of one at the moment).

Ravi -- :rotfl:  :lol: .  It's certainly possible to work real content into such papers, but I think the median science fair project on fractals probably involved close to 0 really mathematical thought (whatever that means) on the part of its creator.  And of course, Meena is one of at least two very mathematically savvy people in her immediate family, so I wouldn't be surprised if science fair projects she produced tended to fall well above the median in quality :)

I think I'll try graph coloring, because that seems like a very interesting topic. But which of the specific topics are hard enough for high school? My teacher doesn't accept ideas that are not challenging enough, and for example, I think the 4-color theorem is a rather simple idea which I first learned about in middle school. Are there a bit more difficult graph coloring topics I can try (or maybe anything related to graph theory in general?)

Well, there are many different types of graph coloring, so you could more than one of them -- some are quite tricky to understand, like vertex list coloring and fractional vertex coloring, so even just understanding how they work and then coming up with examples to re-explain them can be some interesting work.  (One possible source for information is
[url=http://www.emis.de/monographs/Diestel/en/]Diestel's [u]Graph Theory[/u][/url], although it is aimed at a pretty sophisticated audience and so you shouldn't get discouraged if you can't immediately follow everything he writes.)

One important (but hard!) question is how to figure out how to compute the chromatic number of a graph.  One thing to discuss here is possible ways of bounding it -- so, instead of an exact answer, you look for easy-to-compute-but-still-somewhat-informative estimates.

As for planar graphs and the 4-color theorem, of course the [i]statement[/i] of the 4-color theorem is easy, but the [i]proof[/i] is very difficult (and relies on a computer enumeration of some largish number of cases).  However, the [i]5[/i]-color theorem has a proof that's "easy enough," so you could devote much of your project to explaining that proof.

Okay, I think I've given you enough ideas for free now  :P  Go do some research!  :)

[quote="remy1140"]Ok, I'm not sure if this is the right forum to ask about this. But the question is, I have a science fair project in school and I have to think of a specific topic by the end of September. After some consideration, I decided to try researching fractals (e.g. Mandelbrot Set). Are there any ideas for how to proceed my research, and specifically what I should study about?[/quote]

Well :|, Fractals or rather Fractal Image Compression is a very good topic -- I gave a seminar pertaining to fractal image compression in my senior under-graduate year.:D I personally found the topic to be highly enthralling and arresting.

But, just as [b] JBL [/b] had said, this particular topic is becoming rather cliched-- it appears most people nowadays have a predisposition to either choose or select 'fractals' or 'fractal image compression' for their science projects. If you intend to do a highly 'math-oriented project', the topics which [b]JBL [/b] had suggested are pretty good.

I would like to suggest you some topics, but if you insist on doing highly 'math-oriented' projects, the topics which I suggest might not be so worthwhile.The topics which I would like to suggest are: Protonic Memory, Quantum Computing, Augmented Reality, Photonic Networks, etc.

The topics which I had suggested are mostly from a computer science perspective. Between, it appears that you had joined this forum on my birthday. :D

Thank you so much, I'll also consider computer science as well as graph coloring for the project. Also, I am also thinking about psychology. Are there any suggestions on what kind of topics related to psychology (e.g. brain, dreams, behaviors) that I can try?

I really can't suggest any topics in psychology. I'm not so well acquainted with psychology, but you can probably dabble a bit in Cognitive Science. It's a pretty interesting subject and has many interesting topics.

Probably, you could also 'research' a bit on Parapsychology-- which is a subset of Psychology. Parapsychology primarily deals with paranormal events. I would suggest you to do something which suits your interests.

Also, do some "research"! :D


induction
number theory proposed
number theory

Does exist infinitely many numbers $n$ that:
\[ n|3^{n+1}-2^{n+1}\]


Does anyone know any [color=red]good[/color] websites where I could get older mathcounts contests? It seems that all the official websites only have the more recent ones, which creates a problem for those of us who practiced with those for this year's past competition and took this year's at the competition and are looking to practice this summer before their last shot at mathcounts. If anyone knows where I can get older contests free online please respond.
                                        :)  Thanks!

Sorry, but I think you have to buy them. For example, I got mine from online, and some from a friend who was kind enough to lend me the collection he and his sister collected from their years of experience.

The best way to get old math tests for math competitions is to buy them from the program in which they were used. That is "best" because

1) the OFFICIAL solutions are usually more educational than the sketchy solutions you can sometimes find online,

and

2) buying the old materials supports the program, which is always a good cause. Most math competitions are run by nonprofit organizations that don't have a lot of spare money. You can save yourself time and frustration by spending a small amount of your money to buy official practice materials from the math competition programs you like.

Hope this helps!

If I remember correctly, there was a thread in the Mathcounts board that had a website with a lot of practice Mathcounts problems.

What I'm looking for isn't the practice problems right now, I'm trying to really slam down on more real problems (like those from the previous competitions) before school eats my time. My family owns several of the books, but I don't know where most of them are. So I'm down to looking for free tests online that are old rounds. If anyone can find anything I would love to know. Thanks again!

No, I'm pretty sure mathcounts doesn't disclose their tests to any websites, because they make money by selling old tests. There might be some sites with old rounds, but there might not be either. I bought a lot of old rounds and stuff to practice but since I'm too old for mathcounts, I gave them away. You could also have somebody give some to you, like I typed some up for Rep123max last year, but I'm not sure I still have them.

http://mathcounts.saab.org/ is a pretty good site for last minute drills. I used it last year before MathCounts.


geometry
3D geometry
sphere
tetrahedron
circumcircle
geometry unsolved

A sphere is inscribed in tetrahedron $ ABCD$ and is tangent to faces  $ BCD,CAD,ABD,ABC$ at points $ P,Q,R,S$ respectively. Segment $ PT$  is the sphere's diameter, and lines $ TA,TQ,TR,TS$ meet the plane $ BCD$ at points $ A',Q',R',S'$. respectively. Show that $ A$  is the center of a circumcircle on the triangle $ S'Q'R'$.

Consider the stereographic projection on the insphere $ \omega$ with pole $ T$ (the antipode of  P) and stereographic plane $ \pi \equiv BDC.$ This projection maps $ S,Q,R$ into $ S',Q',R',$ $ P$ is a double point and circle $ \odot(SQR)$ on the surface of $ \omega$ is taken into the circle $ \odot(S'Q'R')$ in the plane $ \pi.$ Due to the conformity of the stereographic projection, the sphere $ k$ with center $ A$ and orthogonal to $ \omega,$ which intersects $ \omega$ along the circumference $ \odot(SQR),$ is taken into the sphere $ k'$  containing its image $ \odot(S'Q'R')$ and orthogonal to the plane $ BCD$ $\Longrightarrow$ the center of $ k'$ coincides with the circumcenter of  $ \triangle S'Q'R'.$ Since the pole $ T$ of the stereographic projection (inversion) is collinear with the center of the sphere and its image, it follows that $ A ' \equiv \pi \cap AT$ is the circumcenter of $ \triangle S'Q'R'.$

[hide = Solution]

Here is a solution for those of us who don't know what a stereographic projection is :P.

Let $\phi$ denote the sphere, $\pi$ denote the plane of $\triangle BCD$ and let $T' = AT \cap \phi$. Our first step will be to get rid of the unnecessary things in this problem. Consider the circle $\frak{C}$ consisting of the set of points $Q \in \phi$ with $AP$ tangent to $\phi$. We claim that for any $C \in \frak{C}$, if we define $D = TC \cap \pi$, then $|DA'|$ is fixed. Let $\frak{D}$ be the set of $D$ as described in the previous sentence. First of all, observe that since the inversion about $T$ with radius $TP$ maps $\frak{C}$ to $\frak{D}$, it's clear that $\frak{D}$ is a circle, so hence it has some circumcenter.

It now suffices only to prove that this circumcenter is $A'.$ Let's consider two arbitrary points $C_1, C_2 \in \frak{C}$ so that $A, T, C_1, C_2, T', A'$ are coplanar. Let $D_1, D_2$ be $TC_1 \cap \pi, TC_2 \cap \pi$ respectively. We claim that $A'$ is equidistant from $D_1, D_2$. To see this, take the cross-section of the figure containing $A, T, C_1, C_2, T', A'$. We now have a two-dimensional geometry problem! To solve this new problem, simply project the harmonic quadrilateral $TC_1T'C_2$ onto $\pi$ (which is a line in our new diagram), and note that $TT \cap \pi$ is the point at infinity.

Hence, as we've shown that $A'$ is the midpoint of infinitely many chords of $\frak{D}$ passing through it, and $\frak{D}$ is a circle, it's clear that $A'$ is the center of $\frak{D}$, as desired.

[/hide]


real analysis
function
integration
calculus
derivative
ratio
real analysis unsolved

Assume that the (unbalanced) density of E exists at every point of $\mathbf{R}$. Prove that up to a zero set, $E = \mathbf{R}$ or $E = \emptyset$.

Hint: Topological connectedness of $\mathbf{R}$ is useful in the proof.

The definiton of unbalanced density is similar to Lebesgue density defined at
http://en.wikipedia.org/wiki/Lebesgue_density, with the difference that $B_\epsilon$ is a set of radius that [b]contains[/b] $x$, not centered at $x$.

THanks!

Notation: use $|E|$ to mean the Lebesgue measure of the set $|E|.$

Define a function $f(x)$ as follows:

For $x\ge 0,$ let $f(x)=|E\cap[0,x]|.$

For $x<0,$ let $f(x)=-|E\cap[x,0]|.$

We can combine this together as $f(x)=\int_{0}^{x}\chi_{E}.$

Then for all $x<y,\ f(y)-f(x)=|E\cap[x,y]|=\int_{x}^{y}\chi_{E}.$

By the Lebesgue theorem on the differentiation of the integral, $f'(x)$ exists almost everywhere and $f'(x)=\chi_{E}(x)$ almost everywhere. But the hypothesis of the problem is that $f'(x)$ exists everywhere.

We want to show that either $f'(x)=0$ everywhere or $f'(x)=1$ everywhere.

[Incomplete argument. To be continued later.]

Hmmm, isn't it just the nested interval lemma? Suppose that $E$ is neither essentially empty, nor essentially full. Then there exists a non-degenerate closed interval $I$ such that $\frac{|I\cap E|}{|I|}$ is as close to $0$ as we want and also there is another interval $I$ for which this ratio is as close to $1$ as we want. Since every non-degenerate interval can be continuously transformed into any other non-degenerate interval, we conclude that there exists an interval $I_{1}$ for which the ratio is, say, $2/3$. Note also that such $I$ can be chosen arbitrarily small and lying within any given interval on which $E$ is neither essentially empty, nor essentially full. So, we can find a small subinterval $I_{2}$ of $I_{1}$ for which the ratio is $1/3$. Then we can bring it up to $2/3$ again for some even smaller subinterval $I_{3}$  of $I_{2}$ and so on. Clearly, the point in the intersection of $I_{k}$ will be a point where the unbalanced density does not exist.

Here is another question: suppose that the usual (symmetric) density exists at every point. (This can happen; one simple example is a finite union of intervals). Is it necessarily true then that the boundary of $E$ has zero measure?

An interval minus $\mathbb Q$ has boundary with positive measure.

Erm... :blush: I meant "essential boundary", of course (i.e., the set of points $x$ such that every open interval containing $x$ intersects both the set and its complement by a set of positive measure). Sorry for this stupid inaccuracy.


inequalities
trigonometry
inequalities proposed
algebra

$ a\plus{}bc\equal{}(a\plus{}b)(a\plus{}c)$, so apply that, etc. The inequality is

\[ \sum (b\plus{}c)(a^2\plus{}ab\plus{}ac\minus{}bc)\le \frac{3}{2}(a\plus{}b)(b\plus{}c)(c\plus{}a)\]
\[ \iff \sum_{cyc} a^2b\plus{}b^2a\ge 6abc\]

which is obvious by am-gm.

Another approach:
[hide]The inequality is equivalent to: 4bc/(a+bc) + 4ac/(b+ac) + 4ab/(c+ab)>=3
Using Cauchy-Schwartz's inequality we have: LHS>=4(ac+bc+ab)^2/(b^2c^2+a^2c^2+a^2b^2+3bac)>=3
It suffices to prove that:3(a^2b^2+b^2c^2+c^2a^2)+8abc>=9abc, which is true by AM-GM and a+b+c=1(Done!)[/hide]

Sorry to bump an old post but while preparing for CMO 2011, I found another proof for this. It is in the PDF attachment.

The inequality is equivalent to

$\sum \frac{-2bc}{(a+c)(b+c)} \leq \frac{-3}{2}$ or $\sum \frac{bc}{(a+c)(b+c)} \geq \frac{3}{4}$

Expanding, we must show $4[2,1,0] \geq [1,1,1]+3[2,1,0]$ or $[2,1,0] \geq [1,1,1]$, with a bit less arithmetic than Altheman's straight expansion.

[hide="My Solution"]Notice that: $\frac{a-bc}{a+bc} = \frac{a+bc}{a+bc} - \frac{2bc}{a+bc} = 1- \frac{2bc}{a+bc}$
also notice that: $a+bc = 1-b-c+bc = (1-b)(1-c)$. Then $LHS$ is $= 3-\sum_{cyc}\frac{2ab}{(1-a)(1-b)}$ so the inequality is true $\Leftrightarrow\sum_{cyc}\frac{2ab}{(1-a)(1-b)}\ge\frac{3}{2}\Leftrightarrow\sum_{cyc}\frac{ab}{(1-a)(1-b)}\ge\frac{3}{4}\Leftrightarrow ab+bc+ac\ge\ 9abc$ (Expanding everything)
and this is true because of $AM-HM:$
$\frac{a+b+c}{3}\ge\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$
since $a+b+c=1$ we get the desired result result.[/hide]

\[\frac{a-bc}{a+bc}+\frac{b-ca}{b+ca}+\frac{c-ab}{c+ab}\le \frac{3}{2}\]\[\iff \frac{bc}{a+bc}+\frac{ca}{b+ca}+\frac{ab}{c+ab}\ge \frac{3}{4}\]\[\iff \frac{bc}{(a+b)(c+a)}+\frac{ca}{(b+c)(a+b)}+\frac{ab}{(c+a)(b+c)}\ge \frac{3}{4}\]\[\iff a(b^2+c^2)+b(c^2+a^2)+c(a^2+b^2)\ge 6abc\]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&p=3416434#p3416434]St. Petersburg, Russia 2007[/url]

In $\triangle  ABC$
Let $a=tan\frac{B}{2}tan\frac{C}{2},b=tan\frac{C}{2}tan\frac{A}{2},c=tan\frac{A}{2}tan\frac{B}{2}$,
hance $a,b,c>0,a+b+c=tan\frac{B}{2}tan\frac{C}{2}+tan\frac{C}{2}tan\frac{A}{2}+tan\frac{A}{2}tan\frac{B}{2}=1$,

$\frac{a-bc}{a+bc}+\frac{b-ca}{b+ca}+\frac{c-ab}{c+ab}=\frac{1-\frac{bc}{a}}{1+\frac{bc}{a}}+\frac{1-\frac{ca}{b}}{1+\frac{ca}{b}}+\frac{1-\frac{ab}{c}}{1+\frac{ab}{c}}$

$=\frac{1-tan^2\frac{A}{2}}{1+tan^2\frac{A}{2}}+\frac{1-tan^2\frac{B}{2}}{1+tan^2\frac{B}{2}}+\frac{1-tan^2\frac{C}{2}}{1+tan^2\frac{C}{2}}=  \cos{A} \plus{} \cos{B} \plus{} \cos{C}\leq\frac {3}{2}$.

[quote="April"]Let $ a$, $ b$, $ c$ be positive real numbers for which $ a \plus{} b \plus{} c \equal{} 1$. Prove that
\[ {{a\minus{}bc}\over{a\plus{}bc}} \plus{} {{b\minus{}ca}\over{b\plus{}ca}} \plus{} {{c\minus{}ab}\over{c\plus{}ab}}
\leq {3 \over 2}.\][/quote]
Let $a\rightarrow bc,b\rightarrow ca,c\rightarrow ab ,$  we get

AI3--208( "1691 Algebraic Inequalities - Old and New Problems for Mathematical Olympiads" - AGA, Italy, 2010.) :
Let $ a$, $ b$, $ c$ be positive real numbers for which $  bc \plus{} ca \plus{} ab \equal{} 1$. Prove that
\[ {{1\minus{}a^2}\over{1\plus{}a^2}} \plus{} {{1\minus{}b^2}\over{1\plus{}b^2}} \plus{} {{1\minus{}c^2}\over{1\plus{}c^2}}\leq {3 \over 2}.\]

[quote="sqing"]
Let $ a$, $ b$, $ c$ be positive real numbers for which $  bc \plus{} ca \plus{} ab \equal{} 1$. Prove that
\[ {{1\minus{}a^2}\over{1\plus{}a^2}} \plus{} {{1\minus{}b^2}\over{1\plus{}b^2}} \plus{} {{1\minus{}c^2}\over{1\plus{}c^2}}\leq {3 \over 2}.\][/quote]
$\sum \frac{1-a^2}{1+a^2} \ge \frac{3}{2} \iff \sum \frac{a^2}{1+a^2} \ge \frac{3}{4}$
This is quite obvious by Cauchy Schwarz

I think it is the best constant

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=606069]Generalization of Canada 2008[/url]


modular arithmetic
algorithm
number theory

Find a prime factor of $ 99!\minus{}1$.

[hide]if you are using wilson's thm, can you please prove and explain in easier way,, i have seen the proof and i did not quite understand. [/hide]

Wilson's theorem is the obvious motivation for this problem.  You should instead attempt to better understand the proof of Wilson's theorem.  What part is unclear to you?

I don't understand the part where it says,  the numbers 2,3,4,...,p-2 can form pairs whose multiple $ \equiv 1 \bmod p$.

[b]Definition:[/b]  Given an integer $ a$, an integer $ b$ such that $ ab \equiv 1 \bmod p$ is called a [i]multiplicative inverse[/i] of $ a \bmod p$.

[b]Lemma:[/b]  The multiplicative inverse exists provided that $ a \not \equiv 0 \bmod p$ and is unique.

Proof:  The equation $ ax \plus{} py \equal{} 1$ has a solution in integers $ (x, y)$ by [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=127224]Bezout's lemma[/url].  Thus $ ax \equiv 1 \bmod p$ and $ x$ is a multiplicative inverse of $ a \bmod p$.

Suppose $ b, b'$ are multiplicative inverses of $ a \bmod p$.  Then $ ab \equiv ab' \equiv 1 \bmod p \implies a(b \minus{} b') \equiv 0 \bmod p \implies ba(b \minus{} b') \equiv b \minus{} b' \equiv 0 \bmod p$, hence $ b \equiv b' \bmod p$, so the multiplicative inverse is unique.

So one proof of Wilson's Theorem proceeds by pairing up multiplicative inverses (which are unique).  The only numbers whose multiplicative inverses are themselves ($ \bmod p$) are $ 1, \minus{} 1$, so the others pair up perfectly.

[hide]101 seems to work (yes, I used a calculator, a pretty advanced one)[/hide]

[hide="Solution"]
According to Wilson's theorem: $ 100! \equiv 100 \pmod{101}$
Now $ \gcd(100, 101) \equal{} 1$ so we can divide both sides by $ 100$.
This gives us $ 99! \equiv 1 \pmod{101}$
which is equivalent to $ 99! \minus{} 1 \equiv 0 \pmod{101}$
So $ 101$ is a prime factor (since it is prime).
[/hide]

[quote="wonki_noh"]HOw do they pair up perfectly??[/quote]

It follows directly from the fact that multiplicative inverses are unique.  If you want an algorithm, it works something like this:

- Consider the product $ 2 \cdot 3 \cdot 4 \cdot ... \cdot (p\minus{}2) \bmod p$.  Look at $ 2$ and find its multiplicative inverse and multiply them together.

- Now we have $ 3 \cdot ... \cdot (p\minus{}2) \bmod p$.  Look at $ 3$ and find its multiplicative inverse, which is neither $ 2$ nor its multiplicative inverse (because that contradicts uniqueness) and multiply them together.

- Now we have $ 4 \cdot ...$.  Etc.

Thanks for keep responding to my queries,, tOrajirOu.

I see that there are unique multiplicative inverse pairs. But, how would you know if  for example, the multiplicative pair of 3 is smaller than p-2?

The multiplicative inverse of $ 3 \bmod p$ is some residue $ a \bmod p$, of which some representative exists in $ \{ 0, 1, 2, ... p\minus{}1 \}$.  Moreover, we know it's not $ 0, 1, \minus{}1$.

Thanks tOrajirOu!!

I got it now..

THanks heaps!


On a scale of 1-10 where Meggy-Meg is a 9.5, Im Ji Hye would be about a 36.

Because she gets a default rating of only 12/10, but when you factor in the x3 Korean factor it skyrockets.

No. What would be best for Ubey would be to stalk them obsessively like he did Meggy-meg.

That would be sure to win their affection.

Plus, if that fails, he can always entice them with his lack of math-skills.

(sarcasm shields: AAACK-TIVATE!)

[quote="aspiringchesslete"]Umm what about Hyori Lee? she would get like 999[/quote]

Hyori Lee is pretty hot but Jun Ji Hyun and Im Ji Hye are both hotter in my opinion.

[quote="moogra"][quote="serialk11r"]Albert went from hitting on girls younger than him to hitting on Korean supermodels.[/quote]

How did you know she was a supermodel?
lol so true.[/quote]
Chat from AIM went like:
Ubemaya: dude [whatever the name was] is so hot
trigalg693: huh? who the hell is that
Ubemaya: google

[I type in the name, I see the page excerpts from Google say words like "supermodel" "Korean" blah]

Ok, I'm not going to lie.

When I saw Im Ji Hye, I read it as Im(Ji Hye) as in the imaginary part of the complex number Ji Hye.

I know, I'm not Korean. But I still think that mis-read is pretty funny.

[quote="Ubemaya"][quote="pianoforte"]//When I saw Im Ji Hye, I read it as Im(Ji Hye) as in the imaginary part of the complex number Ji Hye.//

Whoa

But still

I think meggy-meg's pic was hott. Ubemaya is hotter.[/quote]

DANG, I never knew you were in my grade.[/quote]
Huh? His latest blog entry says he's a junior.

[quote="serialk11r"][quote="pianoforte"][quote="Ubemaya"]Wait... I can't add. I saw him being in 8th grade in 06 from his sig and I thought he's in 10th grade now...[/quote]
BECAUSE *THIS* SCHOOL YEAR'S MATHCOUNTS IS IN 2009 DUDE. BUT OK, YOUR HOTT, SO IT DOES NOT MATTER.
DOES NOT MATTER AT ALL.
IN THE SLIGHTEST.[/quote]
Which is why he should start stalking this Im Ji Hye person...[/quote]
Ah yes, google image searching is very "revealing". Very good good sir.

Dude, I read the topic title and I was like what the ****

1) Idk if I was behind the times, but Ubey wrote "> Meggy-Meg"
2) The name you wrote in the topic title i recognized as completely korean. That's really creepy.

Ubey, I think you're going senile. No offense or anything. Just...w/e