Let $ a,b,c>1$ and $ a\plus{}b\plus{}c\equal{}9$. Prove that $ \sqrt{ab\plus{}bc\plus{}ca}\leq \sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c}$

[quote="tdl"]Let $ a,b,c > 1$ and $ a \plus{} b \plus{} c \equal{} 9$. Prove that $ \sqrt {ab \plus{} bc \plus{} ca}\leq \sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c}$[/quote]
Setting $ \sqrt {a} \equal{} x,\sqrt {b} \equal{} y,\sqrt {c} \equal{} z,$ we have $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 9$ and $ 9\min\{x^2,y^2,z^2\} \ge x^2 \plus{} y^2 \plus{} z^2.$ Now, we see that the original inequality is equivalent to
$ x \plus{} y \plus{} z \ge \sqrt {x^2y^2 \plus{} y^2z^2 \plus{} z^2x^2},$
$ (x \plus{} y \plus{} z)^2(x^2 \plus{} y^2 \plus{} z^2) \ge 9(x^2y^2 \plus{} y^2z^2 \plus{} z^2x^2),$
$ 3(x^2 \plus{} y^2 \plus{} z^2)^2 \minus{} 9(x^2y^2 \plus{} y^2z^2 \plus{} z^2x^2) \ge 3(x^2 \plus{} y^2 \plus{} z^2)^2 \minus{} (x \plus{} y \plus{} z)^2(x^2 \plus{} y^2 \plus{} z^2),$
$ 3\sum (x^2 \minus{} y^2)(x^2 \minus{} z^2) \ge 2(x^2 \plus{} y^2 \plus{} z^2)\sum (x \minus{} y)(x \minus{} z),$
$ \sum (x \minus{} y)(x \minus{} z)[3(x \plus{} y)(x \plus{} z) \minus{} 2(x^2 \plus{} y^2 \plus{} z^2)] \ge 0,$
$ \sum (x \minus{} y)(x \minus{} z)[x^2 \plus{} 3x(y \plus{} z) \plus{} 3yz \minus{} 2y^2 \minus{} 2z^2] \ge 0,$
$ \sum (x \minus{} y)(x \minus{} z)[x^2 \plus{} 3x(y \plus{} z) \minus{} yz] \minus{} 2\sum (x \minus{} y)(x \minus{} z)(y \minus{} z)^2 \ge 0,$
$ \sum (x \minus{} y)(x \minus{} z)[x^2 \plus{} 3x(y \plus{} z) \minus{} yz] \ge 0.$
So, we may write the inequality in the form $ X(x \minus{} y)(x \minus{} z) \plus{} Y(y \minus{} z)(y \minus{} x) \plus{} Z(z \minus{} x)(z \minus{} y) \ge 0,$ where $ X \equal{} x^2 \plus{} 3x(y \plus{} z) \minus{} yz,$ and $ Y,Z$ are similar. Because $ 9x^2 \ge x^2 \plus{} y^2 \plus{} z^2,$ we have $ x^2 \ge \frac {y^2 \plus{} z^2}{8} \ge \frac {(y \plus{} z)^2}{16},$ so
$ X \ge \frac{y^2 \plus{} z^2}{8} \plus{} \frac {3(y \plus{} z)^2}{4} \minus{} yz \ge \frac {yz}{4} \plus{} 3yz \minus{} yz > 0.$
Therefore $ X,Y,Z \ge 0.$ On the other hand, assume that $ x \ge y \ge z,$ we have
$ X \minus{} Y \equal{} (x^2 \minus{} y^2) \plus{} 4z(x \minus{} y) \ge 0.$
Thus
$ X(x \minus{} y)(x \minus{} z) \plus{} Y(y \minus{} z)(y \minus{} x) \plus{} Z(z \minus{} x)(z \minus{} y) \ge Y(x \minus{} y)(x \minus{} z) \plus{} Y(y \minus{} z)(y \minus{} x) \equal{} Y(x \minus{} y)^2 \ge 0.$
Our proof is completed. Equality holds if and only if $ a \equal{} b \equal{} c \equal{} 3.$ :)

[quote="can_hang2007"][quote="tdl"]Let $ a,b,c > 1$ and $ a \plus{} b \plus{} c \equal{} 9$. Prove that $ \sqrt {ab \plus{} bc \plus{} ca}\leq \sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c}$[/quote]
Setting $ \sqrt {a} \equal{} x,\sqrt {b} \equal{} y,\sqrt {c} \equal{} z,$ we have $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 9$ and $ 9\min\{x^2,y^2,z^2\} \ge x^2 \plus{} y^2 \plus{} z^2.$ Now, we see that the original inequality is equivalent to
$ x \plus{} y \plus{} z \ge \sqrt {x^2y^2 \plus{} y^2z^2 \plus{} z^2x^2},$
$ (x \plus{} y \plus{} z)^2(x^2 \plus{} y^2 \plus{} z^2) \ge 9(x^2y^2 \plus{} y^2z^2 \plus{} z^2x^2),$
$ 3(x^2 \plus{} y^2 \plus{} z^2)^2 \minus{} 9(x^2y^2 \plus{} y^2z^2 \plus{} z^2x^2) \ge 3(x^2 \plus{} y^2 \plus{} z^2)^2 \minus{} (x \plus{} y \plus{} z)^2(x^2 \plus{} y^2 \plus{} z^2),$
$ 3\sum (x^2 \minus{} y^2)(x^2 \minus{} z^2) \ge 2(x^2 \plus{} y^2 \plus{} z^2)\sum (x \minus{} y)(x \minus{} z),$
$ \sum (x \minus{} y)(x \minus{} z)[3(x \plus{} y)(x \plus{} z) \minus{} 2(x^2 \plus{} y^2 \plus{} z^2)] \ge 0,$
$ \sum (x \minus{} y)(x \minus{} z)[x^2 \plus{} 3x(y \plus{} z) \plus{} 3yz \minus{} 2y^2 \minus{} 2z^2] \ge 0,$
$ \sum (x \minus{} y)(x \minus{} z)[x^2 \plus{} 3x(y \plus{} z) \minus{} yz] \minus{} 2\sum (x \minus{} y)(x \minus{} z)(y \minus{} z)^2 \ge 0,$
$ \sum (x \minus{} y)(x \minus{} z)[x^2 \plus{} 3x(y \plus{} z) \minus{} yz] \ge 0.$
So, we may write the inequality in the form $ X(x \minus{} y)(x \minus{} z) \plus{} Y(y \minus{} z)(y \minus{} x) \plus{} Z(z \minus{} x)(z \minus{} y) \ge 0,$ where $ X \equal{} x^2 \plus{} 3x(y \plus{} z) \minus{} yz,$ and $ Y,Z$ are similar. Because $ 9x^2 \ge x^2 \plus{} y^2 \plus{} z^2,$ we have $ x^2 \ge \frac {y^2 \plus{} z^2}{8} \ge \frac {(y \plus{} z)^2}{16},$ so
$ X \ge \frac {y^2 \plus{} z^2}{8} \plus{} \frac {3(y \plus{} z)^2}{4} \minus{} yz \ge \frac {yz}{4} \plus{} 3yz \minus{} yz > 0.$
Therefore $ X,Y,Z \ge 0.$ On the other hand, assume that $ x \ge y \ge z,$ we have
$ X \minus{} Y \equal{} (x^2 \minus{} y^2) \plus{} 4z(x \minus{} y) \ge 0.$
Thus
$ X(x \minus{} y)(x \minus{} z) \plus{} Y(y \minus{} z)(y \minus{} x) \plus{} Z(z \minus{} x)(z \minus{} y) \ge Y(x \minus{} y)(x \minus{} z) \plus{} Y(y \minus{} z)(y \minus{} x) \equal{} Y(x \minus{} y)^2 \ge 0.$
Our proof is completed. Equality holds if and only if $ a \equal{} b \equal{} c \equal{} 3.$ :)[/quote]
Very nice proof :) I like the part where you use $ \sum (x \minus{} y)(x \minus{} z)(y \minus{} z)^2 \equal{} 0$. I will remember this trick :)

A note: We can also prove it for $ a,b,c \ge \frac {1}{5}$ (or even for $ a,b,c \ge .1169..$) but it is hard using the traditional inequalities..

[quote="can_hang2007"][quote="tdl"]Let $ a,b,c > 1$ and $ a \plus{} b \plus{} c \equal{} 9$. Prove that $ \sqrt {ab \plus{} bc \plus{} ca}\leq \sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c}$[/quote]
Setting $ \sqrt {a} \equal{} x,\sqrt {b} \equal{} y,\sqrt {c} \equal{} z,$ we have $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 9$ and $ 9\min\{x^2,y^2,z^2\} \ge x^2 \plus{} y^2 \plus{} z^2.$ Now, we see that the original inequality is equivalent to
$ x \plus{} y \plus{} z \ge \sqrt {x^2y^2 \plus{} y^2z^2 \plus{} z^2x^2},$
$ (x \plus{} y \plus{} z)^2(x^2 \plus{} y^2 \plus{} z^2) \ge 9(x^2y^2 \plus{} y^2z^2 \plus{} z^2x^2),$
$ 3(x^2 \plus{} y^2 \plus{} z^2)^2 \minus{} 9(x^2y^2 \plus{} y^2z^2 \plus{} z^2x^2) \ge 3(x^2 \plus{} y^2 \plus{} z^2)^2 \minus{} (x \plus{} y \plus{} z)^2(x^2 \plus{} y^2 \plus{} z^2),$
$ 3\sum (x^2 \minus{} y^2)(x^2 \minus{} z^2) \ge 2(x^2 \plus{} y^2 \plus{} z^2)\sum (x \minus{} y)(x \minus{} z),$
$ \sum (x \minus{} y)(x \minus{} z)[3(x \plus{} y)(x \plus{} z) \minus{} 2(x^2 \plus{} y^2 \plus{} z^2)] \ge 0,$
$ \sum (x \minus{} y)(x \minus{} z)[x^2 \plus{} 3x(y \plus{} z) \plus{} 3yz \minus{} 2y^2 \minus{} 2z^2] \ge 0,$
$ \sum (x \minus{} y)(x \minus{} z)[x^2 \plus{} 3x(y \plus{} z) \minus{} yz] \minus{} 2\sum (x \minus{} y)(x \minus{} z)(y \minus{} z)^2 \ge 0,$
$ \sum (x \minus{} y)(x \minus{} z)[x^2 \plus{} 3x(y \plus{} z) \minus{} yz] \ge 0.$
So, we may write the inequality in the form $ X(x \minus{} y)(x \minus{} z) \plus{} Y(y \minus{} z)(y \minus{} x) \plus{} Z(z \minus{} x)(z \minus{} y) \ge 0,$ where $ X \equal{} x^2 \plus{} 3x(y \plus{} z) \minus{} yz,$ and $ Y,Z$ are similar. Because $ 9x^2 \ge x^2 \plus{} y^2 \plus{} z^2,$ we have $ x^2 \ge \frac {y^2 \plus{} z^2}{8} \ge \frac {(y \plus{} z)^2}{16},$ so
$ X \ge \frac {y^2 \plus{} z^2}{8} \plus{} \frac {3(y \plus{} z)^2}{4} \minus{} yz \ge \frac {yz}{4} \plus{} 3yz \minus{} yz > 0.$
Therefore $ X,Y,Z \ge 0.$ On the other hand, assume that $ x \ge y \ge z,$ we have
$ X \minus{} Y \equal{} (x^2 \minus{} y^2) \plus{} 4z(x \minus{} y) \ge 0.$
Thus
$ X(x \minus{} y)(x \minus{} z) \plus{} Y(y \minus{} z)(y \minus{} x) \plus{} Z(z \minus{} x)(z \minus{} y) \ge Y(x \minus{} y)(x \minus{} z) \plus{} Y(y \minus{} z)(y \minus{} x) \equal{} Y(x \minus{} y)^2 \ge 0.$
Our proof is completed. Equality holds if and only if $ a \equal{} b \equal{} c \equal{} 3.$ :)[/quote]

Good job! My solution is use S.O.S technique!


projective geometry
LaTeX
geometry
Euler

[color=red]Καλή αρχή και καλή επιτυχία στην όμορφη αυτή προσπάθεια που ξεκινάμε !!! [/color]

[b]α) [/b]Στο χώρο αυτό θα μαζέψουμε [b]15 ασκήσεις [/b]γεωμετρίας.
[b]β)[/b] Όλες θα αναφέρονται σε ύλη που λέει ο τίτλος : [color=red]Ειδικά θέματα : Συμμετροδιάμεσοι , Πολικές , Αρμονικότητα , Pascal , Desargues , Steiner ,Vecten, Casey  κλπ [/color]. Ας μην πάμε όμως σε πολύ τραβηγμένα πράγματα. Διαβάστε 10 ασκήσεις γεωμετρίας από τις IMO - short list για να έχετε ένα κριτήριο του επιπέδου δυσκολίας  των θεμάτων αυτών ,αλλά και της ύλης στη οποία στηρίζονται οι λύσεις . Επειδή πρόκειται για ειδικά θέματα , δε χρειάζεται να είναι πολύ δύσκολα. Να προτιμήσουμε θέματα που έχουν ήδη τεθεί σε διαγωνισμούς και λύνονται και με αυτά τα θεωρήματα.
 Προτιμάμε αποδεικτικές ασκήσεις. Όχι τόπους ή κατασκευές , εκτός και αν έχουν κάτι το εξαιρετικό.Αν έχουμε χώρο , ίσως τις βάλουμε σε άλλη υποενότητα αυτές.
  [color=blue]Επειδή εδώ οι περισσότεροι χρειάζονται τη σχετική θεωρία, μπορούμε λίγο λίγο να στέλνουμε και τα σχετικά θεωρήματα με τα σχόλιά τους .[/color]
[b]γ)[/b] Αυτός που θα προτείνει μια άσκηση  , θα τη βάζει με τη λύση της ή με τις λύσεις της , αν γνωρίζει περισσότερες.
[b]δ) [/b]Γράψτε τις λύσεις με απλότητα , σαφήνεια και με τα απαραίτητα σχόλια , όπου νομίζετε.Ας μην πλατειάζουμε όμως υπερβολικά για προφανή πράγματα. Αν κάτι είναι λίγο σύντομα γραμμένο , ας το αλλάξει αυτός που θα δυσκολευτεί.
[b]ε) [/b]Θα έχετε τη λύση καθαρογραμμένη με [color=red]ΜΟΛΥΒΙ[/color]  . Μία άσκηση σε κάθε σελίδα.Μόλις τελειώσουν οι παρατηρήσεις , θα ολοκληρώνετε το γράψιμο και θα την κρατάτε στο αρχείο που θα φτιάξετε.
Θα δούμε στη συνέχεια πότε θα μου τις στείλλετε και πώς.
[b]στ) [/b]Τις βοηθητικές γραμμές του σχήματος θα τις βάζετε με κόκκινο.
[b]ζ)[/b] Δε χρειάζεται να πληκτρολογείτε στο LATEX , διότι δεν παίρνει ελληνικά, εκτός και αν κάποιος ξέρει να μας πει πώς βάζουμε ελληνικά.Τα κείμενα θα ξαναπληκτρολογηθούν έτσι και αλλιώς.
[b] η)[/b] Όποιος  μπορεί να φτιάχνει σχήματα σε κάποιο πρόγραμμα , είτε αυτός που προτείνει , είτε κάποιος που διαβάζει την άσκηση , παρακαλείται να το κάνει, για να έχουμε και εποπτεία.Ας είναι και λίγο πρόχειρο.
  [b]θ)[/b] Κάθε άσκηση είναι καλό να την περνάτε σε word ,  να την αποθηκεύετε όλοι σε σχετικό νέο φάκελο , να την τυπώνετε  , να μελετάτε τη λύση και να στέλνετε τις παρατηρήσεις σας . Ακόμα και το : '' την κοίταξα και είναι εντάξει '' είναι σημαντικό.
 [b] ι) [/b]Πρέπει να παρακολουθήσουμε όλοι μας όλες τις ασκήσεις και να διαβάσουμε προσεκτικά τη λύση τους. Δεν έχει νόημα να έχουμε γράψει ένα βιβλίο , να έχουμε μέσα δυο ασκήσεις μας και να μην ξέρουμε τι έχει το βιβλίο ! Όλοι είμαστε συνυπεύθυνοι για το περιεχόμενο.Και θα είμαστε χαρούμενοι να γνωρίζουμε τις ασκήσεις!
   Αυτά για αρχή !   θα μαζέψουμε και άλλες παρατηρήσεις και θα τις βάλουμε και αλλού. !

[i]Μπορείτε ακόμα να χρησιμοποιείτε για ευκολία ,αν θέλετε , και τα ψευδώνυμα. Τα κανονικά ονόματα θα τα φτιάξουμε έτσι και αλλιώς.Θα πρότεινα όμως σιγά σιγά να υπογράφετε με το πραγματικό σας όνομα και για λόγους αρχειακούς.[/i]  [color=blue]ΚΑΛΗ ΔΥΝΑΜΗ ΚΑΙ ΚΑΛΕΣ ΕΜΠΝΕΥΣΕΙΣ  ! !! [/color

[b][color=red]ΕΝΟΤΗΤΑ 3 , ΑΣΚΗΣΗ - 1[/color][/b]

  Θα προηγηθεί μια σύντομη αναφορά στην ένοια της συμμετροδιαμέσου και των βασικών θεωρημάτων , χωρίς απόδειξη.
  ΑΥτή και μια άσκηση που τυχαία κατασκεύασα για το βιβλίο μας πολύ πρόσφατα. Παρακαλώ να τη  ελέγξετε  , όπως εσείς ξέρετε !

 [i]( Ήδη ,καθώς έκανα προεπισκόπηση , είδα δυο σημεία με ελληνική ονομασία και τα διόρθωσα ! )[/i]

Να με συγχωρείτε που δεν γνωρίζω αυτόν τον τομέα καλά :oops: , αλλά θα μπορούσατε να μου πείτε πώς αποδείξατε ότι BD συμμετροδιάμετρος :?: Βασικά γιατί έφτανε να αποδείξουμε ότι DA, DC εφαπτομένες του κύκλου?

Στην πραγματικότητα δεν είμαι σίγουρος. Συμμετροδιάμεσος δεν είναι η συμμετρική της διαμέσου ως προς το ύψος? :blush:
Τι ακριβώς ιδιότητες έχει :?:

[quote="bilot4"]Να με συγχωρείτε που δεν γνωρίζω αυτόν τον τομέα καλά :oops: , αλλά θα μπορούσατε να μου πείτε πώς αποδείξατε ότι BD συμμετροδιάμετρος :?: Βασικά γιατί έφτανε να αποδείξουμε ότι DA, DC εφαπτομένες του κύκλου?

Στην πραγματικότητα δεν είμαι σίγουρος. Συμμετροδιάμεσος δεν είναι η συμμετρική της διαμέσου ως προς το [u]ύψος?[/u] [color=red]ΔΙΧΟΤΟΜΟ [/color]!!! :blush:
Τι ακριβώς ιδιότητες έχει :?:[/quote]

 Αγαπητέ φίλε bilot 4! Μη στενοχωριέσαι ! :) Και εγώ σε πάρα πολλά πράγματα νοιώθω τελείως αδαής! Η μαθηματική επιστήμη είναι τόσο πλούσια , που δε μπορείς ποτέ να τα μάθεις όλα. Απλά , όλοι μας προσπαθούμε να γίνουμε καλύτεροι ! Και αυτό είναι πολύ όμορφο. Το να μαθαίνεις είναι ευχαρίστηση και ανανέωση. Λοιπόν :
   Στο βιβλίο θα τα εξηγήσουμε όλα αυτά ! Πρόκειται για πολύ χρήσιμα πράγματα. Εδώ ας μαζέψουμε πρώτα τις ασκήσεις και με τη θεωρία βλέπουμε.Δεν την έβαλα , για να μη χάνω χρόνο στην πληκτρολόγηση.
  Αυτό που ρωτάς είναι η βασική  μέθοδος κατασκευής της συμμετροδιαμέσου ! Πάρτο έτοιμο αυτό : [color=red]<<  Το σημείο Lemoine  ενός τριγώνου είναι το σημείο Georgone του εφαπτομενικού τριγώνου >>. [/color]Αλλά ,μη χάνεις χρόνο με αυτά. Ψάξε για ωραία θέματα στις άλλες ενότητες και σιγά σιγά, καθώς θα προχωράει και η ενότητα αυτή ,  θα συμπληρώσεις τις θεωρητικές σου γνώσεις . Και  εμείς οι άλλοι θα γινόμαστε μαζί σου σοφότεροι. Ο ένας θα διδάσκει στον άλλο. Γι αυτό είμαστε μια παρέα. Τα ξαναλέμε !
    Ευχαριστώ που μελέτησες το θέμα.-
                           ΜΠΑΜΠΗΣ

Ευχαριστώ κι εγώ πολύ για την απάντηση κ. Στεργίου :)

και θα ακολουθήσω τις συμβουλές σας με χαρά...

Το βιβλίο θα είναι ξεχωριστό για αυτές τις ενότητες  :lol:
Βάζω μια πολύ καλή άσκηση που έχει συζητηθεί στο φόρουμ.
[color=red]ΑΣΚΗΣΗ 2[/color]
Έστω τρίγωνο $ABC$ και $BD$ το ύψος του .Ο κύκλος με διάμετρο την $BD$ τέμνει τις πλευρές $AB$ και $BC$  στα $K,\ L$ αντίστοιχα .Οι εφαπτόμενες στον κύκλο στα $K,\ L$ τέμνονται στο $Μ$ . Να δειχθεί ότι η $BM$ διχοτομεί την $AC$.

[color=red]ΑΣΚΗΣΗ 2-ΛΥΣΗ[/color]
Είναι $\angle{KLB}=\angle{KDB}=90-\angle{ADK}=\angle{KAD}$

άρα το τετράπλευρο $AKLC$ είναι εγγράψιμο και άρα οι ευθείες $KL,\ AB$ είναι αντιπαράλληλες .Αφού η $BM$ είναι συμμετροδιάμεσος της $KL$ και οι $KL,\ AB$ είναι αντιπαράλληλες συνεπάγεται ότι η $BM$ είναι διάμεσος της $AB$ , το ζητούμενο

Προφανώς πριν τη λύση αυτής της άσκησης θα πρέπει να έχει παρουσιαστεί το κατάλληλο θεωρητικό υπόβαθρο

[b][size=109][color=red]ΑΣΚΗΣΗ 3.[/color][/size][/b]

Επί της πλευράς $BC$ τριγώνου $\bigtriangleup ABC$ και προς το εξωτερικό μέρος αυτού, κατασκευάζουμε το τυχόν ορθογώνιο παραλληλόγραμμο $BCEF.$

Εάν $D',$ είναι το σημείο τομής του περιγεγραμμένου κύκλου $(O)$ του $ABC,$ από την ευθεία του ύψους $AD$ και $B',$ $C',$ τα σημεία τομής του από τις ευθείες $AE,$ $AF$ αντιστοίχως, αποδείξτε ότι οι ευθείες $BB',$ $CC'$ και η δια του $D'$ διάμετρος του $(O),$ τέμνονται στο ίδιο σημείο.

Κώστας Βήττας.

ΥΓ. Έχει δημοσιευτεί στο Αγγλικό φόρουμ. http://www.mathlinks.ro/Forum/viewtopic.php?t=122340

Θα καθαρογράψω τη λύση, καθώς και κάποιες σημειώσεις από τη θεωρία περί διπλού λόγου.

[b][size=109][color=red]ΑΣΚΗΣΗ 1 – ΣΧΟΛΙΑ[/color][/size][/b]

Η απόδειξη βασίζεται στο παρακάτω γνωστό θεώρημα της θεωρίας των ισογώνιων ευθειών.

[b][size=109]ΘΕΩΡΗΜΑ Γ3-Θ1[/size][/b]

[b][size=100]Σε κάθε τρίγωνο, το σημείο τομής των εφαπτόμενων του περιγεγραμμένου κύκλου, σε δύο κορυφές του, ανήκει στην ευθεία της συμμετροδιαμέσου που διέρχεται δια της τρίτης κορυφής.[/size][/b]

(η αρίθμηση των θεωρημάτων ξεκινάει αυθαίρετα από αυτό το θεώρημα, αλλά στο τέλος θα μπουν με τη σωστή σειρά).

[b][size=100]1η Απόδειξη.[/size][/b] – (σχήμα Γ3 -1.1)

Έστω $K,$ $L,$ οι προβολές του $P,$ επί των ευθειών $AB,$ $AC$ αντιστοίχως. Για να είναι η $AP,$ συμμετροδιάμεσος του τριγώνου $ABC,$ αρκεί να αποδειχθεί ότι $\frac{PK}{PL}= \frac{AB}{AC}.$

Έστω $B',$ $C',$ τα αντιδιαμετρικά σημεία των $B,$ $C$ αντιστοίχως.

Από τα όμοια ορθογώνια τρίγωνα $\bigtriangleup KBP,$ $\bigtriangleup ABB'$ $($ γιατί $\angle KPB = \angle ABB',$ από $PK\perp AB$ και $PB\perp BB'$ $),$ προκύπτει ότι $\frac{PK}{AB}= \frac{PB}{BB'}$ $,(1)$

Ομοίως, από τα όμοια ορθογώνια τρίγωνα $\bigtriangleup LCP,$ $\bigtriangleup ACC'$ έχουμε ότι $\frac{PL}{AC}= \frac{PC}{CC'}$ $,(2)$

Από $(1),$ $(2)$ και επειδή $PB = PC,$ $BB' = CC',$ συμπεραίνουμε ότι $\frac{PK}{AB}= \frac{PL}{AC}$ $\Longrightarrow$ $\frac{PK}{PL}= \frac{AB}{AC}$ και η πρόταση έχει αποδειχθεί.

[b][size=100]2η Απόδειξη.[/size][/b] – (σχήμα Γ3 -1.2)

Έστω $D,$ $E,$ τα σημεία τομής του περιγεγραμμένου κύκλου $(O),$ του $\bigtriangleup ABC,$ από την ευθεία $PO$ η οποία τέμνει επίσης μεσοκάθετα, την πλευρά $BC.$

Είναι γνωστό, ότι τα σημεία $D,$ $E,$ είναι αρμονικά συζυγή των $P,$ $M$ και άρα η δέσμη $A.DEPM,$ είναι αρμονική.

Επειδή τώρα, δύο ακτίνες της δέσμης αυτής, είναι κάθετες μεταξύ των $($ $AD\perp AE$ $),$ συμπεραίνουμε ότι είναι και διχοτόμοι των γωνιών που σχηματίζουν οι άλλες δύο ακτίνες της δέσμης.

Δηλαδή η διχοτόμος $AE,$ της γωνίας $\angle A,$ του $\bigtriangleup ABC,$ διχοτομεί επίσης και τη γωνία $\angle PAM.$

Αφού η $AM,$ είναι διάμεσος του $ABC,$ η $AP$ είναι συμμετροδιάμεσος και η πρόταση έχει αποδειχθεί.

Κώστας Βήττας.

ΑΣΚΗΣΗ 4 -(Κινέζικη Μαθηματική Ολυμπιάδα 2007)
Έστω $O$ και $I$ το περίκεντρο και το έγκεντρο τριγώνου $\triangle ABC$ αντίστοιχα. Ο εγκεγραμμένος κύκλος του $\triangle ABC$ εφάπτεται των  $BC,CA,AB$ στα  $D,E,F$ αντίστοιχα. Οι ευθείες $FD$ και $CA$ τέμνονται στο $P$,και οι ευθείες $DE$ και $AB$ τέμνονται στο $Q$. Αν $M,N$ είναι τα μέσα των τμημάτων $PE$, και $QF$ αντίσtοιχα,δείξτε ότι  $OI\perp MN$.

ΑΣΚΗΣΗ 4-ΛΥΣΗ (Σιλουανός Μπραζιτίκος)
Αφού οι $AD,\ BE,\ CF$ τέμνονται στο ίδιο σημείο (σημείο Gergonne) θα έχουμε ότι τα $P,\ E$ είναι συζηγή αρμονικά των
$A,\ C$ λόγω του πλήρους τετραπλεύρου που σχηματίζεται.
Για τον ίδιο λόγω τα $F,\ Q$ είναι συζηγή αρμονικά των $A,\ B$
Αφού τώρα $M,\ N$ είναι τα μέσα των $PE,\ QF$ από το θεώρημα Newton θα έχουμε :
$ME^{2}=MC\cdot MA$  και $NF^{2}=NB\cdot NA$

Όμως τα $ME^{2},\ NF^{2}$ είναι οι δυνάμεις των $M,\ F$ ως προς τον εγκεγραμμένο κύκλο του $ABC$ ενώ τα
$MC\cdot MA,\ NB\cdot NA$  είναι οι δυνάμεις των Ν,Μ ως προς τον περιγεγραμμένο κύκλο του $ABC$ . Συνεπώς τα $M,\ N$ ανήκουν στον ριζικό άξονα των δύο κύκλων και αφού η $ΟΙ$ είναι η διακεντρική ευθεία,θα είναι  $OI\perp MN$ , το ζητούμενο

φίλε σιλουανέ εξήγησε μου σε παρακαλώ ποιό είναι το σημείο Gregonne κι τί ιδιότητα έχει??

Φίλε Αμβρόσιε, είμαι σίγουρος ότι ο Σιλουανός θα ξέρει καλύτερα αλλά απ' ότι ξέρω αν σε ένα τρίγωνο φέρεις τον εγγεγραμμένο κύκλο και τις σεβιανές από κάθε κορυφή προς τα σημεία επαφής στην απέναντι πλευρά τότε οι τρεις σεβιανές αυτές συντρέχουν στο σημείο Gergonne. Αποδεικνύεται πανεύκολα με το θεώρημα του Σεβά.
Επίσης είναι το κέντρο του κύκλου Adam.
Για περισσότερες πληροφορίες για το Gergonne και τον Adam circle σε παραπέμπω στο agutie.homestead.com/files/Geoproblem_B.htm

[size=150][color=red]ΑΣΚΗΣΗ 5[/color] [/size]

 [color=blue]Βάζω τώρα ένα θέμα , τελείως στοιχειώδες , αλλά πολύ χρήσιμο στην πολική σημείου. Ο Κώστας , μάλλον θα επωμιστεί το βάρος να γράψει τη σχετική θεωρία με συντομία. Δε χρειάζονται αποδείξεις αλλά τα συμπεράσματα και λίγα σχόλια.
 Τη δίνω σε word για να είναι επεξεργάσιμη.[/color]

ΑΣΚΗΣΗ 2-2η ΛΥΣΗ (Κώστας Βήττας)
Έστω Ε, το σημείο τομής των AB, DL και F, το σημείο τομής των BC, DK.

Θεωρούμε το εγγεγραμμένο τετράπλευρο BKDL, ως ειδική περίπτωση εγγεγραμμένου εξαγώνου του οποίου δύο ζεύγη κορυφών συμπίπτουν στις θέσεις των K, L.

Σύμφωνα τότε, με το θεώρημα του Pascal, τα σημεία E, M, F, είναι συνευθειακά.

Θα αποδείξουμε ότι EF//AC και ότι το M, είναι μέσον του EF.

Από την καθετότητα των ευθειών AB, DL και BC, DK, συμπεραίνουμε ότι το σημείο D, είναι ορθόκεντρο του τριγώνου BEF και επομένως έχουμε EF//AC.

Θεωρούμε ότι M', είναι το μέσον του EF και θα αποδείξουμε ότι συμπίπτει με το M.

Από το ορθογώνιο τρίγωνο KEF => M'K = M'E = M'F, (1).

Από το ορθογώνιο τρίγωνο LEF => M'L = M'E = M'F, (2).

Από (1), (2), => M'K = M'L => το σημείο M', είναι σημείο της μεσοκάθετης του KL και άρα συμπίπτει με το M ( ως σημείο και του EF ).

Από το τραπέζιο EACF, συμπεραίνουμε ότι η ευθεία BM, διέρχεται δια του μέσου του AC και η πρόταση έχει αποδειχθεί.

ΣΗΜΕΙΩΣΕΙΣ.

1) – Μία γενίκευση του προβλήματος είναι η εξής:

Σημείο Κ, κινείται επί ευθείας (e) η οποία διέρχεται από την κορυφή Α, δοσμένης γωνίας <xAz. Ο κύκλος με κέντρο το Κ και ακτίνα KA, τέμνει τις πλευρές Ax, Az, στα σημεία B, C, αντιστοίχως. Ζητείται να ευρεθεί ο γεωμετρικός τόπος του σημείου Μ, τομής των εφαπτομένων του κύκλου (κ), στα σημεία B, C.

Δια του Μ φέρνουμε κάθετη ευθεία επί την (e), η οποία τέμνει τις πλευρές Ax, Az, της δοσμένης γωνίας στα σημεία B, C, αντιστοίχως. Αποδεικνύεται ότι το Μ είναι μέσον του BC και ότι ο ζητούμενος γεωμετρικός τόπος είναι η ευθεία ΑΜ.

2) – Έχω υπόψη μου μία ενδιαφέρουσα επέκταση του προβλήματος ( συνέχεια από την εκφώνηση όπως έχει δοθεί ):

Έστω E, F, τα σημεία τομής της πλευράς AC, από τις εφαπτόμενες στα K, L, αντιστοίχως. Εάν P, είναι το σημείο τομής των KC, LA και Q, το σημείο τομής των KF, LE, να αποδειχθεί ότι η ευθεία PQ, διέρχεται δια του μέσου Ν του AC.

Κύριε Κώστα μπορείτε να βάλετε τις αποδείξεις σας για τις γενικεύσεις για να συμπληρωθεί το πακέτο ?Ευχαριστώ.

[b][size=109][color=red]ΑΣΚΗΣΗ 2 - 1η ΛΥΣΗ  - ΣΧΟΛΙΑ.[/color][/size][/b]

Η απόδειξη αυτή, βασίζεται στη χαρακτηριστική ιδιότητα των αντιπαράλληλων ευθειών, όπου δύο ευθείες που συνδέουν την κορυφή της γωνίας αναφοράς τους, με δύο ομόλογα σημεία πάνω σ’ αυτές, είναι ισογώνιες ως προς τη γωνία αναφοράς ( συμμετρικές, ως προς τη διχοτόμο της ).

$\bullet$  Έστω τρίγωνο $ABC$ και $D,$ $E,$ τα σημεία τομής των ευθειών των πλευρών του $AC,$ $AB$ αντιστοίχως, από τυχόντα κύκλο $(O_{1}),$ χορδής $BC.$

Είναι φανερό ότι τα τρίγωνα $\bigtriangleup ABC$ και $\bigtriangleup ADE,$ είναι όμοια $($ από εγγράψιμο τετράπλευρο $BCDE,$ $\Longrightarrow$ $\angle ADE = \angle B$ και $\angle AED = \angle C$ $).$

Οι ευθείες $BC,$ $DE,$ ονομάζονται [b][size=109]''αντιπαράλληλες ευθείες'',[/size][/b] ως προς τη γωνία $\angle A$ του $ABC$ $($ ή ως προς τη γωνία των ευθειών $AB,$ $AC$ $).$

Εάν $AA',$ είναι η διχοτόμος της γωνίας $\angle A$ $($ $A',$ μέσον του τόξου $BC$ $),$ τότε τα σημεία $B',$ $C',$ συμμετρικά αντιστοίχως των $B,$ $C,$ ως προς αυτήν, κείται επί των ευθειών $AC,$ $AB,$ αντιστοίχως και αποδεικνύεται εύκολα ότι $B'C'\parallel DE$  $($ $\angle AB'C' = \angle B = \angle ADE$ $).$

Επειδή οι ευθείες $BC,$ $B'C',$ είναι συμμετρικές ως προς τη διχοτόμο της γωνίας $\angle A,$ κάθε σημείο της μίας, είναι συμμετρικό με κάποιο σημείο της άλλης και έχει την ίδια ιδιότητα με αυτό. Έτσι το σημείο $M',$ συμμετρικό σημείο του μέσου $M,$ του $BC,$ είναι το μέσο του $B'C'.$  Αυτό αποδεικνύεται εύκολα από τα ίσα τρίγωνα $\bigtriangleup ABC$ και $\bigtriangleup AB'C'.$

Από $DE\parallel B'C',$ συμπεραίνουμε ότι η ευθεία $AM'$ διέρχεται από το μέσον $N,$ του $DE$ και αποδεικνύεται έτσι, ότι οι ευθείες $AM,$ $AN,$ διάμεσοι αντιστοίχως των τριγώνων $\bigtriangleup ABC,$ $\bigtriangleup AB'C',$ των οποίων οι πλευρές $BC,$ $B'C',$ είναι αντιπαράλληλες ως προς τη γωνία $\angle A,$  είναι συμμετρικές ως προς τη διχοτόμο της $($ ισογώνιες ως προς τη γωνία $\angle A$ $).$

Το ίδιο ισχύει για τα ύψη τους $AK,$ $AK',$ καθώς και για κάθε ζεύγος ομόλογων ευθειών τους.

[b][size=109]Σημειώσεις.[/size][/b]

[b]1).[/b] – Όπως είναι γνωστό, σε κάθε τρίγωνο, το ύψος και η διάμετρος του περιγεγραμμένου κύκλου του, από την ίδια κορυφή, είναι ισογώνιες ως προς την αντίστοιχη γωνία.

Έτσι, αφού οι ευθείες $AK,$ $AL\equiv AK',$ είναι ισογώνιες ως προς τη γωνία $\angle A,$ συμπεραίνουμε ότι η ευθεία $AL,$ διέρχεται δια του κέντρου $O,$ του περιγεγραμμένου κύκλου του $\bigtriangleup ABC.$

Επειδή τώρα, $AL\perp DE,$ συμπεραίνουμε ότι η δια του $A$ διάμετρος του $(O),$ είναι κάθετη σε όλες τις αντιπαράλληλες ευθείες της $BC,$ ως προς τη γωνία $\angle A.$

[b]2).[/b] – Η εφαπτομένη του $(O)$ στην κορυφή $A$ του $ABC,$ είναι η ειδική περίπτωση αντιπαράλληλης ευθείας της $BC,$ όταν ο κύκλος χορδής $BC$ ταυτίζεται με τον περιγεγραμμένο κύκλο $(O),$ του $\bigtriangleup ABC.$

[b]3).[/b] – Με τόσο εκτενή σχόλια, είναι σίγουρο ότι θα βγούμε εκτός προϋπολογισμού σελίδων, αλλά και ευρώ...

Σκέφτηκα όμως, ότι καλό είναι να υπάρχουν εδώ και από αυτά ας συμπεριλάβουμε στο τέλος, ότι κριθεί σκόπιμο.

Κώστας Βήττας.

[size=150][color=red]ΑΣΚΗΣΗ 6 [/color][/size]

 Ένα καλό θέμα , εφαρμογή του θεωρήματας pascal. Θα το συναντήσετε σίγουρα ως θέμα  κάποιου διαγωνισμού , όσοι ψάχνετε τη Γεωμετρία στο δίκτυο !

Θα ήθελα να ρωτήσω από πού μπορώ να διαβάσω θεωρία για αυτόν τον τομέα :huh:
Υπάρχει κάποιο βιβλίο που θα μου προτείνατε :?:

[quote="silouan"][quote="stergiu"][size=150][color=red]ΑΣΚΗΣΗ 7[/color][/size]

 Μια επίσης πολύ καλή άσκηση πολύ διδακτική και πολύ χρήσιμη.

   [color=blue]China- 2007 ! [/color][/quote]

Κ. Μπάμπη αν δεν κάνω λάθος είναι ίδια με την άσκηση 4 που έβαλα παραπάνω ! Διορθώστε με αν κάνω λάθος[/quote]

[b]Silouan Sorry ! [/b]

Θυμόμουνα ότι την είχα ξαναδεί την άσκηση , σε σκέφτηκα , αλλά έψαξα στο χώρο με τα θέματα διαγωνισμών(ΙΜΟ κλπ) και δεν ήταν εκεί. Μετά σκέφτηκα να τη βάλω εδώ , γιατί είναι ωραία εφαρμογή στο ριζικό άξονα.
  Δεν πειράζει. Την άσκηση θα τη χρεώσω σε σένα - το ξέρω ότι δεν το λες για αυτό - , αλλά έτσι είναι το σωστό.
 Συνέχισε με την ίδια αρίθμηση και αγνόησε αυτή την άσκηση.
  Σε ευχαριστώ - Μπάμπης

[color=red][size=150][b]ΑΣΚΗΣΗ 8[/b][/size][/color]

Δίδεται τρίγωνο $ABC$ και $D,E,Z$ τα σημεία επαφής του εγγεγραμμένου του κύκλου (κέντρο $O$) με τις πλευρές $BC,CA,AB$, αντίστοιχα. Αν $P$ είναι το αντιδιαμετρικό σημείο του $D$, οι $AP,ZP,EP$ τέμνουν τη $BC$ στα σημεία $M,K,N$, αντίστοιχα. Να αποδειχθεί ότι $KM=MN$.

[size=150][b][color=red]ΛΥΣΗ[/color][/b][/size]

Έστω $S$ το άλλο κοινό σημείο της $AP$ με τον κύκλο κέντρου $O$ και $A'=ZE\cap PS$. Εφόσον η $ZE$ είναι πολική του $A$ ως προς το $O$, η τετράδα $(A,A',P,S)$ θα είναι αρμονική και επομένως και η προβάλλουσα από το $Z$ δέσμη $(ZB,ZA',ZP,ZS)$ θα είναι αρμονική. Όμως και η δέσμη $(PK,PN,PX,PM)$ όπου $PX$ η εφαπτομένη του σημείου $P$ στον κύκλο κέντρου $O$, θα είναι αρμονική λόγω της ισότητας των γωνιών $\hat a=\hat a_{1}, \hat b=\hat b_{1}, \hat c=\hat c_{1}$. Εφόσον πλέον η δέσμη $(PK,PN,PX,PM)$ είναι αρμονική με την ευθεία $PX$ παράλληλη προς τη $ΚΝ$, έπεται ότι το $M$ είναι μέσο του $KN$.

[quote="mostel"][color=red][size=150][b]ΑΣΚΗΣΗ 8[/b][/size][/color]

Δίδεται τρίγωνο $ABC$ και $D,E,Z$ τα σημεία επαφής του εγγεγραμμένου του κύκλου (κέντρο $O$) με τις πλευρές $BC,CA,AB$, αντίστοιχα. Αν $P$ είναι το αντιδιαμετρικό σημείο του $D$, οι $AP,ZP,EP$ τέμνουν τη $BC$ στα σημεία $M,K,N$, αντίστοιχα. Να αποδειχθεί ότι  [color=red]α) [/color]$KM=MN$ [color=red]β) $BD= MC$[/color].

[size=150][b][color=red]ΛΥΣΗ[/color][/b][/size]

α) Έστω $S$ το άλλο κοινό σημείο της $AP$ με τον κύκλο κέντρου $O$ και $A'=ZE\cap PS$. Εφόσον η $ZE$ είναι πολική του $A$ ως προς το $O$, η τετράδα $(A,A',P,S)$ θα είναι αρμονική και επομένως και η προβάλλουσα από το $Z$ δέσμη $(ZB,ZA',ZP,ZS)$ θα είναι αρμονική. Όμως και η δέσμη $(PK,PN,PX,PM)$ όπου $PX$ η εφαπτομένη του σημείου $P$ στον κύκλο κέντρου $O$, θα είναι αρμονική λόγω της ισότητας των γωνιών $\hat a=\hat a_{1}, \hat b=\hat b_{1}, \hat c=\hat c_{1}$. Εφόσον πλέον η δέσμη $(PK,PN,PX,PM)$ είναι αρμονική με την ευθεία $PX$ παράλληλη προς τη $KN$, έπεται ότι το $M$ είναι μέσο του $KN$.[/quote]

[color=red] Είναι πολύ καλό θέμα![/color]

 [color=blue]Συμπλήρωσα ένα ακόμα ερώτημα[/color], γιατί έχει ενδιαφέρον.

[color=red] b) [/color]Θα συμπληρώσω τη λύση με την ομοιοθεσία. Η ομοιθεσία που πάει το Ρ στο Μ , μεταφέρει τον εγγεγραμένο κύκλο στον παρεγγεγραμμένο.Ο παρεγγεγραμμένος  θα εφάπτεται της ΒΓ στο Μ, διότι η εφαπτομένη του εγγεγραμμένου στο Ρ είναι παράλληλη στη ΒΓ και έτσι η χ'χ μεταφέρεται στη ΒΓ. Αλλά από γνωστή πρόταση ΒΔ = ΜC .

 Στέλιο , δες μήπως βγαίνει και το β) με βάση αυτά που έχεις ήδη βρει. Θα το κοιτάξω και εγώ.
 Μπάμπης

[quote="Virgil Nicula"][color=darkblue][b][u]A short proof.[/u][/b] It is well-known the property $\boxed{MC=DB}$, i.e. $DM=|b-c|$.

Therefore, $\left\{\begin{array}{c}ZK\perp ZD\ ,\ BZ=BD\Longrightarrow KB=BD\\\ EN\perp ED\ ,\ CE=CD\Longrightarrow NC=CD\end{array}\right\|$ $\Longrightarrow$

$\left\{\begin{array}{c}KM=KB+BM=BD+DC=BC\\\ NM=NC+CM=CD+DB=BC\end{array}\right\|$ $\Longrightarrow$ $\boxed{MK=MN=BC}$.[/color][/quote]

 [color=red]Dear Virgil ! [/color]

 [b]a)[/b] We are very glad to have you here ! Your help is great ! In this section we try to collect some good problems , to make a little book. We invite you to send us some of your always nice problems in olympiads. Of course , to every problem there will be the name of the proposer and that of the creator , if it is known.
  [b]b)[/b][color=blue]See my post before your's .[/color]
 After your solution,  we have two ways - two proofs for mostels problem. Part a) helps to prove b) and part b) - according to your nice solution - helps to prove part a). Nice work !!!
  Thank you again and we hope to see you here with many ... Niculian problems :) .

  babis

[size=150][color=red][b]ΑΣΚΗΣΗ 9[/b][/color][/size]

Δίδεται τρίγωνο $ABC$ και ο εγγεγραμμένος σε αυτό κύκλος κέντρου $O$. Οι κάθετες στο $O$ στις $OA,OB,OC$, τέμνουν τυχαία εφαπτομένη του κύκλου με κέντρο $O$ στα σημεία $M,N,P$, αντίστοιχα. Να αποδειχθεί ότι οι $AM,BN,CP$ συντρέχουν.

[size=150][color=red][b]ΛΥΣΗ[/b][/color][/size]

Η ευθεία $AM$ περνά από τον πόλο $A$ της $B_{1}C_{1}$, άρα και η $B_{1}C_{1}$ θα περνά από τον πόλο της $AM$. Εξάλλου η πολική του $M$ ως προς τον κύκλο $(O)$ είναι η από το σημείο επαφής $T$ κάθετη επί την $OM$, δηλαδή η  από το $T$ κάθετη επί την $B_{1}C_{1}$, ή $TS_{1}$. Εφόσον η $AM$ περνά από τον πόλο $M$ της $TS_{1}$ και η $TS_{1}$ θα περνά από τον πόλο της $AM$ και συνεπώς ο πόλος της $AM$ είναι το ίχνος $S_{1}$ της από το $T$  κάθετης επί την $B_{1}C_{1}$.

Όμοια ο πόλος της $BN$ είναι το ίχνος $S_{2}$ της από το $T$ κάθετης επί την $A_{1}C_{1}$ και ομοίως ο πόλος της $CP$ είναι το ίχνος $S_{3}$ της από το $T$ κάθετης επί την $A_{1}B_{1}$. Τα σημεία όμως $S_{1},S_{2},S_{3}$ επειδή το $T$ βρίσκεται στην περιγεγραμμένη περιφέρεια στο $A_{1}B_{1}C_{1}$, βρίσκεται επί ευθείας, της ευθείας του Simson του σημείου $T$ ως προς τον κύκλο κέντρου $(O)$ και επομένως οι πολικές των σημείων $S_{1},S_{2},S_{3}$, που είναι αντίστοιχα οι ευθείες $AM,BN,CP$ θα περνούν από το ίδιο σημείο $Z$, πόλο της ευθείας $S_{1}S_{2}S_{3}$ ως προς τον κύκλο κέντρου $(O)$, γιατί οι πολικές των σημείων μιας ευθείας περνούν από τον πόλο αυτής.

[color=red][b]ΑΣΚΗΣΗ 10 [/b][/color]

Θεωρούμε κυρτό τετράπλευρο $ABCD$ περιγράψιμο σε κύκλο και έστω $I$ το κέντρο του κύκλου αυτού.Έστω ότι το $I$ δεν βρίσκεται στη διαγώνιο  $AC$. Οι διαγώνιες  $AC$ και  $BD$  του τετραπλεύρου  $ABCD$  τέμνονται στο  $E$. Η ευθεία που είναι κάθετη στην  $BD$  και διέρχεται από το σημείο  $E$  τέμνει τις ευθείες  $IA$ και  $IC$  στα σημεία $P$ και  $Q$ αντίστοιχα.Να αποδείξετε ότι   $PE=EQ$ .

[b]ΛΥΣΗ[/b]

Έστω ότι η $AB$ τέμνει την  $CD$ στο $J$.Έστω επίσης ότι  ο  κύκλος  $(I)$ εφάπτεται των  $AB,CD$ στα σημεία $N,M$.

Τα σημεία  $N,E,M$  είναι συγγραμμικά ( είναι συνέπεια του θεωρήματος Brianchon). H  $MN$  τέμνει την $AI$ στο  $F$.

Έχουμε ότι :

 $\angle FMD=\angle NMJ=90-\angle NJM/2=(\angle JAD+\angle JDA)/2$$=\angle IAD+\angle IDA=\angle FID$

 οπότε   τα σημεία  $I,M,F,D$  είναι ομοκυκλικά . 'Ετσι :

  $\angle IDM=\angle IFM$ $(1)$ and $\angle IFD=\angle IMD =90=\angle PED$

 οπότε και τα σημεία  $P,E,F,D$ είναι ομοκυλκικά .Επομένως

    $\angle EFP=\angle EDP$  , $(2)$

 Από τις  (1) and (2)  παίρνουμε

                $\angle PDE=\angle IDM=\angle ADJ/2$

και ομοίως  $\angle EDQ=\angle ADJ/2$

ΑΥτό σημαίνει ότι η  $BD$ είναι μεσοκάθετος του τμήματος  $PQ$ , δηλαδή  $EP=EQ$

[b]ΑΛΛΟΣ ΤΡΟΠΟΣ[/b]

Θα τον αποδώσω στην ελληνική άλλη στιγμή.Αν κάποιος θέλει μπορεί να το κάνει και νωρίτερα!

Let intersection of incircle with AB,BC,CD,DA be V,W,T,U in order.and let  $UV\cap WT\equiv S\ \ ,\ AC\cap VW\equiv F$

and  let $IH \perp BD\ \ ,H\in BD$

the polar of $S$ (AC) intersect with polar of $F$ (BD) at $E$ so $SF$ is polar of $E$ hence $IE\perp SF$

we know that $S(WV,EF)=-1$ so perpendicular to these line are also Harmonic.

perpendicular line to SE is IH , to SF is IE, to SV is IA,to SW is IC hence $I(HE,AC)=-1$. and we know that $PQ \parallel IH$ hence $PE=EQ$.

[quote="r_boris"]Μπάμπη , νομίζω ότι η τελειταία άσκηση γεωμετρίας που έστειλα είναι κατάλληλη για εδώ (βιβλίο Ειδικά θέματα...). Αν θες κοίταξέ την και βάλε την ή όχι[/quote]

 Ροδόλφε ,

  αν είναι έτοιμη , με πλήρη και αναλυτική ,  κατά το δοκούν ,  λύση  , θα την βάλω ευχαρίστως ! Το ίδιο θα κάνω και με κάθε άλλη άσκηση που θα στείλλεις και θα σου αρέσει.

 Μπάμπης

[color=red][b]ΑΣΚΗΣΗ 11[/b][/color]

Να δειχθεί ότι, για περιττό $ n$, τα μόνα γνωστά κατασκευάσιμα $ n$-γωνα είναι αυτά για τα οποία ο $ n$ διαιρεί τον $ 2^{32}-1$.

(Τη λύση θα τη βάλλω σε λίγες μέρες αν δεν έχει προταθεί από κάποιον)

[b][color=red][size=150]Βασικό Θεώρημα[/size][/color][/b] (η απόδειξη παραλείπεται καθώς θέλει θεωρία Galois)

[b][color=blue]Αναγκαία και ικανή συνθήκη για να κατασκευάζεται με κανόνα και διαβήτη ένα κανονικό πολύγωνο, είναι το πλήθος των πλευρών του να είναι της μορφής

$ 2^{a_{0}}\left(2^{2^{a_{1}}}+1\right)\left(2^{2^{a_{2}}}+1\right)\cdots \left(2^{2^{a_{m}}}+1\right), \ \ a_{i}\in \mathbb{Z}$ για κάθε $ \ i=0,1,\cdots,m$ με $ \ a_{0}, a_{1},a_{2},\cdots,a_{m}\geq 0$ και μάλιστα οι αριθμοί $ 2^{2^{a_{m}}}+1$ να είναι πρώτοι και διαφορετικοί ανά δύο. (Πρώτοι αριθμοί της μορφής $ F_{m}= 2^{2^{a_{m}}}+1, \ m\in\mathbb{Z}, m\geq 0$ λέγονται πρώτοι του Fermat).[/color][/b]

[b][size=150][color=red]Λύση άσκησης 11[/color][/size][/b]

Δεδομένου του παραπάνω, του ότι οι μόνοι (εως σήμερα) γνωστοί πρώτοι του Fermat είναι οι $ F_{0},F_{1},F_{2},F_{3},F_{4}\ \^{\star}$

και τέλος του ότι $ 2^{32}-1=\left(2^{2^{0}}+1\right)\left(2^{2^{1}}+1\right)\left(2^{2^{2}}+1\right)\left(2^{2^{3}}+1\right)\left(2^{2^{4}}+1\right)=F_{0}F_{1}F_{2}F_{3}F_{4}$

το ζητούμενο είναι άμεσο.

Αλέξανδρος

$ ^{\star}:$ O Fermat πίστευε (είναι αυτό που ονομάζουμε "μικρή πλάνη του Fermat") ότι η ακολουθία των $ F_{m}$ δίνει πρώτους αριθμούς. O Euler aργότερα (1732) έδειξε ότι ο $ F_{5}$ δεν είναι πρώτος. Πάρα πολλοί αριθμοί από τότε εως σήμερα, έχουν αποδειχθεί ότι είναι σύνθετοι, ενώ δεν έχει βρεθεί κανένας πρώτος αυτής της ακολουθίας. Περισσότερα μπορείτε να δείτε και στην ιστοσελίδα http://www.prothsearch.net/fermat.html .

Πολύ ωραία λύση Αλέξανδρε  :lol:

Ξέχασα να πω ότι την άσκηση την άντλησα από το βιβλίο του Στυλιανού Ανδρεαδάκη, "Θεωρία Galois".

[quote="stergiu"]
we know that $S(WV,EF)=-1$ so perpendicular to these line are also Harmonic.

perpendicular line to SE is IH , to SF is IE, to SV is IA,to SW is IC hence $I(HE,AC)=-1$. and we know that $PQ \parallel IH$ hence $PE=EQ$.[/quote]

Μπορεί κάποιος να εξηγήσει λίγο παραπάνω αυτές τις 2 σειρές??


geometry
circumcircle
inradius
geometry theorems

Does anybody know a nice geometrical way of showing that if $a,b,c$ are sides of a triangle and, $s,R,r$ represent the semiperimeter, circumradius and inradius respectively, then

$ab+bc+ca=s^{2}+4Rr+r^{2}$

I think ,i can used Viete's theorem here...

I have a solution :
 $ sinA=\frac{2tg\frac{A}{2}}{1+tg^2\frac{A}{2}} $
$ \iff \frac{a}{2R} = \frac{\frac{r}{(p-a)}}{1+\frac{r^2}{(p-a)^2}} $
$ \iff a^3-2pa^2+(p^2+r^2+4Rr)a -4prR =0 $
 So a,b,c are roots of the equation :
  $ x^3-2px^2+(p^2+r^2+4Rr)x -4prR =0 $
 After that , using Viete ... :)

Otra forma de verlo es reemplazando a=y+z,..... y utilizando que yz=r.ra (semejanza), y ra + rb +rc= 4R+r [te acordás Leo que te conté esa identidad en la IMO con semejanza, área y pedales?].... porque te queda:
ab+bc+ca=(x+y+z)^2+xy+yz+zx=s^2+r(ra+rb+rc)=s^2+r(4R+r).

[u][b]The first solution.[/b][/u]

It is known that $S=sr=\sqrt{s(s-a)(s-b)(s-c)},\ abc=4RS,\ sr^2=(s-a)(s-b)(s-c).$

Therefore, $sr^2=s^3-(a+b+c)s^2$ $+(ab+bc+ca)s-abc\Longrightarrow$

$sr^2+s^3+4Rsr=s(ab+bc+ca)\Longrightarrow ab+bc+ca=s^2+r^2+4Rr.$

[b]Exercise[/b]. Prove that $(s-a)(s-b)+(s-b)(s-c)+(s-c)(s-a)=r(4R+r).$

[u][b]The second solution.[/b][/u]

It is known that $S=sr=(s-a)r_a=(s-b)r_b=(s-c)r_c;$

$r_a+r_b+r_c=4R+r;\ r_a r_b+r_b r_c+r_c r_a=s^2;\ r_a r_b r_c=s^2r$.

 Therefore,$\frac 1 r_a +\frac 1 r_b +\frac 1 r_c =\frac 1 r,$

$a=s\left( 1-\frac rr_a \right),\ a.s.o.\Longrightarrow \sum ab=s^2\cdot \sum \left[ \left( 1-\frac rr_a\right) \left( 1-\frac rr_b\right) \right] \Longrightarrow$

$\sum ab=s^2\cdot \left[ 3-2r\cdot \sum \frac 1r_a +\frac{r^2(r_a+r_b+r_c)}{r_ar_br_c}\right]\Longrightarrow$

$\sum ab=s^2\cdot \left[ 3-\frac{2rs^2}{rs^2}+\frac{r^2(4R+r)}{rs^2}\right]\Longrightarrow ab+bc+ca=s^2+r(4R+r).$

\[ \underline {\overline {\left|\ \{\ a,\ b,\ c\ \}\Longleftrightarrow \{\ r_a,\ r_b,\ r_c\ \}\Longleftrightarrow \{\ R,\ r,\ s\ \}\ \right| }} \]

_________________________________$Farewell,\ my\ dear\ Mathlinks.ro\ !$


How many positive integers less than 2008 have an even number of divisors?

This should be moved to one of the mathematical questions forums I think...

...anyway the only positive integers that don't have an even number of divisors are perfect squares. There are $ 44$ perfect squares below $ 2008$, so there are $ 2007\minus{}44\equal{}\fbox{1963}$ such positive integers.


integration
algebra
polynomial
linear algebra
linear algebra unsolved

If $ S \equal{}\mathbb{R} [X]_{3}$ is a subspace of $ \mathbb{R} [X]$ with
$ \left < f, g \right > \equal{} \int^1_0 f(t) g(t)\,dt$

How do you find the orthogonal complement of the subspace of constant polynomials?

From a theorem I know that $ S\equal{} Q \oplus Q^\bot$.
But $ Q \equal{}\mathbb{R}$ so wouldn't that make the orthogonal complement just  $ Q^\bot \equal{}  \left\{x^3\minus{}x^2 , x^2\right\}$  ?

Its a positive definite form.
$ \left < 1, 1 \right > \equal{}1 > 0$.
$ \mathbb{R}$ is subspace spanned by $ 1$.
So $ \mathbb{R}$ and its orthogonal complement do form a direct sum for $ \mathbb{R}[X]_3$.

For any $ f$, $ f \minus{} \left < f, 1 \right > \in Q^\bot$.
We want a basis for $ Q^\bot$.

Just choose $ f$ accordingly. $ f \in \{x,x^2,x^3\}$.

In that case, $ SPAN \{x\minus{}\frac{1}{2}, x^2 \minus{} \frac{1}{3}, x^3 \minus{} \frac{1}{4}\} \equal{} Q^\bot$.


Binary expansion is writing numbers in base-2 form.

For example, the base-10 number 456 can be written into the base-2 (binary) number:

111001000

think this way

if you were trying to write a number ( let's say 123 ) and wanted to describe it only two symbols, we would use the binary expasian. The binary numbers are made up of only two different numbers. (0 and 1) so we would describe 123 with just 0 and 1

and that would be 1111011
(if i did my math right)


geometry
geometric transformation
reflection
inequalities proposed
inequalities

For $ a,b,c$ positive reals

$ \sum \sqrt{ab(a\plus{}b)} \geq \frac{5}{4}\sqrt{(a\plus{}b)(b\plus{}c)(c\plus{}a)}\plus{}2\sqrt{abc}$

[quote="manlio"]For $ a,b,c$ positive reals

$ \sum \sqrt {ab(a \plus{} b)} \geq \frac {5}{4}\sqrt {(a \plus{} b)(b \plus{} c)(c \plus{} a)} \plus{} 2\sqrt {abc}$[/quote]
It's wrong. Try $ a\equal{}b\equal{}1$ and $ c\rightarrow0^\plus{}.$


function
real analysis
real analysis unsolved

hi,
let $ (x_{n})_{n \geq 1}$ be a sequence with $ x_{1} \equal{} 1$ and for $ x \in \mathbb{R}$ we have
\[ x_{n\plus{}1} \equal{} x^{n} \plus{} nx_{n},\]
$ \forall n \geq 1$. Prove
\[ \prod_{n\equal{}1}^{\infty}{(1\minus{}\frac{x^{n}}{x_{n\plus{}1}})} \equal{} e^{\minus{}x}.\]

This is interesting!  I asked the inverse question about a different function [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=261410]here[/url].

[b]Proposition:[/b]  Let $ A(x) \equal{} \sum_{n \ge 0} a_n x^n$ be a "nice" generating function with $ a_0 \equal{} 1$.  Then there is a unique sequence $ (p_i)$ such that

$ (1 \plus{} p_1 x)(1 \plus{} p_2 x^2)... \equal{} A(x)$,

at least formally.

[i]Sketch.[/i]  Take $ p_1 \equal{} a_1$ and and compute the second-order term of $ \frac {A(x)}{1 \plus{} p_1 x}$, which will give you $ p_2$, and so forth.

I haven't done the computation, but it should give the answer in this case.  It's somewhat disappointing that this particular sequence is so complicated; I was hoping the answer would be nice in simple cases, but it appears as if this product decomposition question is quite deep.


A point $ (x,y)$ is randomly and uniformly chosen inside the square
with vertices (0,0), (0,2), (2,2), and (2,0).  What is the probability
that $ x\plus{}y < 3$?

Graphing it, we get that the shaded area  is the $ 2^2 \minus{} \frac{1}{2} \equal{} 3.5$. SO the probability is $ \frac{3.5}{4} \equal{} \boxed{\frac{7}{8}}$.

If you graph the points, it is quite obvious where to start. But, if you do not, then the best way to get the answer is something that I call, "smart visualization."

Because we are trying to find probability with a graph, it is best that we find the endpoints of our region so that we can do this problem efficiently. So, to begin, our square is a Quadrant 1 square that has sides on the x and y axis. The line mentioned, has x and y intercepts at (3,0) and (0,3) ( You may check for yourself by plugging it in, pretty simple to find ). So, we know that our line intersects the square at two points, both of which we don't quite know. But, because the squares endpoints (2,0) and (0,2) are very close to the intercepts of this line, we can use the same y-values and x-values to get our answer.

Let me explain: Our line here intersects the top side of the square at (1,2), right ( plugging it in, and using the y-value of the top side ). Meanwhile, the line also intersects the right side of the square at (2,1) ( Check with equation ). This may seem hard to do in your mind, but by plugging in x and y values from the squares endpoints, you are doing the problem much more efficiently. Now, since we want to find the area under the line, we know that it is 1- ( the region above the line ). So, by using the square's third endpoint ( 2,2) , we interpret the area of the figure as a representation of a portion of the original square's area, because this is probability

So, we get 1*1/2 ( Using the points (1,2) , (2,1), and (2,2), and since 1/2 is the portion of the figure's that is 4 ( 2*2 = 4), we get that the triangle occupies 1/8 of the square's area.

Solving from there we subtract 1-1/8 to get 7/8, as our final answer. ( Under the line, if you remember )

Hope that helped :)


Find all real pairs $ (x;y)$ of the system

$ x^{2}\plus{}y^{2}\equal{}2$

$ x^{4}\plus{}y^{4}\equal{}4$.

[hide="solution"]
$ x^{2}\plus{}y^{2}\equal{}2$
$ (x^{2}\plus{}y^{2})^{2}\equal{} 4$
$ x^{4}\plus{}y^{4}\plus{}2x^{2}y^{2}\equal{}4$
$ 4\plus{}2x^{2}y^{2}\equal{}4$
$ x^{2}y^{2}\equal{}0$
Thus, a factor of $ x^{2}y^{2}$ must be zero, implying that $ x\equal{}0$ or $ y\equal{}0$. Then the only solutions are $ (\pm\sqrt{2},0)$ and $ (0,\pm\sqrt{2})$
[/hide]


logarithms
calculus
integration
calculus computations

by assuming that a,b >0 can someone help me write a proof using integrals for :
ln(a/b) = ln(a) - ln(b)
ln (a^n) = n[ln(a)]

Define $ \log x: \equal{} \int_1^x \frac {dt}{t}$ for $ x > 0$.  Note that $ \log'(x) \equal{} \frac {1}{x}$ (Fundamental Theorem of Calculus), and $ \log 1\equal{}0$

[u]Theorem[/u]: If $ x,y > 0$, then $ \log(xy) \equal{} \log x \plus{} \log y$.
[u]Proof[/u]: Let $ f(x) \equal{} \log(xy)$ for some $ y > 0$, then $ f'(x) \equal{} \frac {1}{xy}\cdot y \equal{} \frac {1}{x}$.
We see that $ f'(x) \equal{} \log'(x)$, therefore $ f(x) \equal{} \log x \plus{} c$ for some constant $ c$ and all $ x > 0$.
In other words we have $ \log(xy) \equal{} \log x \plus{} c$, and setting $ x \equal{} 1$ we obtain $ \log y \equal{} c$. Conclusion: $ \log(xy) \equal{} \log x \plus{} c \equal{} \log x \plus{} \log y\ \ \ \ \ \ \ \ \square$

Now you should try to prove that $ \log(x^n) \equal{} n\log x$ using induction. You can also extend this to rational $ n$ instead of integer (since you wrote $ n$ I assumed you meant natural $ n$).

[u]Theorem[/u]: If $ x,y > 0$, then $ \log\left(\frac {x}{y}\right) \equal{} \log x \minus{} \log y$.
[u]Proof[/u]: $ \log x \equal{} \log\left(\frac {x}{y}\cdot y\right) \equal{} \log\left(\frac {x}{y}\right) \plus{} \log y$. Now substract $ \log y\ \ \ \ \ \ \ \square$

Or: $ 0 \equal{} \log 1 \equal{} \log\left(\frac {x}{x}\right) \equal{} \log x \plus{} \log\left(\frac {1}{x}\right)$, therefore $ \log\left(\frac {1}{x}\right) \equal{} \minus{} \log x$. It follows that
$ \log\left(\frac {x}{y}\right) \equal{} \log x \plus{} \log\left(\frac {1}{y}\right) \equal{} \log x \minus{} \log y$
[url=http://en.wikipedia.org/wiki/Logarithm#Calculus]here[/url] are other ways to do it.

I meant "n" as an exponent on the left of the equation and "n" as a constant on the right of the equation and I have to prove that equation using integration

Have you looked at my link? Apparently not.

$ \log(x^n)\equal{}\int_1^{x^n}\frac{dt}{t}$. Now set $ u\equal{}t^{1/n}$


[quote="anthony19891019"]Let k be a positive integer.

Find positive integers a,b,c such it is sufficient to show

 \[ ak^{c}k^{b}=bk^{a}k^{c}=ck^{b}k^{a}  \][/quote]
simply the formula
$\frac{a}{b}=k^{a-b}$
$\frac{b}{c}=k^{b-c}$
$\frac{c}{a}=k^{c-a}$
and $\frac{a}{b}=k^{a-b}=(k^{b-c})^{\frac{a-b}{b-c}}=(\frac{b}{c})^{\frac{a-b}{b-c}}$, then
$(\frac{a}{b})^{b-c}=(\frac{b}{c})^{a-b}$
$a^{b-c} \cdot b^{c-a}=c^{b-a}$
$a^b b^c c^a = a^c b^a c^b$
assume $a \ge b$, then $c^{a-b}|b^c a^b$, notice that if b only contains c factor. then $c=b$, $a-b=c$. get $a=2b=2c$. put a,b,c into original formula. it's wrong. if a only contains c factor. then $c=a$, $a-b=b$. it's also wrong. if a and b variable contain c factor. then $a=b=c$(no other case).
if we assume $a \le b$ case, it's similar solution as above.
[b][something is wrong][/b]

[quote="Mathsniper"][quote="anthony19891019"]Let k be a positive integer.

Find positive integers a,b,c such it is sufficient to show

 \[ ak^{c}k^{b}=bk^{a}k^{c}=ck^{b}k^{a}  \][/quote]
simply the formula
$\frac{a}{b}=k^{a-b}$
$\frac{b}{c}=k^{b-c}$
$\frac{c}{a}=k^{c-a}$
and $\frac{a}{b}=k^{a-b}=(k^{b-c})^{\frac{a-b}{b-c}}=(\frac{b}{c})^{\frac{a-b}{b-c}}$, then
$(\frac{a}{b})^{b-c}=(\frac{b}{c})^{a-b}$
$a^{b-c} \cdot b^{c-a}=c^{b-a}$
$a^b b^c c^a = a^c b^a c^b$
assume $a \ge b$, then $c^{a-b}|b^c a^b$, notice that if b only contains c factor. then $c=b$, $a-b=c$. get $a=2b=2c$. put a,b,c into original formula. it's wrong. if a only contains c factor. then $c=a$, $a-b=b$. it's also wrong. if a and b variable contain c factor. then $a=b=c$(no other case).
if we assume $a \le b$ case, it's similar solution as above.[/quote]

if b is the form of b=k*c^(a-b)?


geometry
rectangle
combinatorics unsolved
combinatorics

Consider a $ 2n\times2n$ board.From the $ i$-th line we remove the central $ 2(i\minus{}1)$ unit squares.What is the maximal
number of rectangles $ 2\times1$ and $ 1\times2$  that can be placed on the figure without overlapping or getting outside the
board?

It's REALLY easy ($ 2n^2 \minus{} n \plus{} 1$) :D

Place the rectangles (dominoes) of the next form:

1) Between lines 1 and i-1 place them horizontally, filling completely the side of the board.
2) Between lines i+2 and 2n, place them of the same form.
3) At line i, you just should put two vertical dominoes on the sides and it's done.
4) At line i+1, you have 2n-2 empty squares. Put n-1 horizontal dominoes, and it's done.

Now you have filled completely the board, so you placed $ ((2n)^2 \minus{} (2n \minus{} 2))/2$ = ($ 2n^2 \minus{} n \plus{} 1$) dominoes

I don't think that quite works...some of the rows will be split into two parts of odd size, and you can't cover each part separately by horizontal dominoes.

Here's a sketch of an argument, assuming I understand the question correctly.

I assume that in the original problem lines $ i$ and $ 2n \minus{} i$ both have the central $ 2(i \minus{} 1)$ squares removed, for $ i$ between $ 1$ and $ n$.  We consider the board to be colored like a checkerboard, alternately white and black.

Although there's an equal number of white and black squares, their locations are not evenly distributed throughout the board; if we consider the board as divided into 4 equal $ n \times n$ quadrants, the upper left and lower right will have a large excess of one color, and the upper right and lower left will have an excess of the other.

Each quadrant shares at most 2 dominoes with other quadrants (since we've removed almost all the squares from the boundary of the quadrant).  If follows that if $ A$ is the excess within a single quadrant, at least $ max(A \minus{} 2,0)$ squares in each quadrant must remain uncovered, so you'll fall at least $ max(2A \minus{} 4,0)$ dominoes short of covering the whole board.

Conversely, you can achieve this bound using only horizontal dominoes by first placing dominoes to cover the 4 inter-quadrant locations, then filling each row as much as possible.


ratio geometry circumcircle inradius inequalities trigonometry Euler

Find all triangles which satisfy
$r_a\cdot r_b\cdot r_c=\left(Rr\right)^3$.

thanks...
[u][i][b]fatihôlmez[/b][/i][/u]

[i]unknown source[/i]

It seems that there is a term of degree 3 in the LHS and a term of degree 6 in the RHS.
If the problem is like you posted it, then for any triangle ABC, denote by P the LHS and by Q the RHS.
If you take a triangle similar to ABC, the ratio of similarity being $\left(\frac{P}{Q}\right)^{1/3}$, the new P,Q for this triangle are $P^{\prime}=\frac{P}{Q}\cdot P$ and $Q^{\prime}=\left(\frac{P}{Q}\right)^2\cdot Q$, so P'=Q'.
I don't think this is what you want.

I hope you denote by r, R, $r_a$,... the radii of the inscribed, circumscribed, exscribed circles.

Grobber's version of the problem comes to finding all triangles such that $p^4=R^3r$, where $p=\frac{a+b+c}{2}$ is the semiperimeter of triangle ABC. But at least this is an equation.

Well, I would't take 'my version of the problem', as amfulger calls it, seriously... :D It was just a suggestion. Maybe we should wait for fatih003 to come back.

I am sorry,
the problem was in fact so,

Find all triangles which satisfy
$r_a\cdot r_b\cdot r_c=\left(Rr\right)^{3/2}$.

thanks...

Oh, so "my version of the problem", as amfulger called it, was right :D

[quote="grobber"]Maybe it's $r_a\ ^2\cdot r_b\ ^2\cdot r_c\ ^2=\left(Rr\right)^3$ ?[/quote]

Grobber was right :)

If we denote by $p=\frac{a+b+c}{2}$ the semiperimeter of triangle ABC, then

$r_a\cdot r_b\cdot r_c=\frac{S}{p-a}\cdot\frac{S}{p-b}\cdot\frac{S}{p-c}=\frac{S^3}{\left(p-a\right)\left(p-b\right)\left(p-c\right)}=S\cdot p=r\cdot p^2$.

The equation $r_a\cdot r_b\cdot r_c=\left(Rr\right)^{3/2}$ is equivalent to $S^2\cdot p^2=R^3\cdot r^3$. Or, in other words, $p^4=R^3\cdot r$.

I think the problem was incorrectly stated. I didn't find any nice triangles satisfying it.
What kind of condition should we obtain?  :?

Indeed, there are infinitely many triangles ABC satisfying the condition $r_ar_br_c=\left(Rr\right)^{3/2}$, where R is the circumradius, r is the inradius and $r_a$, $r_b$, $r_c$ are the exradii of triangle ABC. Probably the only halfways interesting thing one can show about such triangles is that they are all obtuse-angled. This is very easy to prove: Assume that an acute-angled or right-angled triangle ABC satisfies the condition $r_ar_br_c=\left(Rr\right)^{3/2}$. As Amfulger showed in post #9, this condition is equivalent to $p^4=R^3\cdot r$, where $p=\frac{a+b+c}{2}$ is the semiperimeter of triangle ABC. But, as I proved in http://www.mathlinks.ro/Forum/viewtopic.php?t=5379 post #3, for any acute-angled (or right-angled, this can be considered as a limiting case) triangle ABC, we have the inequality $\sin A+\sin B+\sin C\geq 2$. By the extended law of sines, in triangle ABC we have a = 2R sin A, b = 2R sin B and c = 2R sin C; thus,

$p=\frac{a+b+c}{2}=\frac{2R\sin A+2R\sin B+2R\sin C}{2}=R\left(\sin A+\sin B+\sin C\right)$
$\geq 2R$

(since $\sin A+\sin B+\sin C\geq 2$). Also, by the Euler inequality, $R\geq 2r$. Thus,

$p^4\geq\left(2R\right)^4=16R^4=16R^3\cdot R\geq 16R^3\cdot 2r=32R^3\cdot r>R^3\cdot r$.

This obviously shows that the equation $p^4=R^3\cdot r$ can impossibly be fulfilled. Hence, we obtain a contradiction, and thus, there is no acute-angled or right-angled triangle ABC satisfying the condition $r_ar_br_c=\left(Rr\right)^{3/2}$. In other words, all triangles satisfying this condition must be obtuse-angled. $\blacksquare$

[b]PS.[/b] I have edited this thread not only to LaTeX the formulas, but also to clear up some misunderstandings based on different use of terminology (some posters used p for the semiperimter of the triangle, while others used s, what was, again, misinterpreted by others as the area S).

  Darij


Prove that $ n! > n^{2}$ for every integer $ n \ge 4$, while $ n! > n^{3}$ for every integer $ n \ge 6$.

[hide="Part One"]

Doing by induction.

For $ n \equal{} 4$ it holds.

Assume now it holds for some $ n \equal{} k$.

Showing that it holds for $ n \equal{} k \plus{} 1$.

$ (n \plus{} 1)! > n^2 \plus{} 2n \plus{} 1$

$ (n \plus{} 1)n! > n^2 \plus{} 2n \plus{} 1$

$ nn! \plus{} n! > n^2 \plus{} 2n \plus{} 1$

$ nn! > 2n \plus{} 1 \minus{} \minus{} \minus{} > nn! > n^2 \plus{} 2 > \equal{} 2n \plus{} 1$

Isnt this problem way to easy for Pre-Olympiad section?[/hide]

Second part goes on the same way.


Suppose $ X$ is a nonempty set, and we want to have certain topology $ \tau$ on it, so $ (X,\sigma(\tau))$ ( as a measure space ), has some propertys.  I want to choose topology $ \tau$ such that $ \sigma(\tau)\equal{}\mathcal{P}(X)$. The obvious way to go, is to choose discrete topology on $ X$. What is the weakest topology $ \tau$ on $ X$, such that $ \sigma(\tau)\equal{}\mathcal{P}(X)$?
I would be grateful is someone could discuss this in a specific setting, that is, if $ X$ were $ \mathbb{R}$.

Can you tell us what you mean by $ \sigma(\tau))$?

Sorry, thought that was a standard notation, if $ \mathcal{F}\neq \emptyset$ is a family of subsets on $ X$,  $ \sigma(\mathcal{F})$ is the smallest $ \sigma$-algebra containing $ \mathcal{F}$. It can be proven that
$ \displaystyle \sigma(\mathcal{F}) \equal{} \bigcap_{\mathcal{G} \sigma \minus{} \textrm{ algebra on } X,\mathcal{F}\subseteq \mathcal{G}}\mathcal{G}$.


geometry
3D geometry
tetrahedron
sphere
ratio
AMC
AIME

1) A tetrahedron is inscribed in a sphere, find the ratio of the volume of the tetrahedron to the volume of the sphere.

2) A sphere is circumscribed around a tetrahedron and another tetrahedron is circumscribed around the sphere. Find the ratio between the volumes of the smaller tetrahedron and the larger tetrahedron.

3) A cube is inscribed in a tetrahedron, find the ratio between the volume of the cube and the volume of the tetrahedron.

#1

Assume :) the side length of tetrahedron is $ 2\sqrt {3}$. The height would be the other leg of a right triangle with hypotenuse = side length and one leg 2, which =  $ 2\sqrt {2}$. The volume of the solid is $ (\frac {1}{3})(2\sqrt {2})(\frac{(2\sqrt {3})^2 \sqrt {3}}{4}) = 2\sqrt {6}$.

To find the radius of the sphere, we need to find the distance from a vertex to the center of the solid. This can be done by setting $ x$ as this distance. By using the length of the height and part of the base, we have $ \sqrt {2^2 + (2\sqrt {2} - x)^2} = x$ which means $ x = \frac {3\sqrt {2}}{2}$. The volume of the sphere is $ (\frac {4}{3})( \pi)( \frac {3\sqrt {2}}{2})^3 = 9\pi\sqrt {2}$. The ratio is
$ \frac {2\sqrt {6}}{9\pi\sqrt {2}} = \boxed{\frac {2\sqrt {3}}{9\pi}}$.

Edit: More like an AIME problem, but I did it within 10 mins :)

2nd Edit: Edited $ \text{\LaTeX}$ mistakes.

If it were an AIME problem, it would've been like:

"The answer can be expressed in the form $ \frac {m\sqrt{n}}{o\pi}$, where $ m, n,$ and $ o$ are relatively prime positive integers and $ n$ is not divisible by the prime of any square. Find $ m \plus{} n \plus{} o$."

[quote="l Bob l"]#1

Assume :) the side length of tetrahedron is $ 2\sqrt {3}$. The height would be the other leg of a right triangle with hypotenuse = side length and one leg 2, which =  $ 2\sqrt {2}$. The volume of the solid is $ (\frac {1}{3})(2\sqrt {2})(frac{(2\sqrt {3})^2 \sqrt {3}}{4}) \equal{} 2\sqrt {6}$.

To find the radius of the sphere, we need to find the distance from a vertex to the center of the solid. This can be done by setting $ x$ as this distance. By using the length of the height and part of the base, we have $ \sqrt {2^2 \plus{} (2\sqrt {2} \minus{} x)^2} \equal{} x$ which means $ x \equal{} \frac {3\sqrt {2}}{2}$. The volume of the sphere is $ (\frac {4}{3})( \pi)( \frac {3\sqrt {2}}{2})^3 \equal{} 9\pi\sqrt {2}$. The ratio is
$ \frac {2\sqrt {6}}{9\pi\sqrt {2}} \equal{} \frac {2\sqrt {3}}{9\pi}$.

Edit: More like an AIME problem, but I did it within 10 mins :)[/quote]

On to the second one. Again, the side length of the tetrahedron (smaller) is $ 2\sqrt {3}$. The distance from the center of the tetrahedron to one vertex is still $ \frac {3\sqrt {2}}{2}$.

Next we find the distance from the center to the middle of one of its faces. This can be done with the pythagorean theroem: $ \sqrt {(\frac {3\sqrt {2}}{2})^2 \minus{} 2^2} \equal{} \frac {\sqrt {2}}{2}$. So the ratio of [the distance from center to the middle of a face] to [the distance from the center to a vertex] is $ \frac {\frac {\sqrt {2}}{2}}{\frac {3\sqrt {2}}{2}} \equal{} \frac {1}{3}.$ :)

Note that the [the distance from the center to a vertex] of the smaller tetrahedron is equal to [the distance from center to the middle of a face]. If the [the distance from the center to a vertex] of the smaller tetrahedron is $ s$ then so is [the distance from center to the middle of a face] of the larger. The [the distance from the center to a vertex] of the larger would be $ 3s$. The ratio of [[the distance from the center to a vertex] of the smaller tetrahedron] to [[the distance from the center to a vertex] of the larger] is $ \frac {1}{3}$. To find the ratio of the volumes, we cube the ratio, giving $ \boxed{\frac {1}{27}}$.

Can someone check my solutions?  :|  :| I doubt I'm the only one here that can solve this mathcounts-level.


Well I randomly decided to make my own mock AIME. It's not done yet, but I only need 3 more problems plus some revision and whatnot. Therefore, I am going to make it available around Dec. 10.

I don't really care if you participate, but if you do the rules are the same as an official AIME. I also don't really care when you send me answers (but hopefully before 2006). I just thought I would give people advance notice :-).

Edit--hmmm seems like the subject thingy doesn't like to have caps: "AIME III." Darn.

I'm in...... but haven't had an opportunity to begin beta's. :blush:

You should release it after Beta's is finished, and I believe he pushed the turn-in date back to the 11th.

Just a heads up, I've started writing a Mock AIME 4 to start off the new year! I'm trying to perfect the difficulty level to as close as possible to something like the 2003/2004 AIME's. We can probably make it to 10 Mocks before the actualy AIME :D

[quote="scorpius119"]

(The geometry on yours should be a bit easier than on beta's. That's an understatement.)[/quote]

LMAO!!! :rotfl:  :heli:  :harhar:  So trueeee....

Yeah, count me in.

Hopefully I'll be able to take both!  (plus Beta's sometime this week, since my essay took way too long, plus the Mock AIME 1, which I still haven't taken)

I'll try to do it on Sunday, make it like paladin8's Mock AIME 1. :D

KNOWING STEPHEN THE GEOMETRY WILL BE A LOT HARDER

AW MAN SO MANY MOCK AIMES SO LITTLE TIME . . .

[quote="MysticTerminator"]
KNOWING STEPHEN THE GEOMETRY WILL BE A LOT HARDER
[/quote]

I agree... expecting to see your geometry problems Alex :) , by the way Alex, how many first name(s) do you have?

i'll do it sometime. i still have to take the 2nd one tho.

[quote="scorpius119"]I'm in!  :D  And i have done beta's...  :stretcher:

(The geometry on yours should be a bit easier than on beta's. That's an understatement.)[/quote]

Haha, you've gotta be kidding me.  :rotfl:

Knowing this kid, he's gonna pound us with all sorts of stuff. Expect it to be very hard. And I'm very serious.

Gee... I think my favorite name for Alex is ZhaiZhai. Haha, that just sounds funny. :rotfl:

:( Don't tell anyone Jarey! But anyways the geo will not be that hard, as I tried to simulate an actual AIME. (But if you want some hard geo I can try to make some :-)) I also tried to balance the topics as much as possible, although I think the first few problems may be slightly too hard.

BTW the problems are all done, but I'm going to ask some random people to check over them.

[quote="white_horse_king88"]

Gee... I think my favorite name for Alex is ZhaiZhai. Haha, that just sounds funny. :rotfl:[/quote]

 :lol: that does sound funny.

Well, if you want a better understanding of the AIME, I would suggest just practicing old AIMEs, and maybe some of Mildorf's problems (but probably not under timed conditions for those). Problem writing is much different from problem solving. Although I think that each helps the other, it seems to me like problem solving helps problem writing much more than the other way around.

But if you just feel like writing a mock AIME anyway, try to think about different scenarios without having a specific problem in mind. Then wander around from there and look for interesting properties. Chances are that you will find something, and you can form a problem from that.

I have all the problems written and texed now. I am only waiting for certain people :mad: to help verify my answers. :-)

[quote="filletwho"]Hmmm...to gain a better understanding of AIME problems, I guess I'll write a mock AIME also, probably over winter break. Can any of you questionwriters give me advice?[/quote]

Try not to write a hard AIME, cuz then nobody would send in answers lol :lol:

This AIME is the bomb. :D

Honestly, I think that all the problems are well-written and enjoyable to solve.  Alex will be releasing it as soon as Arnav and I finish verifying/editing the random nonsense.  All I can say is... you better be good at geometry... like UBERLY good at geometry (though some of it can be brute-forced randomly, that entails a lot of computation).

Yay!!! Geometry is awesome!(hopefully I can solve some of those geometry problems on Alex's AIME) :D

Don't worry, the geometry problems should be pretty close to what AIME geometry is normally like. Right now there are 5 I think, but I'm going to try to replace one today.


Suppose $f(x)$ is a polynomial with integral coefficients. If $f(x)=2$ for three different integers $a,b,c$, prove that $f(x)$ can never be equal to $3$ for any integer $x$

[hide="My solution"]Since $f(\alpha)$ is the remainder that $f(x)$ leaves when divided by $(x-\alpha)$, all $(x-a)$,  $(x-b)$ and  $(x-c)$ divide $f(x)-2$,  so we can write
$f(x)-2=(x-a)(x-b)(x-c)g(x)$ (where $g(x)$ is another polynomial).
Then if $f(x)=3$, we would have that $1=(x-a)(x-b)(x-c)g(x)$, but $1$ cannot be written as a product of three different integers. Q.E.D.[/hide]


1.  At how many minutes past 8:00 will the minute and hour hands form equal angles with the vertical for the first time?

Im just confused what "form equal angles with the vertical" means?

2. What is the number of solutions $ (x,y)$ with $ x$ and $ y$ nonnegative integers, of the equation $ 4^{x}\minus{}9^{y}\equal{}55$

3. In triangle ABC, D is the midpoint of AB, while E lies on BC satisfying BE=2EC.  If angle ADC= angle BAE, then how many degrees are in angle BAC?

4. Find the smallest possible value of $ \frac{x^{2}\plus{}y^{2}}{x\minus{}y}$ for all real numbers $ x$ and $ y$ satisfying $ x>y>0$ and $ xy\equal{}3$

[quote="shawn1"]1.  At how many minutes past 8:00 will the minute and hour hands form equal angles with the vertical for the first time?

Im just confused what "form equal angles with the vertical" means?

2. What is the number of solutions $ (x,y)$ with $ x$ and $ y$ nonnegative integers, of the equation $ 4^{x}\minus{}9^{y}\equal{} 55$

3. In triangle ABC, D is the midpoint of AB, while E lies on BC satisfying BE=2EC.  If angle ADC= angle BAE, then how many degrees are in angle BAC?

4. Find the smallest possible value of $ \frac{x^{2}\plus{}y^{2}}{x\minus{}y}$ for all real numbers $ x$ and $ y$ satisfying $ x > y > 0$ and $ xy \equal{} 3$[/quote]

[hide="Solution for 4"]Note that
\[ \frac{x^{2}\plus{}y^{2}}{x\minus{}y}\equal{}x\minus{}y\plus{}\frac{2xy}{x\minus{}y}\equal{}x\minus{}y\plus{}\frac{6}{x\minus{}y}\]
From AM-GM, we can conclude that
$ \frac{x^{2}\plus{}y^{2}}{x\minus{}y}\equal{} x\minus{}y\plus{}\frac{6}{x\minus{}y}\ge 2\sqrt{6}$

So $ 2\sqrt{6}$ is the minimum, obtained when $ x \equal{} y\plus{}\sqrt{6}$.[/hide]

[quote="erdogankerem123"][quote="vishalarul"]
From AM-GM, we can conclude that
$ \frac{x^{2}\plus{}y^{2}}{x\minus{}y}\equal{} x\minus{}y\plus{}\frac{6}{x\minus{}y}\ge 2\sqrt{6}$
[/quote]
I don't understand how you get that from AM-GM.[/quote]

[hide]Let $ a\equal{}x\minus{}y$, and $ b\equal{}\frac{6}{x\minus{}y}$. By AM-GM,

$ \frac{a\plus{}b}{2}\ge\sqrt{ab}$

$ a\plus{}b\ge 2\sqrt{ab}$

Since $ ab \equal{} 6$,

$ a\plus{}b\ge 2\sqrt{6}$

From our definitions of $ a$ and $ b$,

$ x\minus{}y\plus{}\frac{6}{x\minus{}y}\ge 2\sqrt{6}$

In general, if you have the sum of a few terms with their reciprocals, AM-GM pops out. For example, on this:

$ xy\plus{}x\plus{}y\plus{}\frac{1}{xy}\plus{}\frac{1}{x}\plus{}\frac{1}{y}$

You could use AM-GM (given all the terms are positive) to see that this sum is greater than or equal to $ 6$.

AM-GM even works in stuff like this:
$ x^{2}\plus{}y\plus{}\frac{1}{xy}$.

Just re-write that as:

$ x^{2}\plus{}\frac{y}{2}\plus{}\frac{y}{2}\plus{}\frac{1}{2xy}\plus{}\frac{1}{2xy}$

Try to use AM-GM on your own to solve that now.[/hide]

[hide="Solution for number 2"]By the difference of two squares, we get:
$ 55\equal{}4^{x}\minus{}9^{y}\equal{}(2^{x}\minus{}3^{y})(2^{x}\plus{}3^{y})$

Because $ 3^{y}>0$, $ 2^{x}\plus{}3^{y}>2^{x}\minus{}3^{y}$.  Also, because the second factor is positive, so must the first.  Finally, both factors are integers.  Now, we have only 2 cases...
[hide="1)"]$ 2^{x}\plus{}3^{y}\equal{}55$, $ 2^{x}\minus{}3{y}\equal{}1$
Adding these two equations and dividing by 2 gives:
$ 2^{x}\equal{}28$, for which there are no integer sol'ns.[/hide]
[hide="2)"]$ 2^{x}\plus{}3^{y}\equal{}11$, $ 2^{x}\minus{}3{y}\equal{}5$
The same technique yields: $ 2^{x}\equal{}8$, $ x\equal{}3$[/hide]
Thus, $ (x,y)\equal{}(3,1)$ is our only solution.[/hide]

[quote="vishalarul"][quote="erdogankerem123"][quote="vishalarul"]
From AM-GM, we can conclude that
$ \frac{x^{2}\plus{}y^{2}}{x\minus{}y}\equal{} x\minus{}y\plus{}\frac{6}{x\minus{}y}\ge 2\sqrt{6}$
[/quote]
I don't understand how you get that from AM-GM.[/quote]

[hide]Let $ a \equal{} x\minus{}y$, and $ b \equal{}\frac{6}{x\minus{}y}$. By AM-GM,

$ \frac{a\plus{}b}{2}\ge\sqrt{ab}$

$ a\plus{}b\ge 2\sqrt{ab}$

Since $ ab \equal{} 6$,

$ a\plus{}b\ge 2\sqrt{6}$

From our definitions of $ a$ and $ b$,

$ x\minus{}y\plus{}\frac{6}{x\minus{}y}\ge 2\sqrt{6}$

In general, if you have the sum of a few terms with their reciprocals, AM-GM pops out. For example, on this:

$ xy\plus{}x\plus{}y\plus{}\frac{1}{xy}\plus{}\frac{1}{x}\plus{}\frac{1}{y}$

You could use AM-GM (given all the terms are positive) to see that this sum is greater than or equal to $ 6$.

AM-GM even works in stuff like this:
$ x^{2}\plus{}y\plus{}\frac{1}{xy}$.

Just re-write that as:

$ x^{2}\plus{}\frac{y}{2}\plus{}\frac{y}{2}\plus{}\frac{1}{2xy}\plus{}\frac{1}{2xy}$

Try to use AM-GM on your own to solve that now.[/hide][/quote]

Thanks so much for the detailed reply! However, on the follow-up problem, I'm stumped  :( . I don't see how the x's and y's can cancel out. If I pair the $ \frac{y}{2}$'s with the $ \frac{1}{2xy}$'s, I get $ 2\sqrt{\frac{1}{x}}$, but I don't see how I can do away with the $ x^{2}$.

[quote="erdogankerem123"][/quote][quote="vishalarul"][quote="erdogankerem123"][quote="vishalarul"]
From AM-GM, we can conclude that
$ \frac{x^{2}+y^{2}}{x-y}= x-y+\frac{6}{x-y}\ge 2\sqrt{6}$
[/quote]
I don't understand how you get that from AM-GM.[/quote]

[hide]Let $ a = x-y$, and $ b =\frac{6}{x-y}$. By AM-GM,

$ \frac{a+b}{2}\ge\sqrt{ab}$

$ a+b\ge 2\sqrt{ab}$

Since $ ab = 6$,

$ a+b\ge 2\sqrt{6}$

From our definitions of $ a$ and $ b$,

$ x-y+\frac{6}{x-y}\ge 2\sqrt{6}$

In general, if you have the sum of a few terms with their reciprocals, AM-GM pops out. For example, on this:

$ xy+x+y+\frac{1}{xy}+\frac{1}{x}+\frac{1}{y}$

You could use AM-GM (given all the terms are positive) to see that this sum is greater than or equal to $ 6$.

AM-GM even works in stuff like this:
$ x^{2}+y+\frac{1}{xy}$.

Just re-write that as:

$ x^{2}+\frac{y}{2}+\frac{y}{2}+\frac{1}{2xy}+\frac{1}{2xy}$

Try to use AM-GM on your own to solve that now.[/hide][/quote][quote="erdogankerem123"]

Thanks so much for the detailed reply! However, on the follow-up problem, I'm stumped  :( . I don't see how the x's and y's can cancel out. If I pair the $ \frac{y}{2}$'s with the $ \frac{1}{2xy}$'s, I get $ 2\sqrt{\frac{1}{x}}$, but I don't see how I can do away with the $ x^{2}$.[/quote]
[hide]
Just multiply the addends:

$ x^{2}\cdot y/2\cdot y/2\cdot 1/(2xy)\cdot 1/(2xy) =\frac{1}{16}$.

Take the fifth roots and apply AM-GM:

$ x^{2}+\frac{y}{2}+\frac{y}{2}+\frac{1}{2xy}+\frac{1}{2xy}\ge 5\sqrt[5]{1/16}$.[/hide]

It says that the arithmetic mean of a set of numbers $ \ge$ geometric.

P.S. On your guide to trig, is it necessary to memorize reduction formulas?  :maybe:

[quote="erdogankerem123"]P.S. On your guide to trig, is it necessary to memorize reduction formulas?  :maybe:[/quote]

There's no need to memorize. Just imagine the unit circle in your head or draw it out and it will become apparent. Besides, if you forgot them, you could just use the sum formula. Ex:

$ \sin(90\plus{}\theta) \equal{}\sin(90)\cos(\theta)\plus{}\cos(90)\sin(\theta) \equal{}\cos(\theta)$

BTW, I'm practically done the "guides" and will be submitting them as a single pdf document in a separate thread  :D Just need to edit and get a little help from Stevem.

[quote="Prime factorization"]Don't mean to sound stupid, but what on Earth is AM-GM?[/quote]

[u]A[/u]rithmetic [u]M[/u]ean - [u]G[/u]eometric [u]M[/u]ean rule : $ \frac{x\plus{}y}{2}\geq\sqrt{xy}$ $ \forall x,y\geq 0$


i don't know if this is the right place to post this but i need a help on a SAT writing section question.

[u]Having an exceptionally big living room, the French club chose Lindsay's house[/u] as the best place to hold their end-of-year party.

D. because it has an exceptionally big living room, the French club chose Lindsay's house

E. Based on its exceptionally big living room, Lindsay's house was chosen by the French club

i believe the answer is E, since the modifying phrase is describing Lindsay's house, so Lindsay's house should be closest to it.

but this SAT practice book say it's D because "it is logical and concise; it begins the sentence with an introductory phrase that clearly indicates the cause-and-effect relationship between the attributes of the house and the French club's choice.

could anyone explain to me? thanks

Is this the College Board practice book, or some other brand?

In improving sentence problems, it is always [b]best[/b] to avoid passive voice. The choice E is indeed in a passive voice, so should be eliminated.

This leaves D as our answer.

mws21 is probably right here -- this sentence shouldn't have a passive in it.  Although, in fact, both of those choices are worse than, "The French club chose Lindsay's house [to hold their end-of-year party] because it has an exceptionally big living room," since in that version there's no possibility of conveying that the French club has an exceptionally big living room.  Hopefully you won't come across such a case on the actual exam :)

it's a kaplan sat class book, The Big Purple Book of SAT Practices

i get what mws21 is saying about passive.  but it's just that D makes it seen like the club has a big room.

In my opinion, D ought to be incorrect because of the ambiguous pronoun "it".

If you can, I recommend sticking with official material from the College Board: the Official Study Guide (8 tests), the Official Online Course (7 tests), and released PSATs.

[quote="Ravi B"]If you can, I recommend sticking with official material from the College Board: the Official Study Guide (8 tests), the Official Online Course (7 tests), and released PSATs.[/quote]

I agree with Ravi B's recommendation.

I also agree with the above recommendations about using only officially College Board material. I'd also like to add that, the most important part of doing practice tests isn't doing the questions themselves. You MUST go back over the questions you got wrong and analyze them. I find it pretty hard to believe people can run through all of the college board tests and do analysis and still be left "wanting more tests".

D is definitely wrong. Personally, I say avoid Kaplan because it's full of mistakes, and avoid Princeton Review because it's tests are easier than the real ones.

I remember when I took the SAT 2's, I didn't bother preparing for physics because I got high 800's on 3 straight princeton review practice tests, and then got a 790 on the real thing; if the tests were more representative, I might've studied a bit. I also remember the Kaplan ones having a few incorrect answers, typos, and questions that required the reader to make false assumptions.

GO FOR BARONS!

I've never seen any mistakes in Baron's, and the tests are generally a bit harder than the real ones, so it makes for good practice.


integration
calculus
topology
analytic geometry
derivative
calculus computations

Why isn't Stokes Theorem first taught as follows: $ \int_{\partial S} \omega \equal{} \int_{S} d \omega$? Its much easier to remember.

Can you write down what you would have to do to compute either side? The hard part is [i]defining[/i] the integral of a form and the $ d$ operator.

$ dA \equal{} \frac{\partial A}{\partial x} dx \plus{} \frac{\partial A}{\partial y} dy \plus{} \frac{\partial A}{\partial z} dz$

$ d(A \ dx \plus{} \ldots) \equal{} dA \ dx \plus{} \ldots$

$ d(A \ dy \ dz \plus{} \ldots) \equal{} dA \ dy \ dz \plus{} \ldots$

$ dx \ dx \equal{} 0$

$ dx  \ dy \equal{} \minus{}dy \ dx$

Pedagogically, I tend to be more of a believer in bottom-up, example-driven approaches than top-down, theoretical approaches.

When I was a sophomore in college, I had a course that had as one of its texts Spivak's [i]Calculus on Manifolds[/i]. You can say that that book is beautiful and self-contained, but the truth of it is that I had a hard time making sense of it at the time of the course and only grew to appreciate it in retrospect.

What I do do in an ordinary, engineering-oriented Calculus III course is to write a triangular array of theorems on the board. The coordinates are the dimension of the underlying space and the dimension of the (inner) integral.

In the $ (1,1)$ spot is the Fundamental Theorem of Calculus, in the integral of the derivative form:

$ \int_a^bf'(x)\,dx\equal{}f(b)\minus{}f(a).$

Then on the second row, we have two entries: the fundamental theorem of line integrals and Green's Theorem.

On the third row, we have three entries: the fundamental theorem of line integrals, Stokes' Theorem, and the divergence theorem.

And I mention that we'd have four entries in four-dimensional space, and so on. I try to draw as many parallels as I can among all of these theorems. (I tend to refer to the collection more as the generalization of the FTC rather than as the generalized Stokes' Theorem - again, the message is directed to the class and level.) So I'm putting out the suggestion that these are all instances of one big generalized theorem or principle - but I'm not doing what jmerry is talking about, writing down all of the definitions needed to bind it together. That wouldn't be suitable for that class at that level.

So do you think looking at the problems first and then going back to the relevant sections to learn the material is better then reading the material and then doing the problems?

That's an exaggeration of course - but if we can avoid taking it to an absurd level, yes I do believe that.

Consider the entire notion on which this website rests (whether you call it AoPS or Mathlinks) - the notion of problem solving. That is the idea that the path to interesting mathematics is interesting problems, and theories, theorems, and techniques are things you learn along the way to help you do the problems. In many cases, you learn an idea because you encountered some problem that needed it.


You have one sheet of very thin paper.  One stack of 200 sheets is only one centimeter tall. If you could fold the paper in half 50 times,  how thick,  in kilometers,  would the folded paper be?	  (Select the closest answer.)

A.0.5   B. 5  C .500  D. 50,000   E. 50,000,000

[hide="solution"]Each piece of paper is 0.005cm thick.
Folding in half is doubling that thickness each time.
2^50(.005cm)=5629499534213.12cm.
There are 100,000 cm in a kilometer so the stack is 56294995.3421312 km thick, which is closest to answer E.[/hide]


why not start out by writing out all the primes from 1-20?

Their difference must be less than 20. The only perfect number under 20 is 6.

Also, their sum must be under 40 and greater than the difference. The only perfect number between 6 and 40 is 28.

$p+q=28, \; p-q = 6 \; \implies (p,q)=(17,11)$

[quote="anyone"]what's a perfect number?[/quote]
any number that is the sum of its divisors.

6 has the divisors 3, 2 and 1, consequently 3 + 2 + 1 = 6

the next perfect number is 28 since its divisors are 14, 7, 4, 2, 1 and they add to 28.

[quote="binomial_4eva"]which 2 prime numbers less than 20 have a sum and difference which are both perfect numbers?[/quote]

I modified the problem a little, and removed "less than 20" part... as:

[b]which 2 prime numbers   have a sum and difference which are both perfect numbers?[[/b]

Prove that 11 and 17 are the only answers...
I started a new thread here: Please answer it there ...
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=789734#789734[/url]


[quote="test20"][quote="zhaoli"]I have translated the longlisted problems of IMO 1990 from a Chinese book. The SL problems are following the number listed in the book "The IMO Compendium".[/quote]

Great!

Interestingly, problem (LongList'1990/43) equals problem (IMO'1992/5).[/quote]

I know that shortlist is usually revealed the year after. How about the longlist?

Pan, I suggest you read my post on IMO Shortlists/Longlists [url=http://www.mathlinks.ro/viewtopic.php?p=110248#110248]here.[/url] The IMO jury decided that from 1989 (or maybe though less likely 1990) that IMO Longlists shall no longer be made available. For the years it was provided it usually happened in three different ways. In the early years there was distinction between IMO Shortlists and Longlists. Problems were even not submitted to the IMO problem selection committee apriori but rather directly brought to the IMO jury meetings during the IMO and discussed there. In later years the Longlist was provided with the shortlisted problems marked with an asterix. In a few other later years two lists were produced, i.e. one being the Longlist and the other one being the shortlist.

For 1992 I also got the IMO Longlist but Arkadii Slinko doubts the document is genuine. Apparently the Longlist 1992 has not been distributed during that IMO so it must have come from some other unknown source. For 1990 the IMO Longlist was not distributed at the IMO either as far as I know, and maybe certain people thought publishing it in a Chinese source only is equivalent to not releasing it at all. But luckily due Zhao Li we got the translation.

The motivation for the discontinuation of the IMO Longlist was a lack of good problems. Thus the non-shortlisted problems from the IMO Longlist were returned to the proposer countries such that they can be used for other purposes as national olympiads. You may have realized the rant from more experienced users about recent IMO Shortlists that problems are known, not really new, not very nice etc.

As you can read the in the link above the quality of the Longlist over the years may vary quite a bit.


geometry
Euler
geometric transformation
homothety
circumcircle
ratio
vector

[quote="muks"]Prove that  in any triangle, orthocentre,circumcentre,centroid, all lie in the same straight line.

Thank you.[/quote]

This is the euler line is it not?

Yes, this is the Euler line. You prove that it exists using the homothecy that maps the triangle to its medial triangle (think about it: clearly the centroid maps to itself, but where does the circumcenter go?)

I could not understand that.What is homothecy? Can you please post the solution[partly: a hint would be helpful].

Thank you.

Consider the homothety with center $ G$, where $ G$ is the centroid of $ \triangle ABC$, ratio $ k \equal{} \minus{} \frac {1}{2}$. You will see that $ \mathcal {H}(G,k): \triangle ABC \mapsto \triangle A_0B_0C_0$, where $ A_0,B_0,C_0$ respectively are the midpoints of $ BC,CA,AB$. Now, if $ O$ is the circumcenter of $ \triangle ABC$, note that then $ O$ will be the orthocenter of $ \triangle A_0B_0C_0$, hence $ \mathcal {H}(G,k): H\mapsto O$, where $ H$ is the orthocenter of $ \triangle ABC$. Therefore $ H,G,O$ are collinear. Our proof is completed then.

[b]REMARK-[/b] Vector method can also kill this problem. Try to figure out.

Can you please explain what this homothety is ?Then I will true to understand the problem.

Thanks again.

[url=http://www.pballew.net/eulerln.html]Here[/url] is a link to another proof. (There are a couple of typos, be careful.)  :)

EDIT: From mathVNpro's post, I guess homothety is a similarity relationship, see [url=http://mathworld.wolfram.com/Homothecy.html]here[/url]. So it looks like mathVNpro proved that the centroid mapped to the centroid, and the orthocenter mapped to the circumcenter. Therefore, there is a line that passes through the $ G$ such that $ HG: GO \equal{} 2: 1$, so the Euler line exists. I'm probably wrong... :maybe:

[quote="muks"]Prove that  in any triangle, orthocentre,circumcentre,centroid, all lie in the same straight line.

Thank you.[/quote]
$ \triangle ABC$ with orthocenter, circumcenter,centroid: $ H,O,G$.
Using complex numbers. Let $ h' \equal{} 3g \minus{} 2o$. See that $ (a \minus{} h') \cdot (b \minus{} c) \equal{} ( \minus{} (b \minus{} o) \minus{} (c \minus{} o) ) \cdot (b \minus{} c) \equal{} |c \minus{} o|^2 \minus{} |b \minus{} o|^2 \equal{} CO^2 \minus{} BO^2 \equal{} 0 \iff AH' \perp BC$. Because of symmetry: $ h \equal{} h'$ and hence $ h \minus{} o \equal{} 3(g \minus{} o)$. So $ OH$ is three times as long as $ OG$ and in the same direction. Hence: $ O,G,H$ are collinear.


[quote="turcas_c"]Compute the sum :
\[ S \equal{} \binom{2n \plus{} 1}{1} \plus{} 5 \cdot \binom{2n \plus{} 1}{3} \plus{} 5^2 \cdot \binom{2n \plus{} 1}{5} \plus{} 5^3 \cdot \binom{2n \plus{} 1}{7} \plus{} \dots \plus{} 5^n \cdot \binom{2n \plus{} 1}{2n \plus{} 1}
\]
.[/quote]
[hide="Solution"]
Notice that \[ \frac{(\sqrt{5}\plus{}1)^{2n\plus{}1}}{\sqrt{5}}\equal{}\frac{\binom {2n\plus{}1}{0}\plus{}\sqrt{5}\cdot\binom {2n\plus{}1}{1}\plus{}5\cdot \binom {2n\plus{}1}{2}\cdots \plus{}\sqrt{5}^{2n\plus{}1}\cdot \binom {2n\plus{}1}{2n\plus{}1}}{\sqrt{5}}\] Furthermore, notice that \[ \frac{(\sqrt{5}\minus{}1)^{2n\plus{}1}}{\sqrt{5}}\equal{}\frac{\minus{}\binom{2n\plus{}1}{0}\plus{}\sqrt{5}\cdot \binom {2n\plus{}1}{1}\minus{}5\cdot \binom {2n\plus{}1}{2}\cdots \plus{}\sqrt{5}^{2n\plus{}1}\cdot \binom {2n\plus{}1}{2n\plus{}1}}{\sqrt{5}}\] Thus, \[ \frac{(\sqrt{5}\plus{}1)^{2n\plus{}1}\plus{}(\sqrt{5}\minus{}1)^{2n\plus{}1}}{2\sqrt{5}}\equal{}\binom {2n\plus{}1}{1}\plus{}\binom {2n\plus{}1}{3}\cdot 5\plus{}\binom {2n\plus{}1}{5}\cdot 5^2\cdots \binom \plus{}{2n\plus{}1}{2n\plus{}1}\cdot 5^n\] and thus the answer is $ \boxed{\frac{(\sqrt{5}\plus{}1)^{2n\plus{}1}\plus{}(\sqrt{5}\minus{}1)^{2n\plus{}1}}{2\sqrt{5}}}$ [/hide]


A line through the points $ (2, \minus{}9)$ and $ (j, 17)$ is parallel to the line $ 2x \plus{} 3y \equal{} 21$. What is the value of j?

The slope of 2x+3y=21 is -2/3

Thus, $ \frac{17\minus{}(\minus{}9)}{j\minus{}2}\equal{}\frac{\minus{}2}{3}$
so, $ 78\equal{}\minus{}2j\plus{}4$
$ j\equal{}\minus{}37$


I have downloaded latex and install it.but I can't find the main file(typing formality).maybe because my is not good.can somebody help me?

Have you followed carefully all the steps explained in this tutorial: http://www.mathlinks.ro/LaTeX/AoPS_L_About.php ?

thank you for help.
but I still don't know how to use it.my problem is unknowing how to start it.which file should I chilk it.I can't find the file TeXnicCenter.c
can someone help me again? :lol:

First here are two different things: making up your own $\LaTeX$ document on your computer, and posting in $\LaTeX$ on the forum. Which one is it? :)

well,my problem was I can't find my  own latex document .I don't which one is it?seems a joke.but it's ture.can you tell me what should I do??thanks :)

It seems that this may be a problem of communication. I would recommend in that case posting a help topic in Chinese in the [url=http://www.mathlinks.ro/Forum/index.php?f=279]Chinese Forums[/url], and your Chinese compatriots will help you more quickly that I could.


function
calculus
number theory unsolved
number theory

Let f: R -> R be a function such that, for all x, y belonging to R, both of the conditions f(x + y) = f(x) + f(y) and f(xy) = f(x)f(y) are satisfied. Prove that f is either:

(i) The function which assigns the value 0 to all x belonging to R

or

(ii) The identity mapping, which assigns, to each x belonging to R, the value x.

One hint is that if f(1) does not equal 0, then for each rational number r you have f(r) = r. In addition, if a real number x is > 0, then there exists a non-zero real number y such that x = y^2.

The problem is I can explain and draw pictures of what is happening, but I do not know how to prove that either (i) or (ii) is true.

Thanks.

This is a double post; the problem is also being dealt with in the calculus-analysis forum.

Let's lock this since it clearly isn't number theory.


calculus
integration
trigonometry
logarithms
real analysis
real analysis unsolved

Find the integral    $\int\sin(\ln x)dx$
There should be more than one method we can use

By using substitution $\ln(x) = y$ and partial integral you can get $\frac{e^y(sin(y) - cos(y))}{2}$ and finally substitution $y = \ln(x)$

just want to write out the edigoras's solution.
After doing  subtitution like edigoras said $y= \ln x$, $x=e^y$ , and $dx=x \, dy= e^y \, dy$. So
\begin{eqnarray*} \int \sin \ln x \, dx &=& \int \sin y \, e^y \, dy \\ &=& - e^y \cos y + \int e^y \cos y \, dy \\ &=& -e^y \cos y + e^y \sin y - \int e^y \sin y \, dy \end{eqnarray*}
So $\int \sin y \, e^y \, dy = -e^y \cos y + e^y \sin y - \int e^y \sin y \, dy$ this simply write   $2 \int e^y \, \sin y \, dy = -e^y \cos y + e^y \sin y$. Hence
\[ \int \sin( \ln x) \, dx = \frac{x (\sin (\ln x) - \cos (\ln x))}{2}  \]


the topic basically explains all... what's a sticky topic?

Normally, when someone replies to a message, it moves to the top of the forum.  If no one responds to it, it slowly sinks towards the bottom as other topics move up past it or are created.  Sticky topics always stay at the top, regardless of whether people respond to them or not.

Well...announcements will be above any stickies or threads if there are any, and sticky will be above any threads. If there are many (defined as "two or more") announcements or stickies then they will be ordered from the most recently updated (meaning made or posted upon) to the least.

there is an FAQ section on the top of the forum that will answer these types of questions


Let $p$ be an odd natural number, $p\ge 3$ and a complex number $z$ for which $z^{p}=1\ .$ Prove that :
$1.\blacktriangleright\ \left(1+z\right)\left(1+z^{2}\right)\ldots\left(1+z^{p-1}\right)=1\ .$
$2.\blacktriangleright\ \left(1-z\right)\left(1-z^{2}\right)\ldots\left(1-z^{p-1}\right)=p\ .$
$3.\blacktriangleright\ \sum_{k=1}^{p-1}\frac{1}{1+z^{k}}=\sum_{k=1}^{p-1}\frac{1}{1-z^{k}}=\frac{p-1}{2}\ .$
$4.\blacktriangleright\ \sum_{k=1}^{p-1}\frac{z^{k}}{1+z^{2k}}=2\cdot\sum_{k=1}^{\frac{p-1}{2}}\frac{z^{k}}{1+z^{2k}}=2\cdot \sum_{k=1}^{\frac{p-1}{2}}\frac{1}{z^{p-k}+z^{p+k}}\ .$

You see that $(1-z)(1-z^{2})\dots(1-z^{p-1})=f(1)=p$ which $f(x)=1+x+\dots+x^{p-1}$. Also $z^{2p}=1$ and $(1-z^{2})(1-z^{4})\dots(1-z^{2(p-1)})=p$. Dividing these 2 will conclude the first equation.

[quote="Virgil Nicula"]Let $p$ be an odd natural number, $p\ge 3$ and a complex number $z$ for which $z^{p}=1\ .$ Prove that :

$3.\blacktriangleright\ \sum_{k=1}^{p-1}\frac{1}{1+z^{k}}=\sum_{k=1}^{p-1}\frac{1}{1-z^{k}}=\frac{p-1}{2}\ .$
[/quote]
Using [b]JBL's[/b] idea in an earlier post, for the first one,
$S = \sum_{k=1}^{p-1}\frac{1}{1+z^{k}}= \sum_{k=1}^{p-1}\frac{z^{-k}}{1+z^{-k}}= \sum_{k=1}^{p-1}\frac{z^{k}}{1+z^{k}}= (p-1)-S$
$\Rightarrow S = \frac{p-1}{2}.$

The method is the same for the second one.


geometry
3D geometry
tetrahedron
sphere
probability
circumcircle
trigonometry

(From Mock AMC 1, #21-25)

Here are the last five problems from the first AMC Mock Contest.  Please post your solutions here so that we can hear different methods of doing each problem.

I have also included the source for each problem.

21. What is the length of one edge of a regular tetrahedron inscribed in a sphere of unit radius?
(A) :rt3:
(B) sqrt(6)/2
(C) :rt3:/2
(D) 2*:rt2:/3
(E) 2*sqrt(6)/3

Source: Univ. Maryland 1998 Round 1

22. Fierytycoon has three coins, two of which are fair and one of which is double-headed.  He picks a coin at random and flips it 6 times.  It comes up heads each time.  What is the probability that [u]he[/u] picked the double-headed coin?
(A) 2/3
(B) 15/16
(C) 16/17
(D) 32/33
(E) 63/64

Source: Univ. Georgia 1998 Round 2

23. 223 unbendable sticks with lengths 1776, 1777, 1778, , 1998 are arranged to enclose the maximum possible area.  Which of the following is closest to the radius of the smallest circle that completely encloses the 223-gon?
(A) 25000
(B) 50000
(C) 75000
(D) 100000
(E) 125000

Source: Univ. Georgia 1998 Round 2 (with slight rewording)

24. Two men (Rusczyk and Crawford) and two boys (Rep123max and Syntax Error) need to cross a lake.  Their boat will either carry one man or two boys.  What is the fewest number of one-way trips that it will take to get everybody to the other side of the lake?  [Assume that one person is required to row the boat.]
(A) 4
(B) 5
(C) 8
(D) 9
(E) 11

Source: AASU High School Math Tourney

25. If arctan a + arctan b = 45 degrees, then a + b =
(A) 1/(ab)
(B) ab
(C) 1  ab
(D) 1 + ab
(E) none of the above

Source: MMPC, Round 1, 1962

- mathfanatic

Hokey dokey.. I'm doing these from scratch again since I can't remember what I even submitted for my answers..

21) Ugh, didn't like this one very much. We want to find the "middle" of the tetrahedron which has distance 1 from each vertex. Letting edge length be r, we find the height of the tetrahedron. First find circumcenter of base triangle, which is r:rt3:/3 from each corner. (by pythagoras, then using medians meet 2/3 of the way along thing). Pythagoras again gives height of tetrahedron is (r:sqrt:6)/3. Now we subtract the radius of 1 off that to find how high off the base ourr center point is. (r:sqrt:6)/3 - 1. Finally use pythagoras once again and equate to the radius. End up with r:^2: = (2r:sqrt:6)/3, or [b]E[/b].

22) If he picks a normal coin, the prob it comes up heads 6 times is 1/(2^6) = 1/64. If he picks doubleheaded coin, the prob is 1. So probability of normal and 6 heads is 2/3 of 1/64 = 1/96. Prob of doubleheaded and 6 heads is 1/3. So we get (1/3)/(1/3 + 1/96) = 32/33. [b]D[/b].

23) I don't really know at all if this was the correct way or even a reasonable way of doing this.. I took both extreme cases, one with 223 sticks of length 1776, and one with 223 sticks of length 1998. Splitting these into 223 triangles with angle 360/223 degrees and radius r, use the cosine rule to work out r. One gives a radius of about 63035, the other about 70914. Both are closer to C than any other. [b]C[/b].

24) This is well known.. two boys over, one boy back, one man over, one boy back, and repeat, then both boys over. [b]D[/b].

25) Take tan of both sides. Use the formula for tan (A+B), and we get (a + b)/(1 - ab) = tan 45 = 1. So a+b = 1-ab. [b]C[/b].

21).  I liked this problem and had a different method of solution.  A regular tetrahedron can easily be inscribed in a cube by drawing in six face diagonals, three emanating from one vertex, and three connecting the endpoints of those three lines.  So, instead of considering an inscribed tetrahedron, we can consider an inscribed cube, and its space diagonal is 2R where R is the radius of the circle.  Then if x is an edge of the cube, 3x :^2:  = 4R :^2: , and then 2x :^2:  = 8/3R :^2: , but by the pythagorean theorem, the edge of the tetrahedron is just sqrt(2x :^2: ) since it is a face diagonal.  Then our tetrahedron edge is sqrt(8)/sqrt(3)*R
= 2*sqrt(2)*sqrt(3)*R/3 = 2*sqrt(6)/3*R, and for R = 1 we get 2*sqrt(6)/3 as the edge.

21. in chem we learn the bond angle of a tetrahedral molecule is ~109.5 bust out that law of cosines.

23. just add them all up and assume that's the circumference. then take the next biggest answer choice. (well that's a lower bound. i think i found the upper bound like MMM but i don't remember)

the bond angle of 109.5 depends on one or two lone pairs... H20 will have a bond angle of more like 104.5 sir... :o, I wouldnt trust this method.

water is not tetrahedral.... you are thinking of either trigonal pyramidal (1 lone pair) or angular (2 lone pairs) NOT tetrahedral.

He never said water was tetrahedral, he was just giving examples of cases where the bond angle isn't 109.5 degrees.  However, the point is that all [i]tetrahedral[/i] molecules do have a measure of 109.5 degrees.

Too bad I don't know chemistry so I'm lost.

I just put the tetrahedron inside a cube and inscribed it, pretty much what zscool did.  Luckily I didn't try using 3D vectors for that problem...

hm, zscool's solution is rather savvy. i wouldve done it more of triple m's way


vector
function
linear algebra
linear algebra unsolved

Q#1. Let A be an m x n matrix. Show that if A has linearly independent column vectors, then N(A)={0}.

Now I know that I can use the fact that  Ax=[x1a1 +x2a2 + ... + xnan] for any x element of Rn. But how do I get started?

Q#2 Determine whether the following are subspaces of C[-1,1].
 a) The set of functions f in C[-1,1] such that f(-1)=f(1)
 b) The set of odd functions in C[-1,1]

Again I don't want the answers so much as wanting to understand what the question is asking me so that I will be able to apply it to other similar questions that I may run into.

#1: You have $Ax$ written as a linear combination of the columns of $A$. Suppose $Ax=0$. What does linear independence tell you?

#2: A subset of a vector space is a subspace when it is closed under addition and scalar multiplication. That's two things to check in each case.

maybe you have another notation for things there where you are...
could i know the semnification of those symbols ? $C[-1,1]$ ?

It doesn't really matter, but $C[0,1]$ is the space of continuous functions on $[0,1]$.

This is strictly a linear algebra problem- no analytic facts about this space are relevant.


inequalities
search
function
inequalities proposed

Pretty simple... Note that

\[\frac{a}{5+3a^2} \leq \frac{1}{32}(a+3) \iff 0 \leq (a-1)^2(a+5),\]

which is clearly true.  Now, just sum it up!

\[\sum_{\text{cyc}} \frac{a}{5+3a^2} \leq \frac{1}{32} (a+b+c+d+12) = \frac{1}{2}.\]

[quote="ThAzN1"]Pretty simple... Note that

\[\frac{a}{5+3a^2} \leq \frac{1}{32}(a+3) \iff 0 \leq (a-1)^2(a+5),\]
[/quote]

any good way to find this?

[quote="ThAzN1"]Yes, the technique has been discussed here on the forum... if I remember correctly.  Try a search on "Poland 1996".

[edit] Here's a link:

http://www.mathlinks.ro/Forum/viewtopic.php?highlight=poland+1996&t=180

Specifically, post #5 by [b]hxtung[/b].[/quote]

Thank you.

Ok,It is a special case in my generel problem

Find all positive k such that the inequality satisfy for all positive reals $x_1,x_2,..,x_n$ and $x_1+x_2+...+x_n=n$

$\sum\frac{x_i}{k+x_i^2} \le \frac{n}{1+k}$

This is my solution : We have :
$\frac{x_i}{k+x_i^2}\leq \frac{k-1}{(k+1)^2} .x_i +\frac{2}{(k+1)^2}$.
$=>\sum\frac{x_i}{k+x_i^2}\leq \frac{1}{(k+1)^2}.((k-1)\sum x_i +2n) =\frac{n}{k+1}$ .
Thus , this inequality is true for all positive numbers k .

Checking the case $n=2$ is enough to know you are wrong.

[quote="hungkhtn"]Prove that if $a,b,c,d$ are positive reals and $a+b+c+d=4$ then:

$\frac{a}{5+3a^2}+\frac{b}{5+3b^2}+\frac{c}{5+3c^2}+\frac{d}{5+3d^2} \le \frac{1}{2}$[/quote]

I have another solution. Using AM-GM inequality we have
\[
\frac{a}{5+3a^2}=\frac{1}{3\left(\frac{1}{a}+a\right)+\frac{2}{a}}\leq\frac{1}{6+\frac{2}{a}}=\frac{a}{6a+2}
\]
But the function $f(t)=\frac{t}{6t+2}$ is concave, so by Jensen's inequality, we obtain
\[
\sum_{\text{cyc}}\frac{a}{6+2a}\leq4\frac{(\sum a)/4}{6(\sum a)/4+2}=\frac{1}{2}.
\]

Special solution

For the inequality:
$\sum\frac{a}{2+6a} \le \frac{1}{2}$ we can solve simplyer without convex:

This inequality is equal:

$\sum\frac{2}{2+6t} \le 1$
$\leftrightarrow \sum\frac{1}{1+3a} \le \frac{16}{4+3\sum{a}}=1$

And we are one.

For the best $k_n$ satisfy this inequal..

If $n=2$ then $k=\frac{1}{3}$
If $n=3$ then $k=0.35..$
If $n=4$ then $k=0.37...$

I Wait a complete nice solution.


is it just me, or is the font on this site unreadable?

If you set the font size to anything but medium/middle/normal (depending on what language/browser you use) it's going to be a bit hard to read :). For small adjustments you can click on the As at the right top corner of the Forum main screen.

The whole "temporal font size increase" thing got fixed too, thanks Mr. R :).

you mean too small? if so theres a button near the top of the screen of the forum to do that. or, if you click on view on the very very top, then click text size and pick a size.