Does anyone know any good Mathcounts/amc 10/amc 12 books (besides AOPS)?
Thx

See the book at the top of [url=http://www.agmath.com/152763.html]this[/url] page, as well as the list of books at the bottom. Also, if you click on MATHCOUNTS and AMC on the left sidebar, there are nice lessons and mock/past tests.

At the bottom of the page, where can I buy those books?


A sequence exists such that it is comprised of positive integers including the number $68$, and has an arithmetic mean of $56$. When $68$ is removed, the arithmetic mean of the remaining sequence is $55$. What is the largest number than can occur in such a sequence?

[hide] Solving the simultaneous equations  and , we get  and . Since they must be positive integers, we let 11 of them be  and the final one be our desired value, which is .[/hide]

[quote="4everwise"]A sequence exists such that it is an arithmetic mean of positive integers including the number $68$, and has an arithmetic mean of $56$. When $68$ is removed, the arithmetic mean of the remaining sequence is $55$. What is the largest number than can occur in this sequence?[/quote]
[hide]
Let the total sum of the sequence be $S$ and let the number of elements in the sequence be $n$. Then $S/n = 56$ and $(s - 68)/(n-1) = 55$

so $n = 13$, there are $13$ elements.

$13 \cdot 56 = 728$, $728 - 68 = 660$
in order to average 56 and to fulfill the fact that the sequence consists of only positive numbers, you can have $11$ ones, $1$ really large number, and $1$ 68. so the max is $660 - 11 = 649$

$\fbox{649}$
 [/hide]

[hide="Solution"]Although we don't know the number of numbers that are contained in the sequence, we may still solve it with conventional methods.

From the information given, we see that $\frac{55n+68}{n+1}=56$, where $n$ is the number of elements in the sequence which doesn't contain $68$. Solving this equation, we see that $n=12$.

Because we also know that $\frac{S}{n}=55$, where $S$ is the sum of the second sequence, we see that the sum of the sequence without $68$ is $660$

Now, we find the largest possible value by setting the other $11$ terms in the sequence to $1$. Our answer is therefore $660-11=649$.[/hide]

[hide="Answer"]Say we have $x$ numbers not counting $68$. Then: $\frac{55x+68}{x+1}=56$, so we have $55x+68=56x+56\Rightarrow x=12$. and the sum without $68$ is $12\cdot 55=660$. $11$ of these can be $1$, so the largest possible term in the sequence is $\boxed{649}$.[/hide]


Terra Rossa offers up to 5 different toppings on its pizza bases, how many different toppings can be made?

I don't quite understand the problem...the wording is a little awkward...if you're saying that there are 5 different toppings, so how many different combinations can be made, then, including no toppings, it is 32, because each topping may either be on or off the pizza.

From the wording, I think the no-topping case shouldn't count

If I understood the question correctly you can assume the toppings are $A$, $B$, $C$, $D$, and $E$ and you can choose any combination of any number of toppings (no repeats and you must choose at least 1). Therefore the answer should be $5+10+10+5+1=\fbox{31}$

Because it was designed as a warm up for a warm up.

$\text{That would be the sum of the coefficients of }(x+1)^5-x^5=5x^4+10x^3+10x^2+5x+1\text{.  The sum is }\boxed{\boxed{\color{green}31}}$.


[quote="Silverfalcon"]Since many people are interested on exponent problem, I'll post one.

Find x so that $2^{2^{3^{2^2}}} = 4^{4^{x}}$[/quote]

Is that supposed to be a 3?

[hide="Well if so here is a hint"]
x^y = 2x^(y/2)
[/hide]

Demji -- be careful.  I believe you took some of your exponents from the bottom up.  That is, you read $2^{2^3}$ as $(2^2)^3$.  They are not the same thing!  In this case, to evaluate the number on the LHS, you can only evaluate from the top down, not the bottom up.  In other words, the question could look like this:
$2^{(2^{(3^{(2^2)})})}=4^{(4^x)}$


καλησπερα,
ειμαι στην πρωτη λυκειου και(αν και δεν εινα σιγουρος αν περασα τον ευκλειδη) προετοιμαζομαι για τον αρχιμηδη(πρωτη φορα στους μεγαλους)..αλλα εχω βρεθει μποστα σε ενα τεραστιο βουνο απο θεωρηματα και σχετικα που υποτιθται πως πρεπει να μαθω απ εξω...=S ειδικα ανισωσεις(andrescu,huygens,schur,power means inequality κτλ...)υπαρχει καποιο περιορισμενο συνολο απο τετοια που να ειναι πιο..."sos"...απο τα αλλα?
σημειοτεον οτι οτι βρισκω το αποδεινυω αλλα αυτο δεν αναιρει οτι για να το χρησιμοποιησει κανεις πρεπει να τη θυμαται απ εξω...
ελπιζω να βοηθησετε γιατι τη μνημη μου δεν ειναι να την εμπιστευεται κανεις :(

φίλε darksaga κατ'αρχάς μην απογοητεύεσαι θεωρώντας ότι πρέπει να υπερβείς ένα τεράστιο βουνό. Θα σου δώσω κάποιες συμβουλές, αν και δε ξέρω αν είμαι το καταλληλο άτομο καθώς δεν είχα κάποια αξιόλογη επιτυχία σε διαγωνισμούς της ΕΜΕ αλλά παρακολουθούσα με ιδιαίτερο ενδιαφέρον τα θέματα της. Όσον αφορα τις ανισότητες στις οποίες αναφέρθηκες καλό θα είναι να ξέρεις την BCS,Andreescu,την αριθμητικού-γεωμετρικού-αρμονικού που θεωρούνται οι πιο sos και είναι σχετικα εύκολο να τις θυμάσαι.Η πιο 'sos' όμως είναι η (α-β)^2 >=0  όσο περιεργο κι αν μπορεί να σου φαίνεται αυτό, καθως επίσης και όσες προκύπτουν από αυτή.
Επίσης βρές παλαιότερα θέματα Αρχιμήδη και Ευκλείδη μεγαλύτερων τάξεων ώστε να εξασκηθείς σε πραγματικα θέματα και να καταλαβεις την φιλοσοφία τους.
Επαναλαμβάνω πως καλό θα ήταν να σε συμβουλέψει και κάποια από τα παιδιά που είχαν επιτυχίες σε παλαιότερους διαγωνισμους.

Γεια σου φίλε μου...
Κατ'αρχήν μην αγχώνεσαι (στο λέει πάντως ο πιο αγχώδης τύπος  :P )  .Θα πας πρώτα από όλα να ευχαριστηθείς το διαγωνισμό (μην ξεχνάς ότι είναι Εθνική Ολυμπιάδα) και να γνωρίσεις καινούργια άτομα . Από την πλευρά μου θα σου έλεγα ότι με το τελευταίο που θα πρέπει να ασχοληθείς είναι οι ανισότητες !!! (Εξάλλου δεν έχουν βάλλει τα τελευταία χρόνια αλλά αυτό δε λέει και τίποτα ).Πιστεύω θα πρέπει αρχικά να δώσεις έμφαση στην γεωμετρία και τη θεωρία αριθμών που σχεδόν πάντα μπαίνουν...
Αυτά από εμένα.....Καλό διάβασμα

Υ.Γ. Νομίζω ότι ο πιο αρμόδιος για να σου μιλήσει είναι ο Nick Rapanos αφού αν το καλοσκεφτεί κανείς δεν έχει αποτύχει σε κανένα μαθηματικό διαγωνισμό !!!!!!(στη Β Γυμνασίου Χάλκινο στη Βαλκανιάδα Νέων , τα ίδια και στην Γ Γυμνασίου , στην Α Λυκείου ασημένιο στην Βαλκανιάδα Μεγάλων και Εύφημη Μνεία στην ΙΜΟ ) . Είναι ο μαναδικός ίσως......

σας ευχαριστω και τους δυο παιδια και σας ευχομαι και εγω να περασετε στον αρχιμηδη και καλη επιτυχια σε αυτον

darksaga

Φίλε darksaga,
είμαι κι εγώ πρώτη λυκείου και σε καταλαβαίνω απόλυτα... :(
η προετοιμασία για τον αρχιμήδη είναι πραγματικά ζόρικη καθώς η ύλη είναι απεριόριστη
το καλύτερο που έχεις να κάνεις είναι να δεις τα θέματα προηγούμενων χρόνων(αρχιμήδη και ευκλείδη) για να καταλάβεις την νοοτροπία των θεμάτων :wink:
κατα τα αλλα, καλή σου επιτυχία :first:

ρε παιδια μη σκεφτητε την αποτυχια, δουλευετε με ενα οραμα, απλα ολα ειναι πιθανα  ....
 :)

Φίλε darksaga πέρναμε μου φαίνεται ακριβώς τα ίδια
(και εδώ και στο forum του PCmag)

Και εγώ έχω την εντύπωση ότι έχω περάσει.
Μόνο που δεν ξέρω τι να διαβάσω.

Μπορεί να προτείνει κανείς.....

Το μυστικο για να περασεις τετοιους διαγωνισμους ειναι η ψυχολογικη προετοιμασια (και αυτο ειναι καπως αδικο).Τελικα αν δεις τις ασκησεις των διαγωνσμων σαν απλες ασκησεις που λυνεις και μονος σπιτι σου τοτε πιστευω θα τα πας καλα(εχοντας φυσικα γνωση των βασικων θεωρηματων).Σκεψου μονο οτι υπηρχαν ασκησεις που την ωρα του διαγωνισμου δεν τις ελυνα και σε ελαχιστο χρονικο διαστημα τις ελυνα σπιτι μου.Ξερω ειναι το χειροτερο και πιστευω πολλοι το εχουμε παθει.Και να μην ξεχναμε οτι αυτο που κανετε πρεπει να σας αρεσει πολυ και να μην προσπαθειται πολυ βιαστηκα να καλυψετε κενα με σκοπο να τα μεθετε απλα γρηγορα και τελικα να χασεται την ουσια................Οσο για το τι να διαβασεται.......?Διαβαστε αυτα που σας αρεσουν πραγματικα περισσοτερο.Αν κατι δεν σας αρεσει προσπαθηστε να βρειτε κατι ενδιαφερον σε αυτο και ετσι να συνεχισεται.(φανατικος υποστιρικτης ολων σας:Aρχιμηδης 6)

[quote="Αρχιμήδης 6"]Το μυστικο για να περασεις τετοιους διαγωνισμους ειναι η ψυχολογικη προετοιμασια (και αυτο ειναι καπως αδικο).Τελικα αν δεις τις ασκησεις των διαγωνσμων σαν απλες ασκησεις που λυνεις και μονος σπιτι σου τοτε πιστευω θα τα πας καλα(εχοντας φυσικα γνωση των βασικων θεωρηματων).Σκεψου μονο οτι υπηρχαν ασκησεις που την ωρα του διαγωνισμου δεν τις ελυνα και σε ελαχιστο χρονικο διαστημα τις ελυνα σπιτι μου....[/quote]αυτη για οσους δεν καταλαβαν ειναι σαφης αναφορα στον περσυνο αρχιμηδη των μικρων...btw ηταν αιθουσα αυτη που μας βαλανε να γραψουμε???????? :mad:

[quote="pagio"]Άλλη μια ερώτηση.
και οι τρείς τάξεις του λυκείου γράφουν τα ίδια θέματα????[/quote]
ναι...ειμαστε περιφανοι να γραφουμε πια με τους μεγαλους :P

Καλωσορίζω και εγώ  με τη σειρά μου τα παιδιά της Α΄Λυκείου , που πια από 'μελάλοι' στους 'μικρούς(Junior)' που ήταν , γίνονται τώρα οι 'μικροί' στους 'μεγάλους(senior)'.
 Την ουσία  την ειπαν οι συνομήλικοί σας  παραπάνω. Επισημαίνω όμως ότι :
   [b]α)[/b] Ποτέ δε διαβάζεις κάτι για ένα συγκεκριμένο διαγωνισμό. Όταν ασχολείσαι με τις Ολυμπιάδες , απλά ... διαβάζεις. Διαβάζεις λίγο και λύνεις πολύ ! Αυτή είναι  η μέθοδος ! Και λύνεις επειδη σου αρέσει !
    [b]β)[/b]  Για να μην παιδεύεται όμως κάποιος άσκοπα να λύσει μια άσκηση , είναι καλό να γνωρίζει στην αρχή τα απαραίτητα.Πχ στη   ΓΕΩΜΕΤΡΙΑ : ΤΕΤΡΑΠΛΕΥΡΑ - ΕΓΓΡΑΨΙΜΑ - ΟΜΟΙΟΤΗΤΑ - ΕΜΒΑΔΑ - ΜΕΤΡΙΚΕΣ ΣΧΕΣΕΙΣ
 Δεν πας  πχ να κάνεις αντιστροφή και ριζικό άξονα αν δεν ξέρεις τα κριτήρια ομοιότητας. Διαβάζεις μερικα θεωρηματα και υστερα κάνεις λίγες εφαρμογές.Παρόμοια και στους άλλους κλάδους.
  [b] γ) [/b]Μερικοί διαβάζουν αντίστροφα , αλλά αυτοί έχουν χρόνο : Παίρνουν ένα θέμα διαγωνισμού , πχ την περυσινή γεωμετρία στον Αρχιμήδη και προσπαθούν να τη λύσουν. Ας πούμε ότι δεν την λύνουν. Ανατρέχουν στη λύση και τη διαβάζουν. Βλέπουν τι δεν ήξεραν, ποια θεωρήματα δε γνώριζαν κλπ. Ανοίγουν λοιπόν ένα σχετικό βιβλίο , βλέπουν τα θεωρήματα αυτά , κάνουν και καμιά εφαρμογή και συνεχίζουν.Αυτή είναι καλή μέθοδος , αλλά θέλει πολύ υπομονή και ελεύθερο χρόνο .Δυστυχώς τα ελληνόπουλα δεν έχουν χρόνο , γιατι το πρόγραμμα είναι φορτωμένο.
 Επομένως καλό είναι να διαβάζουν λίγο πιο οργανωμένα. Λίγη θεωρία, λίγες ασκήσεις - ασκήσεις διαγωνσιμών - mathlinks κλπ.
   [b]δ)[/b] Μερικοί έχουν ανθρώπους που ξέρουν μαθηματικά και τους βοηθάνε. Γονείς κλπ. Άλλοι διαβάζουν μόνοι. Τέλος πάντων , είναι μεγάλο το θέμα.
   [b]ε) [/b]Γενικά δεν υπάρχει μεθοδος μελέτης. Κάθε μαθητής βρίσκει το δικό του τρόπο. Ένα είναι σίγουρο : Διαβάζεις όσο μπορείς και όπου ...βγει.
 [color=red] Καλή επιτυχία στον Αρχιμήδη !!![/color]

Φίλε darksaga σε καταλαβαίνω απόλυτα...Ο Σιλουανός τα παραείπε για μένα... δεν ειναι ακριβως έτσι τα πράγματα-για όλους παραμονεύει πάντα μια αποτυχία και όλα είναιμέσα στο πρόγραμμα και όλοι έχουμε αγχος.
Το κρίσιμο ερώτημα είναι γιατί ασχολείσαι με μαθηματικούς διαγωνισμούς, τι θέλεις να πετυχεις, μέχρι πότε θα το κάνεις και τι σκοπεύεις να κάνεις μετα???
Σκέψου πρώτα αυτά και έπειτα βάλε ένα στόχο και διεκδίκησέ τον!Εμένα προσωπικά με καλύπτι η πρόταση (ε) του κύριου Στεργίου και ειδικά το τελευταίο μέρος...Ό,τι διαβάζεις είναι καλό αρκεί να διαβάζεις πράγματα που τα καταλαβαίνεις και μπορείς να το χρησιμοποιήσεις, μην  το βλέπεις σαν ύλη εξετάσεων- τότε είναι σιγουρο ότι θα τρελαθείς από το εύρος των απαραίτητων γνώσεων...Δες το διάβασμα σαν παιχνίδι και το διαγωνισμό σαν μια γιορτή στην οποία θα γνωρίσεις καινούργιους φίλους με τα ίδια ενδιαφέροντα με σένα (όπως εγώ γνωρισα στη Β Γυμνασίου το Σιλουανό και κάποιους άλλους [u]συναγωνιστές[/u] (πρόσεχε τη λέξη γιατί είναι σπουδαία)

Κάτι άλλο για τις ανισότητες που έλεγες πριν... Ποτέ δεν περίμενα ότι στο φετινό Ευκλείδη θα μου χρειαζόταν η Weighted Power Mean Inequality στη μορφή f(1)>=f(0) η οποία σημειοτέων για να αποδειχθεί σρειάζεται γνωσεις μονοτονίας,ορίων, συνέχειας και De L' Hospital...Όμως σε πληροφορώ ότι αν δεν το αντιμετώπιζα έτσι το θέμα δεν ξέρω τι θα γινόταν...
Ούτε πίστευα ποτέ ότι το 4ο θέμα του Αρχιμήδη που δεν έλυσα όταν πήγαινα Β Γυμνασίου, ότι θα ήταν μια απλούστατη εφαρμογή του θεωρήματος Haro-Farey-Cauchy για δύο διαδοχικούς όρους της ακολουθίας  Farey...αλλά να που όλα <<παίζουν>>.

Αυτά για τώρα και δώσε βάση σ' αυτά που προτείνει ο κύριος Στεργίου!

Υ.Γ. Εγώ θα ήθελα να προτείνω δημόσια στο Σιλουανό που τον βλέπω σαν moderator να φτιάξει ένα  topic  με όλες τις απαραίτητε ανισότητες θεωρήμτα και λήματα που μπορούν να φανούν χρήσιμα στον Αρχιμήδη με σκοπό να βοηθηθούμε όλοι..και φυσικά να κάνουμε κατι αντίστοιχο και στους άλλους τομείς...(πχ ο Ράλλης θ.α., εγώ Συναρτησιακές, ο κύριος Βήττας Γεωμετρία και ο κύριος Στεργίου  Συνδυαστική ή κατι τέτοιο που  είναι αρκετά δυσκολο για όλους...)

Αυτα :roll:

[quote="Nick Rapanos"]Φίλε darksaga σε καταλαβαίνω ...
Υ.Γ. Εγώ θα ήθελα να προτείνω δημόσια στο Σιλουανό που τον βλέπω σαν moderator να φτιάξει ένα  topic  με όλες τις απαραίτητε ανισότητες θεωρήμτα και λήματα που μπορούν να φανούν χρήσιμα στον Αρχιμήδη με σκοπό να βοηθηθούμε όλοι..και φυσικά να κάνουμε κατι αντίστοιχο και στους άλλους τομείς...(πχ ο Ράλλης θ.α., εγώ Συναρτησιακές, ο κύριος Βήττας Γεωμετρία και ο κύριος Στεργίου  Συνδυαστική ή κατι τέτοιο που  είναι αρκετά δυσκολο για όλους...)

Αυτα :roll:[/quote]αυτο ακριβως ζητουσα φτιαχνοντας το thread...το νοημα ειναι οτι ειναι πολυ πιο ευκολο να εχεις υπ οψιν σου καποιους τυπους...περσυ και προπερσυ πηγαινα σχεδον τελειως ερημιτζης στον αρχιμηδη...οποτε φετος σκεφτικα να διαβασω λιγουλακι...(οτι κι αν σημαινει αυτο)...η μαγκια ειναι φυσικα να λυνεις "ολα" τα προβληματα χωρις χρηση καμιας ταυτοτητας, με "αγνα" μαθηματικα...δυστυχως ομως το μυαλο δεν παει συνηθως σε καποια πραγματα...ετσι γλυτωνεις χρονο και πααααρα πολλες φορες την ασκηση ξεροντας τον τυπο του "κυρ-μητσου απο το κατω κουτσοχωρι"...πραγματικα με εκνευριζει αυτη η τακτικη αλλα ειναι ενα απαρετητο κακο...χωρις αυτην πολλα προβληματα θα ηταν σχεδον αλυτα...
λυνω κατα μεσο ορο 10 ασκησεις τη μερα και καθε μερα γυρναω τις σελιδες να κοιταξω καποιον τυπο "που νομιζω πως εχω δει καπου και θα βοηθησει στην ασκηση" τουλαχιστον 6 φορες...τι να κανουμε ειναι πααραιτητοι
οσο για το λογο που κανουμε ολο αυτο το "ντραβαλο"...οι λογοι ειναι ποικιλοι...αγαπη για τα μαθηματικα,αναγκη να ξεχωρισεις απο τις μαζες,αφορμη για βολτα στην αθηνα :D  :lol: (ναι ναι καλα γελατε αλλα για εμας τα "χωριατοπαιδα" απο την πατρα.ειναι παραγοντας>0 :P  :D )
αυτο το τελευταιο το ειπα για να δειξω ακριβως οτι οι λογοι ειναι πικοιλοι...ελπιζω να πιασατε το νοημα...ο αρχιμηδης δεν ειναι σκετα μαθηματικα και παρολο που σε καμια απο τις δυο φορες που πηγα δεν γνωρισα ακποιον και μονο η ιδεα οτι βρισκομαι εκει μαζι με τους "ελιτ" με κανει να νιωθω μερικους ποντους ψηλοτερος... :blush:

αυτα απλα γιατι θεσατε το σχετικο προβληματισμο.. :D

Darksaga, όπως και να το κάνουμε, στα μαθηματικά, είτε το θέλεις είτε όχι, καλώς ή κακώς, πρέπει να μάθεις μερικούς τύπους. Δυστυχώς, αν βασίζονταν όλα στα αγνά μαθηματικά όπως το αναφέρεις, η επιστήμη δεν θα είχε νόημα. Δηλαδή αυτό που θέλω να πω είναι το εξής: Η γνώση λειτουργεί [b]συσσωρευτικά[/b]. Κάποιοι συνάνθρωποι μας στο παρελθόν κάναν προόδους στους τομείς των μαθηματικών και πρόσθεσαν και αυτοί ένα μικρό λιθαράκι στον μαγικό κόσμο που αποκαλούμε μαθηματικά (και ό,τι αυτό συνεπάγεται). Εμείς, ως φόρο τιμής, ονομάζουμε το εκάστοτε θεώρημα με το όνομα του δημιουργού του. Από την άλλη, φαντάσου να προσπαθείς από το 0 να φτάσεις στο φεγγάρι την επομένη. Κατά τη γνώμη μου, για να συμμετάσχεις με αξιώσεις σε έναν διαγωνισμό όπως ο Αρχιμήδης, πρέπει να έχεις κάνει "γερή" προετοιμασία, να ασχολείσαι συστηματικά. Προς θεού, δεν λέω ότι είμαι σύμφωνος με την αντίληψη του "κάνω μαθηματικά για να περάσω στον Αρχιμήδη". Κατ' εμέ, τα μαθηματικά τα διαβάζεις πρώτα για 'σένα, επειδή το γουστάρεις, επειδή σε ταξιδεύουν σε έναν κόσμο μαγικό, επειδή σε συναρπάζουν, επειδή είναι η "τζούρα" σου. Στη συνέχεια θα επακολουθήσει και η συμμετοχή στους διαγωνισμούς. Δηλαδή, αυτό που πιστεύω είναι πως δεν θα πρέπει αυτοσκοπός να είναι όπως και δήποτε η διάκριση κάπου. Το κάνουμε πρώτα για μας, όχι για τη μητέρα μας, όχι για τη γειτόνισσα (γενικά αναφέρομαι, όχι συγκεκριμένα. Πολλοί από μας το κάνουμε αυτό καθημερινά). Βέβαια, και η διάκριση είναι σημαντική εφόσον μπορεί να επιτευχθεί, αλλά δεν θα πρέπει να είναι πρώτη προτεραιότητα. Είναι κρίμα που μερικοί βλέπουν τον άλλον ανταγωνιστικά (αυτό ισχύει σε όλες τις πτυχές της καθημερινής μας ζωής). Μακάρι όλοι να μπορούσαμε να δούμε τους διπλανούς μας ως [b]συναγωνιστές[/b], όπως πολύ ωραία ανέφερε ανωτέρω ο Νίκος.

Επειδή ίσως έχω ξεφύγει από το θέμα, να πω πως θα ήταν καλό να ξεσκόνιζες αρχικά το βιβλίο για τις ανισότητες του κυρίου Στεργίου, αφού όπως μου έχεις πει το έχεις. Ποτέ δεν ξέρεις, μπορεί να μπει ανισότητα φέτος. Επίσης, καλύτερα μην ασχοληθείς πολύ με ανάλυση και δούλεψε άλλους τομείς όπως θεωρία αριθμών, γεωμετρία. Δες τις ασκήσεις που έχουν μπει κατά καιρούς στο forum, είναι όλες εξαιρετικές. Αν δεν καταλαβαίνεις κάτι, ρώτα τον συγγραφέα της άσκησης, θα σου απαντήσει.

Αυτά από εμένα. Ίσως σε μερικά σημεία έγινα υπερβολικός, ίσως πάλι κάπου ρομαντικός, ίσως κουραστικός ( :P ). Δεν ξέρω κατά πόσο είναι σωστά αυτά που έγραψα, πάντως αυτή είναι η αποψή μου..

Θα συμφωνήσω με τους προλαλήσαντες..
Διαβάζουμε μαθηματικά γιατί τα γουστάρουμε και επειδή διακρίνουμε μέσα από την κάθε καινούργια πρόταση που μαθαίνουμε έναν κόσμο που μας μαγεύει...Αν προσπαθούμε απλώς να μάθουμε 2-3 πράγματα για πετύχουμε σε έναν διαγωνισμό,για να περάσουμε σε μια σχολή νομίζω πως είναι χάσιμο χρόνου όλη μας η προσπάθεια..
Όσο για τον Αρχιμήδη το μόνο που μπορώ να πω είναι κουράγιο και πως οι ώρες που αφιερώνεται(εγώ είμαι μεγάλος και δεν δίνω...)για μελέτη δεν θα πάνε χαμένες..

Καλή συνέχεια!!!!!!!

[quote="Nick Rapanos"]
Υ.Γ. Εγώ θα ήθελα να προτείνω δημόσια στο Σιλουανό που τον βλέπω σαν moderator να φτιάξει ένα  topic  με όλες τις απαραίτητε ανισότητες θεωρήμτα και λήματα που μπορούν να φανούν χρήσιμα στον Αρχιμήδη με σκοπό να βοηθηθούμε όλοι.[/quote]

Αφού μου το ζητάς Nick δεν μπορώ να αρνηθώ.... :)
Πιστεύω είναι καλύτερο από να συσσωρεύουμε ασκήσεις... :wink:
Δείτε δύο links με αρθράκια που είχα κάνει πέρσυ τέτοιο καιρό με τις δύο πιο βασικές ανισότητες και πείτε μου τη γνώμη σας

http://www.mathlinks.ro/Forum/viewtopic.php?t=80026
http://www.mathlinks.ro/Forum/viewtopic.php?t=78744

Θα φτιάξω πάντως ένα άρθρο μάλλον αύριο ή μεθαύριο με όλες τις χρήσιμες ανισότητες και όλα τα λήμματα συγκεντρωτικά....
Το θέλετε σε μορφή μεθοδολογίας ή με παραδείγματα ?
Πείτε μου σας παρακαλώ....

Silouane enas syndyasmos ton dyo tha itan kalos...alla kalytera na klinei pros ti morfi methodologias...

Οκ....Πάντως σκέφτομαι να μην χρησιμοποιήσω ανάλυση (εκτός από Jensen ίσως....). Τι λέτε ?

Με βρίσκεις σύμφωνο.. καλύτερα άσε την ανάλυση εκτός.. επίσης, θα 'ταν καλό αν δημιουργούσαμε και ένα ξέχωρο section
 με ανάλυση αποκλειστικά.

Θα συμφωνήσω κι εγώ με τον Mostel να δημιουργήσουμε ξεχωριστά ένα topic για προβλήμα απειροστικού λογισμού καθώς είναι ο πιο βασικός τομέας μαθηματικών μαζί με την άλγεβρα και την Ευκλειδια γεωμετρία..
Καθώς και με άλλους τομείς πιο πανεπιστημιακούς όπως η γραμμική άλγεβρα,αναλυτική γεωμετρία,συνολοθεωρία..
Προς το παρόν όμως καλό είναι να βοηθήσουμε τα παιδιά με τον Αρχιμήδη..

[quote]
re paidia mipws kserei kaneis kamia askisi ws efarmogi sta theorimata Ceva kai Menelaou giati psaxnw gia eksaskisi k den vriskw tpt![/quote]
3o 8ema persinu arximidi... :wink: Ceva..proektinis ligo..(Menelao..),8eorima dixotomono+teliosies...pio klasiki efarmogi dn nomizo oti iparxi..

 :D

eyxaristw poly RAKAR!!! an k to eixa dwsei afto to thema perysi den to thimomouna!!!!!


The number of inches in both the length and the width of a
rectangle are prime numbers. The area of the rectangle is
391 square inches. What is the number of inches in the
perimeter of the rectangle?

can't it b 23 and 17?? so then the perimeter would b 68

68? Well, I would not think so.
80 is my answer. :P

391=17*23

(23+17)*2=80

adding is off asiangenius


How many positive integers $n$ satisfy the following condition:

$(130n)^{50}> n^{100}> 2^{200}$

A. 0

B. 7

C. 12

D. 65

E. 125

[hide="It might look clearer if..."] You took the $50^{th}$ root of everything.   :wink: [/hide]

[hide]

so taking the 50th root like suggested we see

$130n>n^{2}>16$

clearly it doesnot hold for $n\ge 130$ so there are $129$ values but we see for $1\le n\le4$ it doesnot hold either therefore there are $\boxed{125}$ possible solutions.[/hide]

[hide]$\implies 130n>n^{2}>2^{4}=16$, so n can range from 129 to 5, so there are 125 solutions. $\boxed{E}$.[/hide]

[hide]

Take the $\text{50}^{th}$ root of everything.

Get $130n>n^{2}>16$. Therefore, $5\le n\le 129\implies\boxed{E}$.
[/hide]


AMC AIME Support FTW blogs Gauss articles

*dannyhamtx (creator??)
*kl2836 + kookamunga - SeventhSecret (mentally insane =])
*thdanh90 - hahagooman - murder (great grand aunt who happens to be male)
*guineapiggies (adopted) + Emton - mathpassion101
*AIME15 - Fantasy Lover + Stargirl240 - superllamel + LHM - cherry_pi + RomanianGenius - mewto55555 + violin321 - pacman2812 - splatyango
*ash-pokemon - weird888 - vamathletes (children of AIME15 =]) isabella2296 - sicarius1029 -mathblitz (children of fantasy and stargirl =]) ernie - xoangieexo - JKowl - xpmath - AIME_is_hard - myyellowducky82 (all children of super and lhm =])

Pi Guy is the Hamburger Helper =]

NON HUMAN FAMILY MEMBERS:
floppy (family pet bunny =])
Mind Storm (family pet turtle =])
Penguin Guy: Family Pet Penguin
Skaldjfhrulez: Family talking dog
Textangle: Family Pet Polar Bear
pinkpiglet: pet pig!!! lol
crazyasianvampire: pet vampire hahaa
asiangenius314: couch ahah
Erabel: family wii hahaha =]
mz94: protoplasmic amoeba that is FantasyLover's comrade that shpeshifts into a DUCK ?
d3mon shad0w: The Penguin Journalist =]

Other ppl in the community:

Pizza guy (lol): 1=2
Well alexhhmun wanted to be something...
SCHOOL BUS DRIVER? that would be convenient for us kids...
uhh (^_^) moderator?? and uhh budi713 wants to be his clone =D
kthxbai: Hobo in the backyard (lives on lawn, on Tuesdays on lawn chair, Thursdays patio floor =] oh yeah and because he stopped asking me to cd he can sleep on the roof on sundays now!)
vallon22: fisherman
myyellowducky82: doctor
iYOA: Plump baker
Indeed: Dentist

- represents siblinghood (is that a word?)
+ represents marriage
* represents one generation

now that made everything easier didn't it?

again: if you want to join pm me or post something. i prefer posts because it makes everything easier. i only have to check one place then... but if you prefer pms... then... pm me.

hahaha thanks for everyone's support =]

erm, excuse me, you missing RG and i! young whippersnappers these days! when i was your age...

[quote="xoangieexo"]thdanh90
guineapiggies (adopted) + Bird
AIME15 - Fantasy Lover  -   superllamel + LHM  - RomanianGenius - mewto55555
                                               Ernie - xoangieexo - (^_^) (CLONE) (all children of super and lhm =])

haha srry this isn't perfect
- represents siblinghood (is that a word?)
+ represents marriage

ahah if you don't want to be up there
pm me.
if you want to get up there
pm me.

 :D[/quote]

ERNIE WANT A LOLLIPOP!

ANGIE STOLE MY COOKIE!

DADDY PUT ME IN THE BASEMENT!- AGAIN!

WAAAAAAAAAAAAAAAAAAAAAAAHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHH!

Let's start a pool on how long it takes the mods to delete this.  :rotfl:  :rotfl:  :rotfl:

hmm... 3 days?  :lol:

we need a last name btw

i think if mods lock this i will send ernie on them  :rotfl:

ernie-WAHHHHH!!
mod-sorry!! i didn't mean to lock it!!
ernie-WAHHH!!!

[quote="guineapiggies"]ERNIE! BEHAVE OR GRANDMA WON'T GIVE U THE WII U WANTED FOR CHRISTMAS!!

mewto is my brother-in-law? meh whatever! :D

o and i LURV this family tree[/quote]

BUT ITS MY BIRTHDAY!

WAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAH!

I WANT A LOLLIPOP!

ANGIE STOLE MY COOKIE!

DADDY PUT ME IN THE BASEMENT!- AGAIN!

WAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAH!

well mewto who do you want to have as a wife?
and do you want children?
haha saying that sounds weirddd

[quote="guineapiggies"]ERNIE! BEHAVE OR GRANDMA WON'T GIVE U THE WII U WANTED FOR CHRISTMAS!!

mewto is my brother-in-law? meh whatever! :D

o and i LURV this family tree[/quote]

No... hes your uhh... son... LOL

I'M HERE!!! BWAHAHAHAHAHA!!!!

hehehe... I wanna join. I don't know why, but...

ill be the family pet bunny. my name will be Floppy.  =] lol

HEY I WANNA JOIN!!!!!!!!!   :bruce:

ERRRR....  I could be some insanely smart baby that wants to take over the world (YEAH YOU CAN BLAME STEWIE FOR THAT)  :diablo:

[quote="zephyredx"]Did you add me as family math tutor yet?   :roll:[/quote]

yes :D

and sure sqrtoit...

and ermm idk what my mommy throws... i'll have to ask her!

Wait, when i'm on your desktop make sure i'm clad in Pink Armor.

[quote="1=2"][quote="Squarerootofisoscelestria"]I could be some insanely smart baby that wants to take over the world.[/quote]

Just because I'm the pizza guy and psychiatrist doesn't mean I don't take down insane scientists regularly.[/quote]

But technically I'm not a scientist, am I?

I'm just abnormally insane.  :D  :D

hi angie! my daughter in law/ ur mommy (LHM) wants to throw random chocolates at ppls. except for like snickers...kitkats(ithink)...and a bunch of other chocolates  :D

yay momma-in-law piggy! can u tell LHM to throw more chocolates at me?  :lol:  :D

/me wants chocolate

lol sure! in the mean time

[hide="click here for a treat"]lollipops!!!!!!! /me throws lollipops at splat[/hide]

yay!

/me says thanks and sucks on a lollipop

wow this is getting off topic xD

replace splatyango with avec_une_h for laa laa

No this thread is becoming dead!

The newspaper needs more publicity because of its awesome writer leoxnlin.  :P
http://www.freewebs.com/xoangieexo/index.htm

[quote="xoangieexo"]awww poor lala always gets replaced =[[/quote]\blargh it's sad :(

[quote="mathblitz"]leo we kinda moved this thread to tmf heres a link: [url]http:themathforum.forumotion.com[/url][/quote]

uh... no.

and yes it's very sad...


algebra
system of equations
binomial theorem
algebra unsolved

Solve in real numbers the system of equations:

         $ (3x\plus{}y)(x\plus{}3y)\sqrt{xy}\equal{}14$
         $ (x\plus{}y)(x^2\plus{}14xy\plus{}y^2)\equal{}36$.

setting $ a\equal{}x\plus{}y,b^2\equal{}xy,a,b>0$ our system is equivalent to
$ (3a^2\plus{}4b^2)b\equal{}14$
$ (a^2\plus{}12b^2)a\equal{}36$
$ \Rightarrow18(3a^2\plus{}4b^2)b\equal{}7a(a^2\plus{}12b^2)$
$ \Leftrightarrow7a^3\minus{}54a^2b\plus{}84ab^2\minus{}72b^3\equal{}0$
$ \Leftrightarrow(a\minus{}6b)(7a^2\minus{}12ab\plus{}12b^2)\equal{}0$
$ \Leftrightarrow a\equal{}6b$ substitute to the first equation we get $ b\equal{}\frac{1}{2},a\equal{}3$
then we can easily find x,y

Nice solution [b]hoangvn.fix[/b] :)
The solution that I encountered used the binomial theorem but your solution is pretty good too.


limit
real analysis
real analysis unsolved

if for every $x \in \left[0,1 \right)$
we have
$\lim_{n \to +\infty}f(x+n)=l$
can we prove
$\lim_{x \to +\infty}f(x)=l$ :)

no.
Let $r_1,r_2,..,r_n,..$ the rational numbers in $[0;1)$ and define $f$ by:
$f(r_k)=k$, $f(r_k+n)=\frac{r_k}{n}$ and $0$ everywhere else.

No we can't. All the sequences $(f(x+n))$ converge independently and with different speeds. For instance consider the case $f(\frac{1}{k}+n)=1$ for $n \leq k$ and $l$ otherwise.

thanks.alekk and julien_santini.
but alekk I think in your example :
$f(r_k+n)=\frac k n$ right?

I still can't understand,where's the right counterexample?


Will the Pumac competition be getting a website outside of wiki?

I don't really think they need one; the wiki works fine.  In fact, it's almost preferential because any staff person can edit it, so info will get updated faster. *cough Duke cough*

[quote="Xevarion"]Yeah, we like the wiki. Ideally, the wiki would have a lot more general math content also. We hope that over time, it will develop into a math resource with a wide range of values. Also, the wiki format is fairly popular and common now, so a lot of Internet users are familiar with it.[/quote]

If that happened, the Wiki would be competing with AoPS, and that probably wouldn't work well for Princeton, unless the Princeton content was at a college level, while AoPS was mostly high school.  Also, for it to be a realistically comprehensive resource, it would have to be open to the public to edit.

[quote="mysmartmouth"]If that happened, the Wiki would be competing with AoPS, and that probably wouldn't work well for Princeton, unless the Princeton content was at a college level, while AoPS was mostly high school.  Also, for it to be a realistically comprehensive resource, it would have to be open to the public to edit.[/quote]
Yeah, I was thinking college-level, and particularly college-related stuff. As you can see, we already have some stuff about summer math opportunities and grad school, but we'd like to expand on that a lot more.

[quote="neelnal"]Well, you guys will post the problems from the pumac exam after the exam right?[/quote]
Yeah I suppose so. We'll also post results and such. It'll probably happen really quickly too, like that evening or the next day.


Hello,

 Determine the valuers of $ a,b,c \in\mathbb{N}$ such as the equation $ ax^2 \minus{} bx \plus{} c \equal{} 0$ accept two distinct roots include in $ ]0,1[$ and the sum $ a \plus{} b \plus{} c$ is smallest possible.

[quote="tonypr"][quote="linda2005"]Hello,

 Determine the valuers of $ a,b,c \in\mathbb{N}$ such as the equation $ ax^2 \minus{} bx \plus{} c \equal{} 0$ accept two distinct roots $ x_1,x_2 \in]0,1[$ and the sum $ a \plus{} b \plus{} c$ is smallest possible.[/quote]

What does the "$ x_1,x_2\in]0,1[$ mean?[/quote]

I edit the post , it's clear now ,  ]0,1[ = the interval (0,1)

[hide="Hopefully correct"]For distinct real roots, we must have $ b^2 > 4ac$. Then $ b - \sqrt {b^2 - 4ac} > 0$ is automatically satisfied (since $ a,b,c$ are naturals). We also must have $ b + \sqrt {b^2 - 4ac} < 2a\iff b^2 - 4ac < 4a^2 - 4ab + b^2\iff b < a + c$. (Note: $ 2a - b > 0$ is guaranteed by Vieta, since $ {b\over a} = x_1 + x_2 < 2$). By Vieta, we have $ {c\over a} = x_1x_2 < 1\iff c < a$.

Therefore we must fulfill a system of inequations

\begin{eqnarray}4ac < b^2 < (a + c)^2 \\
c < a\end{eqnarray}

in natural numbers.

From the LHS of $ (1)$, it follows that $ b^2$ must be strictly greater than a multiple of $ 4$, hence the smallest possible value for $ b$ is $ 3$, which, in the combination with $ (2)$, yields $ (a,c) = (2,1)$, but then the RHS of $ (1)$ isn't satisfied.

For $ b = 4$, LHS of $ (1)$, combined with $ (2)$, yields $ (a,c) = (2,1)$ or $ (a,c) = (3,1)$, but neither combination satisfies the RHS of $ (1)$.

For $ b = 5$, LHS of $ (1)$, combined with $ (2)$, yields $ (a,c)\in\{(2,1), (3,1),(4,1),(5,1),(6,1),(3,2)\}$, but only $ (a,c) = (5,1)$ and $ (a,c) = (6,1)$ satisfy the RHS of $ (1)$. Therefore we choose $ (a,b,c) = (5,5,1)$ as our solution.[/hide]


[quote="bpms"][quote="lord12"]70% of men lost an eye, 80% lost a arm and 85% lost a leg What is the minimum percentage of people who lost all three.

Is there a general way of how to get this without guessing and checking?[/quote]
[hide]70+85+80-200=35[/hide][/quote]

???
[hide]Why do you subtract 200?[/hide]


Ψάχνω την απόδειξη του εξωτερικού γινομένου... Μας έδωσε ο καθηγητής της Φυσικής στην σχολή την αρχή και το τέλος αλλα δεν μπορώ να βρώ το υπόλοιπο... Μάλλον επειδή δεν ξέρω καλά καλά τι ακριβώς είναι το ανάδελτα :)....

Ο καθηγητής μας έδωσε τα εξής:
(θα συμβολίσω το ανάδελτα με το γράμμα Δ)
Δ= (d/dx, d/dy, d/dz)
A= (Ax, Ay, Az)
B= (Bx, By, Bz)
AB= AxBx + AyBy + AzBz

ΔxE = -dB/dt => .... => Δ (Δ Ε) - Δ^2 Ε = -d(Δ+Β)/dE

Καμιά βοήθεια??

Ummm, μπορώ μόνο να υποθέσω τι θέλεις να δείξεις, ελπίζω να έχω μαντέψει σωστά... Ιδίως αφού μιλάς για Φυσική, υποθέτω ότι ξεκινάς με την εξίσωση του Maxwell στο κενό $ \nabla\times E \equal{} \minus{}\frac{dB}{dt}$ και θέλεις να καταλήξεις σε κάποια από τις πολλές δευτεροβάθμιες εξισώσεις που προκύπτουν από τις Maxwell.

Η LHS φαίνεται να είναι: $ \nabla\left(\nabla\cdot E\right) \minus{} \nabla^2 E$, όπου $ \nabla^2$ η Laplacian, εφαρμοσμένη στο διάνυσμα $ E$. Αν όντως αναφέρεσαι στις Maxwell, τότε $ \nabla\cdot E \equal{} 0$ και επιζεί μόνο η Laplacian... hmmm,  μάλλον καταλαβαίνω που το πάς...

Για την RHS, υπάρχει περίπτωση να εννοείς $ \minus{}\frac{d}{dt}\left(\nabla\times B\right)$?

Αν ναι, υπάρχει η ταυτότητα: $ \nabla\times\nabla\times A \equal{} \nabla\left(\nabla\cdot A\right) \minus{} \nabla^2A$ η οποία αποδεικνύεται με αρκετά οδυνηρές πράξεις (εκτός και αν ξέρεις διαφορική γεωμετρία οπότε προκύπτει τουλάχιστον πιο κομψά). Αν θέσεις $ A\equal{}E$ και χρησιμοποιήσεις την Maxwell $ \nabla\times E \equal{} \minus{}\frac{dB}{dt}$ προκύπτει το ζητούμενο (ή, πιο σωστά, αυτό που υποθέτω ότι αναζητάς ;) ). Αν θυμηθείς και ότι το ηλεκτρικό πεδίο είναι ασυμπίεστο, προκύπτει η Laplacian του $ E$: $ \nabla^2 E \equal{} \frac{d}{dt}(\nabla\times B)$

Let me know αν όντως αυτό ψάχνεις.

Cheerio,

Durandal 1707

Ναι νομίζω αυτό θέλω να δείξω...
Μήπως μπορείς να γραψεις την απόδειξη που εγραψες με διαφορική γεωμετρια?

Ok, here goes...

Αν είναι να το προσεγγίσεις από διαφορική γεωμετρία, θεωρείς ότι ο $ \mathbb{R}^3$ έρχεται εφοδιασμένος με μία μετρική $ g$ (στην περίπτωσή σου χρησιμοποιείς την Ευκλείδεια) με την αντίστοιχη volume form $ \Omega$. Τότε, ξεχνώντας τις συντεταγμένες, θα έχουμε για ένα πεδίο $ X$ και μία συνάρτηση $ f$:

0. $ \nabla f \equal{} (df)^\sharp$ όπου $ X^\flat \equal{} g(X,\cdot)$ και $ \sharp \equal{} \flat^{ \minus{} 1}$ οι τελεστές ανόδου/καθόδου που κατεβάζουν διανύσματα σε μορφές και ανεβάζουν μορφές σε διανύσματα. Δηλαδή, αν $ e_j$ ορθοκανονική βάση ως προς την $ g$ και $ X \equal{} X^j e_j$ ένα διάνυσμα, κατεβαίνουμε στη μορφή $ X^\flat \equal{} X^j g(e_j,\cdot)$.
1. $ \nabla\cdot X \equiv \star d(i_X\Omega)$ όπου $ \star$ ο τελεστής Hodge που ορίζεται έμμεσα από την $ \omega\wedge\star\eta \equal{} \left < \omega,\eta\right > \Omega$ για δύο $ k$-μορφές $ \omega, \eta$.
2. $ \nabla\times X \equal{} \left(\star d X^\flat\right)^\sharp$
3. $ \nabla^2$ = $ \delta d \plus{} d\delta$ όπου $ \delta \equal{} (\pm)\star d\star$ και χρησιμοποιείς το $ \minus{}$ για "διανύσματα" (μορφές τύπου $ X^\flat$).

Έτσι έχουμε: $ \nabla\times\nabla\times X \equal{} (d\star d X^\flat)\sharp$ και μετά από λίγα rearrangements προκύπτει το ζητούμενο.

Aν θέλεις περισσότερες λεπτομέρειες, θα βρεις οπωσδήποτε άπειρες πληροφορίες online (πχ. βάλε Hodge star στην Wikipedia).

Τώρα, αν πάλι δε σου λένε τίποτα τα παραπάνω... θα ήταν μάλλον καλύτερο (και πιο διδακτικό!) να κάνεις τις πράξεις. Ούτως ή άλλως, τα παραπάνω απλώς κρύβουν τις πράξεις σε άλλα σημεία που θεωρούνται "γνωστά" αν ξέρεις να τα χειρίζεσαι ως εργαλεία.

Βασικά, εκτός και αν έχεις μία κάποια εμπειρεία στη διαφορική γεωμετρία, θα σου πρότεινα να κάνεις τις πράξεις, δεν είναι πάρα πολύ άσχημες αν θυμάμαι σωστά.

Cheerio,

Durandal 1707


Yep that's right. That is also phi. Funny that some seniors at my school couldn't solve it quickly.

You could also multiply the whole thing through with 2/(x)^2 and then substitute to get a cubic.


A cotton-picking machine picks 60 percent of the cotton each time it is driven over a field. Given that a farmer drives the machine twice over the same field, what percent of the cotton remains in the field?

Assume that there are $ 100$ cottons. At the first drive $ 100(.4) \equal{} 40$ cottons will remain. After the second drive $ 40(.4) \equal{} 16$ cottons will remain.

Hence, $ 16/100 \equal{} .16$ or $ 16\%$

So, if there were x amount of cotton originally, there will be 0.4x left after the farmer drives over it once and (0.4)^2 = 0.16 x left after the farmer drives over it twice.

Therefore there is only 16 percent remaining after the farmer has driven over it twice!


rotation
combinatorics unsolved
combinatorics

[quote="aadil"]consider 2 binary strings $ A$  and $ B$ of length $ 2n$ written one below the other,where each of the 2 strings have equal number of 0,s and 1,s.all cyclic permutations $ B$ are also considered.prove that there exists a cyclic permutation of $ B$ which has atleast n common elements with the corresponding elements of $ B$[/quote]

did you mean
"which has atleast n common elements with the corresponding elements of $ A$" ? Otherwise I don't see how $ A$ comes into play in the problem at all.

But if that's what you meant, this is my solution:
Write $ B$ and all of its cyclic permutations under it, so you end up with an array of $ 2n$ columns and $ 2n$ rows, where row $ i\plus{}1$ is a cyclic permutation of row $ i$.

For each row, we give it a SCORE based on how many common corresponding elements it has with $ A$. So a row's score may range from zero to $ 2n$.
And let $ S$ be the sum of all the row's scores. If there is no row whose score is at least $ n$, then $ S < 2n.n \equal{} 2n^2$.

But for each element of $ A$,  observe the corresponding column in that array. Say this element is 0. Each 0 in that column contributes +1 to the total of $ S$. But that column is also a cyclic permutation of $ B$, because as we rotate $ B$ from row to row, that column is occupied by different elements (ones and zeros) according to $ B$'s movement. So this column contributes a total of $ \plus{}n$ to $ S$.
Thus $ S \equal{} 2n.n \equal{} 2n^2$.

So there must be at least one row whose score is at least $ n$.

I just realized,

your problem might be read as "$ A$ and $ B$ have equal number of ones and zeros to each other but not necessarily $ n$ each"
In that case, my proof still applies, because,
say there are $ k$ ones and $ 2n\minus{}k$ zeros. For each column, if the corresponding element in $ A$ is a one, then that column contributes $ k$ to the sum $ S$, otherwise $ 2n\minus{}k$.

So the total sum of $ S$ is $ k^2 \plus{} (2n\minus{}k)^2 \geq 2n^2 \iff (n\minus{}k)^2 \geq 0$


geometry
3D geometry
sphere
circumcircle

The chords AB and CD of a sphere intersect at X. A, C and X are equidistant from a point Y on the sphere. Show that BD and XY are perpendicular.

[b]Lemma:[/b] The circumcenter of a triangle is an isogonal conjugate of the orthocenter.

Let $O, H$ be the circumcenter and orthocenter of the $\triangle ABC$. The angle $\measuredangle BAH = 90^o - \measuredangle B$. The angle $\measuredangle CAO = (180^o - \measuredangle COA)/2 = (180^o - \measuredangle B)/2 = 90^o - \measuredangle B$, because the triangle $\triangle CAO$ is isosceles and the angle $\measuredangle B$ spans the arc $CA$ with the central angle $\measuredangle COA = 2 \measuredangle B$. Hence, $\measuredangle BAH = \measuredangle CAO$ and similarly, $\measuredangle CBH = \measuredangle ABO$ and  $\measuredangle ACH = \measuredangle BCO$ at the remaining vertices $B, C$.

The 2 intersecting lines $AB, CD$ define a plane and their intersection $X$ is in this plane. The plane cuts the sphere in a circle, which is the circumcircle $(O)$ of the quadrilateral $ACBD$. The locus of points equidistant from $A, X, C$ is a line through the circumcenter $P$ of the triangle $\triangle ACX$ and perpendicular to this plane. This line meets the sphere at 2 points $Y, Y'$. The triangles $\triangle ACX \sim \triangle DBX$, because the angles $\angle AXC = \angle DXB$ are vertical and the angles $\angle CAX = \angle BDX$ span the same arc $BC$ of the circumcircle $(O)$. The sides $AX, DX$ are the corresponding sides of these 2 similar triangles. The line $XP$ passing through their common vertex $X$ and through the circumcenter $P$ of the triangle $\triangle ACX$ forms the same (vertical) angles between the side $AX$ of the triangle $\triangle ACX$ and the side $BX$ of the triangle $DBX$. Hence, it must pass through the isogonal conjugate of the circumcenter $P'$ of the $\triangle DBX$, i.e., through its orthocenter $H'$. Consequently, the line $PX$ is the altitude of the triangle $DBX$ perpendicular to its side $DB$. Thus the 2 perpendicular lines $PX, PY$ define a plane perpendicular to the line $BD$ and any line in this plane is perpendicular to the line $BD$. The line $XY$ is in this plane.

Be more carefull next time yetti :) The problem was caused by you writing $\DEX$, which is a command (because of the "\" slash) and it wouldn't compile (obviously).


absolute value
calculus
calculus computations

Prove that if $ b \geq 0$ then $ |a| \leq b$ if and only if $ \minus{}b \leq a \leq b$. So $ a \leq b$ and $ \minus{}a \leq b$. Thus $ a \geq \minus{}b$ and so $ \minus{}b \leq a \leq b$. Conversely, if $ \minus{}b \leq a \leq b$, then $ a \leq b$ and $ \minus{}a \leq b$. Thus $ |a| < b$.

This is a valid proof? You don't have to go through all the cases?

so you don't after consider the cases if $ a<0$, $ a > 0$ and $ a\equal{}0$?

yes.maybe you should see it in line of Real

what???

So you dont have to consider those cases?

It's a matter of being thorough.

[quote]So $ a \leq b$ and $ \minus{} a \leq b$[/quote]

This is a true conclusion of the statement $ |a| \leq b$. However, if you want to be more thorough, this is where you can inject some casework:

$ |a| \leq b$ means that:

$ a > 0 \; \implies \; \minus{} a < a \leq b$
$ a \leq 0 \; \implies \; a \leq \minus{} a \leq b$

In either case, both $ a \leq b$ and $ \minus{} a \leq b$ are true statements.

[quote="ragingbullrocky"]This is a valid proof?[/quote]  Oddly enough, JRav has given what I consider the most helpful comment so far.  You could explain in more detail which things follow from which others, and doing so could possibly involve invoking the case-definition of absolute value (or more simply, the non-case definition: $ |a| \equal{} \max\{a, \minus{}a\}$), but what you've written is a correct and reasonably clear proof of the desired result.


function conics parabola algebra domain ellipse calculus

please show work and explain how it is done so i can learn from it
thanks

1. determine the vertex of the parabola y = 2x^2 - 12x + 12

2. determine the domain of the relation x^2 + 16y^2 = 16

3. what is the vertex of the parabola f(x) = -2 ( x + 1 )^2 + 3

4.what is the length of the major axis of the ellipse x^2/36 + y^2/81 = 1

You could find the zeros, add them, divide them by 2, set the x value into function, and solve for y to get (x,y).

OR.

You could do the following.
Let $f(x)=ax^2+bx+c$.
$\\f'(x)=2ax+b\\ 0=2ax+b\\ x=\frac{-b}{2a}\\\\ f(\frac{-b}{2a})=a\left(\frac{-b}{2a}\right)^2+b\frac{-b}{2a}+c\\ =\frac{ab^2}{4a^2}-\frac{b^2}{2a}+c\\ =\frac{b^2}{4a}-\frac{b^2}{2a}+c=\frac{-b^2}{4a}+c\\\\ \implies (x,y)=\left(\frac{-b}{2a}, \frac{-b^2}{4a}+c\right)$

If $2a>0$, The vertex is a minimum, and if $2a<0$, it is a maximum.

Teh fourth one is simple.  For the length of each of the axes, take teh square root under the denominator of the x and y containing term and double them for each axis.

i.e.  $\frac{x^2}{36}+\frac{y^2}{9}=1$ gives us axes of length $2\sqrt{36}=12$ and $2\sqrt{9}=6$.
Masoud Zargar

You don't need to use differentiation, just use completing square. :lol:

That also works good, but differentiation decreases the amount of thinking. :D Just kidding; thinking is good.


Well i find these interesting,They are from the GAME :Escape from monkeys island: :D :

These are some of the words that they say before they start the battle:

Pirates:

Every enemy I met I've annihilated!
With your breath, I'm sure they all suffocated.

You're as repulsive as a monkey in a negligee.
I look THAT much like your fiancée?

Killing you would be a justifiable homicide!
Then killing you must be a justifiable fungicide.

I'll skewer you, like a sow at a buffet!
When I'm done with YOU, you'll be a boneless fillet.

Throughout the Caribbean, my great deeds are celebrated.
Too bad they're all fabricated.

You can't match my witty repartee!
I could, if you would use some breath spray.

When your father first saw you, he must have been mortified.
At least mine can be identified.

Would you like to be buried or cremated?
With your around, I'd prefer to be fumigated.

I'll leave you devastated, mutilated & perforated.
Your order alone makes me aggravated, agitated & infuriated.

You're the ugliest monster ever created!
If you don’t count all the ones you've dated.

I have never seen such clumsy swordplay.
You would have, but you were always running away.

Heaven preserve me! You look like something that's died!
The only way you'll be preserved is in formaldehyde.

En garde! Touché!
Oh, that is so cliché!

I can't rest 'til you've been exterminated!
Then perhaps you should switch to decaffeinated.

I'll hound you night & day.
Then be a good dog. Sit! Stay!

Coming face to face with me must leave you petrified!
Is that your face? I thought it was your backside.

My attacks have left the entire island depopulated
With your breath, I'm sure they all suffocated.

Your lips look like they belong on the catch of the day.
When I'm done with YOU, you'll be a boneless fillet.

I have never lost a melee!
You would have, but you were always running away.

You'll find I'm dogged & relentless to my prey!
Then be a good dog. Sit! Stay!

Your stench would make an outhouse cleaner irritated!
Then perhaps you should switch to decaffeinated.

Your mother wears a toupee!
Oh, that is so cliché!

My skills with a sword are highly venerated.
Too bad they're all fabricated.

Never before have I faced someone so sissified.
Is that your face? I thought it was your backside.

Nothing can stop me from blowing you away!
I could, if you would use some breath spray.

I give you a choice. You can be gotted or decapitated.
With your around, I'd prefer to be fumigated.

Your looks would make pigs nauseated.
If you don’t count all the ones you've dated.


floor function
number theory proposed
number theory

We call the mapping $ \Delta:\mathbb Z\backslash\{0\}\longrightarrow\mathbb N$, a degree mapping if and only if for each $ a,b\in\mathbb Z$ such that $ b\neq0$ and $ b\not|a$ there exist integers $ r,s$ such that $ a \equal{} br\plus{}s$, and $ \Delta(s) <\Delta(b)$.
a) Prove that the following mapping is a degree mapping:
\[ \delta(n)\equal{}\mbox{Number of digits in the binary representation of }n\]
b) Prove that there exist a degree mapping $ \Delta_{0}$ such that for each degree mapping $ \Delta$ and for each $ n\neq0$, $ \Delta_{0}(n)\leq\Delta(n)$.
c) Prove that $ \delta \equal{}\Delta_{0}$
[img]http://i16.tinypic.com/4qntmd0.png[/img]

I think there is a typo.It should be $ a\not|\ b$

Perhaps i'm misunderstanding  :oops:

If $ a\equal{}2$ $ b\equal{}4$ $ \Delta\equal{}\delta$

there sholud exist $ r,s$ such that $ 2\equal{}4r\plus{}s$ and $ \delta(s)<\delta(r)$

Now

$ \delta(s)\equal{}\delta(2\minus{}4r)\equal{}\delta(2r\minus{}1)$

But it's easy to show that $ \delta(2r\minus{}1)\ge\delta(r)$

Where is the error?

Oh it must be $ \Delta(s)\leq\Delta(b)$. :blush:

Let $ b\equal{}2^{k},a\equal{}2^{k}\plus{}3,k\ge 3$, then $ s\equal{}3\plus{}(1\minus{}r)2^{k}$ and $ \delta(s)>\delta(b),\ \delta(s)>\delta(r)$.  :lol:

OK, but I don't think it proves the problem.

Official Solution:
For (a), let $ a,b\in\mathbb{Z}$ be such that $ b\neq0$ and $ b\nmid a$. Now divide $ a$ by $ b$ and write $ a \equal{} br' \plus{} s'$ where $ 0 < s' < |b|$. If $ s' < \frac {|b|}2$, then put $ r \equal{} r',s \equal{} s'$; and if $ s'\geq \frac {|b|}2$ then put $ r \equal{} r'\pm1,s' \equal{} s \minus{} |b|$. (the sign $ \pm$ is the same as $ b$'s sign)
One can easily see that $ a \equal{} br \plus{} s$ and $ |s|\leq\frac {|b|}2$ which shows that $ \delta(s)\leq\delta(b) \minus{} 1$.
For (b), let $ \Delta_0(n)$ be the minimum of $ \Delta(n)$ for every degree map $ \Delta$. It suffices to show that $ \Delta_0$ is a degree map. Given $ a,b$, there exists a degree map $ \Delta$ for which	$ \Delta_0(b) \equal{} \Delta(b)$. Now take the $ r,s$ given by $ \Delta$. So $ a \equal{} br \plus{} s$ and $ \Delta(s) < \Delta(b)$. But then $ \Delta_0(s)\leq\Delta(s) < \Delta(b) \equal{} \Delta_0(b)$ which shows that this $ r,s$ work for $ \Delta_0$ too.

Now for (c): For $ n \equal{} \pm1$, $ \delta(n)\geq\Delta_0(n)\geq1 \equal{} \delta(n)$ which shows that $ \delta(n) \equal{} \Delta_0(n)$. Let $ n$ be a number such that $ \Delta_0(n) < \delta(n)$. If there are many, choose one with the minimum $ \Delta_0(n)$. $ n\neq\pm1$ because of the previous result. Now take $ a \equal{} \lfloor\frac n2\rfloor$ and $ b \equal{} n$. Then $ b\nmid a$ because $ a\neq\pm1$.
Let $ r,s$ be such that $ \Delta_0(s) < \Delta_0(n)$ and $ a \equal{} br \plus{} s$. Since $ \Delta_0(s) < \Delta_0(n)$, we have $ \Delta_0(s) \equal{} \delta(s)$. But it can be easily shown that whatever the value of $ s$ is, $ |s|\geq\lfloor\frac {|n|}2\rfloor$ which is at least $ \delta(n) \minus{} 1$. So $ \delta(n)\geq\Delta_0(n) > \delta(n) \minus{} 1$ which shows that $ \Delta_0(n) \equal{} \delta(n)$. A contradiction!


Find the smallest positive integer n that satisfies the following condition: For every finite set of points on the plane, if for any n points from this set there exist two lines containing all the n points, then there exist two lines containing all points from the set.

we have $ min \equal{} 6$
example for  $ n \equal{} 5$ continue sides of tringle $ ABC$ to intersect any arbitary line.now we have 6 point have the condition of problem.
now assume $ n \equal{} 6$
we have 3 point$ a_{1},a_{2},a_{3}$ collinear.name line contains $ a_{1},a_{2},a_{3},l$.now suppose we have 3 another point $ A,B,C$ that are not collinar and also not on the line $ l$.now take$ a_{1},a_{2},a_{3},A,B,C$one of that 2 line should contain two of {$ a_{1},a_{2},a_{3}$}and one of {$ A,B,C$} which contradicts.so all points that are not on line $ l$should be on a line.


Let $p>2$ is a prime and $t$ is a natural number and odd such that $4-p<t<p$.Prove that there isn't exist $m$ such that $(p+t)!-t^m=p^m$.

[quote="ehsan2004"]$t$ is a natural number and odd such that $4-p<t<p$[/quote]

Isn't [tex]t[/tex] an integer rather ?  :?

Well , dear Myth had said that My results were worthless and wrong so i have attached his solution.  (so i will delete my weak arguments)


quadratics
modular arithmetic
algorithm
algebra
polynomial
ceiling function
floor function

Let $ p$ be a prime, and $ a$ is a quadratic residue of $ p$. Determine the number of incongruent solutions to $ x^2\equiv a \pmod {p^n}$, $ n\geq 1$ if
(1) $ \gcd(a,p) \equal{} p$, or
(2) $ \gcd(a,p) \equal{} 1$.

Is there any formula or theorem related to this? Besides, is there any "simple" algorithm to find these solutions?

This might help,

(i) $ x^2\equiv a\pmod{m}$ where $ m\in N$ and $ a\in Z$ such that $ gcd(a,m)\equal{}1$.

(ii) Let $ p$ be an odd prime, and $ gcd(a,p)\equal{}1$. If $ a$ is quadratic nonresidue modulo $ p$, the congruence in (i) has no solution.

(iii) If $ a$ is a quadratic residue modulo $ p$, the congruence in (i) has exactly two incongruent solutions $ x$ modulo $ p$. Also, if $ x_0$ is a solution, then a second incongruent solution is given by $ p\minus{}x_0$.

Solving polynomials $ \bmod p^n$ is done by [url=http://en.wikipedia.org/wiki/Hensel%27s_lemma]Hensel's lemma[/url].  (You may find some statements of it more useful than others.)  See if it leads you to some insight about how the solutions behave as $ n$ changes.  Start with $ n \equal{} 1$.

[quote="JZambrano201"](i) $ x^2\equiv a\pmod{m}$ where $ m\in N$ and $ a\in Z$ such that $ gcd(a,m) \equal{} 1$.[/quote]
We are already given that $ m \equal{} p^n$.  What are you trying to say?

[quote="t0rajir0u"]Solving polynomials $ \bmod p^n$ is done by [url=http://en.wikipedia.org/wiki/Hensel%27s_lemma]Hensel's lemma[/url].  (You may find some statements of it more useful than others.)  See if it leads you to some insight about how the solutions behave as $ n$ changes.  Start with $ n \equal{} 1$.
[/quote]

I think I know the answer to my first question now: there are always exactly 2 incongruent solutions for odd primes in (2), since $ f'(x_0) \equal{} 2x_0$ is always nonzero modulo $ p$. If $ p\equal{}2$, it has 1 solution for $ n\equal{}1$, 2 for $ n\equal{}2$, and 4 for $ n\geq 3$ . (1) is more complicated; it can have a lot of solutions (for example, $ a \equal{} 0$) or no solutions at all.

As for the behavior of solutions...I didn't find much pattern here. I tried $ x^2\equiv 3\pmod {11^k}$; I stopped after my calculator displayed overflow. Am I missing anything?

I'm not talking about the solutions themselves, just counting them.  My point was that Hensel's lemma guarantees that there are exactly $ 2$ solutions for (2).  As for (1), start by writing $ a \equal{} pb$ and see where it leads you.

Okay, here's what I derived:
[hide]Let $ a = b\cdot p^m$, $ m\geq 0$, where $ b = 1,2,...,p - 1$.

[b]Case 1:[/b] $ m\geq n$. The congruence reduces to
\begin{align*} x^2 & \equiv 0 \pmod {p^n} \\
x & \equiv 0 \pmod {p^{\lceil \dfrac{n}{2} \rceil}} \\
x & \equiv k\cdot p^{\lceil \dfrac{n}{2} \rceil} \pmod {p^n} \end{align*}

for $ k = 0,1,...,p^{\lfloor\dfrac{n}{2} \rfloor - 1}$, totally $ p^{\lfloor \dfrac{n}{2} \rfloor}$ solutions.

[b]Case 2: [/b] $ m < n$. We have $ x^2\equiv 0 \pmod {p^m}$, so $ x\equiv 0 \pmod {p^{\lceil \dfrac{m}{2} \rceil}}$. Let $ x = y\cdot p^{\lceil \dfrac{m}{2} \rceil}$,

\begin{align*} y^2\cdot p^{2\lceil \dfrac{m}{2} \rceil} & \equiv b\cdot p^m \pmod {p^n} \\
y^2\cdot p^{2\lceil \dfrac{m}{2} \rceil - m} & \equiv b \pmod {p^{n - m}} \end{align*}

If $ m$ is odd, ${ 2\left\lceil \dfrac{m}{2} \right\rceil} - m = 1$, then the congruence has no solution.

If $ m$ is even, ${ 2\left\lceil \dfrac{m}{2} \right\rceil} - m = 0$, then the congruence reduces to
\[ y^2 \equiv b \pmod {p^{n - m}}
\]
which has a solution if and only if $ b$ is a quadratic residue of $ p$. Furthermore, by Hensel's lemma, if it does have a solution, it has exactly two solutions.[/hide]

Regarding Hensel's lemma, you may be interested in the less known but more general version I mentioned in http://www.mathlinks.ro/viewtopic.php?p=1146101#1146101 :

Let $ f(x)$ be an integer polynomial. Assume there is an integer $ x_0$ and a positive integer $ n$ such that $ f(x_0) \equiv 0 \mod p^{2n\minus{}1}$ and $ f^\prime(x_0) \not\equiv 0 \mod p^n$. Then for all positive integers $ m$ there exists a solution $ x$ to $ f(x) \equiv 0 \mod p^m$.

The case $ n\equal{}1$ is the most common version. But if you are e.g. interested in squares $ \mod 2^m$, this version turns out to be more useful.


probability
geometry
rectangle
trapezoid
perpendicular bisector

A point $ P$ is randomly selected from the rectangular region with vertices $ (0, 0)$, $ (2, 0)$, $ (2, 1)$, $ (0, 1)$. What is the probability that $ P$ is closer to the origin than it is to the point $ (3, 1)$?

$ \textbf{(A)}\ \frac{1}{2} \qquad
\textbf{(B)}\ \frac{2}{3} \qquad
\textbf{(C)}\ \frac{3}{4} \qquad
\textbf{(D)}\ \frac{4}{5} \qquad
\textbf{(E)}\ 1$

The locus of all points that are equidistant from (0,0) and (3,1) is the perpendicular bisector of the segment connecting these two points. We could find this line, but there is an easier way.

Drawing a picture:

[asy]size((500));
draw((0,-3)--(0,3));
draw((-3,0)--(3,0));
draw((0,0)--(2,0)--(2,1)--(0,1)--cycle,red);
draw((1,1)--(1,0));
draw((1,3)--(2,-2));[/asy]

This dividing line will go through the midpoint of the segment from (0,0) to (3,1), that is, (1.5,0.5), the center of the rightmost half of the rectangle, meaning it cuts that square in half. Thus, $ \frac34$ of the points in the rectangle are to the left of the line and therefore closer to the origin.

Our answer is $ \boxed{\text{C}}$

A perpendicular bisector is equidistant from both points.

My solution to this problem was almost exactly the same as Mewto's. Only difference is that I computed the area to the right of the perpendicular bisector by noting that it's a trapezoid (we know that because (2,0) is closer to (3,1) than to the origin, i.e. it must be to the right of the perpendicular bisector) with height $1$, the average of whose bases is $1/2$.

[quote="AlanLiangTheAsian"]So I got everything else but how why is it the perpendicular bisector that seperates closer to 0,0 points? I mean i get the bisector but why does it have to form 90 degrees?[/quote]You could use Pythagoras; but even that's not really necessary if you consider the symmetry.

2002 AMC 12B #18

and this time I think that is right...lol

[hide=Solution]
Drawing the graph,
[asy]
import graph;
xaxis(-1,4,Arrows);
yaxis(-1,4,Arrows);
dot((0,0));
dot((0,1));
dot((2,0));
dot((2,1));
dot((3,1));
draw((0,1)--(3,1));
draw((0,0)--(3,1));
draw((2,1)--(2,0));
[/asy]
Drawing the line perpendicular to the diagonal from $(3,1)$ to $(0,0)$ we get
[asy]
import graph;
xaxis(-1,7,Arrows);
yaxis(-1,7,Arrows);
dot((0,0));
dot((0,1));
dot((2,0));
dot((2,1));
dot((3,1));
dot((0,5));
dot((5/3,0));
draw((0,1)--(3,1));
draw((0,0)--(3,1));
draw((2,1)--(2,0));
draw((0,5)--(5/3,0));
[/asy]
This line has equation, $y=-3x+5$ and intersects $y=1$ at $x=\tfrac{4}{3}.$ So, we want to find everything to the left of this line that lies within the rectangle which becomes a trapezoid, so
$$\left(\frac{4}{3}+\frac{5}{3}\right) \cdot \frac{1}{2}=\frac{3}{2}$$
Dividing by $2$ since that is the area of the rectangle we get $\boxed{\tfrac{3}{4}}.$


The number $85^9-21^9+6^9$ is divisible by an integer between 2,000 and 3,000. Compute that integer.

Is it $2240$? It's divisible by $2^6$, $5$, and $7$ ;) after doing trial and error with mods.

Yes...

[hide="hint for nicer soln"]euclid's algorithm[/hide]


Find the value of x, so that:

$ \frac{(0,2)^{x\minus{}0,5}}{\sqrt{5}}\equal{}5.(0,04)^{x\minus{}1}$

[hide="My Solution"]
Rewriting everything in terms of fractions, we have

$ \frac{\left(\frac{1}{5}\right)^{x \minus{} \frac{1}{2}}}{\sqrt{5}} \equal{} 5\cdot\left(\frac{1}{25}\right)^{x \minus{} 1}\\
\frac{1}{5\sqrt{5}} \equal{} \left(\frac{1}{5}\right)^{2x \minus{} 2 \minus{} x \plus{} \frac{1}{2}}\\
\minus{}\log_{5}{5^{\minus{}\frac{3}{2}}} \equal{} x \minus{} \frac{3}{2}\\
\frac{3}{2} \plus{} \frac{3}{2} \equal{} x\\
x \equal{} \boxed{3}$.
[/hide]

hello, we have $ \frac{\left(\frac{1}{5}\right)^{x\minus{}1/2}}{5^{1/2}}\equal{}5\cdot\left(5^{\minus{}2}\right)^{x\minus{}1}$ this is equivalent with $ 5^{\minus{}x}\equal{}5\cdot5^{\minus{}2x}\cdot5^2$ or $ 5^x\equal{}5^3$ hence $ x\equal{}3$ is the only real solution.
Sonnhard.

Thanks Dr Sonnhard Graubner and sunehra ! ;)

But I don't understand this line:

$ \equal{}\left(\frac{1}{5}\right)^{2x\minus{}2\minus{}x\plus{}\frac{1}{2}}$

How find 2x -2-x+1/2 ?

Thanks.

In sunehra's work:
First write $ \frac{1}{25}$ as $ \left(\frac{1}{5}\right)^{2}$, so you have
$ \frac{\left(\frac{1}{5}\right)^{x\minus{}\frac{1}{2}}}{\sqrt{5}}\equal{} 5\cdot\left(\frac{1}{5}\right)^{2x\minus{}2}$
And then divide both sides by $ 5\cdot\left(\frac{1}{5}\right)^{x\minus{}\frac{1}{2}}$, which will give you the second line in sunehra's solution.


Ross Mathematics Program
ARML
algorithm
pigeonhole principle
geometry
geometric transformation
rotation

Here's an e-mail from Mr. Shapiro (the director of the ROSS program) about the start of the Columbus Math Circle, if anyone's interested in coming

[quote]Hello from the OSU Math Department and from the Ross Math Program!  I hope you're enjoying the start of an excellent school year.

At the request of students, I'm initiating a local Math Circle.  This will involve Sunday afternoon sessions aimed at Central Ohio high school students interested in math (not just for those who love contests).  We hope to provide a relaxed setting that will enable students to learn some accessible mathematical ideas not covered in standard high school math classes.  Along with related math problems for them to solve.

The first meeting of this Math Circle will be on Sunday, September 30 at 2:00 in Davidson Hall on the campus of Columbus State Community College.  (Room number will be posted on the door there.)  At that meeting we will discuss the schedule for future meetings.  Prof. Alan Yang at CSCC is helping with this project.

    Please encourage some of your students to join us.
You are also welcome to participate, and to help with the discussions!

DIRECTIONS for those who will join us on Sunday.

Directions:  http://www.cscc.edu/DOCS/visitor/directions.htm

Campus map of CSCC:  http://www..cscc.edu/Tour/mapscroll.htm

Park in lot 11-S, across from Davidson Hall (labeled DH on map).
Attached is a parking permit which can be printed and placed on the car's dashboard.
If you get lost, call my cell phone at 614-284-7954.

Regards,
   Daniel Shapiro

Professor and Vice Chair, Department of Mathematics
Director, Ross Mathematics Program
http://www.math.ohio-state.edu/ross[/quote]

The parking pass was an attachment, so you'll need to contact either Mr. Shapiro, or Mr. Yang (the Ohio ARML coach) for that.

Hope you come.

The next meeting will be on the 14th, same time and place.  And tell other people too.

I emailed the director of my math club on wednesday and mentioned it, thinking she may mention it on our school's math club meeting the next day, but she didn't even attend that meeting. Nor did she ever respond to my email.

Dang, I'm too far away. It would take too long.

I am not going to spend 3 hours (and then 3 hours back).

eh its just two hours for me

i read a math book there and back

Eh its about 2 hrs for me. Is tjhance going?

i attended yesterday

i will probably attend the one in 2 weeks as well...

Haha

this is becoming more like an... Ohio math circle.

The next Math circle meeting will be in Davison Hall at CSCC on Sunday, October 14 at 2:00 p.m.  In the future, information about Central Ohio Math Circle can be found at http://www.math.osu.edu/ross/MathCircle
For CSCC's campus map and directions, use the links
Campus Map: http://www.cscc.edu/Tour/mapscroll.htm
Directions: http://www.cscc.edu/DOCS/visitor/directions.htm
You are recommended to park in lot 11-S, the big lot at the northwest corner of Mt. Vernon Ave. and Spring St., across from Davison Hall.

1. The topic for the day, included but not limited to "Impossible Scores"

Outline:
In our game there are two ways to score:  A  points for a basket and  B  points for a touchdown.
For example, when  A = 3  and  B = 7,  every positive score is possible, except for 1, 2, 4, 5, 8, and 11.  What happens for other values of  A  and  B?

2.  Here are some possible topics for future meetings.  Which ones seem interesting to you?  What other mathematical topics would you rather talk about?

TOPICS:
  Euclid's algorithm and modular arithmetic.
  Pigeonhole proofs.
  Tiling checkerboards.
  The 4-numbers game.
  The projective plane.
  Unrolling a torus.
  Classifying closed surfaces.
  Geometry of complex numbers.
  Reflections.
  Barycentric coordinates.
  Rotations in space.
  Euler characteristic.
  Finite geometry.
  Geometry of numbers.
  Pick's Theorem.
  Binomial coefficients (mod p).
  Combinatorial games.
  Dealing with determinants.

Anyone interested can contact me at ayang@cscc.edu

Alan Yang

#H34N1,

You asked 'What are the instructions/details about this?'.

I like to know 'this = ?'.

Or, shoot me an e-mail at ayang@cscc.edu.  I'll answer your questions the best to my knowledge.

Columbus Math Circle

This Math Circle is intended for high school and middle school students who enjoy thinking about mathematical ideas, problems, and puzzles.

The next meeting is planned:
        Sunday, April 3,  2:00 - 4:00 PM,  room 724 in the Math Tower.

I expect to send a reminder message when the date gets closer.

At this session we will investigate some properties of integers.

--------------------------------------------------------------------------------
The Math Tower (231 W. 18th Ave) is a building on the OSU campus.  Driving directions are posted at
    http://www.math.ohio-state.edu/ross/directions .

(If the building is locked, or you can't find the room please feel free to call my cell phone at 614-284-7954.)

We hope to get students involved in investigating mathematical questions, talking about their ideas, coming up with guesses and suggestions, and arriving at answers and explanations.

We will also discuss plans for future meetings of our Circle.
Teachers and parents are welcome to join in.

Please feel free to invite other middle school and high school students who would enjoy this session.

If you might join us that Sunday, please respond by email.  (But you are welcome to join us even without a response.)

Sincerely,
   Professor D. Shapiro   shapiro@math.ohio-state.edu
   Professor A. Yang   ayang@cscc.edu

From: Shapiro, Daniel
Sent: Wednesday, May 25, 2011 2:56 PM
To:  . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Subject: RE: Reminder: Math Circle, May 15

Columbus Math Circle

This Math Circle is intended for high school and middle school students who enjoy thinking about mathematical ideas, problems, and puzzles.

The next meeting is planned for the first Sunday in June:
        Sunday, June 5,  2:00 - 4:00 PM,  room 724 in the Math Tower.

Bart Snapp will lead a discussion of numbers and bases.  Other than our familiar "base ten" numbering system, are there other ways to represent numbers with a "place value system"?  Sticking to base ten, we can ask whether the standard list {0, 1, 2, 3, 4, 5, 6, . . . , 8, 9} is the only reasonable set of "digits" to use.

This will probably be our last meeting of this academic year.  I will ask you what went well with this year's Math Circle, what could be improved, and what schedule we should arrange for 2011-2012.

--------------------------------------------------------------------------------
The Math Tower is a building on the OSU campus (at 231 W. 18th Ave).  Driving directions are posted at
    http://www.math.ohio-state.edu/ross/directions.

(If the building is locked, or you can't find the room, please feel free to call my cell phone at 614-284-7954.)

We hope to get students involved in investigating mathematical questions, talking about their ideas, coming up with guesses and suggestions, and arriving at answers and explanations.

Teachers and parents are welcome to join in.
Please feel free to invite other middle school and high school students who would enjoy this session.

If you might join us that Sunday, please respond by email.  (But you are welcome to join us even without a response.)
If you would like to volunteer to bring a snack, please let me know.

Sincerely,
   Professor D. Shapiro   shapiro@math.ohio-state.edu
   Professor A. Yang   ayang@cscc.edu

Do you have handout for the meeting of Columbus Math Circle?

Columbus Math Circle

This Math Circle is intended for high school and middle school students who enjoy thinking about mathematical ideas, problems, and puzzles.

OSU classes will start next week, so things are in place for another year of Math Circle activities.  I hope we can meet every two or three weeks on Sunday afternoons during this academic year.  The first meeting for the 2011-2012 season will be:
        Sunday, October 2,  2:00 - 4:00 PM,  room 724 in the Math Tower.

Topic:  Mass Point Geometry,  led by visiting professor Max Warshauer.

--------------------------------------------------------------------------------
The Math Tower is a building on the OSU campus (at 231 W. 18th Ave).  Driving directions are posted at
    http://www.math.ohio-state.edu/ross/directions.

(If the building is locked, or you can't find the room, please feel free to call my cell phone at 614-284-7954.)

We hope to get students involved in investigating mathematical questions, talking about their ideas, coming up with guesses and suggestions, and arriving at answers and explanations.

Teachers and parents are welcome to join in.
Please feel free to invite other high school and middle school students who would enjoy this session.

If you might join us that Sunday, please respond by email.  (But you are welcome to join us even without a response.)
If you would like to volunteer to bring a snack, please let me know.

Sincerely,
   Professor D. Shapiro   shapiro@math.ohio-state.edu

   Professor Alan Yang ayang@cscc.edu

I went to the math circle yesterday, and there weren't too many people there... :o

Other comments I have:

- hard to work on math problems while sitting on couches and chairs, maybe install tables or desks?
- recruit more high schoolers!

Does any one else plan on going in two or three weeks? How difficult will the topics be in comparison to yesterday's? Thanks for your help.

Dynamite 127,

I've forwarded your comments to Dr. Shapiro.

If you like, I can send you ARML contest related problems for you to work on, or Math articles for you to read.  Just in case, my e-mail address is ayang@cscc.edu
Send me an e-mail.

Alan

Columbus Math Circle

This Math Circle is intended for high school and middle school students who enjoy thinking about mathematical ideas, problems, and puzzles.

The next meeting for the 2011-2012 season will be:
        Sunday, October 23,  2:00 - 4:00 PM,  room 154 in the Math Tower (first floor).

Topic:  Vectors, and their uses in geometry.
Students should have some familiarity with analytic geometry (graphs) and Euclidean geometry (similar triangles).  We might mention sines and cosines.   (We won't discuss any physics).
Location:  Let's try a first-floor classroom rather than a lounge.  This room (154) has about 15 seats.

We hope to get students involved in investigating mathematical questions, talking about their ideas, coming up with guesses and suggestions, and arriving at answers and explanations.

Please feel free to invite other students who would enjoy this session.

--------------------------------------------------------------------------------
The Math Tower is a building on the OSU campus (at 231 W. 18th Ave) in Columbus.  Driving directions are posted at
       http://www.math.ohio-state.edu/ross/directions.
(If the building is locked, or you can't find the room, please feel free to call my cell phone at 614-284-7954.)

If you might join us that Sunday, please respond by email.  (But you are welcome to join us even without a response.)

Parking note:  We can provide attendees with a card to cover the fee for the Tuttle Parking garage.  (It doesn't work for the surface lots.)

Sincerely,
   Professor D. Shapiro   shapiro@math.ohio-state.edu

Columbus Math Circle

This Math Circle is intended for high school and middle school students who enjoy thinking about mathematical ideas, problems, and puzzles.

The next meeting will be:
        Sunday, November 20, 2011,   2:00 - 4:00 PM,   at the OSU Math Department.

Bart Snapp will lead a discussion of:  Numbers and Bases.
In addition to base ten notation as learned in school, some people use "base b" notation, for a positive integer  b.  For instance, binary, octal, and "hex" are notations with bases 2, 8, and 16.
Those base  b  calculations are interesting, but there are other ways to represent numbers with less familiar bases . . .

* Location:  Let's try a different classroom this time.  Signs will be posted at the entrance to the Math Tower.

  (Note: I will bring an extra pair of glasses this time.)

Please feel free to invite other students who would enjoy this session.

--------------------------------------------------------------------------------
The Math Tower is a building on the OSU campus (at 231 W. 18th Ave) in Columbus.  Driving directions are posted at
       http://www.math.ohio-state.edu/ross/directions.
(If the building is locked, or you can't find the room, please feel free to call my cell phone at 614-284-7954.)

If you might join us that Sunday, please respond by email.  (But you are welcome to join us even without a response.)

Parking note:  We can provide attendees with a card to cover the fee for the Tuttle Parking garage.  (It doesn't work for the surface lots.)

Sincerely,
   Professor D. Shapiro   shapiro@math.ohio-state.edu

Columbus Math Circle

This Math Circle is intended for high school and middle school students who enjoy thinking about mathematical ideas, problems, and puzzles.

The next meeting will be:
        Sunday, January 15, 2012,   2:00 - 4:00 PM,   in Cockins Hall, room 232.

Tentative topic:  Combinatorial Games.

Location:  Signs will be posted at the entrance to the Math Tower.  (That is the door we know how to unlock.)

Please feel free to invite other students who would enjoy this session.
Our primitive web page is:   http://www.math.ohio-state.edu/ross/MathCircle

--------------------------------------------------------------------------------
The Math Tower is a building on the OSU campus (at 231 W. 18th Ave) in Columbus.  Driving directions are posted at
       http://www.math.ohio-state.edu/ross/directions.
(If the building is locked, or you can't find the room, please feel free to call my cell phone at 614-284-7954.)

If you might join us that Sunday, please respond by email.  (But you are welcome to join us even without a response.)

Parking note:  We can usually provide attendees with a pass that covers the fee for the Tuttle Parking garage.  (It doesn't work for the surface lots.)

Sincerely,
   Professor D. Shapiro   shapiro@math.ohio-state.edu

Columbus Math Circle

This Math Circle is intended for high school and middle school students who enjoy thinking about mathematical ideas, problems, and puzzles.

The next meeting will be:
        Sunday, February 5, 2012,   2:00 - 4:00 PM,   in Cockins Hall, room 232.

Tentative topic:  Polyhedra and Symmetry.

Location:  Signs will be posted at the entrance to the Math Tower.  (That is the door we know how to unlock.)

Please feel free to invite other students who would enjoy this session.
Our primitive web page is:   http://www.math.ohio-state.edu/ross/MathCircle

--------------------------------------------------------------------------------
The Math Tower is a building on the OSU campus (at 231 W. 18th Ave) in Columbus.  Driving directions are posted at
       http://www.math.ohio-state.edu/ross/directions.
(If the building is locked, or you can't find the room, please feel free to call my cell phone at 614-284-7954.)

If you might join us that Sunday, please respond by email.  (But you are welcome to join us even without a response.)

Parking note:  We can usually provide attendees with a pass that covers the fee for the Tuttle Parking garage.  (It doesn't work for the surface lots.)

Sincerely,
   Professor D. Shapiro   shapiro@math.ohio-state.edu

Columbus Math Circle

This Math Circle is intended for high school and middle school students who enjoy thinking about mathematical ideas, problems, and puzzles.

The next meeting will be:
        Sunday, February 26, 2012,   2:00 - 4:00 PM,   in Cockins Hall, room 232.

Dr. Chris Altomare will meet with the group.  Topic to be announced later.
Location:  Signs will be posted at the entrance to the Math Tower.  (That is the door we know how to unlock.)

Please feel free to invite other students who would enjoy this session.
Our primitive web page is:   http://www.math.ohio-state.edu/ross/MathCircle

--------------------------------------------------------------------------------
The Math Tower is a building on the OSU campus (at 231 W. 18th Ave) in Columbus.  Driving directions are posted at
       http://www.math.ohio-state.edu/ross/directions.
(If the buildings are locked, or you can't find the room, please feel free to call my cell phone at 614-284-7954.)

NOTE:  An OSU basketball game will be at 4:00 that day, so traffic might be an issue.

If you might join us that Sunday, please respond by email.  (But you are welcome to join us even without a response.)

Parking note:  We can usually provide attendees with a pass that covers the fee for the Tuttle Parking garage.  (It doesn't work for the surface lots.)

Sincerely,
   Professor D. Shapiro   shapiro@math.ohio-state.edu

Columbus Math Circle

This Math Circle is intended for high school and middle school students who enjoy thinking about mathematical ideas, problems, and puzzles.

The next meeting will be:
        Sunday, March 11, 2012,   2:00 - 4:00 PM,   in Cockins Hall, room 232.

Topic: Roots of unity.
Complex numbers can be represented as points in the plane, and there are geometric ways to interpret addition and multiplication.  A complex "root of unity" is a number z having some power equal to 1. That is: zn = 1, for some positive integer n.  For instance, solutions when  n = 4  are:  1,  i,  -1,  and  -i.  We will discuss some properties and applications of the roots of unity.  (Exposure to ideas of sines and cosines will be useful for understanding some parts of this discussion.)

Location:  232 Cockins Hall can be reached through the Math Tower.  Signs will be posted at the Math Tower entrance.  (That is the door we know how to unlock.)

Note:  A math conference will be happening in a nearby lecture room, and their people may be milling around the hallways.  But our room in Cockins Hall will be available for us to use.

Please feel free to invite other students who would enjoy this session.
Our primitive web page is:   http://www.math.ohio-state.edu/ross/MathCircle

--------------------------------------------------------------------------------
The Math Tower is a building on the OSU campus (at 231 W. 18th Ave) in Columbus.  Driving directions are posted at
       http://www.math.ohio-state.edu/ross/directions.
(If the buildings are locked, or you can't find the room, please feel free to call my cell phone at 614-284-7954.)

If you might join us that Sunday, please respond by email.  (But you are welcome to join us even without a response.)

NOTES:
Parking:  We can usually provide attendees with a pass that covers the fee for the Tuttle Parking garage.  (It doesn't work for the surface lots.)

Sincerely,
   Professor D. Shapiro   shapiro@math.ohio-state.edu

Columbus Math Circle

This Math Circle is intended for high school and middle school students who enjoy thinking about mathematical ideas, problems, and puzzles.

The next meeting will be in ten days:
        Sunday, March 25, 2012,   2:00 - 4:00 PM,   in Cockins Hall, room 232.

The session will be led by Professor Matthew Kahle.
Topic: Extremal Geometry.
We'll begin with some problems involving length, and then move on to some involving area.
  • Warmup problem #1:  Given ten points in a unit square (side length one), show that some pair of points is no more than distance 1/2 apart.
  • Warmup problem #2:  Given nine points in a triangle of area one, show that some three of the points must form a triangle of area 1/4 or less.
     If you solve that -- can you do better than nine?  What is the smallest number of points for which this is true?

Location:  232 Cockins Hall can be reached through the Math Tower.  Signs will be posted at the Math Tower entrance.  (That is the door we know how to unlock.)

Note:  A math conference will be happening in a nearby lecture room, and their people may be milling around the hallways.  But our room in Cockins Hall will be available for us to use.

Please feel free to invite other students who would enjoy this session.
Our primitive web page is:   http://www.math.ohio-state.edu/ross/MathCircle

--------------------------------------------------------------------------------
The Math Tower is a building on the OSU campus (at 231 W. 18th Ave) in Columbus.  Driving directions are posted at
       http://www.math.ohio-state.edu/ross/directions.
(If the buildings are locked, or you can't find the room, please feel free to call my cell phone at 614-284-7954.)

If you might join us that Sunday, please respond by email.  (But you are welcome to join us even without a response.)

NOTES:
Parking:  We can usually provide attendees with a pass that covers the fee for the Tuttle Parking garage.  (It doesn't work for the surface lots.)

Sincerely,
   Professor D. Shapiro   shapiro@math.ohio-state.edu

Alan

Columbus Math Circle

This Math Circle is intended for high school and middle school students who enjoy thinking about mathematical ideas, problems, and puzzles.

The next meeting will be:
        Sunday, April 15, 2012,   2:00 - 4:00 PM,   in Cockins Hall, room 232.

The session will be led by Bart Snapp.
Topic: Paper Folding Constructions.

--------------------------------------------------------------------------------
In high school geometry, students learn to construct different objects using two "Euclidean tools":  a straight-edge and a compass. For instance,
  • Given three points, can you construct the circle that passes through them?
  • Given a circle and a point outside that circle, can you construct a line passing
     through the point and tangent to the circle?
Meta-question:  Why are such constructions important enough to put into geometry courses?

Today we will use "paper folding" instead of those other tools. You are invited to bring tracing paper, a ruler, and a marker, to help work out some of those new constructions.

--------------------------------------------------------------------------------

Location:  232 Cockins Hall can be reached through the Math Tower.  Signs will be posted at the Math Tower entrance.  (That is the door we know how to unlock.)

Please feel free to invite other students who would enjoy this session.
Our primitive web page is:   http://www.math.ohio-state.edu/ross/MathCircle

--------------------------------------------------------------------------------
The Math Tower is a building on the OSU campus (at 231 W. 18th Ave) in Columbus.  Driving directions are posted at
       http://www.math.ohio-state.edu/ross/directions.
(If the buildings are locked, or you can't find the room, please feel free to call my cell phone at 614-284-7954.)

If you might join us that Sunday, please respond by email.  (But you are welcome to join us even without a response.)

NOTES:
Parking:  We can usually provide attendees with a pass that covers the fee for the Tuttle Parking garage.  (It doesn't work for the surface lots.)

Sincerely,
   Professor D. Shapiro   shapiro@math.ohio-state.edu

It's sad that there's not more people who attend...these math circles are actually really cool.


Dan Brown is beast.

http://www.youtube.com/watch?v=xMjVKuvJw0U&feature=related


invariant
combinatorics unsolved
combinatorics

dans une ile il ya 34 camelions
$17$ de coulours rouge
$10$ de coulours bleu
$7$ de coulours jaune
à la roncontre de deux de different coulrs ils prnd tous les deux la troixieme
à la roncontre de deux de meme coulrs il ne se pass rien
durant une anne les camelions ont la meme couleur là-quelle?
(il faut montre que le cas des autres couleurs est impossible)
et merci pour qui va m'aider à traduire cet exercice

j'espere que tu peut comprener mon francaise. :)

le premier observation est que si on ne se pas que se passe quand 2 cameleons de la meme coleur se rencontre, le problem est insolvable. C'est possible que tous le cameleons d'une coleur se recontre avec cameleons de le meme coleur seulment et ils devient tous le deux rouge par example. Et an final tu as tous les cameleons rouge. Ou le meme c'est possible avel la coleur blue et avec jaune aussi. La conclusion est que tu ne peut dire quel couleur ils se transform an final(ou je n'ai comprende pas le problem correctement).

Si tu suppose que si deux cameleons de la meme coleur quand ils se rencontre ils ne change pas la coleur le problem se resolve comme ci:
Le nombre total de cameleons est 34 qui est de form 3k+1 (ici, k=11). Quand 2 cameleons des couleures distinctes se rencontre ils change leur couleur en la troissieme, la difference modulo 3 entre le nombre de cameleons de chaque couleur est la meme. Par example, le difference initiale entre le nombre de bleu cameleons et les autres couleurs est 17-10=7 et 17-7=10.  7 et 10 sont de la form 3k+1 aussi.
Si, un bleu par example rencontre un rouge, ils se transform en jaune et maintenant le nombre de cameleons blue e 16, de cameleons rouge est 9 est de cameleons jaune est 9 aussi, mais le difference entre le nombre de bleu cameleons et les autres est 16-9=7 et 16-9=7 et les dexu nombres 7 et 7 sont aussi de la form 3k+1 (comment le 10 et le 7 dans le difference initial). Est generallment tous le temp la difference modulo 3 est la meme pars que quand 2 cameleons de la meme couleur se rencontre, les nombres des cameleons de leur couleur descendre avec 1 et le nombre des cameleons de la troissieme couleur ascendre avec 2, et il y a un total gagnement pour la troissieme couleur en comparison avec les autres de 3 cameleons.(la troissieme couluer gagne 2, et les autres perde 1)

Et maintenant le difference initiale est:
blue-rouge=7=3k+1
blue-jaune=10=3k+1
rouge-bleu=-7=3k+2
rouge-jaune=3=3k
jaune-bleu=-10=3k+2
jaune-rouge=-3=3k.
Ces differences il sont le meme(modulo 3) en final quand tu as 34 cameleons de un couleur. Mais si tu as 34 cameleons de la meme couleur, le difference de nombre des cameleons de cette couleur est les autres couleurs est 34 qui est 3k+1. Et le seul couleur qui initialment est different de les autres par nombres de la form 3k+1 est le blue,qui a les differences 10 et 7.
=> bleu e la couleur comune.

bellahsayn:

Can you translate your post into English?

Of course i can :).
So, i was saying that if you cannot say what will happend if two cameleons of the same color meet then the problem is impossible to solve. That's because you can have absolutely any combination possible of colors.
And now we'll suppose that when two cameleons of the same color meet they remain the same color they were before.
Now it's very simeple. The only way cameleons can change colors is if they meet another cameleon of not the sam color as they are. So, let's suppose two cameleons of different colors meet. They will change their color into the third colo. So, the number of cameleons of the third color will increase by two while the number of cameleons of the other 2 colors will decrease by 1. So, totally, the number of cameleons of the third color will increase by 3 comparing to the number of camleons of the other 2 colors. Because this always happends(i mean every time two cameleons of different colors meet, the third color will "gain" 3 comparing to the other two colors), the difference between the number of cameleons of different colors will always remain the same modulo 3.
Now initially we have the differences:

blue-red=7=3k+1
blue-yellow=10=3k+1
red-blue=-7=3k+2
red-yellow=3=3k
yellow-blue=-10=3k+2
yellow-red=-3=3k.
Since in the end we must have 34 cameleons of the same color, it means that in the end the difference between the number of cameleons of that color and the number of cameleons of the other two colors will be 34-0=34=3k+1. So, the only color we could have in the end is the one which initially also (cause as we proved before the differences modulo 3 remain the same) has the differences with the other two colors of the form 3k+1,  and the only color satisfying this is blue which has the differences 7 and 10 both of the form 3k+1.
Therefor the final color is blue.

tank you fdump
let say an other probleme in the invariant quantitises
first we have the 3-uple $(a,b,c)$
the operation  is (trasmettre) this coupl to the 3-uple  to ($\frac{a-b}{\sqrt{2}} $,$\frac{b-c}{\sqrt{2}}$,$\frac{c-a}{\sqrt{2}}$)
begin at $(1,2,0.5)$ we can find ($\sqrt{2}$,1,$\sqrt{2}$-1)
and sorry for  :? zhaoli

doing only that transformation to the tuple
(1,2,0.5) you will only get tuples for which the sum of the three numbers is 0, as it can easily be seen from the transformation. And since the tuple you are trying to obtain has the sum of its numbers 2sqrt(2) and not 0 the answer is no. (i guess on the third position you should have written -sqrt(2)-1; probably it's a typpo. )
anyway, if it's really -sqrt(2) on the last position i will think tomorrow how to solve it...it's late now in romania, so i will go to bed now.)

I can not read and understand the original problem, would you please to translate it into English, bellahsayn? Thanks a lot!


modular arithmetic
number theory
relatively prime
number theory theorems

Take a look!

[url]http://planetmath.org/encyclopedia/ThuesLemma2.html[/url]

About the proof,
[url]http://planetmath.org/encyclopedia/ProofOfThuesLemma.html[/url]

[b]Lemma[/b] (Thue): Let  $p$ be a prime and $a$ an integer relatively prime to $p$. Then the congruence

$ax \equiv y \pmod{p}$

admits a solution $(x,y) = (x_{0},y_{0})$, where

$0 < |x_{0}| < \sqrt{p}$ and $0 < |y_{0}| < \sqrt{p}.$

A more general version of this lemma is Thue's theorem, see [url]http://mathworld.wolfram.com/ThuesTheorem.html[/url]


A train left a station at 12 noon going north at a rate of 50 mph. At 1 pm another train left the same station going 60mph . At what time were the trains 270 miles apart?

By the time the second train started going, the first train traveled $ \frac {50\text{m}}{\text{h}}\cdot1\text{h} \equal{} 50\text{m}$. We can see that the only way for the two trains to be $ 270\text{m}$ apart is for the second train to be $ 270\text{m}$ miles ahead of the first train. Since right now, the second train is $ \minus{} 50\text{m}$ miles ahead of the first train, the second train needs to go a distance of $ 270\text{m} \minus{} ( \minus{} 50\text{m}) \equal{} 330\text{m}$ relative to the first train. Since the second train is moving at $ \frac{60\text{m}}{\text{h}} \minus{} \frac {50\text{m}}{\text{h}} \equal{} \frac {10\text{m}}{\text{h}}$ relative to the first train, this will take $ \frac {330\text{m}}{\frac {10\text{m}}{\text{h}}} \equal{} 33\text{h}$ (or $ 1$ day and $ 9$ hours) after the second train departed, or $ 10$ PM.

[quote="math154"]By the time the second train started going, the first train traveled $ \frac {50\text{m}}{\text{h}}\cdot1\text{h} \equal{} 50\text{m}$. We can see that the only way for the two trains to be $ 270\text{m}$ apart is for the second train to be $ 270\text{m}$ miles ahead of the first train. Since right now, the second train is $ \minus{} 50\text{m}$ miles ahead of the first train, the second train needs to go a distance of $ 270\text{m} \minus{} ( \minus{} 50\text{m}) \equal{} 330\text{m}$ relative to the first train. Since the second train is moving at $ \frac {60\text{m}}{\text{h}} \minus{} \frac {50\text{m}}{\text{h}} \equal{} \frac {10\text{m}}{\text{h}}$ relative to the first train, this will take $ \frac {330\text{m}}{\frac {10\text{m}}{\text{h}}} \equal{} 33\text{h}$ (or $ 1$ day and $ 9$ hours) after the second train departed, or $ 10$ PM.[/quote]

I think he wants an algebraic solution.   :)

Yes i need an algrebaic solution with equations and so fourth  :blush:

or a simpler solution..

Um...
The algebra is boring, but whatever. I won't bother with units this time. :wink:
We will use the fact that $ d=rt$ over and over again.
Let $ (s_1,s_2)=(50,60)$ represent the speeds of the first and second trains, respectively.
Then we know that by the time the second train departed, the first train traveled $ s_1\cdot1=50$.
Then let the time that it takes, after the second train departs, for the two trains to bet $ 270$ miles apart be $ T$. In this time $ T$, the second train will have traveled $ s_2T$ miles, and the first train will have traveled $ s_1(T+1)$ miles.
We can see that this corresponds to the equation $ s_2T-s_1(T+1)=270\implies60T-50T-50=270\implies10T=330\implies T=33$. The rest is the same as above.

[quote="math154"]Um...
The algebra is boring, but whatever. I won't bother with units this time. :wink:
We will use the fact that $ d = rt$ over and over again.
Let $ (s_1,s_2) = (50,60)$ represent the speeds of the first and second trains, respectively.
Then we know that by the time the second train departed, the first train traveled $ s_1\cdot1 = 50$.
Then let the time that it takes, after the second train departs, for the two trains to bet $ 270$ miles apart be $ T$. In this time $ T$, the second train will have traveled $ s_2T$ miles, and the first train will have traveled $ s_1(T + 1)$ miles.
We can see that this corresponds to the equation $ s_2T - s_1(T + 1) = 270\implies60T - 50T - 50 = 270\implies10T = 330\implies T = 33$. The rest is the same as above.[/quote]

i still dont get it  :(

Where do you not get it? I can't exactly help you if I don't know. :lol:

WAIT ONE MORE QUESTION.

At noon, two cards started from the same town and traveled the same direction on the same road. Car a averaged 36mph and car b averaged 31 mph. At what time were the cards 30 miles apart?

Um...why is this so urgent? Have you learned how to do these types of problems before?

As for your question, the answer is $ \frac{30}{36\minus{}31}\equal{}6$, but that doesn't matter if you don't understand these concepts in the first place.

I learned this today in math but my teacher is horrible and only covers concepts she thinks are easy to grasp. SO I havnt done any of these harder problems...

Also for this.

A cargo plane leaves an airport at 180mph. Two hrs later a jet leaves the same airport ant 900mph. How long will it take for the jet to catch up to the cargo?

my answer is 2.5 hrs. is that right?

180x= 900(x-2)

...
x= 2.5

also can you explan the one before this. the one where x= 6

Actually, you have the right idea for these types of problems. However, the cargo plane/jet bla one is $ 2.5\minus{}2\equal{}0.5$, since you want the time AFTER the jet starts. You DO have the right equations, though.

However, say the plane question was:
A cargo plane leaves an airport at 180mph. Two hrs later a jet leaves the same airport ant 900mph. How long will it take for the jet to be 360 miles ahead of the cargo?

Instead of $ 180x\equal{}900(x\minus{}2)\implies180x\minus{}900(x\minus{}2)\equal{}0$, take a guess at what you would have. :)

[quote="math154"]Actually, you have the right idea for these types of problems. However, the cargo plane/jet bla one is $ 2.5 \minus{} 2 \equal{} 0.5$, since you want the time AFTER the jet starts. You DO have the right equations, though.

However, say the plane question was:
A cargo plane leaves an airport at 180mph. Two hrs later a jet leaves the same airport ant 900mph. How long will it take for the jet to be 360 miles ahead of the cargo?

Instead of $ 180x \equal{} 900(x \minus{} 2)\implies180x \minus{} 900(x \minus{} 2) \equal{} 0$, take a guess at what you would have. :)[/quote]

wait the question says when will they catch up or meet

Well, the question is sort of ambiguous. In my experience, it usually would mean what I thought it did. Anyway, I'm asking what IF the question were:
A cargo plane leaves an airport at 180mph. Two hrs later a jet leaves the same airport ant 900mph. How long will it take for the jet to be 360 miles ahead of the cargo?

Umm, Math154, just a simple error, but on the train problem, you said that 270-(-50)=330, however 270-(-50)=320, so the answer to the train problem should be 9pm the next day.


How to to find, with proof, the value of (tan 20)*(tan 40)*(tan 80) ?

[u]Solution.[/u] Using the sum and difference formulae for tangent, we obtain
\begin{eqnarray*}\tan 20^{\circ}\tan 40^{\circ} \tan 80^{\circ} &=& \tan 20^{\circ}\tan (60-20)^{\circ} \tan (60+20)^{\circ}\\ &=& \tan 20^{\circ}\frac{3-\tan^2 20^{\circ}}{1-3\tan^220^{\circ}}\\&=&\tan 60^{\circ}=\sqrt{3},\end{eqnarray*}
where the triple angle formula for tangent was used (easily derivable using the sum formula for tangent). $ \Box$

[u]Remark.[/u] It can easily be proven that $ \cos 20^{\circ}\cos 40^{\circ}\cos 80^{\circ}=\frac{1}{8}$.  From this, one obtains that $ \sin 20^{\circ}\sin 40^{\circ}\sin 80^{\circ}=\frac{\sqrt{3}}{8}$.

General equality: (with a similarly ways of prove)
$ tanx.tan(60^0 \minus{} x).tan(60^0 \plus{} x) \equal{} tan3x$
$ 4sinx.sin(60^0 \minus{} x).sin(60^0 \plus{} x) \equal{} sin3x$
$ 4cosx.cos(60^0 \minus{} x).cos(60^0 \plus{} x) \equal{} cos3x$


Well a easy looking problem it is ,isnt it?
Let $i \in \mathcal N,1 \le i \le 5$,$a_i \in \mathcal R$

1.$\sum_{i=1}^{5} a_i=8$

2.$\sum_{i=1}^{5} a_i^2=16$

Find the maximum value of $a_5$?(or anyone you want :) )

hey guys , what an action.
not even you solved it , even a general problem.
nice ,thanks. :D


what exactly is AP? and do you mean an ordered pair or all 3 numbers?
and you should LaTeX to clarify your expressions

EDIT: *will not make new pointless post*
thx rofler, i just usually dont see those abbreviations.
but when your [u]typing[/u] up a post it shouldnt matter much

so we know that
$ a(\frac1b \plus{} \frac1c),b(\frac1a \plus{} \frac1c),c(\frac1a \plus{} \frac1b)$

are in arithmetic progression. We add 1 to each term which preserves their arithmetic progression

$ 1 \plus{} a(\frac1b \plus{} \frac1c),1 \plus{} b(\frac1a \plus{} \frac1c),1 \plus{} c(\frac1a \plus{} \frac1b)$

$ a(\frac1a) \plus{} a(\frac1b \plus{} \frac1c),b(\frac1b) \plus{} b(\frac1a \plus{} \frac1c),c(\frac1c) \plus{} c(\frac1a \plus{} \frac1b)$

$ a(\frac1a \plus{} \frac1b \plus{} \frac1c),b(\frac1a \plus{} \frac1b \plus{} \frac1c),c(\frac1a \plus{} \frac1b \plus{} \frac1c)$

dividing by $ \frac1a \plus{} \frac1b \plus{} \frac1c$ will be valid since this is equal to zero iff a=0, b=0, or c=0.
after the division, which preserves the arithmetic progression, we get the desired result.

$ a,b,c$

I think 2 is...

2. Prove that if $ \log_y x, \log_y z, \minus{}15\log_z x$ are in arithmetic progression, then $ z^3\equal{}x$.

[quote="stevenmeow"]dividing by $ \frac1a \plus{} \frac1b \plus{} \frac1c$ will be valid since this is equal to zero iff a=0, b=0, or c=0.[/quote]It is never equal to zero... and $ abc\ne0$.

[quote="ayush_2008"]Q1 if a(1/b+1/c),b(1/c+1/a),c(1/a+1/b) are in AP prove that  a,b,c are in Arithmetic Progresion.

Q2 If log x (base y), logz (base y),-15log x (basez)   r in AP then prove that  z^3  =x.[/quote]
For both of these questions you have to use the relation of an A.P. i.e. a+c= 2b where a b c are consecutive terms of an A.P.

If $ (a,b,c)\equal{}(\minus{}2,3,6)$, then those fractions are equal to $ \minus{}1$ which is a valid arithmetic progression, but clearly  $ a,b,c$ are not in an arithmetic progression.