First of all i must say that english is not my native language. Now the problem:

A copper turntable with a ray R= 5cm is set perpendicular to the lines of a steady magnetic field  with induction B = 0.2T. A stream of I = 1A follows the path of the ray of the disc between the tow sliding contacts. The turntable whirls with a velocity of n = 3 turns/sec. Determine:
a) the turning moment that this disc needs.
b) the power of the motor.

What is your native language then?

Did you notice that electromagnetics is not a hot topic. Isn't that nice??.

Anyway the copper disc in a magnetic field . This sets up eddy currents in the disc depending on the conductivity of the disc. I look this up more tonite..........

Where are you ?? I don't talk to an empty audience
Again what is your native language?


Evaluate:
$\sum_{n=1}^{1995} tan(n)*tan(n+1)$

I think I've seen a similar problem somewhere on this site but couldn't find it in the search.

Are the tangents evaluated in degrees or radians?


Let $ n$ be a positive integer such that $ \tfrac{1}{2}\plus{}\tfrac{1}{3}\plus{}\tfrac{1}{7}\plus{}\tfrac{1}{n}$ is an integer. Which of the following statements is [b]not[/b] true?

$ \textbf{(A)}\ 2\text{ divides }n \qquad
\textbf{(B)}\ 3\text{ divides }n \qquad
\textbf{(C)}\ 6\text{ divides }n \qquad
\textbf{(D)}\ 7\text{ divides }n \\
\textbf{(E)}\ n>84$

$ \frac12 \plus{} \frac13 \plus{} \frac17 \plus{} \frac1n \equal{} \frac{41}{42} \plus{} \frac1n$. The value $ n\equal{}42$ kinda jumps out, giving answer $ \boxed E$.


Prove that the sum of the product of four consecutive odd integers and the number 16 results in a square number.


If x is 150% of y, what percent of 3x is 4y? Express your answer to the nearest
whole number.

[hide] Isn't it $\boxed{89}$, or the reciprocal of what you had? :? [/hide]

[hide]set x to be 15 and y to be 10...[/hide]

Umm... [hide] Set x=3, y=2. 3x=9, 4y=8. What percent of 9 is 8? Answer: $\frac{8}{9}$, which is $88.8888...$%. So, the answer is $\boxed{89}$. [/hide]

you are right. Need to edit because I misread it.

[hide]
I think the easiest way is plug in numbers. Let x be 30 and y be 20.
$\frac{80}{90}=.88888888888\approx\boxed{89}$[/hide]


inequalities
number theory proposed
number theory

Find all positive integers $ x,y,z$ satisying $ x^x\plus{}2y^y\equal{}3z^z$

One trivial solution is $ x\equal{}y\equal{}z>0$. Suppose now that $ x\neq y$ (it is quite clear that if $ x\equal{}y$, then necessarily $ x\equal{}y\equal{}z$):

a) $ x>y$. This implies that $ x^x>z^z$, thus $ x\geq z\plus{}1$ and $ x^x\geq (z\plus{}1)^{z\plus{}1}$. Then $ 3z^z\equal{}x^x\plus{}2y^y\geq (z\plus{}1)^{z\plus{}1} \plus{}2y^y> z^zz\plus{}2y^y\,\,\to\,\, 3z^z>z^z z \plus{}2y^y$. This inequality holds only if $ z<3$. It is easy to see that there are no solutions.

b) $ x<y$. The approach is as for point a)

The only solution is $ x\equal{}y\equal{}z>0$  :roll:


Hi everybody

I want to ask whether bacteria(specifically lactobacillus)can be sterlized(just in the case of virus)  :maybe:

well skand i dont have an anser to your question but along wid u ill like to ask the ppl wats kranz type anatomy........???

pyruvic dikinase......???

.............
fossa ovalis.........fossa rotundus
.............???

hey can u plz tell me more about pyruvic dikanase?
thanx
 :)

check this out  :P  :D
  [url]http://en.wikipedia.org/wiki/Pyruvate,_phosphate_dikinase[/url] :P

rituraj.......i have not taken bio.........pcm

evrything went over my head.......... :oops:  :huh:

[quote="rituraj007"]kranz type anatomy........type of arrangement of cells in leaf wrt chloroplast in CAM plants for photosyn
(if u hav bio u'll understand better in due course.otherwise do tell me)

pyruvic dikinase......an enzyme involved in respiration(i don't remember the exact function)

fossa ovalis.........fossa rotundus--openings in the inner ear that help in transfer of sound vibrations

none of these r spot on precise(coz i m not a very good mugger--i remem only after seeing options :P  :P )

as long as i kno,bacillus can be sterilized but not  completely as there spores remain dormant and can be activated under fav circumstances.....[/quote]

I was going to answer exactly that but rituraj beat me to it!!

 :D  :D  :rotfl:

[quote="hell_ever"]

I was going to answer exactly that but rituraj beat me to it!!

[/quote]

You are a bio student?  :o

:P  :P  :P
really didn't knew that   .............


Can  anybody  give  me  a  hint  to this?

a,b,c>  prove  that

ab/c(c+a) +bc/a(a+b) + ca/b(b+c) >= a/(c+a) + b/(a+b) + c/(b+c)

It  was  taken  from  moldavian  mo , 1999 , thanks!!

Note a/b=x, b/c=y, c/a=z and obtain the equivalent form (z-1)/(x+1)+(x-1)/(y+1)+(y-1)/(z+1)>=0. Then, clear denominators and use the fact that sum x^2>=1/3(sum x)^2>=sumx (here we need xyz=1) and sum z^2y>=3xyz=3.

Harazi solution is much better than mine.
After algebraic calculus we have

a^3b^3+b^3c^3+c^3a^3+a^2b^4+b^2c^4+c^2a^4 \geq 3a^2b^2c^2+abc(a^2c+b^2a+c^2b)

We have by AM-GM a^2b^4+b^2c^4+c^2a^4 \geq 3a^2b^2c^2

Then put A=ab   B=bc   C=ca to obtain

A^3+B^3+C^3  \geq C^2A+A^2B+B^2C

to prove again by AM-GM.


Καλή και σύντομη
Μήπως μπορει κάποιος να βάλει τα θέματα .  Η site της ΕΜΕ είναι το κενό σύνολο..........

Το πρώτο θέμα .
Έστωσαν x,y ανήκουν (0,1) μεταβλητοί πραγματικοί αριθμοί . Αν $ a\equal{}x\plus{}ym$ και $ b\equal{}y\plus{}xm$

με α,β, m  ανήκουν Ζ+ .Αν  όλα τα ζεύγη ακεραίων που προκύπτουν καθώς τα x,y ποικίλουν είναι 119 να βρεθεί ο m

To δεύτερο
Να λυθεί στους ακεραίους η εξίσωση $ x^8\plus{}2^{2^x\plus{}2}\equal{}p$ όπου p πρώτος

Το τέταρτο

Αν $ a_1,...,a_n$ είναι θετικοί ακέραιοι  και $ k,t$ ο min kai o max tous ndo

$ (\frac{a_1^2\plus{}...\plus{}a_n^2}{a_1\plus{}...\plus{}a_n})^{\frac{tn}{k}}\geq a_1\cdot a_2\cdot...\cdot a_n$

geia se olous kai sygxaritiria se osous diakrithikan,
tha valw mia lisi(syntomi) mono gia tin proti askisi
loipon oi liseis tou systimatos a=x+ym, b=y+xm einai x= (a+b)/2(m+1)+(b-a)/2(m-1), y=(a+b)/2(m+1)+(a-b)/2(m-1) tora theloume na doume gia poious thetikous akeraious a,b(thewroume to m stathero) ta x,y anoikoun sto (0,1) eukola prokyptei oti oi to synolo ton zeygwn (a,b) gia ta opoia x,y anoikoun sto (0,1) einai ola ta zeygoi (a,b) otan to 1<=a<=m kai 1<=b<=m ektos apo ta zeygoi (1,m) kai (m,1) diladi synolika m^2-2=119 ara m=11.

$ k\le a_1 \le t ... k\le a_n \le t$ τότε $ nk \le nt$ άρα $ n \le \frac {nt}{k}$ άρα αρκεί να δείξουμε ότι $ \frac{{a_1}^2\plus{}...\plus{}{a_1}^2}{a_1\plus{}...\plus{}a_n}\ge \sqrt[n]{a_1...a_n}$ ή $ \frac{{a_1}^2\plus{}...\plus{}{a_1}^2}{a_1\plus{}...\plus{}a_n}\ge \frac{a_1\plus{}...\plus{}a_n}{n}$λόγω Cauchy που είναι η BCS για τους $ a_1,...,a_n$ και τους $ 1,...,1$
(Ισχύει $ \frac{a_1\plus{}...\plus{}a_n}{n} \ge 1$)

Κατ' αρχήν συγχαρητήρια σε όλους του επιτυχόντες και μη:)

Για το τρίτο απλά δύναμη του σημείου $ S$ ως προς τον περιγεγραμμένο του $ ABC$ και μετά $ \angle{BKC}\equal{}BHC\iff \angle{A}\equal{}60$. Σε αυτό το πρόβλημα το πιο δύσκολο κομμάτι ήταν το σχήμα!!! Εχασα πολύ χρόνο γαι να καταλάβω πώς είναι...

Για το 2ο αρκεί να ήξερε κάποιος ότι $ a^4\plus{}4b^4$ δεν είναι ποτέ πρώτος (παραγοντοποιείται :wink: )

Gia $ x\equal{}1$ omws to $ p$ bgainei 17 pou einai prwtos.

[quote="basted"]Gia $ x \equal{} 1$ omws to $ p$ bgainei 17 pou einai prwtos.[/quote]
εκτός και αν το $ a^2\plus{}2b^2\minus{}2ab\equal{}1$ ή $ (a\minus{}b)^2\plus{}b^2\equal{}1$ που σημαίνει ότι ...

Ευχαριστώ Nick για τη λύση... Just respect.

Πάντως η ατάκα ήταν το "μικρό νούμερο", ε;  :rotfl:

[i]PS: Θα παρακαλούσα τον Silouan να βάλει και τη λύση με τους μιγαδικούς που έκανε, έτσι, για να υπάρχει στο φόρουμ. Μερικές φορές η αναλυτική σώζει ...μετάλλια! :)[/i]

[quote="mostel"]Ευχαριστώ Nick για τη λύση... Just respect.

Πάντως η ατάκα ήταν το "μικρό νούμερο", ε;  :rotfl:
[/quote]

Kati den piano...h den thymamai...

][quote]
[i]PS: Θα παρακαλούσα τον Silouan να βάλει και τη λύση με τους μιγαδικούς που έκανε, έτσι, για να υπάρχει στο φόρουμ. Μερικές φορές η αναλυτική σώζει ...μετάλλια! :)[/i][/quote]

nai ki ego ithela na to zitiso auto apo to silouan, an kai pisteuo oti katalava pos akrivos skefthke.

[quote="Nick Rapanos"][quote="mostel"]Ευχαριστώ Nick για τη λύση... Just respect.

Πάντως η ατάκα ήταν το "μικρό νούμερο", ε;  :rotfl:
[/quote]

Kati den piano...h den thymamai...
[/quote]

Όταν σε ρώτησε ο Sil πόσο είναι το αποτέλεσμα και του είπες "μικρό" :p

[quote="Orfikoss"]

.....................................................

Το "=" ισχύει όταν $ a_1 \equal{} a_2 \equal{} ... \equal{} a_n$[/quote]

Περίεργο , αλλά χθες άκουσα ,  μετά την απονομή των βραβείων , ότι η επιτροπή διόρθωσης θεώρησε ότι η ισότητα ισχύει  αν :

 $ a_1\equal{}a_2\equal{}...\equal{}a_n \equal{} 1$

και  ότι έκοβαν μια μονάδα από την παράλειψη του  $ ...\equal{}1$.
 Βλέπω όμως αυτή τη στιγμή ότι η ισότητα των αριθμών αρκεί για το ''ίσον''.Δεν το ρίχνετε και σεις μια ματιά , γιατί είμαι ζαλισμένος ;

 Μπάμπης

Περίεργο μου φαίνεται  :huh:

Γιατί να υπάρχει τέτοιο restriction ? Το ότι λέει πως είναι ακεραίοι, προφανώς η ισότητα ισχύει για $ a_{1} \equal{} a_{2} \equal{} \ldots \equal{} a_{n} \equal{} k , k\in\mathbf{Z^{\plus{}}}$  :wacko:

[quote="stergiu"][quote="Orfikoss"]

.....................................................

Το "=" ισχύει όταν $ a_1 \equal{} a_2 \equal{} ... \equal{} a_n$[/quote]

Περίεργο , αλλά χθες άκουσα ,  μετά την απονομή των βραβείων , ότι η επιτροπή διόρθωσης θεώρησε ότι η ισότητα ισχύει  αν :

 $ a_1 \equal{} a_2 \equal{} ... \equal{} a_n \equal{} 1$

και  ότι έκοβαν μια μονάδα από την παράλειψη του  $ ... \equal{} 1$.
 Βλέπω όμως αυτή τη στιγμή ότι η ισότητα των αριθμών αρκεί για το ''ίσον''.Δεν το ρίχνετε και σεις μια ματιά , γιατί είμαι ζαλισμένος ;

 Μπάμπης[/quote]

Χρησιμοποιώ τις εξής ανισότητες:
-  B-C-S: $ (a_1^2 \plus{} a_2^2 \plus{} ... \plus{} a_n^2)(b_1^2 \plus{} b_2^2 \plus{} ... \plus{} b_n^2) >\equal{} (a_1b_1 \plus{} a_2b_2 \plus{} ... \plus{} a_nb_n)^2$. Το "=" ισχύει όταν $ \frac {a_1}{b_1} \equal{} \frac {a_2}{b_2} \equal{} ... \equal{} \frac {a_n}{b_n}$

- Cauchy ή ανισότητα των μέσων: $ \frac {a_1 \plus{} a_2 \plus{} ... \plus{} a_n}{n} >\equal{} \sqrt [n]{a_1\cdot a_2\cdot...\cdot a_n}$

- Ότι τό μέγιστο είναι μεγαλύτερο ή ίσο από το ελάχιστο, ή $ t >\equal{} k$
Τα παραπάνω συναληθεύουν για $ a_1 \equal{} a_2 \equal{} ... \equal{} a_n$

Στο κάτω-κάτω αν κάνεις τις πράξεις παίρνεις: $ k \equal{} t$ οπότε $ \frac {tn}{k} \equal{} n$
$ \left(\frac {a^2 \plus{} a^2 \plus{} ... \plus{} a^2}{a \plus{} a \plus{} ... \plus{} a}\right)^n \equal{} \left(\frac {n\cdot a^2}{n\cdot a}\right)^n \equal{} a^n \equal{} a\cdot a \cdot ... \cdot a$

Μήπως θεώρησαν οτι ισχυει μονο για n=1???? :huuh: Παντως χτες το ακουσα και εγω οτι ειπαν οτι ισχυει η ισοτητα μονο για κατι τετοιο.Αλλα δεν ισχυει σαφως

Ρε μα αυτο λέω οτι ισχυει για a1=a1=......=aN oxi gia n=1 (Υποθετικά μιλουσα )

[quote="meinstein"]κατά τη γνώμη μου δεν ισχύει μόνο για 1. Κ στο διαγωνισμό εγραψα $ a1 \equal{} a2 \equal{} ... \equal{} an$ Έλεγξέ το Μήτσο να δεις :D[/quote]

Και μένα μου φάνηκε παράξενο , γιατί να ειπωθεί τέτοιο πράγμα. Αλλά δυστυχώς είχα λύσει το πρωί της Κυριακής την άσκηση σε μια χαρτοπετσέτα στο σπίτι της ...πεθεράς στην Αθήνα και μόνο στην περίπτωση για ν= 2 ( δεν είχα χώρο) και δεν ασολήθηκα με την περιπτωση της ισότητας. Ξαφνιάστηκα λοιπόν όταν το άκουσα και δεν προέβαλα αντίρρηση - ήταν και ακατάλληλη η στιγμή.
 Πάντως κάτι έγινε και μάλλον τους ξέφυγε . Ελπίζω να μην χάθηκαν άσκοπα μονάδες και αλλοιώθηκε η βαθμολογία (έστω για λίγο), αν και νομίζω ότι τελικά με 1 μονάδα μάλλον κανένας δεν άλλαζε κλίμακα.Αυτό όμως είναι  δική μου εκτίμηση και εγγυάται τίποτα.Καλό θα είναι να μην έγινε τίποτα από όλα αυτά και όλα να πηγάζουν από ένα κακό παράδειγμα που δόθηκε για τον τρόπο βαθμολόγησης . Πάντως η γνώμη αυτή ακούστηκε πάνω στην κουβέντα ότι κάποια παιδιά περίμεναν περισσότερο και είδαν 2-3 μονάδες παρακάτω από αυτό που περίμεναν. Να σπεύσω όμως να προσθέσω ότι τα γραπτά που βραβεύτηκαν εκτιμήθηκαν από πολλά μέλη της επιτροπής - με διαβεβαίωσαν για αυτό-και θέλω να πιστεύω ότι ακόμα και να συνέβη αυτό που λέμε , μάλλον δεν άλλαξε τίποτα το ουσιαστικό.
   Άντε και:
 [color=red][b] Καλή Μεσογειάδα !!![/b][/color]

Με αφορμή τη συζήτηση γύρω απ το αν ισχύει για α1=α2=....=αν το απαντήσατε ηδη (προφανως και ισχύει)
Επειδή είναι αργά και μπορεί να κάνω λάθος απαντήστε μου ......

Σχόλιο 1.
θεμα 4 Μεγάλοι: Γιατί ο περιορισμός θετικοί ακέραιοι και οχι πραγματικοί μεγαλύτεροι η ισοι του 1;

Σχόλιο 2
θέμα 1 Μεγάλοι: (Εύστοχο σχόλιο διαγωνιζομένου) πως αυτός ο υπολογιστής έκανε καταγραφή απείρων ζευγών σε πεπερασμένο χρόνο

Σχολιο 3
Θεμα 2 Μικροί : Αφού χ^2 +ψ^2+ζ^2=3 (1) γιατί δίνει χ,ψ,ζ<2 μήπως  μπορούν να γίνουν >= του 2 Εξ άλλου εύκολα απ την (1) είναι χ<=3^(1/2) Αν το παρουμε οτι δινεται αυτή η άκομψη ανίσωση για ευκολία μάλον ξεχάσαμε οτι διαγωνίζονται με Μ.Ο.

Σχόλιο 4
Θεμα 3 Μικροί :Πανενωσιακές Τόμος 1 Σελ 38 ασκηση 161 Παλι μας αντιγράφουν οι πρωην Σοβιετικοί......

Για το 2, ο κύριος Τυρλής είπε πως το πρόβλημα δεν ανταποκρίνεται στην πραγματικότητα...

Επίσης, για το πρώτο νομίζω πως δε γίνεται σε εκθέτη να έχουμε π.χ. άρρητο. (Αν μέσα στους πραγματικούς π.χ. ο min ήταν ο $ \sqrt{2}$). Ίσως με ρητούς γινόταν κάτι, αλλά και πάλι δεν αλλάζει η άσκηση.

Σόρι αν κάνω και γω λάθος, αλλά είναι σχεδόν 3 τα χαράματα.........

Στέλιος

Stelio αν ο εκθέτης δεν έπερνε άρρητες τιμές τότε οι εκθετικές συναρτήσεις θα είχαν πρόβλημα παραγωγισιμότητας εξ άλλου ρίξε μια ματια το βιβλίο της Β Λυκείου να δείς πως οριζεται η δύναμη με άρρητο εκθέτη .Υποθέτω οτι ο κατασκευαστής της ανίσωσης είχε στο νού του την λύση της με την ανισότητα των βαρών για αυτό και το ακέραιος και η πατάτα (αν αληθεύει αυτό που γράφεται στα προηγούμενα toric) me to α1=α2=..=αν=1

xp σόρρυ αλλά δεν κατάλαβα το συσχετισμό του λάθους ($ a_1\equal{}a_2\equal{}..\equal{}a_n\equal{}1$) με την ανισότητα των βαρών. Δηλαδή γιατί κάποιος  που χρησιοποιεί αυτήν την ανισότητα να οδηγηθεί σε αυτό το λάθος (το ρωτώ γαιτί εγώ την έλυσα χρησιμοποιώντας την συγκεκριμένη ανισότητα αλλά είπα απλά ότι η ισότητα ισχύει όταν όλα είναι ίσα)

Bilot 4 to ''ακέραιος'' είναι για την ανισότητα και οχι για το 1

Σιλουανέ μπορείς να μου εξηγήσεις πώς μια τυχαία γεννήτρια αριθμών θα δώσει με την πρώτη τις σωστές λύσεις του προβλήματος; Νομίζω ότι πράγματι αυτό θέλανε. (Συγνώμη για τον εκνευρισμό μου αλλά είμαι πολύ καλύτερος στον υπολογιστή από ότι στα μαθηματικά και το λάθος είναι οφθαλμοφανές.)

Το πρόβλημα λέει ότι ο υπολογιστής δημιουργώντας τυχαίους αριθμούς κατέληξε μέσα σε 119 βήματα να προσδιορίσει όλους τους ακεραίους. Ποιος μας λέει ότι μέσα σε αυτά τα βήματα δεν υπάρχουν μη ακέραια ζευγάρια που προσδιορίστηκαν; Σίγουρα όχι η εκφώνηση γιατί τότε έπρεπε να πει ότι ο υπολογιστής χρειάζεται [u]τουλάχιστον[/u] 119 βήματα για να προσδιορίσει τα ζεύγη. :mad: Επομένως, αφού δεν το λέει, θα είναι:

$ m^2 \minus{} 2\le119 \Leftrightarrow m \le 11$

Επίσης για την τελευταία:
η ισότητα μευο 1 είναι ΣΥΛΛΑΘΗ. Αρκεί να βάλουμε στην εκφώνηση όλα τα α ίσα μεταξύ τους και προκύπτει ισότητα.

Συγχαρητήρια πάντως σε όσους τα καταφέρανε με αυτά τα αντιφατικά θέματα! (Για να είμαι δίκαιος, τα 2,3 ήταν ωραία.) Καλλή επιτυχία και στη Μεσογειάδα.

[quote="bilot4"]Κατ' αρχήν συγχαρητήρια σε όλους του επιτυχόντες και μη:)

Για το τρίτο απλά δύναμη του σημείου $ S$ ως προς τον περιγεγραμμένο του $ ABC$ και μετά $ \angle{BKC} \equal{} BHC\iff \angle{A} \equal{} 60$. Σε αυτό το πρόβλημα το πιο δύσκολο κομμάτι ήταν το σχήμα!!! Εχασα πολύ χρόνο γαι να καταλάβω πώς είναι...

[/quote]

Ρίξτε μια ματιά εδώ και πείτε μου τι λέει αυτή η λύση , γιατί δεν προλαβαίνω ούτε να την κοιτάξω  :(

http://www.mathlinks.ro/viewtopic.php?t=190777

 Μπάμπης

[quote="bilot4"]xp σόρρυ αλλά δεν κατάλαβα το συσχετισμό του λάθους ($ a_1 \equal{} a_2 \equal{} .. \equal{} a_n \equal{} 1$) με την ανισότητα των βαρών. Δηλαδή γιατί κάποιος  που χρησιοποιεί αυτήν την ανισότητα να οδηγηθεί σε αυτό το λάθος (το ρωτώ γαιτί εγώ την έλυσα χρησιμοποιώντας την συγκεκριμένη ανισότητα αλλά είπα απλά ότι η ισότητα ισχύει όταν όλα είναι ίσα)[/quote]
Μπορεις να ποσταρεις λιγο την λυση σου  με την ανισοτητα των βαρων???

Σόρρυ που δεν έχω χρόνο να το γράψω αναλυτικά... πάντως θεωρώ τα βάρη $ \frac {a_i}{\sum{a_i}}$ οπότε π.χ. $ \frac{a_1^2}{\sum{a_i}} \equal{} a_1\cdot \frac {a_1}{\sum{a_i}}$ και εφαμρμόζω την ανισότητα...

[quote="math_07"][quote="Nick Rapanos"]Υποσχέθηκα στον mostel το πρωί ότι θα δημοσιεύσω τη λύση μου στο forum...So here it is.[/quote]
Would anyone kindly translate this solution into English? :maybe:  I am really interested to know the solution. :huh:[/quote]

math_07, sorry for the late reply.

Question: H orthocenter of triangle ABC, K circumscribed circle of radius R=1. S intersection of HK and BC with (KS)(KH) = 1. Find the area of ABHCA

Solution (Nick Rapanos):  Let C_1,C_2 be the circumcircles of HSB,HSK. Observe (KS)(KH) = 1 = (KB)^2 = (KC)^2, so KB,KC are tangents to the circles C_1 and C_2. Since angle between chord and tangent is equal to the subtebded angle on the opposite side of the chord we have $ \angle HSB \equal{} \angle HBK$ and $ \angle HSB \equal{} \angle HCK$. It follows that $ \angle HSB \equal{} \angle HCK$ and so the quadrilateral BHKC can be inscribed in a circle. Therefore $ \pi \minus{} \hat A \equal{} \angle BHC \equal{} \angle BKC \equal{} 2 \hat A$, so $ \hat A \equal{} \pi/3$. It now follows easily that KM = R/2 = 1/2 and $ BC \equal{} \sqrt{3}$. The area can now be computed as in the last fomula in the attached pdf of Nick.


Let a,b,c>0 such that $ a^2\plus{}b^2\plus{}c^2\plus{}2abc\equal{}1$. Prove that
$ a^2\plus{}b^2\plus{}c^2\plus{}\frac{9}{2}(a\plus{}b\plus{}c)\leq \frac{15}{2}$

[quote="khanhkhtn"]Let a,b,c>0 such that $ a^2 \plus{} b^2 \plus{} c^2 \plus{} 2abc \equal{} 1$. Prove that
$ a^2 \plus{} b^2 \plus{} c^2 \plus{} \frac {9}{2}(a \plus{} b \plus{} c)\leq \frac {15}{2}$[/quote]

Let $ a\equal{}\sqrt {yz}, b\equal{}\sqrt{zx}, c\equal{}\sqrt{xy}$, We have $ xy\plus{}yz\plus{}zx\plus{}2xyz\equal{}1$,
So we have $ x\equal{}cosA, y\equal{}cosB, z\equal{}cosC(A\plus{}B\plus{}C\equal{}\pi)$.
and we know well-known this inequality:
$ cosA\plus{}cosB\plus{}cosC\le \frac{3}{2}$, and we want to show
$ cosAcosB\plus{}cosBcosC\plus{}cosCcosA$
$ \plus{}\frac{9(\sqrt{cosAcosB}\plus{}\sqrt{cosBcosC}\plus{}\sqrt{cosCcosA})}{2}\le\frac{15}{2}$
But
$ cosAcosB\plus{}cosBcosC\plus{}cosCcosA\plus{}\frac{9(\sqrt{cosAcosB}\plus{}\sqrt{cosBcosC}\plus{}\sqrt{cosCcosA})}{2}$
$ \le \frac{(cosA\plus{}cosB\plus{}cosC)^2}{3}\plus{}\frac{9(cosA\plus{}cosB\plus{}cosC)}{2}\le\frac{15}{2}$,
Done.

[quote="NextPeace"]
Let $ a \equal{} \sqrt {yz}, b \equal{} \sqrt {zx}, c \equal{} \sqrt {xy}$, We have $ xy \plus{} yz \plus{} zx \plus{} 2xyz \equal{} 1$,
So we have $ x \equal{} cosA, y \equal{} cosB, z \equal{} cosC(A \plus{} B \plus{} C \equal{} \pi)$.
[/quote]
Are you sure?

[quote="khanhkhtn"]Let a,b,c>0 such that $ a^2 \plus{} b^2 \plus{} c^2 \plus{} 2abc \equal{} 1$. Prove that
$ a^2 \plus{} b^2 \plus{} c^2 \plus{} \frac {9}{2}(a \plus{} b \plus{} c)\leq \frac {15}{2}$[/quote]
The following stronger inequality is true too.
Let $ a,$ $ b$ and $ c$ are non-negative numbers such that  $ a^2 \plus{} b^2 \plus{} c^2 \plus{} 2abc \equal{} 1.$ Prove that
\[ a^2 \plus{} b^2 \plus{} c^2 \plus{} 3(a \plus{} b \plus{} c)\leq \frac {21}{4}
\]


real analysis
topology
real analysis unsolved

Construct a closed subset $ F \subset [0; 1]$ such that $ F$ does not contain any open interval and $ |F| \equal{} \frac{1}{2}$.

Any help?

Does [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=222669]this[/url] give you any ideas?

Try to construct dense open set of any lebesgue measure:

$ O$ open set of $ [0,1]$ , dense such that $ |O|<x$ for any x.

[hide]enumerate rational numbers of $ [0,1]$, $ r_n$ and $ u_n >0$ with $ \sum u_n \equal{} x$ and let $ O$ = $ \bigcup ]r_n\minus{}u_n/2,r_n\plus{}u_n/2[$[/hide]

then you should conclude with an affine transformation to obtain a measure of exactly 1/2..


limit
real analysis
real analysis unsolved

Let$ \{a_n\}$ be sequence of real numbers such that
$ \displaystyle\lim_{n\to\infty}(a_n \plus{} a_{n \plus{} 1}) \equal{} 0.$
Let $ m$ be the number of limit points of the sequence $ \{a_n\}$.Show that either $ m\leq 2$ or $ m \equal{} \infty$.

Hint:let $ a\equal{}\liminf_{n\to \plus{}\infty}{|a_n|},b\equal{}\limsup_{n\to \plus{}\infty}{|a_n|}$
if $ b>a$ try to prove all the point in $ [\minus{}b,\minus{}a] \cup [a,b]$ is the limit point of $ a_n$


Euler
absolute value
combinatorics unsolved
combinatorics

Let G be a simple connected graph whichh contains a vertex of degree 2005. Prove that it is poossible to colour the edges of the graph red or blue so that for any vertex A the absolute value of the difference between the number of red edges and blue edges emanating from A does not exceed 1.

Bomb

We'll prove that there's a good coloring for any connected graph $G$ with a vertex of odd degree.

A lemma: if a graph has all its vertices of even degree and an even number of edges, we can color it to get all differences equal to $0$. If it has an odd number of edges, we can color it to get all differences equal to $0$ except for one $2$, and we can place the $2$ at any vertex we want. To prove it, just take the Euler cycle (it exists because all vertices are even); if the number of edges is even, color them alternately, and it's clear that we'll get all differences equal to $0$; if the number of edges is odd, color them alternately but with two consecutive edges of the same color at the chosen vertex, then it will have difference $2$ and the rest will have $0$.

Suppose we've proven the required result for all connected graphs with an odd vertex and $1, 2, ..., k-1$ edges, and take a graph with $k$ edges. If it contains a vertex of odd degree $X$, which is joined to a point of even degree $Y$, then removing $XY$ we get a graph with $k-1$ edges, where $X$ is even and $Y$ is odd.

If it is connected, color it using a good coloring. Since $X$ is even, its difference must be $0$, so when we put back $XY$, any way of coloring it won't cause problems with $X$. Then put back $X$ and color it however it's necessary to keep the $Y$ difference smaller than $2$. If removing $XY$ disconnects the graph, color the $Y$ part with a good coloring (it exists because $Y$ is odd). If the $X$ part has another odd vertex, give it a good coloring (possibly switching all edges so that the $X$ and $Y$ differences have the same sign) and then we can color $XY$ without problems; if the $X$ part has no odd vertices, color it so that all vertices different from $X$ have difference $0$, and give the $X$ difference the same sign as the $Y$ difference. Again we can color $XY$.

The case we've left out is when any odd vertex is joined only to odd vertices. In this case, if there's an odd vertex $X$ of degree at least $3$, there will be at least $4$ odd vertices so removing an edge from $X$ will leave at least $2$ odd vertices, and we can apply the same arguments as above. If odd vertices are joined only to odd vertices and odd vertices only have degree $1$, for the graph to be connected it must be just two vertices $X, Y$ joined by an edge, and this case is trivial.


Well,I liked this one ,hope you enjoy it too:
For $x,y,z \ge 0$ prove:

$\sum_{cyc}^{x,y,z} \frac{x}{1+x+xy} \le 1$

Find all the conditions which equality occurs.

One equality case is any $x,y,z$ with $xyz=1$.


Hi I just finished taking pre algebra in college and semester I will take beginners algebra. Are there any good books out there that could help me? I would prefer a book that explains algebra in a very easy to understand way. Also what about computer programs?

Higher Algebra by Hall and Knight
Its not really that "higher" pretty easy stuff in there. By the way, what grade are you in? Just asking, so don't think im a stalker...that would be weird :maybe:

But please take a look at the date of the first post before replying it. ;)

oh.

oops
epic resurrection post :blush:  :D  :blush:


Show that $\frac{x}{y}+\frac{y}{z}+\frac{z}{x}=1$ has no solutions in the natural numbers.

By AM-GM:
$\frac{x}{y}+\frac{y}{z}+\frac{z}{x} \geq 3\sqrt[3]{\frac{xyz}{xyz}}=3$, hence no solution even in positive numbers

[hide]for non negative numbers $\frac{\frac{x}{y}+\frac{y}{z}+\frac{z}{x}}{3} \geqq 1$[/hide]

Ok, now :
Find all $n \in \mathbb{N}$ for which :
$\frac xy + \frac yz + \frac zx = n$ :)

(to be related with this problem : http://www.mathlinks.ro/Forum/viewtopic.php?t=2592 ;))


I have this problem from measure theory which should be relatively easy, but after several attempts, I couldn't solve it. Either I'm overlooking something, or the formulation of the problem is incorrect. Anyway, here it is:

We have $ H: \equal{}I_\mathbb{R}$ (half-open intervals) and $ \mu: H\rightarrow[0,\infty]$ a premeasure with $ \mu(I): \equal{}\infty$ for $ I\not\equal{}{\o}$. Prove that there are uncountably many $ \sigma$-finite extensions $ \nu$ to the Borel-sigma-algebra $ B(\mathbb{R})$.

I can't find even one extension. What bothers me is the following: If the extension is $ \sigma$-finite, there should be a cover of the whole space $ \mathbb{R}$ wih countably many subsets of finite measure, and I don't see how that is possible, because none of those subsets can contain any intervals (because of the monotonicity of the measure and the assumption $ \mu(I): \equal{}\infty$), and if i try to define them somehow differently, I lose the $ \sigma$-additivity. Anybody have any ideas?

[quote="jmerry"]Here's one such extension:

For any set $ X$ of real numbers, $ \mu(X)$ is the number of rational numbers contained in $ X$.[/quote]

 :D I can't believe I haven't thought of that one. Now if we just define $ E_n: \equal{}\{[\minus{}n,n]\setminus\mathbb{Q}\}\cup\{q_1,q_2,...,q_n\}$, where $ \{q_1,q_2,q_3,...\}$ denote the rational numbers, we have $ \mathbb{R}\equal{}\displaystyle\bigcup_{n\equal{}1}^{\infty}E_n$ and $ \nu(E_n)\equal{}n$, so $ \nu$ is indeed $ \sigma$-finite. We get uncountably many such measures by multiplying this one with some b$ \in\mathbb{R}$. Thanks! :)


[quote="April"]Let be given a triangle $ABC.$ Prove that:
$a(p-a)+b(p-b)+c(p-c)\le\frac{3(a^{3}+b^{3}+c^{3})}{2(a+b+c)}$[/quote]

$\Longleftrightarrow \frac{2\sum{ab}-\sum{a^{2}}}{2}\leq \frac{3\sum{a^{3}}}{2\sum{a}}$ , then use chebisev : $3\sum{a^{3}}\geq \sum{a}\cdot \sum{a^{2}}$, the result follows.


ratio
inequalities
trigonometry
geometry unsolved
geometry

Let $ ABCD$ be a parellogram with $ AB>AD$. Suposse the ratio between diagonals $ AC$ and $ BD$ is $ \frac {AC} {BD}\equal{}3$.     Let $ r$ be the line symmetric to $ AD$ with respect to $ AC$ and $ s$ the line symmetric to $ BC$ with respect to $ BD$. If $ r$ and $ s$ intersect at $ P$ , find the ratio  $ \frac {PA} {PB}$

Daniel

See in the topic [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=208213]Hard geo[/url], of [b][size=100]Inequalities Master.[/size][/b]

Kostas Vittas.

Thank's for the link.  Very nice work :)

Daniel

you can try this also... let $ \angle CAD\equal{}\alpha, \angle CAB\equal{}\beta, \angle DBA\equal{}\theta$... we're asked to find $ \sin(\alpha\plus{}\beta\plus{}2\theta)/\sin(\alpha\minus{}\beta)$... from the condition we have that $ \sin(\alpha\plus{}\beta\plus{}\theta)\equal{}3\sin \alpha$ and $ \sin\theta\equal{}3\sin \beta$... if we square this relations and substract them, and we use that $ \sin^2x\minus{}\sin^2y\equal{}\sin(x\minus{}y)\sin(x\plus{}y)$ we obtain that $ \sin(\alpha\plus{}\beta\plus{}2\theta)\sin(\alpha\plus{}\beta)\equal{}9\sin(\alpha\minus{}\beta)\sin(\alpha\plus{}\beta)$, and we're done.  :D

That's nice also! I had tried the trigonometric solution but I had failed to see that $ sin^2 x\minus{} sin^2 y\equal{} sin (x\plus{}y) sin( x\minus{}y)$
Thank you
Daniel


Mhpws gnwrizei kaneis pote dieksagetai o diagwnismos tou Eykleidh?
Pote tha anevoune ta themata?
Eyxaristw.

ps den iksera pou akrivws na to postarw ayto to topic :(

Ο διαγωνισμός έγινε σήμερα το πρωί.
Τα θέματα θα αναρτηθούν σύντομα στην ιστοσελίδα της Ε.Μ.Ε. http://www.hms.gr

Τα θέματα κι οι λύσεις μόλις αναρτήθηκαν στην ιστοσελίδα!

Τα θέματα φέτος ήταν αρκετά τσιμπημένα κατά κοινή ομολογία. Αλλά αυτό δεν έχει σημασία διότι υπήρχαν περισσότεροι διαγωνιζόμενοι και χρειάζεται να κοπούν και περισσότεροι. Θα έλεγα ότι η βάση είναι το 10 χοντρικότατα.

Κατά τη διάρκεια του διαγωνισμού που έκανα επιτήρηση ασχολήθηκα με τα θέματα της Γ Λυκείου.

Το πρώτο ήταν αρκετά εύκολο νομίζω.

Το δεύτερο (γεωμετρία) ήταν ιδιαίτερα δύσκολο για όσους δε γνωρίζουν τις ιδιότητες του σημείου Miquel και το ριζικό άξονα που στον Ευκλείδη θα ήταν σίγουρα πολλοί. Αφήστε που η εκφώνηση του δεύτερου ερωτήματος ήταν ιδιαίτερα ψαρωτική (καλώς ή κακώς)...

Το τρίτο ήταν το δύσκολο θέμα της ημέρας. Λυνόταν ωραία με μιγαδικούς αλλά η λύση που βρήκα εγώ όσο επιτηρούσα είναι καθαρά αλγεβρική. Απλώς ένα hint: Δημιουργήστε και στις δύο εξισώσεις το  με προσθαφαιρέσεις και πολλαπλασιασμούς και εξισώστε.

Το τέταρτο ήταν προσεγγίσιμο πιστεύω. Αν όχι το 5 λόγω δικαιολογήσεων πιστεύω οι περισσότεροι θα έχουν τσιμπήσει μερικά μοριάκια...

Θεωρω οτι ηταν ισως ο πιο δυσκολος Ευκλειδης των τελευταιων ετων (Γ Λυκειου), μονο και μονο για το πολυ εξυπνο 3ο θεμα το οποιο μας ταλαιπωρισε ολους παρα πολυ μεχρι να το λυσει ο Κωστας. Λογο αυτου το θεματος και μονο για το οποιο αξιζει συγχαρητηρια η επιτροπη, ειναι πολυ πιθανον το Αριστα στον φετινο Ευκλειδη της Γ Λυκειου να ειναι το 15

Το 1ο θεμα της Γ ηταν κλασικο και ευκολο για οποιον εχει διαβασει λιγο για ολυμπιαδες (η συγκεκριμενη ακολουθια ειναι πολυ γνωστη). Οπως επισεις ευκολο ηταν και το 4ο, που η μονη δυσκολια του ηταν να δει καποιος οτι οι ενδιαμεσες τιμες λαμβανονται ακριβως 2 φορες ενω οι δυο ακραιες ακριβως μια φορα, ωστε να μη βγει λαθος η μετριση.

Το 2ο της Γ ειχε ενα πολυ απλο 1ο ερωτημα, και ενα πολυ ψαρωτικο δευτερο (η ιδεα του σημειου που ανηκει στο ριζικο αξονα δυο κυκλων ειναι παρα πολυ γνωστη, ετσι οπως ηταν διατυπωμενο ομως το συγκεκριμενο υποερωτημα, αρκετοι που ασχοληθηκαμε λιγακι με το θεμα δεν σκευτηκαμε καν οτι τα σημεια Α, Κ1, Μ1 θα μπορουσαν να ειναι συνευθειακα για την τυχαια επιλογη του Μ1).

Παρα τη δυσκολια του 3ου θεματος, βλεπω τη βαση για τη Γ Λυκειου φετος να βρησκεται γυρω στο 9, λογο της ευκολιας του 1ου και του 4ου θεματος και του 1ου ερωτηματος του 2ου.

Καλα αποτελεσματα σε ολους.


There's a  :!: sticky stating "PLEASE do not post calculus problems in this forum!" Please move it to the appropriate forum.

Please do not double post.
[hide] Substitute $ u\equal{}\sin x$. $ \int_{0}^{\frac{\pi}{2}}\frac{du}{\sqrt{1\plus{}u^{2}}}\equal{}\sinh^{\minus{}1}(u)]_{u\equal{}0}^{u\equal{}1}\equal{}\ln(1\plus{}\sqrt2)$.[/hide]
You forgot the $ dx$.

[quote="lotrgreengrapes7926"]Substitute $ u \equal{}\sin x$. $ \int_{0}^{\frac{\pi}{2}}\frac{du}{\sqrt{1\plus{}u^{2}}}\equal{}\sinh^{\minus{}1}(u)]_{u \equal{} 0}^{u \equal{} 1}\equal{}\ln(1\plus{}\sqrt2)$.
You forgot the $ dx$.[/quote]
If FOURRIER doesn't like hyperbolic ones, then set $ v\equal{}u\plus{}\sqrt{1\plus{}u^{2}}$

[quote="lotrgreengrapes7926"]Please do not double post.
[hide] Substitute $ u \equal{}\sin x$. $ \int_{0}^{\frac{\pi}{2}}\frac{du}{\sqrt{1\plus{}u^{2}}}\equal{}\sinh^{\minus{}1}(u)]_{u \equal{} 0}^{u \equal{} 1}\equal{}\ln(1\plus{}\sqrt2)$.[/hide]
You forgot the $ dx$.[/quote]

Thanks

U forgot the : $ sin(\frac{\pi}{2}) \equal{} 1$  :D


hey here's one cute little prob from tournment of towns
Show that there are infinitely many pairs of positive integers (m, n) such that m divides n^2 + 1 and n divides m^2 + 1.
i have a very short solution for this one but i'm interested in knowing other types of solution(my solution uses the  even terms of the fibonacci series,i mean F_(2n).
best regards

I thought I had a different type of solution, but I didn't. Here's what I did: if m and n check the 2 relations, then assume n<m, and m and (m^2+1)/n also check the relations. But by considering the initial pair (n,m)=(1,2), by this method we get.. guess what? exactly what you got   :D, so it's not a different type of solution. How did you get the pairs (F_2n, F_(2n+2)) ?

nice work grobber .my solution is xactly the same.i just checked out some values and found they were terms of the fibonacci series and the rest was smooth sailing :D
cheers

Hi grobber,

yesterday I was trying to solve the following pb:
**
let a,b>0 so that ab divides a^2+b^2+1:prove that  (a^2+b^2+1)/ab = 3
**
and i found a nice proof using your solution.

Try it ;-)

Let k=(a^2+b^2+1)/ab=f(a,b).. Assume, like before, a < b (if a=b then a=b=1). We can see (we just calculate) that k=f((a^2+1)/b, a). But because b>a (a^2+1)/b >= a, with equality iff a=1, b=2.. So if we make the transformation (a, b)->((a^2+1)/b, a) we will eventually get the pair (1, 2) (and then (1, 1) if we make one more step). For each pair (a, b) obtained like this f(a, b)=k, so k=f(1, 2)=3. Another important thing we can observe is that the pairs (F_(2n-2), F_2n), which Galois found, are, in fact, THE ONLY solutions.

Sorry about the writing, I don't know why it came out like that

Okay.. when you wrote a < b "he" thought  you wanted to use bold so that is why it appeared like that! But I edited your post and  now it looks good!

when you write a < b use spaces between a, <, and b. I think that was it!!

cheers! :D


[hide="hint"]one of the factors is (c+a)  :wink: [/hide]

Hmmm I was going to quote your post for the problem but now I can't from copyright issues :P

Okay...

it is $a^{3}+b^{3}+ca^{2}+cb^{2}=a^{2}(a+c)+b^{2}(b+c)=(a-c)(a+c)^{2}+(b-c)(b+c)^{2}$

... never mind my method does nothing :P

But I verified your solution... it is correct for (a,b,c)=(3,4,5), and (5,12,13) :)

SM4RT - this was on an older mandelbrot and I think they asked for three distinct factors, although I think your solutio nworks

another hint

[hide]

is there really any difference between c+a and c+b?

[/hide]

[quote="me@home"]Of course there is a difference... $a\not = b$, unless you are implying something else...[/quote]

Yes but..

[spoiler]

Is there any difference between the pythagorean triple (a,b,c) and (b,a,c)? So suppose c+a is a factor...than so much c+b be.

[quote="me@home"]But that is of no rigorous help since we don't rigorously know that $a+c$ divides it... it might help to find the factorization but won't supply any (formal, proof writing able) motivation

EDIT: I am finding a contradiction...

Assume $(a+c)|f(c)=a^{3}+b^{3}+c^{3}$
Then, $f(-a)=0\implies b^{3}=0$

hmm....

perhaps $a-c$?[/quote]

if a=c, then b=0  :wink:

[quote="me@home"]But that is of no rigorous help since we don't rigorously know that $a+c$ divides it... it might help to find the factorization but won't supply any (formal, proof writing able) motivation

EDIT: I am finding a contradiction...

Assume $(a+c)|f(c)=a^{3}+b^{3}+c^{3}$
Then, $f(-a)=0\implies b^{3}=0$

hmm....

perhaps $a-c$?[/quote]

It's perfectly rigorous, there's no fault in you trying out random factors until it factors correctly (though janiel has the most elegant, direct factorization).

I meant for explaining the motivation. The most elegant solution explains its motivation. :)

For instance, janie's solution is elegant partly because it uses the product of $(c^{2}-b^{2})(a+c)$ to introduce symmetry

[quote="me@home"]But that is of no rigorous help since we don't rigorously know that $a+c$ divides it...[/quote]

By symmetry (which is completely rigorous) if $b+c$ is a factor then $a+c$ must be a factor, and vice versa.

Of course that part is rigorous but it is not rigorous from the fact that the motivation for using the (a+c) in the first place was given in a hint :wink:

I just like to know the motivations behind what I'm doing. However after seeing janie's proof, motivations are more clearly supplied...

Just forget what I said about all that stuff about motivation. It doesn't really matter


geometry
circumcircle
incenter
geometry solved

Prove that

OH \geq IH \sqrt 2

where O,I and H are the circumcenter,the incenter and the orthocenter of a triangle.

Source:Matematika v skole 1966 N°3,60

[color=red][Moderator edit: Also discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=38700 .][/color]

Well...it is known that :
OH 2  = 9R 2  - 2s 2  + 2r 2  + 8Rr.
And :
IH 2  = 4R 2  - s 2  + 3r 2  + 4Rr.
Thus :
OH 2  - 2IH 2  = R 2  - 4r 2 , which holds from Euler's inequality.

Pierre.


If $ a,b>0$, show that

\[ \lim_{n\rightarrow \infty} \sqrt{(n\plus{}a)(n\plus{}b)}\minus{}n\equal{}\frac{(a\plus{}b)}{2}\]
 :)

This is usually done by multiplying with $ \frac {\sqrt {(n \plus{} a)(n \plus{} b)} \plus{} n}{\sqrt {(n \plus{} a)(n \plus{} b)} \plus{} n}$, but we can do this as well:

$ \text{L} \equal{} \lim _{n\to 0} \frac {\sqrt {(1 \plus{} a\,n)(1 \plus{} b \, n)} \minus{} 1}{n} \equal{} ^{\text{H}} \lim_{n \to 0} \equal{} \frac {2n \plus{} a \plus{} b}{2 \sqrt {(1 \plus{} a\,n)(1 \plus{} b \, n)}} \equal{} \frac {a \plus{} b}{2}$

(The $ \equal{} ^{\text{H}}$ means the usage of L'Hospital's theorem)

EDIT: How about this

$ \lim _{n \to \infty} \left(\sqrt [n]{\prod^{n \minus{} 1}_{i \equal{} 0}(n \plus{} c_i)} \minus{} n\right)$

In fact, I need the solution for one of my friend in the $ \epsilon$ language without using L'Hosopital's.

this question in [3.2, 12, pg 67 , Bartle and Sherbert , 3rd edition] ]Basic real analysis.

---------

Is the answer of your question is the mean on $ c_i$'s .?

 :)

[quote="Hidden Scofield"]If $ a,b > 0$, show that
\[ \lim_{n\rightarrow \infty} \sqrt {(n \plus{} a)(n \plus{} b)} \minus{} n \equal{} \frac {(a \plus{} b)}{2}\]
:)[/quote]
$ T \equal{} \sqrt {(n \plus{} a)(n \plus{} b)} \minus{} \frac {a \plus{} b \plus{} 2n}{2} \equal{} \frac {4(n \plus{} a)(n \plus{} b) \minus{} (a \plus{} b \plus{} 2n)^2}{4\sqrt {(n \plus{} a)(n \plus{} b)} \plus{} 2(a \plus{} b \plus{} 2n)}$
$ \equal{} \minus{} \frac {(a \minus{} b)^2}{4\sqrt {(n \plus{} a)(n \plus{} b)} \plus{} 2(a \plus{} b \plus{} 2n)}$
$ 4\sqrt {(n \plus{} a)(n \plus{} b)} \plus{} 2(a \plus{} b \plus{} 2n) \ge 4n\plus{}2a\plus{}2b\plus{}4n > n$
So: $ |T| < \frac {(a \minus{} b)^2}{n}$. And from here it is easy to see that $ T \to 0$ as $ n \to \infty$

[quote="Hidden Scofield"]Is the answer of your question is the mean on $ c_i$'s .?[/quote]

Yes it is..

hello, you can rewrite your term as
$ \frac{(n\plus{}a)(n\plus{}b)\minus{}n^2}{\sqrt{n^2\plus{}na\plus{}nb\plus{}ab}\plus{}n^2}\equal{}$
$ \frac{a\plus{}b\plus{}\frac{ab}{n}}{\sqrt{1\plus{}\frac{a}{n}\plus{}\frac{b}{n}\plus{}\frac{ab}{n}}\plus{}1}$
and this tends to $ \frac{a\plus{}b}{2}$.
Sonnhard.

In fact if $ p_1 \plus{} ... \plus{} p_m \equal{}1$, then $ \prod_{k\equal{}1}^m (n\plus{}a_k)^{p_k} \minus{} n \rightarrow p_1a_1 \plus{} ...\plus{} p_ma_m$ as $ n \to \infty$. You can allow any real numbers for $ a_k$ and $ p_k$ as long as $ p_1 \plus{} ... \plus{} p_m \equal{}1$. To prove this, factor out an $ n^{p_k}$ from each term in the product to get $ n (\prod_{k\equal{}1}^m (1\plus{}\frac{a_k}{n})^{p_k}\minus{}1)$, then use $ (1\plus{}x)^p \equal{} 1 \plus{} px \plus{} O(x^2)$ on each factor in the latter product. After multiplying this out the expression looks like $ n(1 \plus{} (p_1a_1 \plus{} ...\plus{} p_ma_m)/n \plus{} O(1/n^2) \minus{} 1)$, which has the desired limit.


Given that $ a,b$ are unequal positive real numbers satisfying the equation $ a^{3}\plus{}b^{3}\plus{}9ab\equal{}27$,then find the minimum possible value of $ \frac{9}{ab}\plus{}\frac{ab}{9}$

OMG factor it...

$ a^3\plus{}b^3\plus{}(\minus{}3)^3\minus{}3(a)(b)(\minus{}3)\equal{}(a\plus{}b\minus{}3)(...)\equal{}0$ where $ ...$ is not allowed since $ a\ne b$. Then the problem is the same as: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=291399


Let $ a$ be a real number satisfying $ 0 \leq a \leq 1$
Find min and max of $ A=\sqrt[4]{1-a^{2}}+\sqrt[4]{a}$

:oops: Let $ a=sin(x)$
Then $ A(x)=\sqrt{cos(x)}+\sqrt[4]{sin(x)}$
$ A^\prime (x)=-\frac{1}{2}sin(x) \frac{1}{\sqrt{cos(x)}}+\frac{1}{4}cos(x) \frac{1}{\sqrt[4]{sin^{3}(x)}}=0$
So: $ 16sin^{7}(x)=cos^{6}(x) (*)$
$ (*)$ has one solution. It should be an extremum point.
$ A(0)=A(1)=1$
$ A(\frac{\pi}{4})=\frac{1}{\sqrt[4]{2}}+\frac{1}{\sqrt[8]{2}}\approx 1.757>1$
So.. $ (*)$ it will be maximum and minumum will be $ A(0)=A(1)=1$ :oops:
Interesting to saw another solution...


I use Kile and have Fedora 10(if that makes a difference). How can I get Asymptote?

Download Asymptote here: http://sourceforge.net/projects/asymptote/ (they have an .rpm file for FC.. I'm not sure, but doesn't that make things automatic? Anyway..)

If you download the source, follow steps in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1291733#1291733]this past post[/url].

Hope that helps!


8. How many 3-digit numbers between 100 and 300 are there for which one of the digits equals the sum of the other two?

I don't know how to solve this besides going through each number?  Can anyone help?

Going through every number isn't that bad.  Since the hundreds digit would have to be 1 or two, the other two digits would have a difference of 1 or 2, or the other two digits would add up to the hundreds digit.

101-2 (110, 101)
112-2 (112, 121)
123-2
134-2
145-2
...
189-2
9*2=18

then

202-2
213-2
...
279-2
8*2=16

18+16=34

then there is 211

so 34+1=35

If I messed up somewhere, just ignore me...

According to the answers, you got it right.  Thanks. :D

[hide]
The hundreds digit is either 1 or 2.

For 1,
The other two digits have a difference of 1
0,1
1,2
2,3
...
8,9
That is 9 pairs. You can order in two ways

For 2,
The other two digits have a difference of 2
0,2
1,3
2,4
...
7,9

That is 8 pairs. You can order in two ways.

So total it is $9\cdot2+8\cdot2=18+16=34$

There's also 211
35 total[/hide]


linear algebra
matrix
trigonometry
function
complex numbers

I did a problem in class today that evaluated $ f(t)\equal{}e^{At}$ for $ A_{2,2}\equal{}\begin{bmatrix}2&1 \\\minus{}1&4 \end{bmatrix}$ to a matrix form.

The answer I got was:

$ f(t)\equal{}\begin{bmatrix}e^{3t}\minus{}te^{3t}&te^{3t} \\\minus{}te^{3t}&e^{3t}\plus{}te^{3t} \end{bmatrix}$

Factoring we have:

$ f(t)\equal{}e^{3t}\begin{bmatrix}1\minus{}t&t \\\minus{}t&1\plus{}t \end{bmatrix}$

My question is, is there some simple general expression for simplifying $ e^{At}$ to a matrix form? Maybe something that resembles $ e^{tA_{2,2}}\equal{}e^{\lambda t}\begin{bmatrix}1\minus{}t&t \\\minus{}t&1\plus{}t \end{bmatrix}$

but for any size matrix.

No. This is a very special case, in that the eigenvalues are equal. Most of the time (for a $ 2\times 2$), you'll get $ e^{At} \equal{} e^{\lambda_1 t}A_1 \plus{} e^{\lambda_2 t}A_2$ for some distinct complex numbers $ \lambda_1$ and $ \lambda_2$ and rank-one matrices $ A_1$ and $ A_2$.

For an $ n\times n$ matrix, there are $ n$ eigenvalues, and we get cases based on how many of them coincide.

Once upon a time (a differential equations class, a long time ago) I actually did write out a complete formula for the exponential of any $ 2\times 2$ real matrix. For better or worse, here is that formula:

[b]Formula:[/b] The general formula for the exponential of the real $ 2 \times 2$ matrix
\[ A \equal{}\begin{bmatrix}  a&b\\c&d\end{bmatrix} :\]
[i]Preliminaries:[/i] Let  $ r_0 \equal{}  \frac{a \plus{} d}2  ;$ let $ \Delta \equal{} ad \minus{} bc;$ and let $ D \equal{} r_0^2 \minus{}\Delta.$
($ \Delta$ is the [i]determinant[/i] and $ D$ is the [i]discriminant[/i].)

[i]Case 1:[/i] $ D > 0$	 [Two real eigenvalues.]

Let  $ r_1 \equal{} r_0 \plus{} \sqrt{D}$   ; and let $ r_2 \equal{} r_0 \minus{} \sqrt{D}   .$  Then:
\[ e^{tA}  \equal{}  \frac1{r_1 \minus{} r_2}\begin{bmatrix}  (r_1 \minus{} d)e^{r_1t} \minus{} (r_2 \minus{} d)e^{r_2t} &b(e^{r_1t} \minus{}e^{r_2t})\\c(e^{r_1t} \minus{}e^{r_2t})&(r_1 \minus{} a)e^{r_1t} \minus{} (r_2 \minus{} a)e^{r_2t}\end{bmatrix}\]

[i]Case 2:[/i] $ D \equal{} 0$	[Repeated eigenvalues.]
\[ e^{tA}  \equal{}  e^{r_0t}\begin{bmatrix}  1 \plus{} \left(\frac{a \minus{} d}2\right) t&bt\\ct&1 \minus{} \left(\frac{a \minus{} d}2\right) t\end{bmatrix}\]

[i]Case 3:[/i] $ D < 0$	 [Complex eigenvalues.]

Let  $ \omega  \equal{} \sqrt{\minus{}D}$
\[ e^{tA}  \equal{}  e^{r_0t} \begin{bmatrix} \cos \omega t \plus{} \left(\frac{r_0 \minus{} d}{\omega}\right) \sin\omega t&\left(\frac{b}{\omega}\right) \sin\omega t\\\left(\frac{c}{\omega}\right) \sin\omega t&\cos\omega t \plus{} \left(\frac{r_0 \minus{} a}{\omega}\right) \sin\omega t\end{bmatrix}\]

=========================

epkid08's example falls into Case 2, with $ r_0\equal{}3,\Delta \equal{} 9,$ and $ D\equal{}0.$

Again, from a differential equations class: an interesting tidbit, if you've developed Laplace transforms as a tool.

The Laplace transform of the scalar function $ e^{at}$ (on $ [0,\infty),$ by default) is $ \frac1{s\minus{}a}.$

The Laplace transform of the matrix-valued function $ e^{tA}$ for $ t\in [0,\infty)$ is $ (sI\minus{}A)^{\minus{}1}.$


AMC
USA(J)MO
USAMO
trigonometry
ratio
geometry
geometric transformation

Two circles intersect at two points, one of them $X.$ Find Y on one circle and $Z$ on the other, so that $X, Y$ and $Z$ are collinear and $XY \cdot XZ$ is as large as possible.

Let A be the other intersection of the circles $\mathcal C_{1},\ \mathcal C_{2}$ with radii $r_{1},\ r_{2}.$ The angles $\angle ZYA,\ \angle YZA$ are fixed, hence the remaining angle $\alpha = \angle YAZ$ of $\triangle AYZ$ is also fixed. Denote $\phi = \angle YAX,$ then $\angle ZAX = \alpha-\phi.$

$XY \cdot XZ = 2r_{1}\sin \phi \cdot 2r_{2}\sin (\alpha-\phi)$

is maximum when $f(\phi) = \sin \phi \sin (\alpha-\phi)$ is maximum,

$0 = \frac{\text{d}f}{\text{d}\phi}= \cos \phi \sin (\alpha-\phi)-\sin \phi \cos (\alpha-\phi) = \sin(\alpha-2\phi),\ \ \ \phi = \frac{\alpha}{2}.$

$\frac{\text{d}^{2}f}{\text{d}\phi^{2}}\left|_{\phi = \frac{\alpha}{2}}=-2\cos (\alpha-2 \phi)\right|_{\phi = \frac{\alpha}{2}}=-2 < 0,$

hence $f(\phi)$ has a maximum at $\phi = \frac{\alpha}{2}.$ Thus $XY \cdot XZ$ is maximum, when AX bisects the angle $\angle YAZ.$ The angles $\angle YAX = \angle ZAX$ are then equal, which means that the chords XY, XZ of $\mathcal C_{1},\ \mathcal C_{2}$ are in the ratio of the circle radii, $\frac{XY}{XZ}= \frac{r_{1}}{r_{2}},$ so that the line YXZ passes through the external homothety center of $\mathcal C_{1},\ \mathcal C_{2}.$

Differentiating !,Show some Merci at them. :D

For a more[i] gentle[/i] solution take a look at mathx (at the end )

http://www.mathlinks.ro/Forum/viewtopic.php?highlight=max&t=48171

[color=darkblue] This problem is equivalently with the next three classical (old) problems :[/color][color=darkred]

[b][u]Problem 1[/u][/b] ([u]very nicely[/u]). Given are a line $CD$ and a point $A\not\in CD\ .$ Denote the point $B\in CD$ for which $AB\perp CD\ .$

Project the points $C$ , $D$ on a [u]mobile line[/u] $d$ , $A\in d$ in the points $E$ , $F$ and denote  the intersection $L\in CD\cap EF\ .$

Prove that the product $p=AE\cdot AF$ is maximum if and only if the points $E$ , $F$ are (harmonical) conjugate w.r.t. the points $A$ , $L\ ,$

i.e. the ray $[BA$ is the bisector of the angle $\widehat{EBF}\ .$ Construct the $p$-maximum position of the line $d\ .$[/color]

[color=darkblue][b]Proof.[/b] Denote the foot $L$ of the $B$-bisector in the triangle $EBF\ .$ The triangle $EBF$ is imilarly to itself for any position of the line $d\ .$

Thus, the ratio (homogeneous) $\frac{LE\cdot LF}{LB^{2}}$ is constant. From the next problem obtain $\frac{AE\cdot AF}{AB^{2}}\le \frac{LE\cdot LF}{LB^{2}}$

and the product $AE\cdot AF$ is maximum iff $A: =L\ ,$ i.e. the ray $[AB$ is the $B$- bisector in the triangle $EBF\ .$

[b]Construct.[/b] Denote the fixed circles $w_{1}$ , $w_{2}$ and $E\in w_{1}$ , $F\in w_{2}\ .$ Let the points $C\in w_{1}$ , $D\in w_{2}$ for which $B\in CD$ and $CD\perp AB\ .$ Denote

the point $L_{0}\in\overline{CBD}$ which belongs to the $A$-bisector of the triangle $ACD\ .$ Then, the required $p$-maximum position of the line $d$ is the perpendicular line $d_{0}$

raised from the point $A$ on the line $AL_{0}\ .$ Indeed, if denote the point $P\in d_{0}$ for which $BP\perp d_{0}\ ,$ then $\widehat{PBA}\equiv\widehat{BAL_{0}}\ ,$ i.e. $BP\parallel AL_{0}\ ,$ $\implies$ $d_{0}\perp AL_{0}\ .$[/color]

[color=darkred][b][u]Problem 2[/u][/b] ([u]easily[/u]). Let $ABC$ be a fixed triangle and let $M\in [BC]$ be a mobile point.

Prove that the ratio $\frac{MB\cdot MC}{MA^{2}}$ is maximum if and only if the point $M$ is the foot of the $A$- bisector.[/color]

[hide="Here is proof."][color=darkblue]Apply the [u]Stewart's theorem[/u]  : $MA^{2}+MB\cdot MC=\frac{1}{a}\cdot\left(c^{2}\cdot MC+b^{2}\cdot MB\right)\ .$ Therefore, the ratio $\frac{MB\cdot MC}{MA^{2}}$ $\mathrm{\--\ max.\ }\Longleftrightarrow$ $\frac{MA^{2}}{MB\cdot MC}$ $\mathrm{\--\ min.\ }\Longleftrightarrow$

$\frac{c^{2}}{MB}+\frac{b^{2}}{MC}$ $\mathrm{\--\ min.\ }$ It is achieved (see the next problem) when $\frac{MB}{c}=\frac{MC}{b}=\frac{a}{b+c}\ ,$ i.e. the point $M$ is the foot of the $A$- bisector.[/color][/hide]

[color=darkred][b][u]Problem 3[/u][/b] ([u]commonplace[/u]). Given are the positive constants $a$ , $b$ , $c\ .$

If $x>0\ ,\ y>0\ ,\ x+y=a\ ,$ then the sum $\frac{c^{2}}{x}+\frac{b^{2}}{y}$ is minimum if and only if $\frac{x}{c}=\frac{y}{c}=\frac{a}{b+c}\ .$[/color]

[hide="Here is proof."][color=darkblue] From the C.B.S.' inequality $(c+b)^{2}=\left(\frac{c}{\sqrt x}\cdot\sqrt x+\frac{b}{\sqrt y}\cdot\sqrt y\right)^{2}\le \left(\frac{c^{2}}{x}+\frac{b^{2}}{y}\right)\cdot (x+y)=a\left(\frac{c^{2}}{x}+\frac{b^{2}}{y}\right)$

obtain $\frac{c^{2}}{x}+\frac{b^{2}}{y}\ge \frac{(b+c)^{2}}{a}\ .$ The equality achieves iff $\frac{\sqrt x}{\frac{c}{\sqrt x}}=\frac{\sqrt y}{\frac{b}{\sqrt y}}\ ,$ i.e. $\frac{x}{c}=\frac{y}{b}=\frac{a}{b+c}\ .$[/color][/hide]

See the nice synthetical [b][u]Grobber's proof[/u][/b] at the topic http://www.mathlinks.ro/Forum/viewtopic.php?t=48171 .

[quote="lomos_lupin"][color=darkred]Differentiating ! Show some Merci at them. For a more[i] gentle[/i] solution take a look at mathx (at the end )[/color][/quote]
[quote="Yetti"][color=darkred]You had helluva time with the thing  :!: [/color][/quote]
[color=darkblue]This exchange of kindess between $ L.L.$ ("Duke of Lanchester") and $ Y.$ ("Duke of York") suggest mine a question : what is the problem solving (I"ll refer to geometry) ? For example, I don't commit problem solving. I like geometry and I found in it an entertainment, as tennis, chess a.s.o. In the my opinion, any method is good if it is correctly and quickly. Lomos Lupin offered here a quick proof of this problem ; Grobber presented a complete synthetical proof ; Yetti used calculus with derivatives ; I solved easy metrically. What is the problem ? Please, common on to renounce use the estimates down or up ! [u]If somebody solved correctly and quickly a problem then the whole is O.K.[/u] I propose to open a topic with this subject : [b][u]what is problem solving ?[/u][/b]

[i]With friendship, Virgil Nicula.[/i]

[b][u]P.S.[/u] Yetti[/b], what means "halluva time" ?[/color]

I could not agree more with what you said. Lomos Lupin was just teasing me and I teased him back, no ridicule was involved on either side.

[quote="Virgil Nicula"]
[color=darkblue][b][u]P.S.[/u] Yetti[/b], what means "halluva time" ?[/color][/quote]

Helluva time is slang for hell_of_a_time. To have helluva time means to have a lot of fun, go wild, do things which seem to be a good idea only at a party, etc.

Yetti


That only picks one group.  Instead, it's $ \frac{\binom{9}{3}\binom{6}{3}}{3!}\equal{}280$ - the top chooses the way to create groups, and the denominator accounts for the arrangements of the group (e.g. ABC, CBA are not counted separately).

[quote="Thunder365"]Nats 2009 sprint #22
22. A tennis coach divides her 9-player squad into
three 3-player groups with each player in only
one group. How many different sets of three
groups can be made?

why isnt the answer just 9C3 (9*8*7/3*2*1)= 84? instead the answer is 280 .[/quote]

The answer isn't 9C3 because the coach is dividing up the squad into different groups--she isn't picking people for 3 different positions.

Thank you for reading my 1600th post. Have a nice day.

Oh wow im so dumb how did i not notice that :oops:

And i just noticed that this problem is almost the same as the one in states with the science teacher!

Yeah, except with bigger numbers.

Figures, they just like wasting our time.


Three circles are mutually externally tangent. Two of the circles have radii 3 and
7. If the area of the triangle formed by connecting their centers is 84, then the
area of the third circle is kπ for some integer k. Determine k.

[quote="The Pink Panther"]Three circles are mutually externally tangent. Two of the circles have radii 3 and
7. If the area of the triangle formed by connecting their centers is 84, then the
area of the third circle is k*pi for some integer k. Determine k.[/quote]

Lets see...hmmm.....connect the radii of the circles...
You'll get a triangle with sides of length $ 10, 3 \plus{} x$, and $ 7\plus{} x.$
Find the area of the circle in terms of x, and that equals 84.
Just solve for x!


geometry
analytic geometry
calculus
trigonometry
ratio
area of a triangle

Let ABC be a right triangle such that AB=AC. Let P be the midpoint on the side AC. Let Q be the point on the hypotenuse BC for which AQ is perpendicular to BP. Let R be the point of intersection of the segments AQ and BP. What percentage of the area of the triangle ABC is the area of the triangle PQR? :?

This is very inelegant, but you could just introduce a coordinate plane and solve it that way.

Ya...but the teacher prefer us to solve this in a geometrical way... ie...use some transformation properties...rather than doing in a calculus way...

The answer i get is:
[hide]$\frac{1}{15}$[/hide]

Here is a simple geometrical i've found(it even does not cantain trig)
[hide]$BQ:CQ=\frac{[BAQ]}{[CAQ]}\cdot\frac{[APB]}{[APB]}$
$=\frac{AB\cdot AQ}{AP%Error. "dcotBP" is a bad command.
}\cdot\frac{AB\cdot BP}{AQ\cdot AC}$
$=\frac{AB}{AP}$
$=\frac21$
and:
$BR:PR=\frac{[ABQ]}{[APQ]}$
Since $AP=PC$, we have $[APQ]=\frac12[ACQ]$
$BR:PR=2\cdot\frac{[ABQ]}{[AQC]}=2\cdot\frac{BQ}{CQ}=4$

Hence we have: $\frac{[PQR]}{[BPQ]}=\frac15$
Because $[BQP]:[BQA]=1:2$ and $[BQA]:[ABC]=2:3$, we shall obtain:
$[PRQ]:[ABC]=1:15$

[/hide]

I don't think JBL was telling you to use calculus.

[quote="math beginner"]hi Shobber,
can u kindly explain the first part of your solution?
I don't understand how you get from BQ:CQ to 2:1

Thank you![/quote]

Since $ABQ$ and $ACQ$ has the same attitude, so the ratio of their area is equal to the ratio of their base, that is $[ABQ]:[ACQ]=BQ:CQ$, and also, <BAR=<APB, <PBA=<CAQ, so we just add $[APB]$, then, by using the theorem of the areas, we can get 2:1

There seem to be many theorems about the triangle areas......
For two triangles ABC and A'B'C', if <A=<A', then we will have:
$[ABC]:[A'B'C']=\frac{{AB}\cdot{AC}}{{A'B'}\cdot{A'C'}}$

I got this from a Chinese math book so i don't know its english name. Can someone tell me it? And can you post more about the area theorem of the triangles??

[quote="JBL"]That formula you posted follows immediately from the fact that the area of a triangle is given by $\frac{ab\sin(C)}{2}$ where a and b are two sides of the triangle and C is the angle between them.[/quote]

Yes, it does. But we can get that without any trig. Just draw two triangles with the same angle and compare their areas.


linear algebra
matrix
invariant
linear algebra unsolved

A nice problem:
Let $V$ be an $n$-dimensional space, and $V_1,V_2   \subseteq V$ such that $V_1 \oplus V_2 = V$.
Let $A$ be a diagonizable matrix, $B$ - a base in $V$ and let $f \in End(V)$ such that $M_B(f)=A$.

Prove that: $f(V_1) \subseteq V_1 \Rightarrow f(V_2) \subseteq V_2$.

sorry for a typo, it should be "basis" not "base" , $M_B(f)$ is the matrix of $f$ in basis $B$.

So, basically, you're saying that if $f$ is a diagonalizable endomorphism, and $V_1$ is an invariant subspace of $f$, then every complement $V_2$ of $V_1$ is also invariant. This is not true:

Take a linear space of dimension $2$, with basis $e_1,e_2$. Define $f$ by $f(e_1)=e_1,f(e_2)=0$, and take $V_1=\langle e_1\rangle,V_2=\langle e_1+e_2\rangle$. $V_1$ is invariant under $f$, but the element $e_1+e_2$ of $V_2$ goes to $e_1$, which does not belong to $V_2$ (and, of course, it's obvious that $V=V_1\oplus V_2$).

Am I missing something? :?

yes, something's wrong in the text, but soon I'll give a proper version.


geometry
abstract algebra
induction
perpendicular bisector
geometry unsolved

Let $n$ be an odd integer. In a regular n-agon take two adjacent vertices, $V_{1}$ and $V_{2}$, and consider the perpendicular bisector of side $V_{1}V_{2}$. Define $V_{k}$ as the vertex of the $n$-agon through which the line, so defined for $V_{k-1}$, $V_{k-2}$, passes. Find all $n$ for which the sequence $\{V_{i}\}$ contains all vertices, at least once.

quite a few geometry eh

Daniel

This is solved at
http://www.mathlinks.ro/Forum/viewtopic.php?t=112895

Hmm... can somebody translate into English?

Enumerate the vertices of the polygon (in a consecutive way) with the residues module $n$ such that $\overline{i}$ is the number at the vertex $V_{i}$, for $i=1,2$.
Let $a_{k}$ be the residue corresponding to the vertex $V_{k}$.
It's easy to see that
$2a_{k+1}\cong a_{k}+a_{k-1}mod(n)$ for all $k>1$
We have,
$2a_{k+1}+a_{k}\cong 2a_{k}+a_{k-1}mod(n)\cong 2a_{2}+a_{1}mod(n)\cong 5mod(n)$.
Multiplying by -(-2)^k, we have
$(-2)^{k+1}a_{k+1}-(-2)^{k}a_{k}\cong-(-2)^{k}5 mod(n)$
The sum since $k=1$ until $k=r-1$ give us:
$(-2)^{r}a_{r}+2\cong 10\frac{(-2)^{r-1}-1}{-3}mod(n)$
or, equivalently,
$(-2)^{r}(3a_{r}-5)\cong 4 mod(3n)$.
If $n$ have a prime divisor $p>3$, there exist a residue $l$ module $n$  such that $p$ divides $3l-5$, and then we never have $a_{r}\cong l mod(n)$.
Then, $n=3^{m}$ for some positive integer $m$.
Now, we have to prove that such $n$ satisfies the condition.
We have to prove that for all $0\leq l\leq n-1$, there exists an nonnegative integer $r$ such that $(-2)^{r}(3l-5)\cong 4 mod(3n)$.
Or, equivalently, for all $0\leq t\leq 3n-1$ with $t\cong 1 mod(3)$, there exists $r$ such that $(-2)^{r}\cong t mod(3n)$
If $a,b$ are nonnegative integers, we can prove (by induction over $b$) that
$(-2)^{a.3^{b}}\cong-a.3^{b+1}+1 mod(3^{b+2})$
Now, we can afirm that the numbers $(-2)^{3^{s}}$ for $0\leq s\leq n-1$ have distinct residues module $3n$, and moreover, such residues are of the form $1 mod(3)$.
So, we have between them all residues module $3n$ with the form $1 mod(3)$.


Measuring activity of uranium in a Geiger counter pulses arrive according to a Poisson process with rate 2 arrivals per second. Measured plutonium the arrival rate is 5 seconds.

For 0.2 seconds if there were no arrivals, what is the probability that uranium was being measured?

I think such problems can be solved with the Erlang distribution or truncated Poisson,

is this correct, or how they do it?

Thank you very much

[quote="extranger"]
For 0.2 seconds if there were no arrivals, what is the probability that uranium was being measured?
[/quote]
To be honest, I do not really know what it means ? Are you using a Bayesion approach (you then need a prior distribution) ?

Hello, no, it simplmenete a problem that must be done through a conditional Poisson:

Here's a better rewording:

By measuring pulses of uranium with a Geiger counter, those arriving to a rate of 2 per second. Measuring Plutonium, the arrival rate is 5 per second. Si durante dos segundos no se producen arrivos,

For two seconds if there are no arrivals,

What is the probability that uranium had been measuring?

I hope you or someone can help me, I still think that a conditional Poisson distribution is Erlang, but I'm not sure.

A greeting and thanks

I think he means: assuming that no pulse arrived in the last two seconds, what is the probability that it is uranium you are measuring, rather than plutonium?

exactly

I think I have the solution, but I want to know the opinion of you and if anyone knows what that is the Truncated Poisson or Erlang distribution.

How to solve this problem?

Greetings

The statement of the problem is clear. This is not the problem. What alekk said is that some information is missing to get a definite answer. Depending on what assumptions you make on the "a-priori" probabilities of what you are measuring, you will get different answers.

I guess what the problem implicitly assumes is that before we make any observation, we have no bias whatsoever about what we measure. This corresponds to an a-priori 50-50 distribution, of whether it is uranium or plutonium. It is easy to compute the probabilities for no pulse arriving in the first to seconds both for uranium and plutonium, using Poisson distributions with the given parameters. Then you can use Bayes' formula (and the above "a-priori" assumption) to calculate the reverse conditional probability. This is fairly elementary and does not require any knowledge of a Truncated Poisson (or Erlang) distribution, just the usual Poisson (or exponential) distribution.

[quote="solyaris"]I guess what the problem implicitly assumes is that before we make any observation, we have no bias whatsoever about what we measure. This corresponds to an a-priori 50-50 distribution, of whether it is uranium or plutonium. It is easy to compute the probabilities for no pulse arriving in the first to seconds both for uranium and plutonium, using Poisson distributions with the given parameters. Then you can use Bayes' formula (and the above "a-priori" assumption) to calculate the reverse conditional probability. This is fairly elementary and does not require any knowledge of a Truncated Poisson (or Erlang) distribution, just the usual Poisson (or exponential) distribution.[/quote]
Thank you Solyaris: this is exactly what I meant, and you perfectly explained it.  :)

Yes, that is, assuming it 50-50, and using Bayes theorem,

Thank you very much


$ \frac{7!}{4!3!} \equal{} \frac{4! \cdot 5 \cdot 6 \cdot 7}{4! \cdot 6} \equal{} 5 \cdot 7 \equal{} \boxed{35}$.


geometry perimeter calculus integration ARML Harvard college

nobody seems to be posting questions today, so i'll post some.

1.  How many terms are there in the expansion of (x+y+z)^10.  You can also generalize, e.g how many terms are there in (x_1 + x_2 + x_3 + ... x_k)^n ?

2.  How many right triangles, whose sides are all positive whole numbers, have the property that the area is numerically equal to its perimeter, and what are their measures?

3.  Point P is 6 inches from the center of a circle with radius 10.  Compute the number of chords with integral length that pass through P.

4.  If the three numbers 796, 1157, and 1594 are divided by the positive integer q, they all leave a remainder of r.  Compute r.

5.  If log_a b + log_b c + log_c a = 0,  find the value of (log_a b)^3 + (log_b c )^3 + (log_c a) ^3.

These seem incredibly familiar . . . what are they from?

the first 4 were questions from an arml practice yesterday, i dont know where they got that from though, a lot of times i think they like to rip questions out of the nyciml contests, so thats why they might be familiar.  The 5th is from nyciml 2003, and the last one is from Harvard-Mit.

Not today, I guess.  Eventually someone will.

Finally someone will attempt these problems, or at least some of them.

[b]1.[/b] 12C2 = 66.  The reason of this is because the terms are all in the form of x^a * y^b * z^c where a + b + c = 10 and a,b,c :ge: 0.  Finding all nonnegative integer solutions to a + b + c = 10 will get the answer of 66.

[b]3.[/b]
The shortest chord containing point P will be when the radius is a perpendicular bisector.  That length of the chord will be 16.  Then the diameter of the circle is 20.  There are 2 chords each of lengths 17, 18, and 19 since it's symmetric, meaning there are six of those.  Adding the diameter and the perpendicularly bisected chord, we have 8 chords in all.

[b]4.[/b]
x < y < z
r < q
x,y,z > 0
796 = qx + r
1157 = qy + r
1594 = qz + r
Subtracting equations results in
q(y-x) = 361
q(z-y) = 437
(361,437) = 19
q = 19
19x + r = 796
r = 17 since the only time x>r is when (x,r) is (41,17).

we need to solve a+b+sqrt(aa+bb)=ab/2 which is equilvalent to ab/4-a-b+2=0
ie a=4+8/(b-4)
i am sure u can finish it off by trying a few numbers

you could also [hide]isolate the square root, square, and factor[/hide], which is how i think bugzpodder got the equivalent form. i like this factorization better: [hide](a-4)(b-4)=8[/hide] but yours works too.

im in a hurry, but i took a look at the solutions and they're all right.

But for 2.

[hide]

we could use the algorithm to generate pythag. triples.

m^2-n^2, 2mn, m^2+n^2.  the first 2 are the legs, last is hypotenuse.

Perimeter is 2m^2+2mn = 2m(m+n)

Area is (m+n)(m-n)*mn

So (m-n)*n = 2.  Now we only have to work with 2 cases, and it's easy from there.[/hide]

ugh. i still haven't seen a proof that that algorithm generates ALL pythagorean triples. i know it generates primitive triples, but what about the rest?

Actually, the best way to express it is merely to toss in a "d" in each of the terms (to account for non-primitive triples).  Otherwise, I'm not sure the formula does in fact generate all triples.  (At the very least, it won't generate them uniquely.)

you can actually see from my deriviation that gcd(a,b,c) could be 1 or 2 (but nothing else) but that means you have to do the derivation first ;)


Let $ O$ be the center of an ellipse an $ P$ a point in that ellipse. Consider point $ Q$ in the semimajor axis such that $ PQ$ is perpendicular to the tangent to the ellipse by $ P$. Find the maximum posible value of $ \angle OPQ$. Semimajor axis is $ a$ and eccentricity is $ e$.

$ E, E'$ are the ellipse foci. Assuming that the given excentricity is $ e \equal{} OE \equal{} OE'$ (absolute). The ellipse semiminor axis is then $ b \equal{} \sqrt{a^2 \minus{} e^2}.$ $ OP$ cuts the ellipse again at $ P'$ and parallel through $ O$ to the ellipse tangent at $ P$ cuts the ellipse at $ R, R'$ so that the angle $ \angle POR$ is acute. $ PP', RR'$ is a pair of the ellipse conjugate diameters; if the ellipse is projected to a circle by a parallel projection, they become perpendicular diameters of this circle. The angle $ \angle OPQ \equal{} 90^\circ \minus{} \angle POR$ is maximum, when the angle $ \angle POR$ is minimum, which happens when $ OP \equal{} OR,$ $ \tan \frac{\widehat {POR}}{2} \equal{} \frac{b}{a},$ $ \tan \widehat{POR} \equal{} \frac{2ab}{a^2 \minus{} b^2},$ $ \tan \widehat{OPQ} \equal{} \frac{a^2 \minus{} b^2}{2ab}.$


inequalities
LaTeX
induction
trigonometry
calculus
derivative
strong induction

[quote="farbod_mp2002"]prove that:

1 / ( 2^(n-1) ) =< Sin^(2n) + Cos^(2n) =< 1

and do sorry for this no-latex thing
(even an induction-like approach if appreciated...)[/quote]

welcome in the forum farbod_mp2002. The LaTeX version is:

$\frac{1}{2^{n-1}}\le\sin^{2n}{x}+\cos^{2n}{x}\le1$

you can learn some stuff about LaTeX here:

http://artofproblemsolving.com/LaTeX/AoPS_L_About.php

thanks alot mr/mrs/miss conceive!
but i still want answerss...

Write $a = \sin^{2}{x}$ and $1-a = \cos^{2}{x}$. Then we want to prove $a^{n}+(1-a)^{n}\leq 1$ where $a \in [0,1]$. Take the second derivative with respect to $a$ and you will get something that's always positive. So the LHS is convex and has a maximum on an endpoint or endpoints. We check to see that $a = 0$ and $a = 1$ give $1 \leq 1$, which is sufficient to prove it for all $a \in [0,1]$.

We'll do the other side by induction. For $n = 1$ we have a basic pythagorean identity.
Now suppose $2^{1-n}\leq \sin^{2n}{x}+\cos^{2n}{x}$.
Multiply the RHS by $\cos^{2}{x}+\sin^{2}{x}= 1$. $2^{1-n}\leq \sin^{2n+2}{x}+\sin^{2n}{x}\cos^{2}{x}+\sin^{2}{x}\cos^{2n}{x}+\cos^{2n+2}{x}=$
$\sin^{2n+2}{x}+(\sin^{2}{x}\cos^{2}{x})(\sin^{2n}{x}+\cos^{2n}{x})+\cos^{2n+2}{x}$ or $2^{1-n}-(\sin^{2}{x}\cos^{2}{x})(\sin^{2n}{x}+\cos^{2n}{x}) \leq \sin^{2n+2}{x}+\cos^{2n+2}{x}$.
Working with the LHS, we get $2^{1-n}-(\sin^{2}{x}\cos^{2}{x})(\sin^{2n}{x}+\cos^{2n}{x}) =$
$2^{1-n}-\frac{\sin^{2}{2x}}{4}(\sin^{2n-2}{x}+\cos^{2n-2}{x}) \leq$
$2^{1-n}-\frac{2^{2-n}}{4}= 2^{-n}$ and we're done.
(note that for $n = 1$, $\sin^{2n-2}{x}+\cos^{2n-2}{x}= 2$, and $n > 1$ the principle of strong induction applies, so there is no problem substituting $2^{2-n}$ in).

[hide="right hand side..."]We know that $\sin^{2}x+\cos^{2}x=1$, and notice that $\sin^{2}x,\cos^{2}x$ are always positive (trivial inequality).
$\implies 1^{n}=(\sin^{2}x+\cos^{2}x)^{n}=\sin^{2n}x+\text{(a bunch of non-negative terms)}+\cos^{2n}x\ge\sin^{2n}x+\cos^{2n}x$.
$\implies \sin^{2n}x+\cos^{2n}x\le1$.[/hide]

use binomial theorem with $\left(\sin^{2}{x}+\cos^{2}{x}\right)^{n}= 1$ on the left-hand side. :)

should work.


A small blob of putty of mass m falls from the ceiling and lands on the outer rim of a turnable of radius R and moment of inertia $I_{0}$ that is rotating with angular speed $w_{i}$about its vertical fixed symmetry axis. After several turns, the blob flies of the edge off the turnable . What is the angular speed of the turnable after the blob flies off?

$w_{f}= \frac{I_{0}}{I_{0}+mR^{2}}w_{i}$

I think this is correct. If so, i got it from equating
iitial angular momentum to final, since there is no external torque on the system   :maybe:

Your answer is the same as mine. I think it is correct too. I use the law of conservation of angular momentum in the axis of rotation. When the mass is ready to fall down, I calculate L (m, radius, and the tangential velocity). What is your method?

my method is more or less the same. Equating the initial (bfr the ball hits the table) to the final (after it leaves), we get this expression.


Let $ABC$ be a triangle with area $S$, and let $P$ be a point in the plane. Prove that $AP+BP+CP\geq 2\sqrt[4]{3}\sqrt{S}$.

Minimum by $P$ of $AP+BP+CP$ is attained either for $P$ inside $ABC$ such that $\angle APB=\angle BPC=\angle CPA=2\pi/3$, either for a vertex $A$, if $\angle BAC\ge 2\pi/3$. In the first case,

\[S=\frac{\sqrt{3}}4(AP\cdot BP+BP\cdot CP+CP\cdot AP)\le \frac{(AP+BP+CP)^2}{4\sqrt{3}}.\]

In the second case, we have $S\le \frac{\sqrt{3} BA\cdot CA}4< \frac{(BA+CA)^2}{4\sqrt{3}}$.

$AP+BP+CP\geq\frac{h_a+h_b+h_c+\sqrt3\cdot\frac{a+b+c}{2}}{3}$
$\frac{1}{h_a}+\frac{1}{h_b}+\frac{1}{h_c}=\frac{1}{r}$,    $\Longrightarrow$    $h_a+h_b+h_c\geq 9r$
finally, $\frac{9r+\sqrt3\cdot p}{3}\geq 2\sqrt{pr\sqrt3}$

Nice solutions. Particularly I find Salamitrosky's solution interesting, although it took me some time to realize how he proved the inequality $AP+BP+CP\geq\frac{h_a+h_b+h_c+\sqrt3\cdot\frac{a+b+c}{2}}{3}$. In fact, this inequality can be proven as follows:

Erect an equilateral triangle BCX outwardly on the side BC of triangle ABC. As shown in http://www.mathlinks.ro/Forum/viewtopic.php?t=19316 post #4, we have $AP+BP+CP\geq AX$. Now, of course, the segment AX is longer or equal to its projection onto the A-altitude of triangle ABC. But the projection of the segment AX onto the A-altitude of triangle ABC equals to the sum of the A-altitude of triangle ABC with the X-altitude of triangle BCX. The A-altitude of triangle ABC has length $h_a$, and the X-altitude of triangle BCX has length $\frac{\sqrt3}{2}a$ (since the triangle BCX is equilateral and has the sidelength BC = a). Thus, the projection of the segment AX onto the A-altitude of triangle ABC equals $h_a+\frac{\sqrt3}{2}a$. Since the segment AX is longer than this projection, we thus have $AX\geq h_a+\frac{\sqrt3}{2}a$. Hence, $AP+BP+CP\geq AX\geq h_a+\frac{\sqrt3}{2}a$. Similarly, $AP+BP+CP\geq h_b+\frac{\sqrt3}{2}b$ and $AP+BP+CP\geq h_c+\frac{\sqrt3}{2}c$. Adding up these three inequalities, we get

$3\left(AP+BP+CP\right)\geq \left(h_a+\frac{\sqrt3}{2}a\right)+\left(h_b+\frac{\sqrt3}{2}b\right)+\left(h_c+\frac{\sqrt3}{2}c\right)$
$=h_a+h_b+h_c+\sqrt3\cdot\frac{a+b+c}{2}$,

so that $AP+BP+CP\geq\frac{h_a+h_b+h_c+\sqrt3\cdot\frac{a+b+c}{2}}{3}$, as we wanted to prove.

Here is my solution of the original problem:

According to http://www.mathlinks.ro/Forum/viewtopic.php?t=19316 post #4 Theorem 2, we have $AP+BP+CP\geq W$, where $W=\sqrt{\frac12\cdot\left(a^2+b^2+c^2+4\sqrt3S\right)}$. On the other hand, Weitzenboeck's inequality (see http://www.mathlinks.ro/Forum/topic-14388.html or http://www.mathlinks.ro/Forum/topic-8059.html ) yields $a^2+b^2+c^2\geq 4\sqrt3S$. Thus,

$AP+BP+CP\geq W=\sqrt{\frac12\cdot\left(a^2+b^2+c^2+4\sqrt3S\right)}\geq\sqrt{\frac12\cdot\left(4\sqrt3S+4\sqrt3S\right)}$
$=\sqrt{4\sqrt3S}=2\sqrt[4]{3}\sqrt{S}$.

And the problem is solved.

  Darij


analytic geometry
combinatorics proposed
combinatorics

Let $\displaystyle n \in \mathbb N$, $\displaystyle n \geq 2$. Determine $\displaystyle n$ sets $\displaystyle A_i$, $\displaystyle 1 \leq i \leq n$, from the plane, pairwise disjoint, such that:

(a) for every circle $\displaystyle \mathcal C$ from the plane and for every $\displaystyle i \in \left\{ 1,2,\ldots,n \right\}$ we have $\displaystyle A_i \cap \textrm{Int} \left( \mathcal C \right) \neq \phi$;

(b) for all lines $\displaystyle d$ from the plane and every $\displaystyle i \in \left\{ 1,2,\ldots,n \right\}$, the projection of $\displaystyle A_i$ on $\displaystyle d$ is not $\displaystyle d$.

You can simply take them to be dense (this is actually what (a) says) and countable (so we're sure that (b) is satisfied).

If $S$ is the set of points in the plane with both coordinates rational, for example, and $a_1,\ldots,a_n$ are reals which are linearly independent over the rationals, then we can take $A_i=S+(a_i,a_i)$.


let (a,b,c) denote the GCD of a,b,c
let [a,b,c] denote the lowest common multiple of a,b,c

prove: xyz=(xy,yz,xz)[x,y,z]

Suppose that $x=p^ax'$, $y=p^by'$, $z=p^cz'$, $a\geq b\geq c$, numbers $x',y',z'$ are not divisible by $p$. Then $(xy,yz,zx)[x,y,z]$ is divisible by $p^{a+b+c}$ and not divisible by $p^{a+b+c+1}$. It proves the assertion.


vector
superior algebra
superior algebra solved

Let V_1, V_2, ... V_s are non-trivial linear subspaces of V. Prove that there exists a vector \alpha \in V s.t. \alpha does not belong to V_i for i=1,2,...,s.

I think it must be simple... but I don't find a solution.... :(

Perhaps there is a condition missing ?

If you take V1=V2=...Vs=V such alpha does not exist ???

sorry. they are non-trivial. i've modified the description and i worked it out a little later than the post.

Probably the problem is to find alpha such that alpha does not belong to any Vi  i.e.  \cup Vi != V ??

I think I have something If V is a K-vectorial space with K infinite (the usual case !).

The case n=2 is easy : x1 in V1\V2, x2 in V2\V1, then x1+x2 in (wolog) V1 = y1 then x2=y1-x1 in V1 contradiction

Wolog, Vn not included in  \cup V1 .. V(n-1).
Let x in Vn \ ( \cup V1 .. V(n-1)), y not in Vn.

Then a.x + y not in Vn for all a (same as above).
If a1.x+y and a2.x+y in the same Vi (i < n) for a1!=a2 then x in Vi, contradiction. But since K is infinite this is not possible.

please do not post sources as [i]textbook[/i]. write the name of the book and author if possible. :) see http://www.mathlinks.ro/viewtopic.php?t=3468 for more details.

of course, the dimension should be definite.

the textbook is in Chinese and written by a group in a department of a school.


logarithms
calculus
calculus computations

How would you isolate L in this situation?

$I*exp(-\frac{r+R}{L}*t)=i$

I have tried changing the letters around and got this one as the best result among others. I doubt it is correct though.

$L=\frac{t}{\frac{r+R}{-ln(\frac{i}{I})}}$

$I\cdot e^{-\frac{r+R}{L}t}=i\Longleftrightarrow e^{-\frac{r+R}{L}t}=\frac{i}{I}.$

$-\frac{r+R}{L}t=\ln \frac{i}{I}\Longleftrightarrow L=\boxed{-\frac{r+R}{\ln \frac{i}{I}}t}.$

Great, thx a lot. Moving around with letters and numbers has always been a problem for me. This example can surely help in the future as it has many aspects involved, so if only i can remember to go back and check this topic for help i should be able to do the most moving :)


function
induction
combinatorics proposed
combinatorics

Let $S=\{a_{1},a_{2},...,a_{n}\}$ be a set with elements from $\mathbb{N}$ in increasing order.
Define the function $g(A)=\max\{a_{k+1}-a_{k}\}$
Show that for any finite colouring of $\mathbb{N}$ there exists a natural $d\geq 1$ (depending only on the colouring) and arbitrarily large monochromatic sets $S$ with $g(S)=d$

[hide]i was told that one can derive van der Waerden from this  :huh: [/hide]

Let A be a set of natural numbers.
First, we prove that the following two propositions are equivalent:
(P1) There exists a positive integer $d$ such that $g(S) = d$ for arbitrary large subsets S of A
(P2) There exists a positive integer $d$ such that $g(S) \le d$ for arbitrary large subsets S of A
(P1) => (P2) is obvious, and (P2) => (P1) is obvious too, take the maximum m of g(S) when S is a subset such that $g(S) \le d$, take two consecutive elements a<b with difference m, then if $g(S) = k \le d$, then $g((S \cap \{x \in A : x > b\}) \cup \{a,b\}) = m$.

We call a set A "fat" if it has the property P1.
We call a set A "slim" if it is not fat.
We call a set A "antislim" if its complement (in N) is slim.

Suppose we have a partition of N in finitely many sets.
The proposition "there exists a d such that for arbitarly large monochromatic sets $S$, $g(S) = d$" is equivalent to "there exists a d such that for arbitarly large monochromatic sets $S$, $g(S) \le d$" and, since there are finitely many sets, it is equivalent to "there is a monochromatic fat set".

We can prove that if the union of finitely many sets is N, then at least one of them is fat.
We wil prove the stronger assertion: if the union of finitely many sets is fat, then at least one of them is fat.
This is equivalent, by induction, to "if the union of two sets is fat, then at least one of them is fat".
This is equivalen to "the union of two slim sets is slim".
This is equivalen to "the intersection of two antislim sets is antislim"
And this is what we shall prove  :D   :rotfl:

How is an antislim set?
- a fat set is "there exist a d such that for arbirarly large subsets S, $g(S) \le d$"
- a slim set is "for each d, there exists a k such that among any k consecutive elements, there is a hole of length $> d$ betweet two of them"
Given an antislim set, A we say that k consecutive naturals belonging to A form a k-island. So, a set A is antislim if and only if, for each d, there exists a k, such that there infinitely many d-islands, and the difference between two consecutive d-islands is less than k"

Now, after all this chatting, comes the proof.
We are given two antislim sets A,B and a natural d. We want to find d-islands in their intersection. There is a k such that the distance between two consecutive d-islands of A is less than k. There is a k' such that the distance between two consecutive (2d+k)-islands of B is less than k'. Now, a (2d+k)-island of B must contain a d-island of A.  And this is also a d-island of B, of course! Therefore, we can find consecutive d-islands in $A \cap B$ with distance at most k'. Here ends the proof.

Still I can't find the link with Van der Waerden's theorem.
Nice problem indeed!  :)


Consider an acute angle triangle ABC and le P be an interior point of ABC. Suppose the lines BP and CP, when prodice, meet AC and AB in E and F respectively. Let D be the point where AP intersects the line segment EF and K be the foot of perpendicular from D on BC. Show that DK bisects angle EKF

Search for "a tuff nut" in the "Solved Problems" section of the "Geometry" forum (I'd post the link, but I can't copy it from my browser's address bar, because it seems to be a bit messed up right now).

[color=red][Moderator edit: Indeed, it's http://www.mathlinks.ro/Forum/viewtopic.php?t=17 .][/color]


I have taken a look at Engel's solution, however I would like to see some very creative solutions to the following problem.

[i][b]Problem:[/b] There is a positive integer in each square of a rectangular table.  In each move, you may double each number in a row or subtract 1 from each number of a column.  Prove that you can reach a table of zeros by a sequence of these permitted moves.[/i]

Masoud Zargar

It's enough to make an arbitrary column 0.

Consider the following algorithm:

1.  If there are any "ones", then we double that row so that they become "twos".

2.  Let's subtract 1 from the whole column.

3.  If all numbers are not ones, repeat step 1.

4.  Subtract 1 from the whole column, so that it is all zero.

This algorithm can only be non-terminating if all numbers can never be ones.  However, by considering the largest integer in the column, it is strictly decreasing on every repetition of step 2, but no number ever becomes less than 1.  Thus, eventually the largest integer must become 1.  Then all integers are 1, and we don't repeat step 1 during step 3.  Contradiction.  So the algorithm always terminates, and the column becomes zero.  The result follows.

[quote="t0rajir0u"]In each corner?  What is there in the rest of the table?[/quote]
Sorry, I meant square.


a ball of density D and volume V is totally immersed in water (density d). A rope on the bottom of the container (tension T) is connected with the ball.
The container moves upwards with an acceleration a= g/4.
a) Find the density D of the ball

The rope is cut away, the container keeps on moving with acceleration a

b) Find the acceleration of the ball relative to a.

enjoy!

maybe the text is not clear? i've my "solution" but i'm waiting for yours...surely there'll be better

unfotunately your results are wrong. i have numeric values for:
T= 4*10^-2 N
V= 4*10^-6 m^3
d=1000 Kg/m^3

matbe you didn't consider that when the container moves, on the ball there's a force F1=d*V*a

yes...i found these results...but how to describe the force F1=d*V*a...i'm searching for a theoric explaining

I think you should just introduce $ g'=g+a$ in place of $ g$ in the container frame of reference...

[quote="Ummagumma"]when the container moves, on the ball there's a force F1=d*V*a[/quote]

What is the direction of that force?

up, like the direction of the motion...it's a problem similar to the weight in a lift moving upwards with an acceleration a


Use Gauss’s Lemma to prove that $(\frac{-2}p) = +1$ or -1 according as p = 1,3 mod 8  or p = 5,7 mod 8

The conditions are $p \equiv \pm 1 \mod 8$ for the first and the others for tha second. In total it's straightforward and counting only.


What are some great math programs for a PC?

Also, what are some great programs to run on a PC?

Great math programs: Maxima or another Macsyma derivative, KSeg or Kig.

Great other programs: Firefox <1.5 - Deer Park Alpha 2, about FF1.4, is the best; Google Pack takes care of search and antispyware; and OpenOffice for all your document needs. :)

If you run it as a Windows program, Linux is. (Take screensaver Linux for instance :D)

I'd call Texniccenter a great application (so is Texshop if you're using a Mac - it might even be a bit better).


let $ x$ , $ y$ ,$ z$ $ \in$ $ {\mathbb R}^{ \plus{} }$ , prove that :

$ \frac {x^2 \plus{} y^2}{x \plus{} y} \plus{} \frac {y^2 \plus{} z^2}{y \plus{} z} \plus{} \frac {z^2 \plus{} x^2}{z \plus{} x} \geq \sqrt {xy} \plus{} \sqrt {yz} \plus{} \sqrt {zx}$

$ \frac {x^3 \plus{} y^3 \plus{} z^3 }{(x \plus{} y)(y \plus{} z)(z \plus{} x)}\geq \frac {3}{8}$

$ \frac {x^3 \plus{} y^3}{x \plus{} y} \plus{} \frac {y^3 \plus{} z^3}{y \plus{} z} \plus{} \frac {z^3\plus{}x^3}{z \plus{} x} \geq x\sqrt {yz} \plus{} y\sqrt {zx} \plus{} z\sqrt {xy}$

[quote="y-a-s-s-i-n-e"]let $ x$ , $ y$ ,$ z$ $ \in$ $ {\mathbb R}^{ \plus{} }$ , prove that :

$ \frac {x^2 \plus{} y^2}{x \plus{} y} \plus{} \frac {y^2 \plus{} z^2}{y \plus{} z} \plus{} \frac {z^2 \plus{} x^2}{z \plus{} x} \geq \sqrt {xy} \plus{} \sqrt {yz} \plus{} \sqrt {zx}$

[/quote]
The following inequality is true too:
$ \frac {x^2 \plus{} y^2}{x \plus{} y} \plus{} \frac {y^2 \plus{} z^2}{y \plus{} z} \plus{} \frac {z^2 \plus{} x^2}{z \plus{} x} \geq \sqrt{3(x^2\plus{}y^2\plus{}z^2)}.$

[b]2[/b].   $ LHS\geq \frac {(x \plus{} y \plus{} z)^3}{9(x \plus{} y)(y \plus{} z)(z \plus{} x)} \equal{} \frac {x^3 \plus{} y^3 \plus{} z^3 \plus{} 3(x \plus{} y)(y \plus{} z)(z \plus{} x)}{9(x \plus{} y)(y \plus{} z)(z \plus{} x)} \equal{} \frac {LHS}{9} \plus{} \frac {1}{3}$

  3    $ \sum {\frac {x^3 \plus{} y^3}{x \plus{} y}}\geq\sum {\frac {(x \plus{} y)xy}{x \plus{} y}} \equal{} \sum {(\sqrt {xy})^2}\geq \sum {\sqrt {xy}\sqrt {yz}}$

but the following is wrong already:

$ \frac {x^2 \plus{} y^2}{x \plus{} y} \plus{} \frac {y^2 \plus{} z^2}{y \plus{} z} \plus{} \frac {z^2 \plus{} x^2}{z \plus{} x} \geq \sqrt [3]{9(x^3 \plus{} y^3 \plus{} z^3)}.$ :wink:

As for the third inequality we notice that $ \frac{x^3\plus{}y^3}{x\plus{}y}\equal{}x^2\minus{}xy\plus{}y^2$ so we have to prove that $ 2(x^2\plus{}y^2\plus{}z^2) \minus{}xy\minus{}yz\minus{}zx \ge RHS$ But then we know that $ x^2\plus{}y^2\plus{}z^2 \ge xy\plus{}yz\plus{}zx \ge x \sqrt{yz} \plus{}y\sqrt{zx}\plus{} z\sqrt{xy}$.

By Cauchy Schwarz :
$ \sum \frac{x^2\plus{}y^2}{x\plus{}y} \ge \sum \frac{x\plus{}y}{2} \ge \sum \sqrt{xy}$

$ \frac {x^3 \plus{} y^3 \plus{} z^3 }{(x \plus{} y)(y \plus{} z)(z \plus{} x)}\geq \frac {3}{8}$
$ \frac {x^3 \plus{} y^3}{x \plus{} y}\geq \frac {(x \plus{} y)^2}{4}$ then $ \frac {x^3 \plus{} y^3 \plus{} z^3 }{(x \plus{} y)(y \plus{} z)(z \plus{} x)}\geq \frac {1}{8} \cdot \left( \frac {(x \plus{} y)^2}{(x \plus{} z)(z \plus{} y)} \plus{} \frac {(z \plus{} x)^2}{(x \plus{} y)(z \plus{} y)} \plus{} \frac {(y \plus{} z)^2}{(x \plus{} z)(x \plus{} y)} \right)$.

Applying AM-GM this last is $ \geq 3$

Ronald Widjojo: How do you use Cauchy-Schwarz? (And why :)

For Mathias_DK

$ (1\plus{}1)(x^2\plus{}y^2) \ge (x\plus{}y)^2$
$ \frac{x^2\plus{}y^2}{x\plus{}y} \ge \frac{x\plus{}y}{2}$


polynomial
group theory
abstract algebra
Galois Theory
superior algebra

:(  can anyone help me with this one? it seems elementary
but i can't solve it

it is known that an irreducible polynomial over K of degree 3 (with K not characteristic
2 or 3 ) has a splitting field of degree six, if and only its discriminant is a
square in K

however, does it also hold for degree four if the poly is irreducible, that
the degree of the splitting field is four iff the discriminant is square in K
??

i worked on the case x^4+a*x*x+b  and found that if it was irreducible
it had a splitting field of degree four iff b or (a*a-4*b)*b is a square

now it gets even more confusing : i can't find a single example of numbers
with b and (a*a-4*b) not square, but b*(a*a-4*b) square

and also? the discrimnant of such polynomials is 16*b*(a*a-4*b)^2
which is square iff b is square and no mentioning of a*a-4*b

so can anyone make this a bit more clear? plz?

Let's see..

The splitting field of a quadratic field has degree 1 if and only if the discriminant is a square.
The splitting field of a cubic polynomial has degree dividing three if and only if its discriminant is a square.
Both of these are quite trivial to prove, and the general result that follows is that the Galois group is a subgroup of the alternating group if and only if the discriminant is a square. (Consider a square root of the discriminant as the product of the differences of the roots, taken with a particular choice of sign. An even permutation preserves the overall sign, while an odd permutation reverses it. Therefore, the square root of the discriminant is in the base field if and only if none of the odd permutations which would change it are in the Galois group.)

Your conjecture seems to be that the degree of the field is at most four for a fourth degree polynomial with a square discriminant.
The alternating group on four elements has order twelve.
I'm looking for a counterexample, and I'm sure I'll find one.

Here's the example I was looking for:
$f(x)=x^4+x+\frac34$, over the rational numbers. We can show that the discriminant of a polynomial $x^4+ax+b$ is $4^4b^3-3^3a^4$. For $a=1,b=\frac34$, this becomes $3^4=9^2$.

$f$ is irreducible, so 4 divides the degree of the splitting field.
Now to prove that the splitting field has degree 12 rather than 4, we show that the resolvent cubic is irreducible, and its roots also lie in the splitting field. This is, of course, part of the procedure for solving a quartic.

Let the roots of $f$ be of the form $r+s+t,r-s-t,-r+s-t,-r-s+t$. The pairwise sums of roots of $f$ are $\pm2r,\pm2s,\pm2t$, so $r,s,t$ are in the splitting field. This form is general for any quartic missing its $x^3$ term.

Now let $g(y)=(y-r^2)(y-s^2)(y-t^2)=y^3-ay^2+by-c$. We can show
that $f(x)=x^4-2ax^2 \pm 8\sqrt{c}x+a^2-4b$. Matching to the original coefficients, we have that $a=0,b=\frac{-3}{16},c=\frac{1}{64}$.

This gives the resolvent cubic as $y^3-\frac3{16}y-\frac1{64}=0$, or $(4y)^3-3(4y)-1=0$. This cubic is obviously irreducible, and we are done; the quartic has a splitting field of degree 12.

Incidentally, you were looking in the wrong place; if you leave out both odd degree terms, the resolvent cubic has a trivial factor, and the whole thing is of degree dividing 8. In that case, the degree divides four when the discriminant is a square, but this only happens when the polynomial factors.


for $ n\ge 2$ let $ d_i\equal{}(1\plus{}\dots\plus{}i)\bmod{n}$. Find all $ n$ s.t. $ \{d_1,\dots,d_{n\minus{}1}\}\equal{}Z_n\setminus\{0\}$.

It equavalent $ (2k\plus{}1)^2\not \equal{}1\mod 8n,k\equal{}1,...,n\minus{}1$. It may be only if $ n\equal{}2^k$.


linear algebra
matrix
vector
analytic geometry
linear algebra unsolved

Let $A$ be a row stochastic matrix such that $||A||=1$, where the norm is that induced by the euclidean norm. Show that $A$ is doubly stochastic.

I think I can prove it, but there's a step I didn't exactly prove. It's very intuitive though :).

harazi told me via PM that the entries of $A$ are in $[0,1]$. However, I think it suffices to assume the entries are real s.t. the sum on every row is $1$, and the conclusion is that the sum on every column is $1$ as well.

Since $\|A\|=1$, it means that $A$ is a convex combination of orthogonal matrices (this has been shown before on the forum), so let's write it as $A=\sum_{i=1}^ka_iU_i$, where $a_i\in[0,1],\sum a_i=1$, and $U_i$ are orthogonal. Let $x$ be the column vector which has all entries equal to $1$. We have $Ax=x$, so $\sum a_i(U_ix)=x$. All the vectors $U_ix$ have the same norm as $x$, and, since the Euclidean norm on $\mathbb R^n$ is strictly convex, it means that $U_ix=x,\ \forall x$, or, in other words, the sum of the entries of all the rows equals $1$ in all $U_i$'s.

Based on the only cases I can imagine: $n=2,3$, I think that from the fact that $U_i$ are orthogonal and have sum $1$ on every row we can conclude that they have the following special form: the points representing the rows are the vertices of a regular $n-1$-dimensional simplex in the hyperplane $x_1+\ldots+x_n=1$ ($x_i$ are coordinates of a point in $\mathbb R^n$), with center $t=\left(\frac 1n,\frac 1n,\ldots,\frac 1n\right)$. From this it follows that the sum on every column is also $1$ in all $U_i$, because for each $U_i$ the sum on each column is $n$ times the respective coordinate of $t$, the centroid of the $n-1$-dimensional simplex formed by the points representing the row vectors of $U_i$, and all the coordinates of $t$ are $\frac 1n$.

Since $A$ is a convex combination of the $U_i$'s, and the sum of the entries in each column equals $1$ in all the $U_i$'s, the same must hold for $A$.

Could this work?

Why cannot we  prove it directly  :?  ?
 Let's consider everything over the reals ( vector space $
\mathbb{R}^{n} $, euclidean norm of it, ...).
 We know that for a vector $ \alpha = (1,1,
\ldots ,1) $ we have $ A \alpha = \alpha $. Now take some vector $
v $, which is orthogonal to $ \alpha $. Then we have for every
real $ t $, that $ || \alpha + tv ||^{2} = || \alpha ||^{2} +
t^{2} || v ||^{2} $, and $ || A (\alpha + tv) ||^{2} = || A \alpha
+ tAv ||^{2} = || \alpha + tAv ||^{2} = || \alpha ||^{2} + 2t(
\alpha , Av) + t^{2}||Av||^{2} $, so we know that $ || \alpha
||^{2} + 2t( \alpha , Av) + t^{2}||Av||^{2} \leqslant || \alpha
||^{2} + t^{2} || v ||^{2} $ for every real $ t $, therefore we
have $ ( \alpha, Av ) = ( A^{*} \alpha,v) = 0 $ for every $ v $,
such that $ ( \alpha, v) = 0 $. This means that $ A^{*} \alpha $
is proportional to $ \alpha $, so sums at every column is equal to
some constant $ a $. But then if we compute sum of all elements in
matrix $ A $ by summing through rows, we get $ n $, and by summing
through columns, we get $ na $, so we get that $ a = 1 $, so sum
of elements in each column is $ 1 $.

I hope it is correct.

Of course, this is the natural solution I was expecting for.


limit
logarithms
real analysis
real analysis unsolved

Hi
The sequence $ p_{n}$ is non increasing sequence of integer.
If $ f(x) \equal{} \sum_{n \equal{} 0}^{\infty }x^{p_{n}}$
can you help me to prove that if $ \lim_{x\rightarrow 1^{ \minus{} } }(1 \minus{} x)f(x) \equal{} 0$
then $ \lim_{n\rightarrow \plus{} \infty } \frac {p_{n}}{n} \equal{} \plus{} \infty$

Do you mean nondecreasing? Nonincreasing makes no sense.
WLOG, the $ p_n$ are all positive; we can just multiply $ f$ by some $ x^k$ if there are negative exponents.

Consider $ 1-x=\frac1n$. We have $ f(1-\frac1n)>\sum_{k=0}^{n}(1-\frac1n)^{p_k}\ge \sum_{k=0}^{n}(1-\frac1n)^{p_n}=(n+1)(1-\frac1n)^{p_n}$
Thus $ \frac1n f(1-\frac1n)>(1+\frac1n)(1-\frac1n)^{p_n}$. Since this quantity goes to zero as $ n\to\infty$ by hypothesis, $ (1-\frac1n)^{p_n}\to 0$ as well. Taking logarithms, $ p_n\ln(1-\frac1n)< -p_n\cdot\frac1{n+1}\to -\infty$. Negate that, and we have the desired result $ \frac1n p_n\to\infty$.

sorry $ p_{n}$ is nondecreasing, $ p_{n\plus{}1} \geq p_{n}$
ans $ p_{n} \in N$


modular arithmetic
function
AMC
AIME
number theory
totient function

Hello. Im just a new comer and Im very poor in math. Can anybody explain how to get the answer of this question:

What is the remainder on dividing 6^83 + 8^83 by 49?

Thanks in advance

[hide]$6^3 = 216 \equiv 20 \pmod {49}$

$6^6 \equiv 20 \cdot 20 = 400 \pmod {49} \Longrightarrow 6^6 \equiv 22 \pmod {49}$

Just keep multiplying

$6^{18} \equiv 512 \pmod {49} \Longrightarrow 6^{18} \equiv 22 \pmod {49}$

$6^{36} \equiv 484 \pmod {49} \Longrightarrow 6^{36} \equiv 43 \pmod {49}$

$6^5 \equiv 720 \pmod {49} \Longrightarrow 6^5 \equiv 34 \pmod {49}$

$6^{41} \equiv 34 \cdot 42 =1462 \pmod {49} \Longrightarrow 6^{41} \equiv 41 \pmod {49}$

By Euler's totient function, $6^{42} \equiv 1 \pmod {49}$
$6^{83} \equiv 41 \pmod {49}$

$8^3 \equiv 22 \pmod {49}$

$8^6 \equiv 484 \pmod {49} \Longrightarrow 8^6 \equiv 43 \pmod {49}$

$8^{12} \equiv 1849 \pmod {49} \Longrightarrow 8^{12} \equiv 36 \pmod {49}$

$8^{24} \equiv 1296 \pmod {49} \Longrightarrow 8^{24} \equiv 22 \pmod {49}$

$8^36 \equiv 22 \cdot 36 = 792 \pmod {49} \Longrightarrow 8^36 \equiv 8 \pmod {49}$

$8^5 \equiv 22 \cdot 15 = 330 \pmod {49} \Longrightarrow 8^{5} \equiv 36 \pmod {49}$

$8^{41} \equiv 36 \cdot 8 = 288 \pmod {49} \Longrightarrow 8^{41} \equiv 43 \pmod {49}$

Again by Euler's totient function, $8^{42} \equiv 1 \pmod {49}$ hence $8^{83} \equiv 43 \pmod {49}$ so

$8^{83} + 6^{83} \equiv 36 + 43 \pmod {49} \Longrightarrow 8^{83} + 6^{83} \equiv 35 \pmod {49}$[/hide]

Your way is going to take a pretty long time and there may be a mistake. Here is a well-known answer.

[hide="View"]

Notice the relationship of 6 and 8 as 7. 6 = 7-1 and 8 = 7+1.

Rewriting this, we have:

$(7-1)^{83} + (7+1)^{83}$, and expanding:

$(7-1)^{83} = \binom{83}{0} \cdot 7^{83} - \binom {83}{1} \cdot 7^{82} + ... \binom {83}{82} \cdot 7^1 - \binom {83}{83} \cdot 7^0$

$(7+1)^{83} = \binom{83}{0} \cdot 7^{83} + \binom {83}{1} \cdot 7^{82} + ... \binom {83}{82} \cdot 7^1 + \binom {83}{83} \cdot 7^0$

All terms are divisible by 49 except last two ones in each and $\binom{83}{83} \cdot 7^0$ part cancels because one is positive and one is negative.

So, the problem is now rather simple.

$\binom{83}{82} \cdot 7^1 + \binom{83}{82} \cdot 7^1 = 2(83)(7) \equiv r \mod 49$

where $r$ is the remainder we're trying to find. $2(83)(7) = 1162$ and dividing that by 49 gives remainder of 35, or $r = 35$.

Therefore, our answer is 35.[/hide]

Please cite your problem if possible. This one was AIME 1983 #6.

I might have thought a little more if I didn't use my handy dandy calc.exe.  :P

Silverfalcon's answer was nice and clean.  :10:


anybody get an invitation to the NSLC? anybody going?

[quote="kheg_k"]what is NSLC?[/quote]
I'm guessing National Student Leadership Conference. See the subject line. ;)

I've heard excellent things from people who have gone to things like that. The thing is though (not to sound conceited) everyone gets one of those invitations, and it costs a lot to attend. So I dunno...it's not like a prestigous camp or anything where the "environment" is actually something special. I dunno if you get what I'm saying...lol...but anyway...


Too hard..........................I spent one hour to solve it ,but failed

[quote="ductrung"]Let $a,b,c$ and $d$ be positive real numbers such that $a^{2}+b^{2}+c^{2}+d^{2}=4$. Prove that
\[\frac{a+b+c+d}{2}\le \sqrt[3]{(1+abcd)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)}\]
[/quote]
The solution is:
From $a^{2}+b^{2}+c^{2}+d^{2}=4$, then
\[abcd \le 1,a+b+c+d \le 4 \]
Applying AM - GM inequality, we have
\[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\ge \frac{4}{\sqrt[4]{abcd}}\ge \frac{4}{\sqrt{abcd}}\ge \frac{8}{abcd+1}\]
Hence,
\[RHS \ge 2 \ge LHS \]
We are done. :)

[quote="Rzeszut"][quote="toanhocmuonmau"]
$\frac{4}{\sqrt[4]{abcd}}\ge \frac{4}{\sqrt{abcd}}$
[/quote]
Correct me if I'm wrong, but I think you reversed the sign.[/quote]
Sorry for my stupid mistake. I will try it again :)

[quote="ductrung"]Let $a,b,c$ and $d$ be positive real numbers such that $a^{2}+b^{2}+c^{2}+d^{2}=4$. Prove that
\[\frac{a+b+c+d}{2}\le \sqrt[3]{(1+abcd)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)}\]
[/quote]
The main ideas in my proof is prove that
\[3(a+b+c+d) \le abcd+11 \]
:)

[quote="ductrung"]Let $ a,b,c$ and $ d$ be positive real numbers such that $ a^{2} \plus{} b^{2} \plus{} c^{2} \plus{} d^{2} \equal{} 4$. Prove that
\[ \frac {a \plus{} b \plus{} c \plus{} d}{2}\le \sqrt [3]{(1 \plus{} abcd)\left(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \plus{} \frac {1}{d}\right)}\]
[/quote]
A new and nice solution, see: http://can-hang2007.blogspot.com/2010/01/inequality-48-p-h-duc.html

With the same conditions $ \frac{a\plus{}b\plus{}c\plus{}d}{2}\le\sqrt{\frac{(1\plus{}abcd)\left(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\plus{}\frac{1}{d}\right)}{2}}$  is also true.


The formula $ t \equal{} \frac{n(n\plus{} 1)}{2}$ will determine a triangular number, where $ n$ is a natural number. For what value of $ n$ will the triangular number 325 be found?

$ n(n \plus{} 1) \equal{} 650 \implies n^{2} \plus{} n \equal{} 650 \implies n^{2} \plus{} n \minus{} 650 \equal{} 0 \implies (n \plus{} 26)(n \minus{} 25) \equal{} 0 \implies n \equal{} \boxed{25}$