i)x,y mh midenikoi pragmatikoi ari8moi kai  (1-x^2)(1-y^2)=1 Na upologistei to A=(1+|x|)*(1+|y|)!

ii) x^2+y^2+z^2=x+y+z+1 na apodeixth oti dn exei lush stous rhtous

iii)P(x^2)=x^2(x^2+1)P(x) , P(2)=12 na bre8ei to poluwnhmo (P(x) me pragmatikous sidelestes..)

Για το τρίτο είναι $ p(x)\equal{}x^4\minus{}x^2.$ αν δεν έχω κάνει κάπου λάθος στις πράξεις....

swsta tis ekanes..! to prwto einai to pio wraio try it

Το 3ο ειναι πρωτο θεμα απο παλιο Αρχιμηδη, ευκολουτσικο

Στο 1ο οι αριθμοι $ |x|^2, |y|^2$ ειναι ριζες καποιας εξισωσης $ x^2 \minus{} ax \plus{} a$ με $ a \geq 4$, για καθε τετοιο $ a$ η παρασταση ισουτε με $ f(a) \equal{} \sqrt {a \plus{} 2\sqrt {a}} \plus{} \sqrt {a} \plus{} 1$ (αν δεν μου ξεφυγε κατι). Η f τωρα ως συνεχης και γνησιως αυξουσα συναρτηση του $ a$, σαρωνει ολο το διαστημα $ [2\sqrt {2} \plus{} 3, \plus{} \infty)$, οταν ο $ a$ σαρωνει το διαστημα $ [4, \plus{} \infty)$. Δεν καταλαβαινω τι αλλο μπορει να ζηταει το προβλημα...

Edit: Οποτε η ελαχιστη τιμη ειναι $ 2\sqrt {2} \plus{} 3$

Το δευτερο ευκολα γινετε ισοδυναμο με το να δειχτει οτι πανω στη σφαιρα με κεντρο την αρχη των αξονων και ακτινα $ \sqrt {7}$ στον $ R^3$, δεν υπαρχει σημειο με ρητες συντεταγμενες, προβλημα το οποιο εχω την εντυπωση πως ειναι αρκετα γνωστο αλλα ποτε δεν το εχω μελετησει και για την ωρα δεν μπορω να φανταστω πως μπορει να αντιμετοπιστει, με αντικαταστασεις παντως των ρητων με πηλικα ακεραιων δεν νομιζω να βγαινει διοτι γινεται χαμος απο μεταβλητες ενω οι διαιρετοτητες δεν οδηγουνε πουθενα, ισως να θελει infinite descent, το μονο σιγουρο παντος ειναι οτι δεν προκειτε περι σχολικης ασκησης....

Καλα σχολικο δεν το λες αλλα δεν ειναι ιδιαιτερα δυσκολο το 2ο αν και θελει προσοχη.
Ουσιαστικα ειναι ισοδυναμο με την εξισωση $ (2x \minus{} 1)^2 \plus{} (2y \minus{} 1)^2 \plus{} (2z \minus{} 1)^2 \equal{} 7$ $ x,y,z$ στο Q.
Ετσι το προβλημα ειναι σαν να θελουμε να δειξουμε οτι δεν εχει ακεραιες λυσεις η εξισωση $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 7s^2$ ως προς $ a,b,c,s$ (δεν ειναι  ισοδυναμο αλλα επαρκες σιγουρα)
Ευκολα βλεπουμε οτι μπορουμε να κανουμε την τετραδα περιττους($ \mod 4$) και αρα μετα $ \mod 8$ εχουμε ατοπο.
Για την σφαιρα που λες nick μπορει καποιος να παρατηρησει οτι η το προβλημα 2 γενικευεται για οποιοδηποτε αριθμο $ 8k \plus{} 1$ στο RHS.

btw leftmast καλο ειναι γενικα οταν εχεις λυσει κατι να βαζεις που και που εστω, μια περιπου ολοκληρωμενη λυση.
Η ιδεα ειναι σημαντικη αλλα και η διατυπωση οπως και οι λεπτομερειες εχουν σημασια  για την λυση ενος προβληματος.

[quote="Athinaios"]
Ευκολα βλεπουμε οτι μπορουμε να κανουμε την τετραδα περιττους($ \mod 4$) και αρα μετα $ \mod 8$ εχουμε ατοπο.
[/quote]

Πλακα κανεις τωρα.... και παλευα 5 ωρες τωρα με 6 μεταβλητες και διαιρετοτητες και ηταν τοσο απλο... εχω αρχισει και ξεχναω βασικα πραγματα κατι μου φαινεται....

Zητω συγγνωμη .....! :blush: Το εκανα ασυναισθητα!!!Παντως για την δευτερη ασκηση, αφου κανουμε τισ αντικαταστασεις αν το παμε mod3 και μετα mod 9 θα βγει,αν και μου επιασε 2 σελιδες περιπου...Μολις βρω χρονο θα βαλω και την λυση μου!!!!Πολυ καλη η λυση σου Athinaios παντως!!! :)

[quote="NickNafplio"][quote="KapioPulsar"][quote="lefmast"]το δευτερο με μια προχειρη ματια βγαινει με απειρη καθοδο.....Εχω καποιες αμφιβολιες για το πρωτο ομως...Αν θεσουμε καποιες τιμες στο χ και λυσουμε ως προσ y δεν θα εχει την ιδια τιμη το Α.... :(  :maybe:Τι εννοεις να υπολογιστει????Αν μπορεις εξηγησε το μου λιγο σε παρακαλω....... :)[/quote]

lipon katarxas sorry eprepe na diefkrinisw oti zitw thn elaxisth timh tou A , kai episis ola afta ta 8emata lunode polu wraia me gnwseis lukeiou (ulh panelinion) ara kai xwris mod hlia :P[/quote]

Τα παραλες τωρα... το πρωτο και το τριτο προβλημα ειναι οντως σχολικα, ενω το τριτο μπορει να μπει και σε εναν διαγωνισμο μικρου επιπεδου (οπως ειπα και παραπανω, ειχε μπει σαν ευκολο θεμα σε Αρχιμηδη παλια). Το δευτερο ομως δεν εχει καμια σχεση με την σχολικη υλη, ειδικα με αυτη των πανελληνιων, αφου μετα το μετασχηματισμο αναγεται σε μια πολυπλοκη διοφαντικη εξισωση με τα τετραγωνα 6 ακεραιων μεταβλητων, οποτε οπως και να γινει χρειαζονται γνωσεις και τεχνασματα απο τη θεωρια των αριθμων, εγω οταν ειμουν στο λυκειο (στο δικο μου λυκειο τουλαχιστον) δεν θυμαμε να αναφερθηκε ποτε ουτε καν ο ορος "διαιρετοτητα" στην ταξη...[/quote] emeis ton anaferame persi sthn 2 lykeiou kai malista kaname kai eisagwgi sto mod alla mexri kai tis diofadikes kaname akoma kai duskoles

gia thn prwth askhsh exw sto mualo mou mia polu wraia trigwnometrikh lush!  :D

[quote="KapioPulsar"][/quote][quote="NickNafplio"][quote="KapioPulsar"][quote="lefmast"]το δευτερο με μια προχειρη ματια βγαινει με απειρη καθοδο.....Εχω καποιες αμφιβολιες για το πρωτο ομως...Αν θεσουμε καποιες τιμες στο χ και λυσουμε ως προσ y δεν θα εχει την ιδια τιμη το Α.... :(  :maybe:Τι εννοεις να υπολογιστει????Αν μπορεις εξηγησε το μου λιγο σε παρακαλω....... :)[/quote]

lipon katarxas sorry eprepe na diefkrinisw oti zitw thn elaxisth timh tou A , kai episis ola afta ta 8emata lunode polu wraia me gnwseis lukeiou (ulh panelinion) ara kai xwris mod hlia :P[/quote]

Τα παραλες τωρα... το πρωτο και το τριτο προβλημα ειναι οντως σχολικα, ενω το τριτο μπορει να μπει και σε εναν διαγωνισμο μικρου επιπεδου (οπως ειπα και παραπανω, ειχε μπει σαν ευκολο θεμα σε Αρχιμηδη παλια). Το δευτερο ομως δεν εχει καμια σχεση με την σχολικη υλη, ειδικα με αυτη των πανελληνιων, αφου μετα το μετασχηματισμο αναγεται σε μια πολυπλοκη διοφαντικη εξισωση με τα τετραγωνα 6 ακεραιων μεταβλητων, οποτε οπως και να γινει χρειαζονται γνωσεις και τεχνασματα απο τη θεωρια των αριθμων, εγω οταν ειμουν στο λυκειο (στο δικο μου λυκειο τουλαχιστον) δεν θυμαμε να αναφερθηκε ποτε ουτε καν ο ορος "διαιρετοτητα" στην ταξη...[/quote][quote="KapioPulsar"] emeis ton anaferame persi sthn 2 lykeiou kai malista kaname kai eisagwgi sto mod alla mexri kai tis diofadikes kaname akoma kai duskoles[/quote]

Η θεωρια αριθμων που υπαρχει στο βιβλιο της Β Λυκειου ισως να καλυπτει 2-3 σελιδες απο ενα βιβλιο θεωριας αριθμων για διαγωνισμους, οποτε αν κανατε δυσκολα προβληματα αριθμοθεωριας οπως λες τοτε σιγουρα βγηκατε πολυ εξω απο την υλη του σχολειου, αν ειναι ετσι τα πραγματα τοτε αξιζει συγχαρητηρια ο καθηγητης σας διοτι σπανια καθονται οι καθηγητες να διδαξουν πραγματα που δεν χρειαζονται στις εξετασεις... παντος θα ηθελα πολυ να δω τη λυση χωρις τετραγωνικα υπολοιπα που λες οτι εχεις βρει.

[quote="NickNafplio"][/quote][quote="KapioPulsar"][/quote][quote="NickNafplio"][quote="KapioPulsar"][quote="lefmast"]το δευτερο με μια προχειρη ματια βγαινει με απειρη καθοδο.....Εχω καποιες αμφιβολιες για το πρωτο ομως...Αν θεσουμε καποιες τιμες στο χ και λυσουμε ως προσ y δεν θα εχει την ιδια τιμη το Α.... :(  :maybe:Τι εννοεις να υπολογιστει????Αν μπορεις εξηγησε το μου λιγο σε παρακαλω....... :)[/quote]

lipon katarxas sorry eprepe na diefkrinisw oti zitw thn elaxisth timh tou A , kai episis ola afta ta 8emata lunode polu wraia me gnwseis lukeiou (ulh panelinion) ara kai xwris mod hlia :P[/quote]

Τα παραλες τωρα... το πρωτο και το τριτο προβλημα ειναι οντως σχολικα, ενω το τριτο μπορει να μπει και σε εναν διαγωνισμο μικρου επιπεδου (οπως ειπα και παραπανω, ειχε μπει σαν ευκολο θεμα σε Αρχιμηδη παλια). Το δευτερο ομως δεν εχει καμια σχεση με την σχολικη υλη, ειδικα με αυτη των πανελληνιων, αφου μετα το μετασχηματισμο αναγεται σε μια πολυπλοκη διοφαντικη εξισωση με τα τετραγωνα 6 ακεραιων μεταβλητων, οποτε οπως και να γινει χρειαζονται γνωσεις και τεχνασματα απο τη θεωρια των αριθμων, εγω οταν ειμουν στο λυκειο (στο δικο μου λυκειο τουλαχιστον) δεν θυμαμε να αναφερθηκε ποτε ουτε καν ο ορος "διαιρετοτητα" στην ταξη...[/quote][quote="KapioPulsar"] emeis ton anaferame persi sthn 2 lykeiou kai malista kaname kai eisagwgi sto mod alla mexri kai tis diofadikes kaname akoma kai duskoles[/quote][quote="NickNafplio"]

Η θεωρια αριθμων που υπαρχει στο βιβλιο της Β Λυκειου ισως να καλυπτει 2-3 σελιδες απο ενα βιβλιο θεωριας αριθμων για διαγωνισμους, οποτε αν κανατε δυσκολα προβληματα αριθμοθεωριας οπως λες τοτε σιγουρα βγηκατε πολυ εξω απο την υλη του σχολειου, αν ειναι ετσι τα πραγματα τοτε αξιζει συγχαρητηρια ο καθηγητης σας διοτι σπανια καθονται οι καθηγητες να διδαξουν πραγματα που δεν χρειαζονται στις εξετασεις... παντος θα ηθελα πολυ να δω τη λυση χωρις τετραγωνικα υπολοιπα που λες οτι εχεις βρει.[/quote]

loipon gia to deftero kai egw me mod to elusa alla mou eipe o ka8igiths mou (pou mou thn eixe pei o idos ) oti lunete kai me aples gnwseis giafto kai thn anarthsa bas kai thn brei kapios afthn thn polupo8eth lush :S

gia to prwto twra h lush mou einai trigwnometrikh :to arxiko ginete $ \frac {1}{x^2} + \frac {1}{y^2} = 1$ara yparxei gwnia b anoikoi (0,π/2) = >1/|x|=sin(b) 1/|y|=cos(b) ara A=(1+1/sin(b))(1+1/cos(b))= 1+(sin(b)+cos(b)+1)/(sin(b)cos(b))=1+2(2^(1/2)*sin(π/4+b)+1)/(sin(2b))=1+2^(3/2)*(sin(π/4+b))/sin(2b))+2/sin(2b)
pou pernei thn elaxisth timh otan to sin2b einai megisto dld gia b=π/4 opou $ A=3+2\sqrt{2}$

[quote="silouan"]Το δεύτερο θέμα είναι από τη Μεσογειάδα του 2003 (πρώτο θέμα). Το τρίτο θέμα είναι όντως από τον αρχιμήδη του 2005 πρώτο θέμα και πάλι. Αλλά μην τρελαθούμε, το ότι λύνονται με σχολικές γνώσεις δεν σημαίνει ότι είναι σχολικά ! Πχ το 6ο θέμα της φετινής ΙΜΟ επαγωγή ήθελε, που διδάσκεται στο λύκειο ... άρα ήταν σχολικό ?
Και μια παρατήρηση ακόμα... Ποστ του style "πιστεύω ότι λύνεται με cauchy-swartz εύκολα", τα θεωρώ ανούσια και spam[/quote]  oxi da3i , egw eipa apla oti 8elei mexri gnwseis sxoleiou ade kai tetragwnika upolipa afto apla

[quote="KapioPulsar"] oxi da3i , egw eipa apla oti 8elei mexri gnwseis sxoleiou ade kai tetragwnika upolipa afto apla[/quote]

Oπως πολύ εύστοχα παρατήρησε ο Σιλουανός παραπάνω το γεγονώς ότι στη λύση μιας άσκησης ''χρειάζονται'' γνώσεις λυκείου δεν σημένει ότι είναι και λυκειακού επιπέδου.....Π.χ. στη γεωμετρία αν έχεις παρατηρήσει οι πιο πολλές ασκήσεις στους διαγωνισμούς στυρίζονται στα ''σχολικού επιπέδου''...(πόσο μάλον αν αναλογιστείς ότι μέσα στο σχολικό βιβλίο αναφέρονται τα θεωρήματα stewart,ceva κλπ).Αυτό όμως δεν σημένει ότι οι ασκήσεις που μπένουν στους διαγωνισμούς είναι ''σχολικού'' επιπέδου....

Η μοναδική άσκηση από αυτές που ανέβασες που θα μπορούσε να χαρακτιριστεί έτσι είναι η τρίτη :)

Φιλικά
Δημήτρης


trigonometry
function
calculus
calculus computations

All I know is arcsin(sin x)=x. How to calculate arcsin(sin x/2), arctan(tan(15*t), etc.?
The result should not contain trigonometrical functions.

And I've learned calculus.

If you know that $ \arcsin\left(\sin\left(x\right)\right)\equal{}x$, then how is $ \arcsin\left(\sin\left(\frac{x}{2}\right)\right)$ different? Think of $ \frac{x}{2}$ as a single unit. So...$ \arcsin\left(\sin\left(\frac{x}{2}\right)\right)\equal{}\frac{x}{2}$.

But be careful: these things are only true within certain limited domains. $ \arcsin\left(\sin\left(\frac{x}{2}\right)\right)\equal{}\frac{x}{2}$ is only true for $ \minus{}\pi\le x\le\pi,$ for instance. That function is defined for all real $ x,$ and gives a periodic ziz-zag line graph.

sorry I meant to say arcsin(cos x/2), things like that
help


calculus
integration
algebra
polynomial
real analysis
real analysis unsolved

$f(x)$ is a polynomial with the degree of $n$,such that $\int_0^1 x^kf(x)dx=0(k=1,2,\cdot,n)$
prove:$\int_0^1f^2(x)dx=(n+1)^2(\int_0^1f(x)dx)^2$

hmm. Let $f(x)=a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0$

Thus
$\int_0^1 x^k f(x)dx = \frac{a_n}{n+k+1} +\cdots+ \frac{a_1}{k+2}+\frac{a_0}{k+1} = \frac{Q(x)}{(k+1)(k+2)\cdots(n+k+1)}$ (1)
, where $Q(k)$ is a polynomial of degree $n$.

And we have $Q(k)=0$ for $k=1,2,\dots,n$. Hence

$Q(k)=c(k-1)(k-2)\cdots(k-n)$ (2)

Substitute (2) into (1), and multiply $k+1$ in both sides. Let $k=-1$. We obtain that
$a_0 = (-1)^n(n+1)c$ (3)
Substitute (2) into (1) and ket $k=0$ then we have $\int_0^1 f(x)dx = \frac{(-1)^n}{n+1}c$ (4)

Combine (3) and (4), eliminating $c$, we get $a_0=(n+1)^2\int_0^1 f(x)dx$.

Now we are going to prove that $\int_0^1 f^2(x)dx = a_0\int_0^1 f(x)dx$. This is quite straight-forward.
$\int_0^1 f^2(x)dx = \int_0^1 (a_nx^n +\cdots +a_1x + a_0)f(x)dx = a_0\int_0^1f(x)dx$.


Compute the limit of x_n, where
x_n=([sqrt(1)]+[sqrt(3)]+[sqrt(5)]+..+[sqrt(4n^2+4n-1)])/n^3.

where [x]=x-{x}.

cheers! :D

Take a_n=  \sum (k=1, 2n^2+2n) [ \sqrt 2k-1]. We would like to apply Cesaro-Stolz. But a_(n+1)-a_n=  \sum (k=2n^2+2n+1, 2n^2+6n+4) [ \sqrt 2k-1]. Observe that each such integer part is 2n+1 or 2n+2. Thus, a_(n+1)-a_n is between (4n+4)(2n+1) and (4n+4)(2n+2). Thus, a_(n+1)-a_n divided by n^2 is about 8. Thus, a_n/n^3 is 8/3 when n is large.


geometry
analytic geometry
circumcircle
symmetry
parameterization
2009 USAMO
USAMO

Given circles $ \omega_1$ and $ \omega_2$ intersecting at points $ X$ and $ Y$, let $ \ell_1$ be a line through the center of $ \omega_1$ intersecting $ \omega_2$ at points $ P$ and $ Q$ and let $ \ell_2$ be a line through the center of $ \omega_2$ intersecting $ \omega_1$ at points $ R$ and $ S$.  Prove that if $ P, Q, R$ and $ S$ lie on a circle then the center of this circle lies on line $ XY$.

[hide="Solution"]Let $ PQ$ and $ RS$ meet at $ Z$, let the centers of $ \omega_{1}$ and $ \omega_{2}$ be $ O_{1}$ and $ O_{2}$, respectively, and let the center of the circle containing $ P,Q,R,S$ be $ O$.

First, observe that by power of a point, $ (PZ)(ZQ)\equal{}(RZ)(ZS)$. But the LHS is the power of $ Z$ with respect to $ \omega_{2}$, while the RHS is the power of $ Z$ with respect to $ \omega_{1}$. Since these are equal, $ Z$ is on the radical axis of $ \omega_{1}$ and $ \omega_{2}$, that is, $ X,Y,Z$ are collinear.

It is well known that $ XY\perp O_{1}O_{2}$ (this is easy to prove with congruent triangles). Note that $ O$ and $ O_{1}$ are both on the perpendicular bisector of $ RS$, so $ OO_{1}\perp RS$ and $ OO_{1}\perp O_{2}S$. Similarly, $ OO_{2}\perp O_{1}Q$. Thus, $ O_{1}Q\cap O_{2}S$ is the orthocenter of $ \Delta OO_{1}O_{2}$, but this is just $ Z$. Thus, $ OZ\perp O_{1}O_{2}$. But we have $ O_{1}O_{2}\perp XY$ and $ Z$ is on $ XY$, so thus $ O$ is also on $ XY$, as desired.[/hide]

Let w be circle passing through PQRS. Let radius of $ w_1,w_2,w$ be $ r_1,r_2,r$ respectively, center be $ O_1,O_2,O$ respectively. Now power of $ O_1$ according to circle w is $ O_1P$*$ O_1Q$=$ O_1O^2\minus{}r^2$. But $ O_1P$*$ O_1Q$=$ O_1O_2^2\minus{}{r_2}^2$, so $ O_1O^2\equal{}r^2\minus{}{r_2}^2\plus{}O_1O_2^2$. So the power of O according to $ w_1$ is $ O_1O^2\minus{}{r_1}^2\equal{}r^2\minus{}{r_2}^2\plus{}O_1O_2^2\minus{}{r_1}^2$. Similarly, O has same power to $ w_2$. So O lies on radical axis XY.

let the centres of the circles be O1,O2,O3. (O3 being the circle around PQRS)

let O1 be (x1,y1), O2 be (x2,y2), and O3 (x3,y3)
let the radiuses of the three circles be r1,r2,r3.

we construct 3 equations, for each circle.

by subtracting the equation of O3 from O1, we get the equation of line RS.  -(1)
by subtracting the equation of O3 from O2, we get the equation of line PQ.  -(2)
by subtracting the equation of O2 from O1, we get the equation of line XY.  -(3)

but we have another condition: O1 is on PQ and O2 is on RS.

we put in the coordinates of O1 into (1), and the coordinated of O2 into (2). then we add them (or subtract them)

we get an equation that says O3 must be on line (3), which is XY.

CAN SOMEBODY VERIFY MY SOLUTION?

[quote= 'CatalystOfNostalgia']But the LHS is the power of Z with respect to w_2, while the RHS is the power of Z with respect to w_1. Since these are equal, Z is on the radical axis of w_1 and w_2.
[/quote]

[quote= 'nine34']we construct 3 equations, for each circle.

by subtracting the equation of O3 from O1, we get the equation of line RS[/quote]

How would you prove either of these assertions? Thanks for the help.

hmm I thought this was an odd choice for a #1 .. my writeup was done within 15 minutes of the start, but I'm pretty sure I would not have been able to solve this last year.

My general method:
Suppose $ Z$ was the intersection of $ PQ$ and $ RS$, and $ D$ was the intersection of $ XY$ and $ O_1O_2$, where $ O_1$ and $ O_2$ are the centers of $ \omega_1$ and $ \omega_2$ respectively. Also suppose $ C$ is the center of the circle through $ PQRS$.

Step 1: Prove $ C,Z,D$ are collinear. Check.
Step 2: Prove $ Z$ is on $ XY$. Uhhhhhh..... Nope, I had no idea what a radical axis was during the test :( Too bad, it would've been trivial otherwise.

And plus I missed the parallel case like so many other people T-T. So at most a 2 for me on problem 1.

yeah, a easy one, just a radical axis and an orthocentre, solved within 40 minutes.

[quote="bzauzmer"]

[quote= 'nine34']we construct 3 equations, for each circle.

by subtracting the equation of O3 from O1, we get the equation of line RS[/quote]

How would you prove either of these assertions? Thanks for the help.[/quote]

Well, since R and S are both on O3 and O1, by subtracting the equation of O3 from O1, we get a equation with both R and S on them.

but this equation is linear, so it is a line, and there is only one line that can go thru R and S. (which is line RS)

[quote="Yongyi781"]My general method:
And plus I missed the parallel case like so many other people T-T. So at most a 2 for me on problem 1.[/quote]

What was the parallel case?  :o

Let the center of $\omega_1$ be $ A$, and the center of $\omega_2$ be $ B$. Let the circumcenter of $ PQRS$ be $ C$. Let $ r_1, r_2, r_3$ be the radii of $ A, B, C$ respectively. $ PQ$ is the radical axis of circles $ A$ and $ C$, $ RS$ is the radical axis of circles $ C$ and $ B$, and $ XY$ is the radical axis of circles $ A$ and $ B$, because the circles intersect and it is a well known fact that the radical axis of intersecting circles goes through the intersection points.
By definition of radical axis, the power of $ A$ with respect to circles $ B$ and $ C$, and similarly for the points $ B$ and $ C$ with respect to the other 2 circles, we get:
$ AC^2 \minus{} r_3^2 \equal{} AB^2 \minus{} r_1^2$
$ AB^2 \minus{} r_2^2 \equal{} AC^2 \minus{} r_3^2$
Adding the 2 equations gives us:
$ AC^2 \minus{} r_3^2 \plus{} AB^2 \minus{} r_2^2 \equal{} AB^2 \minus{} r_1^2 \plus{} AC^2 \minus{} r_3^2$
$ BC^2 \minus{} r_2^2 \equal{} AC^2 \minus{} r_1^2$
Then by defintion,$ C$ has equal power with respect to circles $ A$ and $ B$, so $ C$ lies on the radical axis of circles $ A$ and $ B$, which is $ XY$, so we are done.

Basically what greentreeroad did, use the powers being equal to find the third circle center having the same power with respect to the other 2 circles.

That takes care of the parallel case since the lengths aren't affected if triangle ABC is degenerate, and the powers being equal still holds.

Letting the circle containing $ P,Q,R,S$ have center $ C$, I proved that $ C$ had a power of a point equal with respect to $ \omega_1$ and $ \omega_2$. But instead of citing radical axes (I only have a vague idea about them, mostly that it is the line $ XY$, so I was careful), I said to consider the extension of line $ CX$ and its two intersections with $ \omega_1$ and $ \omega_2$ be $ X_1, X_2$, respectively. Proving the power of a point is equal for both circles proves that $ CX_1\equal{}CX_2$, implying that they are the same point, i.e. $ Y$. Then this implies that $ C,X,Y$ are collinear.
Does this logic work?

Yeah that looks like it works (and it's pretty much how you prove the existence of radical axes [in the case of intersecting circles]).

[quote="math10"]a problem very easily, I do not think this is the problem USAMO.[/quote]

What do you mean?  This was problem #1 on the 2009 USAMO.

Here I noticed that the intersection of $C_2S$ and $C_1Q$ is the radical axis of $\omega_{1,2,3}$, but here I don't think the Radical Axis Theorem says that the center of $\omega_3$ lies on $XY$  :wacko:

Let $T = PQ \cap RS$, $(PRQS) = \omega_3$, the center of $\omega_3$ be $O_3$, and $K = XY \cap O_1O_2$. We first assume $O_1$, $O_2$, and $O_3$ are non-collinear. Notice $$Pow_{\omega_1}(T) = TR \cdot TS = Pow_{\omega_3}(T) = TP \cdot TQ = Pow_{\omega_2}(T) \implies T \in XY$$ (the Radical Axis of $\omega_1$ and $\omega_2$). Since $RS$ is the Radical Axis of $\omega_1$ and $\omega_3$, and $PQ$ is the Radical Axis of $\omega_2$ and $\omega_3$, we know the intersection of the perpendicular from $O_1$ to $RS$ and the perpendicular from $O_2$ to $PQ$ is $O_3$.

Looking at $\triangle O_1O_2O_3$, it's obvious that $T$ is its orthocenter. Now, let the foot of the altitude from $O_3$ to $O_1O_2$ be $K'$. Then, $\angle TK'O_1 = 90^{\circ}$. Using Radical Axis properties, however, we know $\angle TKO_1 = 90^{\circ}$. Thus, $K = K'$ because they both lie on $O_1O_2$. Hence, $O_3, T, K$ are collinear or $O_3 \in XY$ as desired.

If $O_1$, $O_2$, and $O_3$ are collinear, then we know $PQ \parallel RS \parallel XY$.

[b]Claim:[/b] If $O_1$, $O_2$, and $O_3$ are collinear, then $O_3 = K = XY \cap O_1O_2$.

[i]Proof.[/i] The claim is equivalent to showing $KP = KQ = KR = KS$ (in this configuration). This is trivial via the Pythagorean Theorem, however, so we're done. $\square$
$\blacksquare$

[b]Remark:[/b] In order for the problem to hold when the centers are collinear, $O_3 \in O_1O_2$ and $O_3 \in XY$ has to hold, implying $O_3 = XY \cap O_1O_2 = K$ must be true, motivating the last claim.

[hide=Solution]Clearly $XY$ is the radical axis of $\omega_1, \omega_2$. Define the circumcircle of $PQRS$ to be $\omega$. Similarly, $SR$ is the radical axis of $\omega, \omega_1$. Let $O$ be the center of this circle. Since $O_2$ lies on $RS$, we have $\text{Pow}_{\omega} O_{2} = \text{Pow}_{\omega_1} O_2$, or $OO_2^2-OP^2=O_1O_2^2-O_1P^2$. Also, we get $OO_1^2-OP^2=O_1O_2^2-O_2Q^2$ from similar methods. Subtracting the first equation from the second gives $OO_1^2-OO_2^2=O_1P^2-O_2Q^2$, or $OO_1^2-O_1P^2=OO_2-O_2Q^2$. This is all that's necessary to finish and we are done.[/hide]

Let the center of the circle $\omega$ passing through $P,Q,R,S$ be $O$, and let the length of its radius be $r$. Let the center of $\omega_1$ be $O_1$ and let the center of $\omega_2$ be $O_2$, and let the lengths of their radii be $r_1,r_2$ respectively. It suffices to show that $O_1O^2-r_1^2=O_2O^2-r_2^2$. But
\[O_1O^2-r^2=O_1P\cdot O_1Q=O_1O_2^2-r_2^2\]
\[O_2O^2-r^2=O_2R\cdot O_2S=O_1O_2^2-r_1^2\]
so we are done.

By Radical Center theorem we know O1Q and O2S and XY meet at single point Z.
Let O3 be center of SPRQ. we have O1O3 ⊥ SR and O2O3 ⊥ PQ and O1O2 ⊥ XY so Z is orthocenter of O1O2O3 and as it lies on XY we have O3 lies on XY as well.
we're Done.

Let $\omega_3$ and $O_3$ be $(PQRS)$ and its center, respectively. Notice $$O_1O_2^2-r_2^2=\textit{pow}_{\omega_2}(O_1)=\textit{pow}_{\omega_3}(O_1)=O_1O_3^2-r_3^2$$ and $$O_2O_3^2-r_3^2=\textit{pow}_{\omega_3}(O_2)=\textit{pow}_{\omega_1}(O_2)=O_2O_1^2-r_1^2.$$ Adding yields $$\textit{pow}_{\omega_2}(O_3)=O_2O_3^2-r_2^2=O_1O_3^2-r_1^2=\textit{pow}_{\omega_1}(O_3)$$ so $O_3$ lies on $\overline{XY}.$ $\square$

[hide=for fun.]
Let $O$ be the center of $(PQRS)$, let $O_1,O_2$ are centers of $\omega_1,\omega_2$, respectively. By radical axis theorem, $Z=\overline{PQ}\cap\overline{SR}$ lies on $\overline{XY}\perp \overline{O_1O_2}$. Observe that by perpendicular bisectors, $Z$ is the orthocenter of $\triangle OO_1O_2$. Hence, $\overline{OZ}\perp\overline{O_1O_2}\perp\overline{AX}\implies Z\in \overline{XY}$, as desired.
[/hide]

imagine doing amo problems "for fun" orz

Let $O_1, r_1$ and $O_2, r_2$ be the centers and radii of $w_1$ and $w_2$, respectively. Also, let $w_3$ be the circle containing $P, Q, R, S$ with a center at $O_3$ and a radius of $r_3$. Since $O_1$ is on the radical axis of $w_2$ and $w_3$, we have:
$$Pow_{w_2}(O_1)=Pow_{w_3}(O_1)$$
$$O_1O_2^2-r_2^2=O_1O_3^2-r_3^2$$
Similarly, since $O_2$ lies on the radical axis of $w_1$ and $w_3$:
$$Pow_{w_1}(O_2)=Pow_{w_3}(O_2)$$
$$O_1O_2^2-r_1^2=O_2O_3^2-r_3^2$$
Subtracting these two equations gives the following equation, which we continue to simplify.
$$(O_1O_2^2-r_2^2)-(O_1O_2^2-r_1^2)=(O_1O_3^2-r_3^2)-(O_2O_3^2-r_3^2)$$
$$r_1^2-r_2^2=O_1O_3^2-O_2O_3^2$$
$$O_2O_3^2-r_2^2=O_1O_3^2-r_1^2$$
$$Pow_{w_2}(O_3)=Pow_{w_1}(O_3)$$
Therefore, $O_3$ lies on the radical axis (line $XY$) of $w_1$ and $w_2$. $\square$

Let $O_1, O_2$ be the centers of the circles $w_1,$ and $w_2$ with radius $r_1$ and $r_2$ respectively. Furthermore, let $O_3$ be the center of the circle $w_3$ passing through $P, Q, R,$ and $S$ with radius $r_3$. Then, the problem is equivalent to proving that $P(O_3, w_1) = P(O_3, w_2) \Leftrightarrow O_3O_1^2 - r_1^2 = O_3O_2^2 - r_2^2$.

Note that $P(O_1, w_2) = P(O_1, w_3) \implies O_1O_3^2 -r_3^2 = O_1O_2^2 - r_2^2$ and $P(O_2, w_1) = P(O_2, w_3) \implies O_2O_3^2 - r_3^2 = O_2O_2^2 - r_1^2$. Subtracting the two equations gives $O_1O_3^2 - O_2O_3^2 = r_1^2 - r_2^2 \implies O_3O_1^2 - r_1^2 = O_3O_2^2 - r_2^2$ which is what we wanted to prove.

Let $(PQRS)=\omega_3$ and let $r_i$ and $O_i$ denote the radius and center of circle $\omega_i$ for $i\in[1,2,3]$. Since $O_1$ is on the radical axis of $O_2$ and $O_3$ we have $$\text{Pow}_{\omega_2}(O_1)=\text{Pow}_{\omega_3}(O_1) \iff O_1O_2^2-r_2^2=O_1O_3^2-r_3^2.$$ Since $O_2$ is on the radical axis of $O_1$ and $O_3$ we have $$\text{Pow}_{\omega_1}(O_2)=\text{Pow}_{\omega_3}(O_2) \iff O_2O_1^2-r_1^2=O_2O_3^2-r_3^2.$$ Subtracting yields $$r_1^2-r_2^2=O_1O_3^2-O_2O_3^2 \iff \text{Pow}_{\omega_1}(O_3)=\text{Pow}_{\omega_2}(O_3)$$ and we are done.

Let $O_1,O_2,O_3$ be the centers of $\omega_1,\omega_2$, and $(PQRS)$. Let $C$ be the intersection of $PQ$ and $RS$. Note that $C$ also lies on $XY$ since the pairwise radical axes of the three circles are concurrent. Now, note that $O_3$ lies on the perpendicular bisector of both $PQ$ and $RS$. Thus, we have that $O_1O_3$ is perpendicular to $RS$, and $O_2O_3$ is perpendicular  to $PQ$. Therefore, $C$ is the orthocenter of $\triangle O_1O_2O_3$. This means that $O_3C$ is perpendicular to $O_1O_2$. Since $C$ lies on $XY$, we are done.

Let $M=\overline{SR} \cap \overline{PQ}$. Let $O_1,O_2,O_3$ be the centers of $\omega_1,\omega_3$ and $(SPRQ)$. Now, it is well known that the radical axis of two circles is perpendicular to the line joining their centers. Then, $SR \perp O_1,O_2$ and $PQ \perp O_2O_3$. This means that $M$ must be the orthocenter of $\triangle O_1O_2O_3$. This means, $O_3M\perp O_1O_2$ as well.  Clearly $M$ lies on $XY$ ($MP\cdot MQ = MR\cdot MS$ so it lies on the radical axis of $\omega_1$ and $\omega_2$). But then, $XY$ is the line passing through $M$ and perpendicular to $O_1O_2$. This means that indeed $O_3$ lies on the line $XY$ and we are done.

Let $O_1$ be the center of $\omega_1$ and let $O_2$ be the center of $\omega_3$. Denote the center of $(PRQS)$ to be $O_3$ and let the intersection of $PQ$ and $SR$ be $C$. Now note that $SR$, $PQ$, and $XY$ are the three pairwise radical axes of $\omega_1$, $\omega_2$, and $(PRQS)$.

Denote the length of $O_1O_2$ to be $a$ and let the radii of $\omega_1$, $\omega_2$, and $(PRQS)$ be $r_1$, $r_2$, and $r_3$, respectively. Since $O_2$ is on the line $PQ$, by the Power of a Point definition of the radical axis, we have that
\[a^2-r_1^2=O_2O_1^2-r_1^2=Pow_{\omega_1}(O_2)=Pow_{\omega_3}(O_1)=O_2O_3^2-r_3^2,\]
which gives that $O_2O_3=\sqrt{a^2-r_1^2+r_3^2}$. Similarly, since $O_1$ is on $RS$, using Power of a Point on $O_1$ with respect to $O_2$ and $O_3$, we get that $O_1O_3=\sqrt{a^2-r_2^2+r_3^2}$. Finally, using these lengths and Power of a Point one last time,
\[Pow_{\omega_1}(O_3)=O_1O_3^2-r_1^2=a^2+r_3^2-(r_1^2+r_2^2)=O_1O_2^2-r_2^2=Pow_{\omega_2}(O_3),\]
which implies that $O_3$ is on the radical axis of $\omega_1$ and $\omega_2$, which is $XY$, finishing the problem.


Do we have a new administrator nsato, or has he always been here? I understand he's an instructor for some classes.

Naoki Sato joined the staff of Art of Problem Solving as of yesterday.  He will be a regular instructor.  He was an IMO medalist and worked with the Canadian IMO team for several years, so he is an experienced teacher of top students.


function
limit
algebra
domain
calculus
derivative
analytic geometry

Prove that the function $ \sqrt{x}$ is uniformly continuous for $ x\in[1,\infty)$.

The hint given was to use the facts that the function continuous on a segment is also uniformly continuous on it, and the fact if the function is continuous for $ x\in[a,\infty)$, and the $ \lim_{x\to\infty}$ exists, then it's uniformly continuous. But still I do believe that the word "finite" (limit) was dropped out of this last theorem, and that it cannot be applied here.

It's an unbounded function on an unbounded domain-it's quite surprising for me that it's uniformly convergent... (but again I have to say that I lack knowledge in more than few topics in analysis...)

That's the wrong hint at $ \infty$; what's going on here is that the derivative is bounded for large $ x$. On the other side, you only need a second argument if you extend all the way to zero.

Oh, I didn't think of the fact that if the derivative is bounded on the interval (and the function is continuous), then the continuity is uniform (that's "sufficient Lipschitz", right?)-the theorem is not in the curriculum I have to work by as a UTA (and it would surely be fun to prove it to my students using MVT they've never heard of-oh well, they don't know derivatives yet, so the argument will be a bit misty for them).

So I'll try an intuitive-slope-approach, and hope they'll understand  :)

I just wonder is there a nice $ \epsilon \minus{} \delta$ proof? It seems to me that $ \epsilon\equal{}\delta$ should do the trick, as $ x_1\minus{}x_2>\sqrt{x_1}\minus{}\sqrt{x_2}$ for $ x_1>x_2$.

If $ |f'(x)|\le k$ for all $ x$, $ f$ satisfies the Lipschitz condition $ |f(x)-f(y)|\le k|x-y|$ by the Mean Value Theorem. That implies uniform continuity; if $ |x-y|<\frac{\epsilon}{k}$, $ |f(x)-f(y)|<\epsilon$.

In this case, the derivative is no more than $ \frac12$, so we use $ \delta=2\epsilon$ or smaller.

Introducing technical notions like uniform continuity before we know about derivatives and the MVT sounds very weird to me.

Similarly, the function $ \sqrt{|x|}$ is uniformly continuous.
$ \left|\sqrt{|x|}-\sqrt{|y|}\right|\leq \sqrt\left\parallel{}x|-|y|\right|}\leq\sqrt{|x-y|}$ $ \forall x,y\in\mathbb{R}$.
Given $ \varepsilon>0$, we can be sure that $ |f(x)-f(y)|<\varepsilon$ as long as $ |x-y|<\varepsilon^{2}$.

[quote="jmerry"]
Introducing technical notions like uniform continuity before we know about derivatives and the MVT sounds very weird to me.[/quote]

This Mathematics course is extremely wide, so some things are put upside down /but de facto they're there just pro formae/. First semester has to cover single-variable Calculus and the Analysis basics while in the other semester: multivariable Calculus, DE, Laplace & Fourier, Field theory... Third semester: Complex Analysis, Fourier, Laplace (again)... Quite a blend for young engineers  :)

[quote="hsiljak"]
I just wonder is there a nice $ \epsilon \minus{} \delta$ proof? It seems to me that $ \epsilon \equal{} \delta$ should do the trick, as $ x_1 \minus{} x_2 > \sqrt {x_1} \minus{} \sqrt {x_2}$ for $ x_1 > x_2$.[/quote]

This proof is correct on the given interval. Of course, it won't work for $ x \leq 1$ because then the above inequality won't hold.


function
real analysis
real analysis theorems

I have always wondered whether there is a geometric interpretation for a function which is uniform continuity.   For instance, the geometric interpretation for a continuous function from $ R$ to $ R$ is the classic "a function is continuous if its graph can be drawn without lifting the pencil from the paper".  Another geometric interpretation of a definition is the one for uniform convergence (I like that because it is very helpful) that says that a sequence of functions converges uniformly to a function $ f$ if I can draw a "tube" ,of size epsilon, around $ f$ such that in some moment the tail of the sequence get inside this tube (I hope I explained this well :P).

I appreciate you can help me with my question and if you want you can add another nice geometric interpretation of some definition.

Thank you!

My friend and I are in an advanced analysis course and when we first got to uniform continuity and tried to do some proofs, we concluded that a uniformly continuous function can't go "Blaaah!"  Now that phrase probably doesn't mean anything to you, but when my friend attempted to draw a function that was not uniformly continuous, he basically drew a line shooting upwards.  In other words, the function can't blow up without control.  For example, $ f(x)=x^{2}$ can't be uniformly continuous because it's getting steeper and steeper so we can't pick a $ \delta$ that always works.  On the other hand, $ f(x)=\sqrt{x}$ (and even $ \sqrt{|x|}$) are both uniformly continuous because with the exception of the cusp at the origin, it's very easy to control the function.

Did that help at all?

I guess an o.k. geometric analogy is that a function is uniformly continuous on an interval $ I$ if when you pick a horizontal strip of size $ \epsilon$, there exists a $ \delta$ such that no matter where you slide your $ \epsilon$ strip across the function (as long as it stays in $ I$) any points that are $ \delta$ units away from each other or less will have images that fit into that epsilon strip.

It's easier to visualize if you draw it out  :P.

I agree with [b]max_tm[/b]...and with a little searching look what I found...[youtube]oGkgIzSppik[/youtube]

That's basically the idea behind what I wrote in the first post: the function never blows up without control.


topology
real analysis
real analysis unsolved

Need some help.  Consider the lower limit topology defined on $ \mathbb{R}$. That is, the topology generated by the basis, $ \{[a,b): a,b\in\mathbb{R}\}$.

Give a covering of $ [0.1]$ to show that $ [0,1]$ is not compact.

The obvious covering to try is something about upper limits; how about $ V_n \equal{} \left[ 0, 1 \minus{} \frac {1}{n} \right)$?  A finite subcover of this cover is of the form $ \left[ 0, 1 \minus{} \frac {1}{N} \right)$ and hence cannot cover $ [0, 1]$.

[quote="t0rajir0u"]$ V_n \equal{} \left[ 0, 1 \minus{} \frac {1}{n} \right)$.[/quote]

But it is not an open cover of $ [0, 1]$, neither. You may add $ [1, 2)$ to get around this problem.

Nope. All compact subsets are at most countable, but all open sets are uncountable (as they are intervals), so no compact set can contain an open set.


The lengths of the tangents from vertices A, B, C to the incircle are x, y and z respectively. If $a\geq b\geq c$, prove that $az+by+cx \geq \frac{a^2+b^2+c^2}{2} \geq ax+by+cz$

WE have obviously $x+y=c$
                            $x+z=b$ and
                            $y+z=a$
Solving this little system we determine $x,y,z$ and the ineq is then easily proved!
Do not forget the triangle ineq!!!! :)  :D  :D


search
geometry
cyclic quadrilateral
geometry open

[b]
Let $ ABCD$ be a cyclic quadrilateral and $ I$ denote the point of intersection of the diagonals
$ AC$ and $ BD$, $ M$ is the midpoint of side $ BC$ and $ N$ is the midpoint of side $ AD$.
If $ P \in AB$ such that $ IP \bot AB$ and $ R \in CD$ such that $ IR \bot CD$ prove that $ MN \bot PR$.
[/b]
------------------------------------------------------------------------------------------------------------------
[img]http://img.photobucket.com/albums/v218/hudrea/metrutest.jpg[/img]

I have seen this problem here ---> http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1549164493&t=55836
yetti's very nice answer  :)

Thank you very much for your valuable opinion.

this problem will be solve very easy by the down lemma and congruity triangles. :wink:

lemma: if $ ABC$ is a triangle and  $ x ,y$  are the points that $ ABX , ACY$ are rectangular triangles  . ($ BXA \equal{} CYA \equal{} 90$)  then if $ M$ is the midpoint of $ BC$  and   $ XBA \equal{}YCA$   then we have

[size=150]$ XMY \equal{} 90$ and $ mx \equal{} my$[/size]

It seems this problem is very well known and many people tried it in advance and there are lots of skilled people in mathlinks.
Do you know what is the source of the problem?

I like finding interesting math problems and I post most many of them here.

no. i don't know  . i solve this 8month  ago when i was 16. :o  :roll:

I cant understand u. please speak more cleraly :wink:  :arrow:

I want to say - if people solve such problems when they are 16 years old - they have good teachers that propose such problems and prepare them to solve these problems.

Thank you very much. I said that because when I was 16 old a had no good teacher. But she said me that the Bulgarian Math Olympiad exist and there was good problems in the competitions in their earlier stages. It is the reason for me to invent math problems and to try to find interesting problems and to propose them in forums, magazines and comeptitions.


For $n>1$ , prove that  $\sum_{j=1}^{n}\frac{1}{2j-1}$ can not be integer.

Analogously with a similar proof for the harmonic series, we can show that the greatest power of $3$ that divides the common denominator is greater than the greatest power of $3$ that divides the common numerator.

There is a formulation with $p$-adic numbers, but so far I have been unable to tease it out.  Could somebody who knows about such things sketch it out for me?

I wonder that are there other solution except this classical solution? Can anybody help me?

From Bertrand's Postulate, there exists $k\le n$ such that $n\le 2k-1=p\le 2n-1$ for some prime $p$. Then the fraction (the sum) in lowest terms has denominator divisible by $p$. on the other hand the numerator is equal to $\text{(some terms divisible by }p)+\text{(a term not divisible by }p)$. thus it cannot be an integer.

We can prove the following more general result.

Let $a_{1},a_{2},...,a_{n}(n>1)\in\mathbb{Z}-\{0\}$ . Suppose there is a prime $p$ and positive integer $h$ such that $p^{h}|a_{i}$ for some $i$ and $p^{h}\not |a_{j}\forall j\not =i$. Then show that $\frac{1}{a_{1}}+\frac{1}{a_{2}}+...+\frac{1}{a_{n}}\not\in\mathbb{Z}$.


What is the value of $ 501^2 \minus{} 499^2$?


number theory
prime numbers
arithmetic sequence
algebra unsolved
algebra

proof that $\xi=0,p_{1}p_{2}p_{3}....$is irrational
$p_{n}$is the $n$ th prime number.

Same reasoning than in http://www.mathlinks.ro/Forum/viewtopic.php?t=104311 , noting that there are for example abitrary long sequences of zero's in the (decimal) representation because there are primes $\equiv 1 \mod 10^{k}$ for all $k$.

i want a proof and i don't that it's so simple.
may be the first is simple but this no :D

Well, if you don't ask for a proof of Dirichlet's theorem (even in the special case mentioned by ZetaX), I don't see what you are wanting...

Pierre.

I didn't just say it is simple but gave a sketch of a proof...  :maybe:

A bit more explicit:
Assume it is rational. Then it's (decimal) representation is periodic. By that, there can't be arbitrary long sequences of zeros in it (this could only happen for a periodic representation only consisting of zeros from some point, but this is not the case).
So if we proof that there are arbitrary long sequences of zeros in it, we are done. So we look for prime numbers of type $a\cdot 10^{k}+1$ since they have (at least) $k-1$ consecutive zeros in them (they look like $...x000...0001$).
So it suffices to prove that we can find such a prime for all given $k$, meaing to prove there is a prime $\equiv 1 \mod 10^{k}$. But such prime exists by Dirichlets theorem on primes in arithmetic progression, so we are done (note that this case of Dirichlet can be done on an elementary way; for example all prime divisors $\neq 2,5$ of $x^{4 \cdot 10^{k-1}}-x^{3 \cdot 10^{k-1}}+x^{2 \cdot 10^{k-1}}-x^{10^{k-1}}+1$  work).

well, I can post (or link since it surely is on MathLinks) the proof of Dirichlet's theorem for that case if you want...


$n$ is integer $>0$.let $A={1,2,...n}$.find $f(n)_{max}$ such that when $A$ is divised to $f(n)$ any subset  $A_1,A_2,...A_{f(n)}$st :
1)$A_{i}\bigcap\(A_{j}=\emptyset$ for all $i,j$
2)$A_{1}\cup\(A_{2}\cup\...\cup\(A_{n}=A$
  there exist $A_{k}$ include three elements $x<y<z$ st $x+y\geq z$

[quote="dieuhuynh"]noone .i think it is very easy :roll:[/quote]

Maybe the problem is easy. But without $\LaTeX$ it is hard to read and nobody bothers reading it. Think about that.  :)


inequalities
induction
inequalities unsolved

Let $ x_i \in \mathbb{R}^\plus{}$ for $ i\equal{}1,2,\ldots ,n$ such that $ x_1 \le x_2 \le \ldots \le x_n$. Prove the inequality:
\[ \frac{x_1}{x_2}\plus{}\frac{x_2}{x_3}\plus{}\ldots\plus{}\frac{x_{n\minus{}1}}{x_n}\plus{}\frac{x_n}{x_1} \geq \frac{x_2}{x_1}\plus{}\frac{x_3}{x_2}\plus{}\ldots\plus{}\frac{x_n}{x_{n\minus{}1}}\plus{}\frac{x_1}{x_n} \]

Denote    $ {a_k} \equal{} \frac{{{x_k}}}{{{x_{k \plus{} 1}}}}$ for    $ 1 \le k \le n \minus{} 1$    then      $ 0 < {a_k} \le 1$
and the inequality became
$ \sum\limits_{k \equal{} 1}^{n \minus{} 1} {{a_k}}  \plus{} \frac{1}{{\prod\limits_{k \equal{} 1}^{n \minus{} 1} {{a_k}} }}\ge \sum\limits_{k \equal{} 1}^{n \minus{} 1} {\frac{1}{{{a_k}}}}  \plus{} \prod\limits_{k \equal{} 1}^{n \minus{} 1} {{a_k}}$
If $ n\equal{}2$  the inequality is trivial
Let us assume that the inequality is true for order less than (n-1)
By induction  we have  for numbers $ {a_1},{a_2},...,{a_{n \minus{} 2}},{a_{n \minus{} 1}}{a_n}$
 $ \sum\limits_{k \equal{} 1}^{n \minus{} 2} {{a_k}}  \plus{} {a_{n \minus{} 1}}{a_n} \plus{} \frac{1}{{\prod\limits_{k \equal{} 1}^n {{a_k}} }} \ge \sum\limits_{k \equal{} 1}^{n \minus{} 2} {\frac{1}{{{a_k}}}}  \plus{} \frac{1}{{{a_{n \minus{} 1}}{a_n}}} \plus{} \prod\limits_{k \equal{} 1}^n {{a_k}}$
Then $ \sum\limits_{k \equal{} 1}^n {{a_k}}  \plus{} \frac{1}{{\prod\limits_{k \equal{} 1}^n {{a_k}} }} \ge \sum\limits_{k \equal{} 1}^n {\frac{1}{{{a_k}}}}  \plus{} \prod\limits_{k \equal{} 1}^n {{a_k}}  \plus{} {a_{n \minus{} 1}} \plus{} {a_n} \plus{} \frac{1}{{{a_{n \minus{} 1}}{a_n}}} \minus{} {a_{n \minus{} 1}}{a_n} \minus{} \frac{1}{{{a_{n \minus{} 1}}}} \minus{} \frac{1}{{{a_n}}}$
 i.e  $ \sum\limits_{k \equal{} 1}^n {{a_k}}  \plus{} \frac{1}{{\prod\limits_{k \equal{} 1}^n {{a_k}} }} \ge \sum\limits_{k \equal{} 1}^n {\frac{1}{{{a_k}}}}  \plus{} \prod\limits_{k \equal{} 1}^n {{a_k}}  \plus{} \frac{{\left( {1 \minus{} {a_{n \minus{} 1}}{a_n}} \right)\left( {1 \minus{} {a_{n \minus{} 1}}} \right)\left( {1 \minus{} {a_n}} \right)}}{{{a_{n \minus{} 1}}{a_n}}}$
thus   $ \sum\limits_{k \equal{} 1}^n {{a_k}}  \plus{} \frac{1}{{\prod\limits_{k \equal{} 1}^n {{a_k}} }} \ge \sum\limits_{k \equal{} 1}^n {\frac{1}{{{a_k}}}}  \plus{} \prod\limits_{k \equal{} 1}^n {{a_k}}$


geometry
incenter
inradius
geometry proposed

Let $ P$ be an arbitrary point inside a given triangle $ ABC$. Let $ A',B',C'$ be the orthogonal projection of $ P$ on $ BC,CA,AB$ respectively. Let $ I$ be the incenter and $ r$ be the inradius of the triangle $ ABC$. Find the least value of the expression"

$ PA'\plus{}PB'\plus{}PC'\plus{}\frac{PI^2}{2r}$

Dear Mathlinker,
Where does this problem come from?
In Vietnam, it was in [b]Mathematics and Youth[/b] magazine two months ago and still not out of date. So to mod: please lock this topic!


A merchant has determined that if the selling price of a peach is 15 cents, he will sell 400 each day. He has also determined that for each cent the price is increased, he will sell 16 fewer peaches each day. At what price in cents, should he sell each peach in order to maximize the daily income from peaches?

I got the answer from guessing and checking, but when I read the formula in the solutions: p(640-16p) = 640p - 16p^2, I didn't completely understand.

Help, please and thanks :]

If he increases the price by $ x$ cents, then the amount of income the merchant will get is $ (15\plus{}x)(400\minus{}16x)$.  We wish to maximize this quadratic, which occurs at $ x\equal{}5$, so $ 15\plus{}5\equal{}\boxed{20}$.

Someone wish to explain?

the reason the maximum occurs at x=5 is when you expand the quadratic you get

$ \minus{}16x^2\plus{}160x\plus{}6000$

the maximum of a negative parabolic function (or the minimum of a positive one) occurs at $ x\equal{}\minus{}\frac{b}{2a}$ so $ x\equal{}\minus{}\frac{160}{(\minus{}16)}\equal{}\boxed{5}$


function
superior algebra
superior algebra solved

In a comutative ring with n^2 elements, n>=2 there exists at most n-2 divisors of 0.Show that the ring is a field.

Unfortunately, I published this problem in Recreatii Matematice when I was in the 12-th form and I didn't know allmost anything about superiour algebra. So, I didn't know it is well-known and I didn't know it is easy. I just had an idea I thought is nice. But I was wrong. So, the source should be: well-known.

Does this work?

We know that in a finite ring an element is not invertible iff it's a divisor of 0 (even if the ring isn't commutative, but that's just a little bit harder to prove), so in our ring there are at least n2-n+1 invertible elements. Let's look at the set x*U(A), where x is a divisor of 0. All the elements of the set are divisors of 0, so |x*U(A)|<=n-2. This means that there is a set M included in U(A) s.t. |M|>=(n2-n+1)/(n-2)>n+1 (so |M|>=n+2) and x*M has only one element. Let the elements of M be ai with i from 1 to |M|. The elements ai-a1 with i from 2 to |M| are distinct divisors of 0, but there are at least n+1 such elements, so we have a contradiction (because there are at most n-2<n+1 divisors of 0).

This shows that there are no divisors of 0 (x, in our case), so the ring is a field.

I think that with some effort, this could have been proved for non-commutative rings as well, because in all finite rings we have the equivalence a non-invertible iff a divisor of 0 and in all such rings if a is a divisor of 0 and b is invertible then ab is a divisor of 0.

What is the problem for the non-commutative rings? :)

What part of your proof is not valid if the ring is not commutative?

Oh, they're all valid, but the equivalence a non-invertible <=> a divisor of 0 is a bit trickier.

We consider the function f:A->A, f(x)=ax. Let's assume a is not a divisor of 0. This means that f is injective, but f is injective iff it?s surjective, so there is an x\in A s.t. ax=1. Let M be the set of all x\in A s.t. ax=1. If x\in M then x(a+1)-1\in M. Since f is injective, x=x(a+1)-1, so xa=1=ax, so a is invertible. Conversely, if a is invertible, then it's clearly not a divisor of 0.

This shows that the proof works fine for any ring with n2 elements.


For every n positive reals $a_i (i=1,..,n)$ such that $a_1^2+a_2^2+..+a_n^2=1$ prove that

$a_1+a_2+...+a_n +\frac{1}{a_1a_2....a_n} \geq \sqrt{n} +(\sqrt{n})^n$

First of all I will prove a beautiful lemma (which I cheated from Romanian magazine):

We have  e ≤ a ≤ b ≤ f  and we want to prove e+f ≥ a+b.

(e-a)(f-a) ≤ 0  => a2+ef ≤ a(e+f)
(e-b)(f-b) ≤ 0   => b2+ef ≤ b(e+f)

Summing : a2 + b2 + 2ef  ≤ (e+f)(a+b)

Hence if we prove ef  ≥ ab then (e+f) ≥ (a+b)
 and if we prove e+f ≤ a+b then ef ≤ ab.

So we have by AM-GM inequality we have:
(a1*a2 an) ≤ (∑ai)^n / n^n ≤ 1/ (√n)^n

(∑ai) ≤  (√n) ≤ (√n)^n ≤ 1/(a1*a2 an)

By our lemma it is enough to prove:
(∑ai)/(a1*a2 an) ≥ (√n)^(n+1)

But (∑ai)/(a1*a2 an) ≥ n/(a1*a2 an)^[(n-1)/n] ≥ (√n)^(n+1)

Or again (√n) ≤ 1/(a1*a2 an)^n q.e.d

I have another proof
I will use this Theorem
${p_1}{a_1}+{p_2}{a_2}+..+{p_n}{a_n} \ge (\sum p_i)({a_1}^{p_1}+{a_2}^{p_2}+..+{a_n}^{p_n})^\frac{1}{\sum p_i}$  for positive reals $a_i,p_i$
So rewrite the inequality
$a_1+a_2+..+a_n+\sqrt{n}^{n+1}\frac{1}{a_1a_2..a_n\sqrt{n}^{n+1}}  \ge   \sqrt{n}+\sqrt{n}^n$
Using the Theorem in the left hand side of the inequality we have
$a_1+a_2+..+a_n+\sqrt{n}^{n+1}\frac{1}{a_1a_2..a_n\sqrt{n}^{n+1}} \ge (n+\sqrt{n}^{n+1})\left(\frac{\prod a_i}{(\prod a_i)^{\sqrt{n}^{n+1}}\sqrt{n}^{(n+1)\sqrt{n}^{(n+1)}}}\right)^\frac{1}{n+\sqrt{n}^{(n+1)}}$
Now we have to prove that
$\left(\frac{\prod a_i}{(\prod a_i)^{\sqrt{n}^{n+1}}\sqrt{n}^{(n+1)\sqrt{n}^{(n+1)}}}\right)^\frac{1}{n+\sqrt{n}^{(n+1)}}$  $\ge\left(\frac{1}{\sqrt{n}}\right)$
Using the fact that $\left(\frac{1}{\sqrt{n}}\right)^n \ge \prod a_i $ which holds right away from the condition of the problem  one can easily prove the remaining part of the inequality. Of course Prowler's solution is much more easier but i really hope anybody reads mine     ;)

I hope first inequality is
 p1*x1 + p2*x2 +...+pn*xn ≥ (∑pi)*(x1^p1 +x2^p2+...+xn^pn)
Solution is very beautiful, my congratulations.

Andrei proof is a very nice application of AM-GM generalized inequality.

Sorry, Prowler but I don't understand your post.

I think that Andrei initial theorem is correct.

I am sleeping :blush:
I meant p1*x1 + p2*x2 +...+pn*xn ≥ (∑pi)*(x1^p1*x2^p2*...*xn^pn)^1/(∑pi)
because
[quote="Andrei"]I have another proof

${p_1}{a_1}+{p_2}{a_2}+..+{p_n}{a_n} \ge (\sum p_i)({a_1}^{p_1}+{a_2}^{p_2}+..+{a_n}^{p_n})^\frac{1}{\sum p_i}$  for positive reals $a_i,p_i$

and uses it in a such way:

$a_1+a_2+..+a_n+\sqrt{n}^{n+1}\frac{1}{a_1a_2..a_n\sqrt{n}^{n+1}} \ge (n+\sqrt{n}^{n+1})\left(\frac{\prod a_i}{(\prod a_i)^{\sqrt{n}^{n+1}}\sqrt{n}^{(n+1)\sqrt{n}^{(n+1)}}}\right)^\frac{1}{n+\sqrt{n}^{(n+1)}}$
[/quote]
   Or I misunderstand solution.

[quote="prowler"]First of all I will prove a beautiful lemma (which I cheated from Romanian magazine):

We have  e ≤ a ≤ b ≤ f  and we want to prove e+f ≥ a+b.[/quote]
If $ef \geq ab$, of course.


calculus
integration
inequalities
real analysis
real analysis theorems

I am looking for a simple proof of the following inequality:

${ \left\lgroup\int_{x=a}^{b}\left\lgroup \int_{y=c}^{d}f(x,y)\,dy\right\rgroup^{p}\,dx\right\rgroup^{1/p}\leq \int_{y=c}^{d}\left\lgroup \int_{x=a}^{b}f(x,y)^{p}\,dx\right\rgroup^{1/p}\,dy}$

where $p>1$. If summation sign is used instead of integral sign, this is Minkowski Inequality. Where can I find a proof?  :oops:

I had this on a problem set in analysis -- the proof there went via Holder, although of course then you would have to prove the Holder integral inequality first.  (All together, than took about 2 pages of official solution from the professor.)

It goes by the name "Minkowski integral inequality."

One way to prove it: "linearize" it by making usue of the following fact:

$\|f\|_{p}=\sup_{\|g\|_{p'}=1}\int fg$

where $p'$ is the dual exponent, $\frac1p+\frac1{p'}=1.$

What is the equality condition for the inequality?
The following link gives the prove without the equality condition:

http://calclab.math.tamu.edu/~sivan/math663_06a/appendix2.pdf
 :roll:

Here's the complete result:

$(a)~~{ \left\lgroup\int_{x=a}^{b}\left\lgroup \int_{y=c}^{d}f(x,y)\,dy\right\rgroup^{p}\, dx\right\rgroup^{1/p}\leq \int_{y=c}^{d}\left\lgroup \int_{x=a}^{b}f(x,y)^{p}\,dx\right\rgroup^{1/p}\,dy }$

$(b)~~{ \left\lgroup\int_{x=a}^{b}\left\lgroup \int_{y=c}^{d}f(x,y)\,dy\right\rgroup^{p}\, dx\right\rgroup^{1/p}= \int_{y=c}^{d}\left\lgroup \int_{x=a}^{b}f(x,y)^{p}\,dx\right\rgroup^{1/p}\,dy ~~\Leftrightarrow~~\frac{f(x,y)}{f(x,z)}=\frac{f(u,y)}{f(u,y)}}$

 :P


Find $ f:R--->R $ satisfying :
 $ f(f(x-y)) = f(x)-f(y)+f(x)f(y)-xy $

Put x=y,
f(x)^2 - x^2 = f(f(0)) = C (a constant)
Then f(x) = sqrt(x^2+C) or -sqrt(x^2+C)
Then substitute some values of x,y and it can be found that C=0 or 2

Who are you , Romano ?
 This problem is rather easy :
  Take $ x=y --> f(f(0)) = f^2(x)-x^2 $
  $ ----> f(f(0))=f^2(0) $
  Take $ x=f(0) ---> f(0)=0 $ or $ f(0)=\sqrt2 $
 After that , this problem will be easy !!!!

But if for all $x$, we have $f(x) = \sqrt {x^2 + 2}$ or $f(x) = x$, it does not mean that we have
$f(x) = x$ for all $x$
or
$f(x) = \sqrt {x^2 + 2}$ for all $x$.

Pierre.

So is the problem completely solved? ;)  :D

[quote="pbornsztein"]No.

Pierre.[/quote]

do u mind posting ur solution..?

I didn't search this problem. I was only reading the proposed answers.

Pierre.

I hope this is ok.
Assume f(0)=0, then
f(x)^2=x^2
Putting y=0 we have
f(f(x))=f(x)
so f(x-y)=f(x)-f(y)+f(x)f(y)-xy
Put x=0 we have f(-y)=-f(y)
So f(x+y)=f(x)+f(y)-f(x)f(y)+xy
Sum them up and squaring
f(x-y)^2+f(x+y)^2+2f(x-y)f(x+y)=4x^2
f(x-y)/(x-y) * f(x+y)/(x+y)=1
If we take f(x-y)=x-y, then f(x+y)=x+y
So f(x)=x for all reals, when we take y tends  to 0.
Similarly we can have f(x)=-x for all reals too.

for this case is it ok..?

Case 1: Suppose $f(x) = 0$.  It's easy to check that $f(x) = -x \forall x \in \mathbb{R}$ is not a solution, and that $f(x) = x \forall x \in \mathbb{R}$ is.  Select $a \neq 0$ such that $f(a) = a$, and suppose there is $b \neq 0$ such that $f(b) = -b$.  Let $x = a, y = b$ in the original equation, then we find that $a=1$.  This means that $f(x) = -x \forall x \neq 0, 1$.  But now, setting $x \neq 0, 1, y = 1$, it's easy to check that this does not work.

The other case looks...ugly :( .

Case 2: $f^2(x) = x^2+2$.

Suppose that there exists an $a > 0$ such that $f(a) = f(-a)$.  Substituting $x = -y = a$ into the original equation, we obtain $\sqrt{4a^2 + 4} = 2a^2+2$ (observe that the LHS cannot be $- \sqrt{4a^2 + 4}$.  But this implies that $a = 0$.  Hence $f(-x) = -f(x) \forall x \neq 0$.

Case 2.1: Suppose that $f(a) = \sqrt{a^2+2}$ for some $a \neq 0$.  Let $x = -y = a$, then we get $\pm \sqrt{a^2+1} = \sqrt{a^2+2} - 1$.  It's easy to verify that there are no solutions for both $+$ and $-$.

Case 2.2: Suppose that $f(a) = -\sqrt{a^2+2}$ for some $a \neq 0$.  Using the same method, we get $\sqrt{a^2+1} = \sqrt{a^2+2} + 1$, which is even more impossible.

this case is really ugly..... anyway u solved it :)


geometry
geometric transformation
reflection

particle man is located at $ (2,1)$ and wants to get to the lines $ x\equal{}3, x\equal{}0,$ and $ y\equal{}0$. what is the length of the shortest path he can take?

Graph and it will seem much, much easeier.

[hide]Since the lines $ x \equal{} 0$ and $ x \equal{} 3$ are $ 3$ units apart and particle man is located $ 1$ unit away from the nearest of these lines, he must move a total of at least $ 4$ units in the $ x$-direction.
Since he is located $ 1$ unit away from $ y \equal{} 0$, he must move a total of at least $ 1$ unit in the $ y$-direction.
Thus, the path must be at least $ \sqrt {4^2 \plus{} 1^2} \equal{} \sqrt {17}$ units long.

Reflect $ (2,1)$ over $ x \equal{} 3$ to $ (4,1)$. Connecting $ (0,0)$ to $ (4,1)$ and reflecting the part such that $ x > 3$ over the line $ x \equal{} 3$.
This gives $ (0,0)$ to $ (3,\frac {3}{4})$ to $ (2,1)$ which has a total length of $ \sqrt {17}$.[/hide]


probability
counting
derangement
search
function
probability and stats

A person writes $ n$ mails and he also writes addresses on $ n$ envelopes. What is the probability that neither one mail will reach its destination? What becomes this probability when $ n \to \infty$.

Read [url=http://en.wikipedia.org/wiki/Derangement]here[/url]. You can find also useful links at the end.

Or put the word "derangement" into the search function of this site to find many past discussions.


Ramsey Theory
combinatorics unsolved
combinatorics

There are $ n$ points on the plane, no three of which are collinear. Each pair of points is joined by a red, yellow or green line. For any three points, the sides of the triangle they form consist of exactly two colours. Show that $ n<13$.

[quote="KenHungKK"]There are $ n$ points on the plane, no three of which are collinear. Each pair of points is joined by a red, yellow or green line. For any three points, the sides of the triangle they form consist of exactly two colours. Show that $ n < 13$.[/quote]

I can certainly show that $ n < 17$.  Suppose, to the contrary, that there exists a set of $ n \equal{} 17$ points such that any triangle is colored with exactly $ 2$ colors.  Consider a given point $ A$, and assume without loss of generality that the color most commonly used in lines connected to $ A$ is red.  Then at least $ 6$ points must be joined with a red line to $ A$.  Any two of these points must be joined with yellow or green (lest we have a triangle with all red lines).  Consider a given point in this subset, $ B$.  Assume without loss of generality that the color most commonly used in lines, within this subset, connected to $ B$ is yellow. Then at least $ 3$ points must be joined with a yellow segment to $ B$.  Any two of these points must be joined with green (lest we have a triangle with all yellow lines).  This gives us a triangle with all green lines, which provides a contradiction.

NoYoo-hooForMe, what you have proved is essentially the Ramsey number $ N(3,3,3,2) \le 17$. (But here is it is equal to 17 due to the additional condition that any triangle will have two same colored edges.

I think this problem is quite hard(atleast for me). Can someone give a hint?


A right circular cone whose base has a radius of 4 cm and whose height is greater than 2 cm is inscribed in a sphere with a radius of 5 cm, as shown. What is the ratio of the volume of the cone to the volume of the sphere? Express your answer as a common fraction

Draw the radius from the center of the sphere to a point on the circumference of the base of the cone. Then, the distance from the center of the sphere to the center of the base of the cone is $ \sqrt {5^2 \minus{} 4^2} \equal{} 3$, so the height of the cone is 3+5=8. The volume of the sphere is $ \frac {4}{3}5^3\pi \equal{} \frac {500\pi}{3}$, and the volume of the cone is $ \frac {1}{3}4^2\times 8\pi \equal{} \frac {128\pi}{3}$, so the desired ratio is $ \frac {\frac {128\pi}{3}}{\frac {500\pi}{3}} \equal{} \frac {32}{125}$


Find all natural numbers $n$ less than 200 such that $n^2+(n+1)^2$ is a perfect square.

$n=k^2-m^2$ and $n+1=2km$
or
$n=2km$ and $n+1=k^2-m^2$

and in each of these cases solving it like the Pell's equation.


induction
LaTeX
strong induction
combinatorics unsolved
combinatorics

[quote="hollandman"]There are two strips. On the first is written letter $ A$, and on the second letter $ B$. In each minute, John writes all the letters presented on one strip on another (in the same sequence), either to the right or to the left of the existing letters. (For example, if at a moment the two strips are $ BAB$ and $ ABB$, John can write to the right of the first strip to make the two strips $ BABABB$ and $ ABB$.) Prove that at any moment, the letters on each strip form two palindromes, one of which could be empty. (For example, $ BBA$ is the palindrome $ BB$ and the palindrome $ A$.)[/quote]

Posted before, but nobody wrote a solution.

To make the problem statement clearer, John doesn't need to copy the two strips one after another. That is, he can copy the second one to right or left of the first strips multiple times successively.

This problem is VERY VERY tricky and hard. I had it solved but can only supply some basic ideas. Several suggestions.

1.Use strong induction on number of steps
2.At the beginning, Let's say John copy B to the first strip p time left and q time right. If p=q, then the thing on the first strip is still a palindrome and we can look at it as a whole. (i.e. BBABB can be simply denoted as C) If p>q, we have on the first strip like BBBAB. Make it into a palindrome and several B's either to left or right. So BBBAB becomes BBC, or BBABBBB becomes CBB. After that, John should start to copy the thing on the first strip to the other one at least once. At this point, things are getting complex. However, the main goal is to show that several steps later, you can always reach a situation that can be achieved using less steps if you change A and B at the beginning to some other palindromes. This is when the strong induction applies.

Hmm.. I would appreciate it if you could write a complete solution. I'm trying to understand your basic ideas, but they seem complicated.

Sorry, I will try to explain some detail when I have time. However, since I have few knowledge regarding Latex, it will be very hard for me to draw a diagram.

Have you at least understand all the thing before the "After that" in my second point? If so, just start copy things on the first strip to the second strip and you might see some magic happening.


modular arithmetic
algebra
function
domain
quadratics
Ring Theory
algorithm

Let $p$ be a prime number, and suppose that $x$ is an integer such that $x^2 \equiv -2 \ mod p$.By cnsidering $u+xv$ for various pairs of $(u,v)$.show that at least one of the equations $a^2+2b^2=p, a^2+2b^2=2p$has a solution.

First, note that $x^2 + 2\cdot 1^2 = mp$ for some positive integer $m$.

Choose $u,v$ from the interval $[-m/2,m/2]$ so that $u\equiv x \pmod m$ and $v\equiv 1\pmod m$. Then $mr=u^2+2v^2$ where \[r=\frac{1}{m}(u^2+2v^2)\leq \frac{1}{m}(\frac{m^2}{4}+\frac{m^2}{2})<m.\] Combine $mp=x^2 + 2\cdot 1^2$  and $mr=u^2+2v^2$ to get $m^2rp=(xu+2v)^2 + 2(xv-u)^2$, where both summands are divisible by $m$. Hence,
\[rp=\left(\frac{xu+2v}{m}\right)^2 + 2\left(\frac{xv-u}{m}\right)^2.\] We decreased $m$ to $r$. Similarly we can decrease $r$ farther until $r=1$ producing desired representation of $p$.

Nice, these quadratic problems are given all the time repeatedly and the solutions are always the same :).

[quote="maxal"]We decreased $m$ to $r$. Similarly we can decrease $r$ farther until $r=1$ producing desired representation of $p$.[/quote]
I think there is wrong wth you because you say that always $a^2+ 2b^2=p$ has solution but the problem say another thing. :?

Maxal and grobber are right! Even if you only know the statement of the problem, on can prove that there are already $a,b$ with $a^2 +2 b^2=p$:

Let be $c^2 + 2a^2 =2p$ for some $a,c$, then $c(=2b)$ is even. But this gives immediatly $4b^2+2a^2=2p$ or equivalently $a^2+2b^2=p$.

[quote="grobber"]Actually, $a^2+2b^2=p$ always has a solution.

This is because $A=\mathbb Z[\sqrt{-2}]$ is a unique factorization domain, and since $p|x^2+2=(x+\sqrt{-2})(x-\sqrt{-2})$, but it doesn't divide either one of $x+\sqrt{-2},x-\sqrt{-2}$, it means that $p$ is not a prime in $A$, which, in turn, means that it's decomposable: we can write it as $(a+b\sqrt{-2})(c+d\sqrt{-2})$.

Now, since $a+b\sqrt{-2}|p$, we also have $a-b\sqrt{-2}|p$, and the only possibility left is $a-b\sqrt{-2}=c+d\sqrt{-2}$, because both $a+b\sqrt{-2}, c+d\sqrt{-2}$ must be primes in $A$ (their norm is $p$, which is a prime in $\mathbb Z$).[/quote]

how do you know if $\mathbb{Z}[\sqrt{D}], D \in \mathbb{Z}$ is a UFD? Do you have to try to prove a Euclidean algorithm for each particular case?

[quote]Do you have to try to prove a Euclidean algorithm for each particular case?[/quote]

No, because most numbers fields with class number 1, that is, with ring of integers which is a UFD, are not Euclidean.  For example, Barnes and Swinnerton Dyer in two papers in [i]Acta Math.[/i] (1952) showed that $Q(\sqrt{97})$ is not Euclidean.  Davenport, Hua and Inkeri proved that there are only 22 quadratic Euclidean fields.

A way of showing that a quadratic field has class number 1 is by the theory of binary forms.  An ideal $A$ in a number ring $R$ in a quadratic field is a rank 2 $Z$-module, generated by $a_1,a_2$ say, and $N(a_1X + a_2Y)$ is a quadratic form.  The $SL_2(Z)$-classes of forms correspond to the ideal classes, and the theory of reduction gives an effectively computable way of listing the reduced forms, which are canonical representatives of the classes.  If there is only one reduced form, then the ring is a PID.


[quote="mdk"]In how many ways can the letters of the word AFFECTION be arranged, keeping the vowels
in their natural order and not letting the two F’s come together?[/quote]

when you mean vowels in natural order do you mean A has to be 1st or a vowel has to be first?

I have the solution :D

[hide="Solution"]

Ok, the first thing I will consider is the two F's:

There are 9 different places for these F's to be placed. First consider the placing of the first F:

There 2 different types of places, the ends, and all the places in between. For the 2 ends, there are 7 valid places left over to place the second F, for the 7 places in between, there are 6 places left over to place the second F. This gives the following amount of combinations:

$ 2*7+7*6 = 56$

(This is the edited bit:)
HOWEVER, here the 2 F's have been considered as different letters! This means that this calculation has counted each combination twice. So I will simply divide by 2:

$ (2*7+7*6)/2 = 28$

After the 2 F's are placed, there are 7 spaces left to arrange the remaining letters, and for each arrangement in the seven places, there will be 28 corresponding combinations for application to all the different F placements.

Next I'll consider the vowels in the 7 remaining positions:

If I first consider every combination that the 4 vowels can be placed, there will be:

$ \frac{7!}{(7-4)!}$

combinations. However for each different selection of 4 places to place each vowel, there will be $ 4!$ different ways of placing the vowels in those 4 places, but only 1 of them is right. It follows that only $ \frac{1}{4!}$ of the first combinations is valid. This means there remains the following amount of combinations:

$ \frac{7!}{4!(7-4)!}= 35$

This means there are:

$ 28*35 = 980$

Different ways of arranging the F's and the vowels, and there are 3 places left for placing the 3 remaining letters. Now the number of ways of placing the 3 remaining places in the 3 remaining positions is simple:

$ 3! = 6$

This means the total number of combinations is:

$ \frac{(2*7+7*6)}{2}* \frac{7!}{4!(7-4)!}* 6$
$ = 28*35*6$
$ = 5880$

[/hide]

I made a mistake!

In the above post, I forgot to divide by 2 in the F's section, as the method I described considers the 2 F's as different letters, counting each combination twice.

PS: I've now edited the post above.


function
parameterization
algebra unsolved
algebra
Fixed point

Suppose that $ u$ is a real parameter with $ 0<u<1$. Define $ f(x)\equal{}0$ if $ 0 \le x \le u$, and $ f(x)\equal{}1\minus{}(\sqrt{ux}\plus{}\sqrt{(1\minus{}u)(1\minus{}x)})^2$ if $ u \le x \le 1$, and define the sequence $ u_n$ recursively by $ u_1\equal{}f(1)$ and $ u_n\equal{}f(u_{n\minus{}1})$ for all $ n >1$. Show that there exists a positive integer $ k$ for which $ u_k\equal{}0$.

[quote="moldovan"]Suppose that $ u$ is a real parameter with $ 0 < u < 1$. Define $ f(x) \equal{} 0$ if $ 0 \le x \le u$, and $ f(x) \equal{} 1 \minus{} (\sqrt {ux} \plus{} \sqrt {(1 \minus{} u)(1 \minus{} x)})^2$ if $ u \le x \le 1$, and define the sequence $ u_n$ recursively by $ u_1 \equal{} f(1)$ and $ u_n \equal{} f(u_{n \minus{} 1})$ for all $ n > 1$. Show that there exists a positive integer $ k$ for which $ u_k \equal{} 0$.[/quote]

The key here is to see that $ f(x)<x$ $ \forall x>0$. So $ u_n$ is a positive non increasing sequence, so converging towards the only fixed point of $ f(x)$, so $ 0$.
So, for $ n$ great enough, $ u_n\leq u$ and $ u_{n\plus{}1}\equal{}0$


[b]In a class 54 % of the student are girls and 16 % are boys.If the number of girls is 27 , find the total number of student ?[/b]

where did the other 30% go?
As if there is none, this is a FAILED question  :(

The (fixed) problem (I'm guessing) is here http://www.artofproblemsolving.com/Forum/viewtopic.php?t=302062

Didn't you make this thread twice ra ny?

Uh 30% are neither girls nor boys??


Prove :  equation $ x^2\minus{}2y^2\plus{}8z^3\equal{}3$ don't have solution $ x,y,z\in N$

Check both sides $ mod \;\; 4$ we have $ x\equal{}2a\plus{}1, y\equal{}2b\plus{}1$ are odd.
So the equation becomes $ 4a^2\plus{}4a\minus{}8y^2\minus{}8y\equal{}4 \Leftrightarrow a(a\plus{}1)\minus{}2y^2\minus{}2y\equal{}1$
$ L.H.S.$ is even but $ R.H.S.$ is odd. So there is no solution.

please post this in the pre olympiad section


linear algebra
matrix
linear algebra unsolved

Solve the following matrix system: $\{\begin{array}{cc}AX-BY=C \\ BX-AY=C \end{array}$ where $A,B,C,X,Y\in\mathcal l M_{n}(R)$ and $A+B,A-B$ are not singular matrix.

So  let`s see .... by adding the two equations we get $(A+B)(X-Y)=2C \rightarrow X-Y=2C(A+B)^{-1} \rightarrow X=Y+2C(A+B)^{-1}$ and by substracting one from another we get:
$(A-B)(X+Y)=O_n\rightarrow (A-B)[2Y+2C(A+B)^{-1}]=O_n$
$\rightarrow (A-B)Y=-(A-B)C(A+B)^{-1}$
$\rightarrow Y=-C(A+B)^{-1} \rightarrow X=C(A+B)^{-1}$


I'm not so smart, so I did it in probably the longest way possible. :D

[hide]Anyway, I listed all of the factors and came out with these:

1800: 1, [color=red]1800[/color], 2, [color=red]900[/color], 3, [color=red]600[/color], 5, [color=red]360[/color], 6, [color=red]300[/color], 9, [color=red]200[/color], [color=red]10[/color], [color=red]180[/color], 18, [color=red]100[/color], [color=red]20, 90, 30,  60, 50[/color], 36, [color=red]60, 30, 90, 20, 100,[/color] 18, [color=red]200[/color], 9, [color=red]300[/color], 6, [color=red]600[/color], 3, [color=red]900[/color], 2. That was all I came up with. :(

Then I counted up the factors:
1800, 900, 600, 360, etc.

Final Answer: 23 factors are multiples of ten.

Please tell me if this is wrong, thank you...also tell me any shortcut way[/hide]

YOu counted numbers like 600 and 900 twice, and there's definetly a better solution (should just take 35 seconds max) with prime factorization.

Oops...sorry.  :blush:

[hide]Here's my answer now:

After a close examination of my solution, I found out I had repeated a LOT of numbers.

Now my answer is 14 factors.

And also, I seriously need to know the 35 seconds way.[/hide]

No, it is incorrect. This is one of two ways to solve this that I know (without listing). Contribute other ways.

1800 = 180*10.
So any factor of 180 times 10 is a factor of 1800 (and a multiple of 10)!.
We need to find the number of factors in 180.

180 = $2^2\cdot3^2\cdot5$. That means 3*3*2 = 18 factors.

[quote="236factorial"]How many factors of 1800 are multiples of 10?[/quote]

Using prime factorizaton, i get 18

[hide]Well $1800=2^3\cdot3^2\cdot5^2$  We have to have at least one $2$ and one $5$, so from the exponents we have $3\cdot3\cdot2=18$[/hide]

[quote="math92"][quote="236factorial"][quote="math92"][hide="woot!!!"]
i found all the factors of 180 and then multiplied them by 10.  There are 18 factors. ( i don't know how to do this by prime factorization).  So the final answer is [b]28[/b][/hide][/quote]

Did you make a typo? I can't see where 28 came from.[/quote]
oops, :blush:
that was a typo
[hide]18 is the real answer[/hide][/quote]

i dont think so, 1800 has 18 different factors , but the factors thats a multiple of 10 is not 18 ( 6 for example isnt a multiple of 10 :s)

[quote="Mo²"]i dont think so, 1800 has 18 different factors , but the factors thats a multiple of 10 is not 18 ( 6 for example isnt a multiple of 10 :s)[/quote]
No, $1800=2^33^25^2$ $4\cdot3\cdot3=36$  There has to be a $2$ and a $5$ to make a $10$, so you can have $1,2,3,$ 2's, $0,1,2$ 3's, and $1,2$ 5's.  $3\cdot3\cdot2=18$

[quote="loohcsnuf"]I got it!:)[/quote]
then what did you get?

[quote="loohcsnuf"]I got it!:)[/quote]
what did you get?


algebra
polynomial
trigonometry
algebra solved

$ f(x)=a_1cos x + b_1sinx + a_2 cos (2x) + b_2 sin (2x) $
 Prove that : If $ f(x) \leq \sqrt{a_2^2+b_2^2} $ then $ a_1=b_1=0 $

Let $\theta$ satisfies $a_{2}\cos 2\theta+b_{2}\sin 2\theta=\sqrt{a_{2} ^2+b_{2} ^2}$.
Substitute $x=\theta$ ,and $x=\theta+\pi$ and you will get it.

No war

[i]Edited by Myth[/i]

:D Sorry, the proof is not enough  :D
 :cool:     Try again !  :cool:

Exist $a,b$ st:
$f(x)=usin(x+a)+vcos(2x+b)$
with $u=\sqrt{a_1^2+b_1^2}$
       $v=\sqrt{a_2^2+b_2^2}$
$f(x)\le v<->v(1-cos(2x+b))-usin(x+a) \ge 0$ for all $x$
$<->2vsin^2(x+\frac{b}{2})+usin(x+a+pi) \ge 0$ for all $x$
then exist $c$:
$2vsin^2(x-c)+usin(x) \ge 0$ for all $x$.
we can assume that $-pi \le c \le pi$
For $x=c->sinc \ge 0 ->0\le c \le pi$
For $x=c-pi->sin(c-pi)\ge0->0\le c-pi\le pi$
Then $c=0,pi,-pi$
$->2vsin^2(x)+usinx \ge 0 for all x$
$->u=0$


prove that  for prime $ p > 2$ , $ p|[(2 \plus{} \sqrt {5})^p] \minus{} 2^p *2^1$

We have the question
$ \Leftrightarrow p|(2\plus{}\sqrt{5})^p\plus{}(2\minus{}\sqrt{5})^p\minus{}2^p \times 2$
$ \equal{}2(_p C_2 2^{p\minus{}2}(5)\plus{}_p C_4 2^{p\minus{}4}(5^2)\plus{}...\plus{}_p C_{p\minus{}1} 2^{1}(5^{\frac{p\minus{}1}{2}})$

Obviously, $ _p C_k\equal{}\frac{p!}{(p\minus{}k)!(k)!}$,
so for $ 1 \le k \le p\minus{}1$, $ p|nominator$ but $ p \nmid denominator$ $ \Rightarrow p|_p C_k$,
So $ p|[(2 \plus{} \sqrt {5})^p] \minus{} 2^p *2^1$, done.

stephencheng i think this is what you meant
$ (2 \plus{} \sqrt{5})^p \equal{}$ integer say $ I$ + some number $ k$ where $ k$ is a number $ >0,<1$
so gif of this $ (2 \plus{} \sqrt{5})^p$  is $ I$ and
$ (2 \minus{} \sqrt{5})^p$ is a negative fraction say $ h$
adding both we get
$ (2 \plus{} \sqrt{5})^p \minus{} (2 \minus{} \sqrt{5})^p \equal{}I \plus{} k \plus{} h$ but
$ (2 \plus{} \sqrt{5})^p \minus{} (2 \minus{} \sqrt{5})^p$ is an integer (by the binomial expansion we get so)
so $ I\plus{}k \plus{}h$ is an integer which implies $ k \plus{} h$ is an integer
we have $ 1>k>0 ,\minus{}1<h<0$ so the only integer satisfying this condition is $ k\plus{}h\equal{}0$
which implies  $ [(2 \plus{} \sqrt{5})^p] \equal{} [(2 \plus{} \sqrt{5})^p \minus{} (2 \minus{} \sqrt{5})^p ]$ ,so if we prove that $ [(2 \plus{} \sqrt{5})^p \minus{} (2 \minus{} \sqrt{5})^p ]$ is congruent to $ 2^p *2$
for which you have given the proof


function
Pascal\u0027s Triangle
algebra
binomial theorem

Find a combinatorial argument to explain the recurrence
\[\binom{n+1}{k}= \binom{n}{k-1}+k\binom{n}{k}\]

uuuh... $\binom{3}{2}$ does not equal $\binom{2}{1}+2\binom{2}{2}$   ...typo?

Presumably, they mean $\binom{n+1}{k}=\binom{n}{k}+\binom{n}{k-1}$.

This can be explained as follows. Suppose you have to pick k beads from a box, of which 'n' are black and 1 is white (but all are distinguishable). Therefore, the number of ways of doing this is the LHS. However, you can split it into two cases: EITHER you pick the white bead in which case there are $\binom{n}{k-1}$ ways to choose the remaining black beads, OR you don't pick the white bead, in which case there are $\binom{n}{k}$ ways to choose your k beads from the black beads. Therefore the LHS = the sum of these two numbers!

[quote="Fraxia"]Find a combinatorial argument to explain the recurrence
\[\binom{n+1}{k}= \binom{n}{k-1}+\binom{n}{k}\]
[/quote]

Or you can use [hide="explosives"] Pascal's Triangle. [/hide]

Hmm, I copied as it is from ACoPS, it's 6.4.17. No wonder I couldn't find a argument... Not sure if Pascal's identity was meant here, but w/e  :P

There are many ways to show that $\binom{n+1}{k}=\binom{n}{k}+\binom{n}{k-1}$.

[hide="algebraic method"]
$\binom{n}k+\binom{n}{k-1}=\frac{n!}{k!(n-k)!}+\frac{n!}{(n-k+1)!(k-1)!}$
$\frac{n!(n-k+1)+n!\cdot k}{(n-k+1)!k!}$
$\frac{(n+1)!}{(n-k+1)!k!}=\binom{n+1}{k}$[/hide]

[quote="cincodemayo5590"]Or you can use [hide="explosives"] Pascal's Triangle. [/hide][/quote]

But then you need to prove that the entries in Pascal's triangle are given by ${n \choose k}$, which would be equivalent to proving this lemma in another way.

find the $x^{k}$ coefficient of $(x+1)^{n+1}=x*(x+1)^{n}+(x+1)^{n}$

obviously this relies on the same proof for the binomial theorem, but the function proofs are always cool...this is a fast way to remeber it at least...


A simple pendulum is suspended from a peg on a vertical wall. THe pendulum is pulled away from the wall to the horizontal position and released. Rhe ball hit the wall  and the coeffecient of restitution is 2/sqrt(5) . WHat is the minimum number of collissions after which the amplitude of oscillation becomes less than 60 degrees?

By the definition of coefficient of restitution, the velocity after each impact is $ \frac{2}{\sqrt{5}}$ of the velocity before it. Hence, the Kinetic energy after impact is $ \frac{4}{5}$ of that before.

To make amplitude $ \leq 60^0$, the maximum potential energy (and hence, the total energy) should be halved.

Observe that $ \left( \frac{4}{5}\right)^4<\frac{1}{2}$, and so, $ N\equal{}4$

One request to all of u visiting this forum just hide the answers u post so that everyone gets equal chance :)

[quote="valeriummaximum"]Yeah right....

Nihat i would expect u to post the detailed solution when u post ur answers...

@shreyas-- nice proof[/quote]

I don't like to risk.It could also be a wrong answer.  :)

But Nihat, the reason you should post your solution too, is that if its wrong then we can point out the mistake in it.

It's not at all embarassing to post a wrong solution. :)

This is a forum for discussion after all :D


geometry
circumcircle
trigonometry
projective geometry
geometry theorems

At the vertices of a triangle $ \triangle{ABC}$, we draw three tangent lines to its circumcircle. If $ AA$ meets $ BC$ at $ D$, if $ BB$ meets $ AC$ at $ E$, and if $ CC$ meets $ AB$ at $ F$, show that $ D$, $ E$, $ F$ are collinear.

[i]Remark.[/i] Here, if $ P$ is some point on some circle $ k$, then $ PP$ is an abbreviation for the tangent to the circle $ k$ at the point $ P$.

[i]Alternative formulation.[/i] Prove that the tangents to the circumscribed circle of a triangle, passing through its vertices, intercept the opposite sides in three collinear points.

i  dont  know  if  its a  theorem  or not  but this is easy  proved  with  pascal  on  AABBCC ,  so even  if it has a name or not  it doesnt take much  time to  write the proof.  i  think  i saw this problem in a  moldova contest  or something like that though  :lol:

Well, I got this "theorem" from a Chinese website, the name seems Morlain line theorem...... :?

Actually, I didn't think of using Pascual...... I used Menelause on $\triangle{ABC}$.  :lol:

[color=blue] [b]Lemma.[/b] Given are a triangle $ABC$ and a point $P$ of the sideline $BC$.
Then there is the following relation:$\boxed {\frac{PB}{PC}=\frac{AB}{AC}\cdot \frac{\sin \widehat {PAB}}{\sin \widehat {PAC}}}\ .$[/color]

Here is a ([color=red]very easily[/color]) solution of the proposed problem ([u]Shobber[/u] is right : without ... [color=brown]atomic bombs[/color]).

$D\in BC\cap AA\Longrightarrow \frac{DB}{DC}=\frac{AB}{AC}\cdot \frac{\sin \widehat {DAB}}{\sin \widehat {DAC}}=\frac cb\cdot \frac{\sin C}{\sin (C+A)}=$$\frac cb\cdot \frac{\sin C}{\sin B}=\frac cb\cdot \frac cb=\left(\frac cb\right)^2\ .$
$\frac{DB}{DC}=\left(\frac cb\right)^2\ ;\ \frac{EC}{EA}=\left(\frac ac\right)^2\ ;\ \frac{FA}{FB}=\left(\frac ba\right)^2\Longrightarrow$ $\frac{DB}{DC}\cdot \frac{EC}{EA}\cdot \frac{FA}{FB}=1\Longrightarrow F\in DE\ .$

[b]Exercise.[/b] Prove the [color=green]Steiner's theorem[/color]:

Given are a triangle $ABC$ and two isogonal rays $[AM$, $[AN$, where the points $M$ and $N$ belong to the sideline $BC$, i.e. the angles ${\widehat BAC}$ and $\widehat {MAN}$ have the same angle-bisector.
Then exists the [color=green]Steiner's relation[/color]: $\boxed {\frac{MB}{MC}\cdot \frac{NB}{NC}=\left(\frac{AB}{AC}\right)^2}\ \ (1).$
[hide="Here is a solution."] $\frac{MB}{MC}=\frac{AB}{AC}\cdot \frac{\sin \widehat {MAB}}{\sin \widehat {MAC}}\ ;\ \frac{NB}{NC}=\frac{AB}{AC}\cdot \frac{\sin \widehat {NAB}}{\sin \widehat {NAC}}\ .$ The product of these two relations $\Longrightarrow (1)\ .$[/hide]

Finally I found the name of it. It is called the Lemoine Theorem, the line is called the Lemoine line of triangle ABC.

This line is a polar line of a Lemoine Point. You can easily proof that these tree points are collineral using Pascal Theorem (degenerate case)

Look at the points $A,A,B,B,C,C$ and apply Pascal.

[quote="N.T.TUAN"]Use ''http://en.wikipedia.org/wiki/Menelaus'_theorem''
If you can't then I will post my solution  :wink:[/quote]
Can you please post your solution?  :blush:

[b]Problem from different viewpoint, with different notations.[/b] Let triangle $ABC$ and $D,E,F$ be points where the incircle touches with sides($D\in BC,E\in CA,F\in AB$). Suppose $A_{1}=EF\cap BC, B_{1}=FD\cap CA, C_{1}=DE\cap AB$. Prove that $A_{1},B_{1},C_{1}$ lies a line.

[b]Solution.[/b]

By Menelaus's theorem we have
$\frac{A_{1}B}{A_{1}C}.\frac{EC}{EA}.\frac{FA}{FB}=1$( $A_{1},F,E$ lies a line).

$\frac{C_{1}A}{C_{1}B}.\frac{DB}{DC}.\frac{EC}{EA}=1$($C_{1},D,E$ lies a line).

$\frac{B_{1}C}{B_{1}A}.\frac{FA}{FB}.\frac{DB}{DC}=1$($B_{1},F,D$ lies a line).

From these we obtain $\frac{A_{1}B}{A_{1}C}.\frac{C_{1}A}{C_{1}B}.\frac{B_{1}C}{B_{1}A}=1$ and have done also by Menelaus's theorem . :)


inequalities
floor function
inequalities unsolved

There are $ n \geq 2$ permutations $ P_1, P_2, \ldots, P_n$ each being an arbitrary permutation of $ \{1,\ldots,n\}.$ Prove that

\[ \sum^{n\minus{}1}_{i\equal{}1} \frac{1}{P_i \plus{} P_{i\plus{}1}} > \frac{n\minus{}1}{n\plus{}2}.\]

So cute :) First, the statement is altogether wrong; there are [b]not[/b] $n$[b] permutations [/b]$P_1,P_2,\ldots,P_n$, each being an arbitrary permutation of $\{1,2,\ldots,n\}$, since $\dfrac {1} {P_i + P_{i+1}}$ makes no sense as a number, but rather $n$ [b]numbers[/b] $P_1,P_2,\ldots,P_n$, being an arbitrary[b] permutation [/b]of $\{1,2,\ldots,n\}$. Second, the proof is quite nice, up to a patently reversed inequality; the right form is
\begin{align*}
\sum^{n-1}_{i=1} \frac{1}{P_i + P_{i+1}}&>\frac{(n-1)^2}{2(P_1+P_2+\cdots+P_n)-P_1-P_n}
\\&=\frac{(n-1)^2}{2(1+2+\cdots+n)-P_1-P_n}
\\&=\frac{(n-1)^2}{n(n+1)-P_1-P_n}
\\&\ge\frac{(n-1)^2}{n(n+1)-1-2}
\\&=\frac{(n-1)^2}{n^2+n-3}
\\&>\frac{(n-1)^2}{n^2+n-2}
\\&=\frac{n-1}{n+2}.\end{align*}

Most likely the best such permutation is $1,n,2,n-1,\ldots, \lfloor n/2\rfloor +1$.

Could you use a smoothing argument to turn all the fractions into 1/(n+1)? I turned $p_i < p_j < p_k < p_l $ into $ \frac{1}{p_i+p_j}+\frac{1}{p_k+p_l} > \frac{1}{p_i+p_l} + \frac{1}{p_j+p_k}$ but I'm not sure if one can treat this as an invariant and solve the problem.

*Also Cauchy-Schwarz in Engel-Form = Titu's Lemma for those who read the solutions above and were a bit confused on what that was. (#me)

You can see also here for a very similar problem. http://artofproblemsolving.com/community/c6h17331p118699

$\noindent \textit {Sketch}$: use the HM-AM inequality, in which your terms are $P_i+P_{i+1}$, for $i\in \{1, 2\ldots n-1\}$ (a total of $n-1$ terms).


How will the twin impacts of the Lunar Crater Observation and Sensing Satellite, or LCROSS conducted by NASA affect science?

Only five and a half hours remain.Check this out and do express your views on the subject!

[hide][url]http://www.nasa.gov/mission_pages/LCROSS/main/index.html[/url][/hide]

The mission seems to have been successfull. Now we just have to wait for the results.

I totally want to see that being played on the news, and then have someone hack in and edit the video so the moon blows up at the end. "NASA invented a weapon capable of destroying THE MOON!"


function
trigonometry
logarithms
integration
calculus
calculus computations

(1) Find the general solution of the following differential equation.

$ \frac {dy}{dx} \plus{} P(x)y \equal{} Q(x)$ where $ P(x),\ Q(x)$ are functions of $ x$.

(2) Solve the following differential equation.

$ x\frac {dy}{dx} \plus{} 2y \equal{} \sin x$

For later easier notation, suppose we have:

$ \frac {dy}{dx} + y p(x) = q(x)$

Then

$ \frac {dy_h}{dx} + y_h p(x) = 0$

$ \frac {dy_h}{ y_h} = - p(x) dx$

$ \ln y_h = - \int p(x) \,dx + C = - P(x) + C$

$ y_h = C e^{ - P(x)} \Rightarrow y = C(x) e ^{ - P(x)}$

$ \frac {dy}{dx} = C'(x) e ^{ - P(x)} - C(x) p(x) e ^{ - P(x)}$

Therefor:

$ C'(x) e ^{ - P(x)} - C(x) p(x) e ^{ - P(x)} + C(x) e ^{ - P(x)} p(x) = q(x)$

$ C'(x) e ^{ - P(x)} = q(x)$

$ C'(x) = q(x) e ^{P(x)}$

$ C(x) = \int q(x) e ^{P(x)} \,dx + K$

And finally:

$ y = e^{ - P(x)} (\int q(x) e ^{P(x)} \,dx +K)$ with $ P(x) = \int p(x) \,dx$

Solving the second problem is easy. We have $ p(x) = \frac {2}{x}$ and $ q(x) = \frac {\sin x}{x}$. This gives us

$ P(x) = \int \frac {2}{x} \,dx = 2 \ln x$

and with this:

${ y = \frac {1}{x^2} (\int \frac {\sin x}{x} x^2} \,dx + C) = \frac {1}{x^2} (\sin x - x \cos x + C) = \frac {\sin x}{x^2} - \frac {\cos x}{x} + \frac {C}{x^2}$

That's correct.

Alternative Solution for (1):

Multiply by integerating factor $ e^{\int P(x)dx}$ both sides of the differential equation, we have

$ \frac {d}{dx} e^{\int P(x)dx}y \equal{} e^{\int P(x)dx}Q(x)$

$ \therefore y \equal{} e^{ \minus{} \int P(x)dx}\left(e^{\int P(x)dx}Q(x)dx \plus{} C\right)$.

I have a feeling these are going to get harder real quick.

[quote="kabi"]HINT for 2::
divide the eq. by x.Find Integrating factor which is equal to x^2.

Answer of integral ::
y*(x^2)=sin x - x cos (x) + C[/quote]

That's correct. :)

[quote="JRav"]$ \boxed{1}$
Set $ \mu(x) \equal{} \exp\left(\int P(x)\ dx\right)$.  Then some algebra reveals that
$ y(x) \equal{} \frac {1}{\mu(x)}\int\mu(x)Q(x)\ dx \equal{} \exp\left( \minus{} \int P(x)\ dx\right)\int\left(\exp\left(\int P(x)\ dx\right)Q(x)\ dx\right)$

Did I handle this correctly?[/quote]

Yes, you did.


group theory
abstract algebra
superior algebra
superior algebra solved

I like this one:

For any finite group G define n(G) as the number of elements of G and s(G) as the number of subgroups of G. Determine whether the following sentences are true or false:

a) For any a>0 there is a finite group G s.t. n(G)/s(G)<a.
b) For any a>0 there is a finite group G s.t. n(G)/s(G)>a.

[i]Romanian NMO/final round/Alba Iulia 1999/2'nd problem/12'th grade/proposed by Barbu Berceanu[/i]

For any finite group G define n(G) as the number of elements of G and s(G) as the number of subgroups of G. Determine whether the following sentences are true or false:

a) Take a commutative G where the order of every element is 2 (such groups exist: Z2*Z2*...*Z2 is one example).
let k=n(G)
For any x<>y in G-{e} we have that e,x,y,xy is a subgroup of G. There are (k-1)(k-2)/6 such subgroups.
So s(G)>=(k-1)(k-2)/6 -> n(G)/s(G) <= k/( (k-1)(k-2)/6 ) which goes to 0 as n goes to infinity.

b) Take (Zp,+) for p prime and get n(G)/s(G) = p/2>a

Thx! This was the official solution as well. There's a similar problem, only using the number of elements (n) and the number of multiplicative operations which can be defined on Zn s.t. it's a ring. Try it (it's also composed by Barbu Berceanu).

Can there be defined more than one multiplicative operation on Zn?

Of course not :D. Here's the real problem:

(G,+) is a commutative group with n(G) elements. i(G) is the number of laws (G,*) s.t. (G,+,*) is a ring. Find out if there exists a sequence of groups Hk (commutative and finite groups) s.t.

(a) n(Hk)/i(Hk)->0 as k->00;
(b) n(Hk)/i(Hk)->00 as k->00.

b) This is the easy part. you only need to take Zn and use the fact that only one multiplicative law makes it a ring and get n/1->00  :)

a) I will consider the group Gn=(Z2,+)x(Z2,+)x...x(Z2,+).
Denote:
  e1=(1,0,...,0)
  e2=(0,1,...,0)
  ...
  en=(0,0,...,1)
Every element of Gn can be written uniquely as a sum of some of this elements. So it is enough to define the multiplicative law for this elements.
In order to have a unit element, define e1ei=eie1=ei
The other n2-2n+1=(n-1)2 products may take any of the 2n values in Gn.
So n(Gn)/i(Gn)=2n/((n-1)2*2n)=1/(n-1)2->0  :)

I guess it's Ok. This is the 4'th problem from the final round of the RMO in 2003, 12'th grade

There is something wrong in my proof: for Zn we may choose fi(n) multiplicative laws, because the unit element is not necesarily 1. (fi(n) is Euler's function).

But the result is still OK as pn/fi(pn)=p and the set of primes goes to 00. :D

This is easier than what I tried. I used the fact that \pi{i=1->n}pi/phi(\pi{i=1->n}pi)->00 as n->00, but it's a bit harder to prove (not really hard, but your idea is definitely nicer).


calculus
integration
logarithms
function
calculus computations

Let $ f(x)\equal{}[x(1\plus{}(\ln x)^2)]^{\minus{}q/p}$.
Fix $ p \in (0,\infty)$.
For what  $ q \in (0,\infty)$ will the integral of $ f(x)$ over $ (1,\infty)$ exsists(finite)?
For what  $ q \in (0,\infty)$ will the integral of $ f(x)$ over $ (0,1)$ exsists(finite)?
Thanks for any help.

Hint: $ \minus{} \frac {d}{dx}\frac {1}{\log x} \equal{} \frac {1}{x(\log x)^2}$, and the presence of the $ 1$ has negligible effect on the function near $ \infty$ and near $ 0$ (which is where convergence or divergence of this integral is determined).

I still don't know how to do the second part.
Could someone explain more?

This is simply a matter of comparison tests for the integrability of a positive function.

For large $ x,$ $ f(x)\approx \frac1{x^{q/p}(\ln x)^{2q/p}}.$

For $ \frac{q}{p}\ne 1,$ it's the power of $ x$ that matters, and we have convergence for $ \frac{q}p>1$ and divergence for $ \frac qp<1.$

The borderline case is $ \frac qp\equal{}1,$ in which case we have essentially $ \int_?^{\infty}\frac1{x(\ln x)^2}\,dx.$ That's a convergent integral (which is what JoeBlow was saying).

For $ x$ near zero, we also have $ f(x)\approx \frac1{x^{q/p}(\ln x)^{2q/p}}.$

This time, we have convergence for $ \frac qp<1$ and divergence for $ \frac qp>1.$

Looking at the borderline case again, and let $ u\equal{}\minus{}\ln x$ :

$ \int_0^?\frac1{x(\ln x)^2}\,dx\equal{}\int_?^{\infty}\frac{du}{u^2},$ which converges.

(Note: the use of a question mark as a limit of integration simply means some value which doesn't cause us any unnecessary problems.)

[color=green][Unneeded quote removed by moderator.][/color]

Why we can neglect the effect  of $ (\ln x)^{2q/p}$ near 0?

Ok,I see.
Thanks for your help.

If $ x$ has any exponent other than $ \minus{}1,$ than we can peel of a sliver of a power of $ x$ to overwhelm any logarithm. That's why we have divergence near zero for $ \frac qp>1.$ On the other hand, if $ \frac qp\equal{}1,$ then we can't neglect the logarithm at all; it really matters. That last line of mine is a change of variables.

Also: please don't quote the entire previous post; that's what you're assumed to be replying to anyway. Use the "reply" button in most cases; only use "quote" to quote a specific portion of a post not the whole thing.


$a,b\in\mathbb{N}$ and $p=\frac{b}{4}\sqrt{\frac{2a-b}{2a+b}}$ is a prime,
find the maximum value of $p$(with proof! :) )

[hide=" "]$p=\frac{b}{4}\sqrt{\frac{2a-b}{2a+b}}$ is prime.
$p=\frac{b}{8a+4b}\sqrt{4a^2-b^2}$ is prime. Make c=2a:
$p=\frac{b}{4c+4b}\sqrt{c^2-b^2}$ is prime for $b\in\mathbb{N}$ and c is even.
Multiply by 4 and square:
$16p^2=\frac{b^2(c+b)(c-b)}{(b+c)^2}=\frac{b^2(c-b)}{c+b}$.
Since $16p^2$ has 15 factors, so does $\frac{b^2(c-b)}{c+b}$...[/hide]

Nevermind. Just being my ordinary stupid self. I ought to get better at olympiad problems before I use this forum...

hmm, i don't know where you're going with this, but i'm interested... :)

it is also on the forum .thats is for iran olympiads.

yes it is correct, it is from the iran's 16th, (i think 1998) olympiad, second round

[hide="hint"]
$p^2=\frac{b^2}{16}\frac{2a-b}{2a+b}$
$b$ must be an even number[/hide]

I was able to do a little simplification, but not much more.

[hide]
b obviously must be even.  Let $b = 2b_0$.  Then it simplifies to
\[
p = \frac{b_0}{2} \sqrt{\frac{a-b_0}{a+b_0}}
\]
Now because $\frac{a-b_0}{a+b_0} = \frac{a^2 - b_0^2}{(a+b_0)^2}$, we can rewrite it as
\[
p = \frac{b_0}{2} \cdot \frac{\sqrt{a^2 - b_0^2}}{a+b_0}
\]
$(\sqrt{a^2-b_0^2},b_0, a)$ must be a Pythagorean Triple, so let $a = r^2+s^2$ and $b_0$ can be either $2rs$ or $r^2-s^2$.  No matter which $b_0$ we chose, the equation for $p$ simplifies to the same thing, so for now let $b_0 = 2rs$.  ($r$ and $s$ are integers.)  We have $a^2-b_0^2 = (r^2-s^2)^2$ and $a + b_0 = (r+s)^2$, so it simplifies to
\[
p = rs \cdot \frac{r^2-s^2}{(r+s)^2} = rs \cdot \frac{r-s}{r+s}.
\]
(note how b_0/2 = 2rs/2 = rs) This looks much nicer, but I have no clue what to do with it :P
[/hide]

I didn't get any farther than that...

i dont think that would help you, don't  do the $r$ and $s$  thing... ;)

since no one answered this, i wont hold this anymore, here is my proof( i will do half of the problem and leave the other half to you! :) ):
$ p^2=\frac{b^2}{16}\frac{2a-b}{2a+b}$
let $b=2c$ so:
$p^2=\frac{c^2(a-c)}{4(a+c)}\Rightarrow 4p^2=c^2(\frac{a+c-2c}{a+c})=c^2(1-\frac{2c}{a+b})\Rightarrow 4p^2=c^2-\frac{2c^3}{a+b} \Rightarrow c^2-4p^2=\frac{2c^3}{a+b}\Rightarrow a+c=\frac{2c^3}{c^2-4p^2}\Rightarrow c^2-4p^2|2c^3$
$(c^2-4p^2)(-2c)+(2c^3)=8p^2c-2c^3+2c^3=8p^2c \Rightarrow c^2-4p^2=(c^2-4p^2, 2c^3)|8p^2c$
now there is two situations, first: $p\not |c$ and second: $p|c$
if $p\not |c$ then $c^2-4p^2$ and $p^2$ are primitive thus:
$c^2-4p^2|8p^2c \Rightarrow c^2-8p^2|8c$
$(c^2-4p^2)(-8)+(8c)(c)=-8c^2+32p^2+8c^2=32p^2 \Rightarrow c^2-4p^2=(c^2-4p^2, 8c)|32p^2 \Rightarrow  c^2-p^2|32$ (because $c^2-4p^2$ and $p^2$ are perimitive)
$c^2-4p^2=(c-2p)(c+2p)|2^5$ because $c-2p$ and $c+2p$ are the same, so both must be even, also $c-2p<c+2p$ so:
$p=3\Leftarrow\{
\begin{array}{cc}
c-2p=2
\\
c+2p=4
\end{array} $
$p=5\Leftarrow\{
\begin{array}{cc}
c-2p=2
\\
c+2p=8
\end{array} $
$p=4.5\Leftarrow\{
\begin{array}{cc}
c-2p=2
\\
c+2p=16
\end{array} (unacceptable)$
$p=3\Leftarrow\{
\begin{array}{cc}
c-2p=4
\\
c+2p=8
\end{array} $
so we can write $p=3,5$ in form of $(a,b\in\mathbb{N})p=\frac{b}{4}\sqrt{\frac{2a-b}{2a+b}}$ so now all you have to do is to see is there another prime bigger than $5$ that can be written in this form if not then the answer to the problem is $p=5$
now you do the second part let $p|c$...
so .... :)


Why would this be?
[img]http://img189.imageshack.us/img189/2840/theoot2.jpg[/img]

Thanks a lot.

yes but If the expresion that is red circeled then the numerator must also negatief and why is this so?

Thanks.


w/o a calculator find
a) intercepts
b) vertex
c) max/min

-4x^2 +16x-15

i completely forget how to do this stuff, dont really care about the answers, just looking for the steps that I need to take in order to get to the answers
tanks

I dont know if this is correct but as far as i can remember, Quadratic Function in vertex form is

  f(x) = a(x-h)^2 + k

and the vertex is ( (-b/2a), ((c - b^2)/4a).

When of a > 0, the parabola opens upward, therefore the vertex is minimum. When a < 0, the parabola opens downward, so the vertex is maximum. I hope this helps.

To find the intercepts, if you want to find x, substitute y to 0, the reverse (of course) if otherwise.

hERES A PROBLEM GIVEN TO US AT mATHpATH...FOR SOME REASON, ITS NO LETTIN ME WRITE IS LOWERCASE

ONE STORMY NIGHT, A PIRATES SHIP GETS CAUGHT IN A STORM.  THE SHIP IS SHIPRECKED BAND EVERYONE DIES...EXCEPT 5.  THEY SWIM TO SHORE SAFELY AND START GATHERING AS MANY COCONUITS AS THEY CAN FIND.  AT THE END IOF THE DAY, THEY PILE ALL THE COCONUTS TOGETHER AND DECIDE TO SPLIT THE PILE TOMMOROW.  THE FIRST SAILER DOESNT TRUST THE  OTHERS SO HE TAKES ONE-FIFTH OF THE PILE AND BURYS IT WILE EVERYONE IS SLEEPING.  THERE IS ONE LEFT OVER SO HE GIVES IT TO THE MONKEY.  SAILER 2 DOES THE SAM...SO DOES 3,4, AND 5

IN THE MORNING, THEY START TO SLIT THE PILE...NONE OF THEM SAY ANYTHING BECAUSE THEY ALL TOOK SOME..AFTER SPLITTING THE PILE, THERE IS ONE LEFT OVER SO THEY GIVE IT TO THE MONKEY.

WHAT IS THE SMALLEST POSSIBLE NUMBER OF COCONUTS THEY COULD HAVE FOUND.

p.s.  TRY TO ALSO FIND A NEGATIVE SOLUTION


For second use $ sin \theta \equal{} \cos(\frac {\pi}{2} \minus{} \theta)$
and $ \sin^{2}\theta \plus{} \cos^{2}\theta \equal{} 1$

For third use $ 4\sin A\sin (60 \minus{} A)\sin (60 \plus{} A) \equal{} \sin 3A$
so answer will be $ \frac {\sin60^\circ}{4} \equal{} \frac {\sqrt {3}}{8}$

and 1^{st }is very trivial

$ 4\sin A\sin (60 \minus{} A)\sin (60 \plus{} A) \equal{} \sin 3A$ is a useful identity, but back in the day (hah,  :roll: ) when I was going through high school, people didn't promote that identity much.

In general, a useful approach is to try to write the expressing using as few "weird" angles and as many "common" angles as possible, and then try to simplify the expression. Oftentimes, the "weird" angles will cancel out, or sometimes, such as in this case, you'll end up with "common" angles.

Let $ S_\theta \equal{} \sin \theta$ and $ C _\theta \equal{} \cos \theta$, for shorthand.

Let's see how this plays out in this case:

[hide="Weird angle is 10"]$ S_{20} S_{40} S_{80}$

$ \equal{} S_{(30 \minus{} 10)} S_{(30 \plus{} 10)} C_{10}$

$ \equal{} (S_{30} C_{10} \minus{} C_{30} S_{10})(S_{30} C_{10} \plus{} C_{30} S_{10} ) C_10$

$ \equal{} \left( S^2 _{30} C^2 _{10} \minus{} C^2 _{30} S^2 _{10} \right) C_{10}$

$ \equal{} \left( \frac {1}{4} c^2 _ {10} \minus{} \frac {3}{4} s^2 _{10} \right) C_{10}$

$ \equal{} \frac {1}{4} \left( C^3 _{10} \minus{} 3 S^2 _{10} C_{10} \right)$

$ \equal{} \frac {1}{4} C_{(3 \cdot 10)} \equal{} \frac {\sqrt {3}}{8}$[/hide]


One side of a square is 12 + 3x centimeters and another side is 32 + x centimeters. How many centimeters is the length of the diagonal of the square? Express your answer in simplest radical form.

[hide] We have $ 12\plus{}3x\equal{}32\plus{}x \implies x\equal{}10$. Therefore, the side length of the square is $ 32\plus{}10\equal{}42$ and the diagonal is $ \boxed{42\sqrt{2}}$.[/hide]

Just set the two expressions equal.  12+3x=32+x

2x=20, x=10.

Therefore, the length of the diagonal is $ 42\sqrt{2}$.


function
inequalities
inequalities solved

Show that if $x,y,z>0$ we have:

$ \frac{x}{x+2y+2z} + \frac{y}{2x+y+2z} + \frac{z}{2x+2y+z} \geq \frac{3}{5}$

I have found this solution:

WLOG $x+y+z=1$

we know that the function $f(x)= \frac{1}{x}$ is convex (you can show this using derivates) and you apply Jensen's inequality with weights $x,y,z$.

$ \frac{x}{x+2y+2z} + \frac{y}{2x+y+2z} + \frac{z}{2x+2y+z} \geq \frac{1}{1+2( \sum(xy))} \geq \frac{3}{5}$ because $ ( \sum(x))^2 \geq 3 \sum(xy)$

what is your solution?

cheers

If you put a=x+2y+2z,b=2x+y+2z,c=2x+2y+2 you obtain

x=(2b+2c-3a)/5
y=(2c+2a-3b)/5
z=(2b+2c-3a)/5

so that your inequality is tranformed in

b/a +a/b +c/a + a/c +b/c + c/b \geq 6

which is true as a/b +b/a \geq 2   by AM-GM.


geometry
3D geometry
algebra
factorization
difference of cubes
special factorizations

[hide]Multiplying both sides by $ \sqrt[3]{2}\minus{}1$ gives $ \sqrt[3]{4}\minus{}\sqrt[3]{2}$[/hide]

How do I know what to multiply to the denominator to get rid of the radicals?

[hide="sub"]We let $ \sqrt[3]{2}\equal{}x$. Thus we want $ \dfrac{x}{1\plus{}x\plus{}x^2}$. We recognize the denominator from the difference of cubes: $ x^3\minus{}1\equal{}(x\minus{}1)(x^2\plus{}x\plus{}1)$. Thus we can multiply the numerator and the denominator by $ \sqrt[3]{2}\minus{}1$ to get $ \dfrac{\sqrt[3]{4}\minus{}\sqrt[3]{2}}{2\minus{}1}\equal{}\sqrt[3]{4}\minus{}\sqrt[3]{2}$.[/hide]

Mmm okay a new one

Simplify

$ \frac{1}{1\plus{}\sqrt{2}\plus{}\sqrt{3}}$

You'll notice the trick used above no longer works.  What to do?

[hide="Idea"] Get rid of one radical first and then the other.  In other words, let's try multiplying by $ (1 \plus{} \sqrt{2} \minus{} \sqrt{3})$.  This gives

$ (1 \plus{} \sqrt{2})^2 \minus{} 3 \equal{} 2 \sqrt{2}$.

Now we just have to multiply by $ \sqrt{2}$. [/hide]
[hide="Generalization"] Rationalize $ a \plus{} b \sqrt{c} \plus{} d \sqrt{e}$ by multiplying by

$ (a \plus{} b \sqrt{c} \minus{} d \sqrt{e})(a \minus{} b \sqrt{c} \plus{} d \sqrt{e})(a \minus{} b \sqrt{c} \minus{} d \sqrt{e})$.

These three quantities are the (generalized) [url=http://en.wikipedia.org/wiki/Conjugate_element_%28field_theory%29]conjugates[/url] of $ a \plus{} b \sqrt{c} \plus{} d \sqrt{e}$. [/hide]