calculus
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calculus computations

I don't know how to solve the following problem.

I thought if f(x) = a sin (nx), then the integral becomes zero and hence the minimum, why is that incorrect?

To prove:

If
\[ f \ is \ integrable \ on \[\left [ \pi , - \pi \right ]\]

\[ the \ minimum \ value \ of \ \[\int_{ - \pi}^{\pi}[f(x) - a \ sin \ nx]^{2}dx\]
occurs when
\[ a = \frac {1}{\pi }\int_{ - \pi }^{\pi } f (x)\ sin (nx)\ dx\]

That's not correct; that shouldn't be $ \sin nx$ in one place and $ \cos nx$ in another - it should be sine in both places or cosine in both places.

This is an orthogonal projection in an inner product space.

The analogous problem in $ \mathbb{R}^n$ would be this: given a vector $ x$ and a nonzero vector $ v,$ find $ a$ that minimizes $ \|x\minus{}av\|^2.$ The answer will be that $ a\equal{}\frac{x\cdot v}{v\cdot v}.$

[quote="Kent Merryfield"]That's not correct; that shouldn't be $ \sin nx$ in one place and $ \cos nx$ in another - it should be sine in both places or cosine in both places.

This is an orthogonal projection in an inner product space.

The analogous problem in $ \mathbb{R}^n$ would be this: given a vector $ x$ and a nonzero vector $ v,$ find $ a$ that minimizes $ \|x \minus{} av\|^2.$ The answer will be that $ a \equal{} \frac {x\cdot v}{v\cdot v}.$[/quote]

You are correct. It should be sine in both places or cosine in both places. The wrong $ \cos nx$ in post #1 have been corrected.

But I still have no clue how to prove the $ a$ which gives the minimum value for $ f(x)$.

We have the following:

$ G(x,a) \equal{}\left(f(x) \minus{} a\sin nx\right)^2$

$ \equal{} f(x)^2 \minus{} 2af(x) \sin nx \plus{} a^2 \sin^{2} nx$

$ \equal{} f(x)^2 \minus{} 2af(x) \sin nx \plus{} a^2\cdot \frac{1 \minus{} \cos 2nx}{2}$.

Set $ I(a) \equal{} \int_{\minus{}\pi}^{\pi}G(x,a)dx$

We get:

$ \frac{dI}{da} \equal{} \int_{\minus{}\pi}^{\pi} a dx \minus{} 2a\int_{\minus{}\pi}^{\pi}\frac{1 \minus{} \cos 2nx}{2} dx \minus{} 2\int_{\minus{}\pi}^{\pi}f(x) \sin nx dx$

$ \equal{} 2a\pi \minus{} 2\int_{\minus{}\pi}^{\pi}f(x) \sin nx dx$

Now, $ \frac{dI}{da} \equal{} 0 \implies a \equal{} \frac{1}{\pi}\int_{\minus{}\pi}^{\pi}f(x) \sin nx dx$

akech, you made that harder than it had to be.

Here's a way that does use differentiation, although since this is ultimately a linear algebra problem, that's not as illuminating as it could be.

Let $ I(a)\equal{}\int_{\minus{}\pi}^{\pi} (f(x)\minus{}a\sin nx)^2\,dx$

$ \frac{dI}{da}\equal{}\int_{\minus{}\pi}^{\pi}(\minus{}2\sin nx)(f(x)\minus{}a\sin nx)\,dx$

This derivative equals zero when that integral equal zero, which is when

$ \int_{\minus{}\pi}^{\pi}f(x)\sin nx\,dx\equal{}a\int_{\minus{}\pi}^{\pi}\sin^2 nx\,dx.$

We do need to know that $ \int_{\minus{}\pi}^{\pi}\sin^2 nx\,dx\equal{}\pi.$ Once we've made that calculation, we have the answer.

Just for completeness, $ \frac{d^2I}{da^2}\equal{}\int_{\minus{}\pi}^{\pi}2\sin^2x\,dx>0,$ so the critical point is a minimum.

I'll also add that the whole statement is rather strange in that you haven't done anything with the $ n.$ By implication, you haven't yet done anything with the fact that $ \{\sin nx: n\in\mathbb{N}\}$ is an orthogonal set with respect to the inner product

$ \langle f,g\rangle\equal{}\int_{\minus{}\pi}^{\pi}f(x)g(x)\,dx.$

(Or, for that matter, the fact that $ \{\cos nx,\sin nx\}$ is an orthogonal set.)

Of course, there are many textbooks that develop Fourier series.

Now I realize the required condition, which have not been stated on the corresponding question in my textbook:

$ a$ must be a constant, cannot be a variable dependent on $ x$

The question plagued me earlier because I attempted to put
\[ a \equal{} \frac {f(x)}{sin(nx) }\]

My question posted in this thread is the part (a) of the corresponding question on my "integral calculus" textbook. In part (b) of the corresponding question, we have to use the orthogonality relations to prove an identity.


Επειδη το χθεσινο ζεσταμα δεν φανικε να συγκινησε αρκετους εχουμε και λεμε....
      [b]ΓΙΑ ΜΕΤΑ ΤΟ ΖΕΣΤΑΜΑ.......[/b]
     1)Να δειχθει οτι η εξισωση x^ν+y^ν=z^ν,  ν εΝ,ν>,δεν εχει λυση (x,y,z)στους ακεραιους με
        0<x<=ν  , 0<y<=ν
     2)Νδο 20/60<sin20<21/60  (Yποδ:Θ.Bolzano για την f(x)=8x^3-6x+3^(1^2)
     3)Nδο η εξισωση x^ν+y^ν=z^(ν+1) εχει απειρες λυσεις (x,y,z) στους ακεραιους οπου ν=θετ.ακεραιος

       [b]ΛΙΓΟ ΠΡΙΝ ΤΟ ΚΑΨΙΜΟ!......[/b]
      1)Να βρεθουν οι θετ.ακεραιοι x,y για τους οποιους ισχυει
        x^y-y^x=x-y
      2)Για καθε φυσικο αριθμο x,εστω S(x) το αθροισμα των ψηφιων του και P(x) το γινομενο των ψηφιων του.
        Νδο για καθε φυσικο αριθμο ν,υπαρχουν απειροι φυσικοι αριθμοι x ωστε
        S(S(x))+P(S(x))+S(P(x))+P(P(x))=ν

       [b]ΦΩΤΙΑ ΣΤΑ ΜΠΑΤΖΑΚΙΑ ΜΑΣ........!!![/b]
       1)Εστω ακολουθια α(ν) που οριζετε ως:
         α(1)=α(2)=1    και
         α(ν+2)=14α(ν+1)-α(ν)-4  για ν=1,2,....
         Νδο καθε ορος της ακολουθιας ειναι τελειο τετραγωνο ακεραιου αριθμου.
       2)Για την ακολουθια α(ν) ισχυει:
         α(1)=2  και
         α(ν+1)=(2α(ν))/(((((α(ν)/(2^ν))^2+1)^(1/2))+1) δηλαδη ο α(ν) δια του 2^ν, ολο στο τετραγωνο ,μετα συν 1
         ολο αυτο σε ριζα,και μετα συν 1 και ολο αυτο στον παρανομαστη.
        Νδο α(ν)>π/2    (π=3,14..)για καθε ν=1,2,....
      3)Αν x,y,z θετ.αριθμοι με xy+yz+zx=1
        Nδο(x^4+x^2)(y^4+y^2)(z^4+z^2)<=(2/3)^6

        ΚΑΛΟ ΒΟΛΙ.....

        ΡΑΛΛΗΣ :D

[b]  ΠΑΡΑΛΕΙΨΗ!!!!!!.....[/b]
ΓΙΑ ΤΟ 1 ΑΠΟ ΤΟ ΖΕΣΤΑΜΑ ΕΙΝΑΙ <.....ν>1...>

Ραλλη το χεις παρακάνει με τις φωτιές

[quote="RAKAR"]Επειδη το χθεσινο ζεσταμα δεν φανικε να συγκινησε αρκετους εχουμε και λεμε....
      [b]ΓΙΑ ΜΕΤΑ ΤΟ ΖΕΣΤΑΜΑ.......[/b]

     $1)$Να δειχθει οτι η εξισωση $x^{n}+y^{n}=z^{n}, n \in \mathbb{N} \, ,n>1$ ,δεν εχει λυση $(x,y,z)$ στους ακεραιους με
          $0<x\leq n$  ,   $0<y\leq n$

     $2)$Νδο $\frac{20}{60} < sin20< \frac{21}{60}$  (Yποδ:Θ.Bolzano για την $f(x)=8x^{3}-6x+\sqrt{3}$)

     $3)$Nδο η εξισωση $x^{n}+y^{n}=z^{n+1}$ εχει απειρες λυσεις $(x,y,z)$ στους ακεραιους οπου $n \in \mathbb{Z}_{+}$

       [b]ΛΙΓΟ ΠΡΙΝ ΤΟ ΚΑΨΙΜΟ!......[/b]

      $1)$Να βρεθουν οι θετ.ακεραιοι $x,y$ για τους οποιους ισχυει $x^{y}-y^{x}=x-y$

      $2)$Για καθε φυσικο αριθμο $x$,εστω $S(x)$ το αθροισμα των ψηφιων του και $P(x)$ το γινομενο των ψηφιων του.
        Νδο για καθε φυσικο αριθμο $n$,υπαρχουν απειροι φυσικοι αριθμοι $x$ ωστε $S(S(x))+P(S(x))+S(P(x))+P(P(x))=n$

       [b]ΦΩΤΙΑ ΣΤΑ ΜΠΑΤΖΑΚΙΑ ΜΑΣ........!!![/b]

       $1)$Εστω ακολουθια $a_{n}$ που οριζετε ως:

         $a_{1}=a_{2}=1$   και   $a_{n+2}=14a_{n+1}-a_{n}-4$  για $n=1,2,...$.

         Νδο καθε ορος της ακολουθιας ειναι τελειο τετραγωνο ακεραιου αριθμου.

       $2)$Για την ακολουθια $a_{n}$ ισχυει:

         $a_{1}=2$  και

         $a_{n+1}=\frac{2a_{n}}{1 + \sqrt{{ \left( \frac{a_{n}}{2^n} \right)}^2+1}}$

        Νδο $a_{n}>\frac{\pi }{2}$   για καθε $n =1,2,...$.

      $3)$ Αν $x,y,z$ θετ.αριθμοι με $xy+yz+zx=1$

        Nδο $(x^4+x^2)(y^4+y^2)(z^4+z^2) \leq {\left( \frac{2}{3}\right) }^6$

        ΚΑΛΟ ΒΟΛΙ.....

        ΡΑΛΛΗΣ :D[/quote]

     [u]Αυτές είναι οι ασκήσεις[/u] ??    :)

Ναι αυτες ειναι οι ασκησεις,ευχαριστω.Με δυσκολευει αρκετα η latex αλλα θα την μαθω που θα παει!!!  :wallbash:
Κανεις,καμια ιδεα ....... :whistling:  :juggle:

   ΡΑΛΛΗΣ :D

Στη πρώτη του ζεστάματος .....
[i]Απόδειξη.[/i] Υποθέτουμε ότι υπάρχουν ακέραιοι x, y, z ώστε  $x^n+y^n=z^n$ και  $0<x\leq n$  και 0 < y.Υποθέτουμε ΧΒΤΓ ότι  $x\leq y$  λόγω συμετρίας ως προς x,y
Αφού  0 < x, έχουμε  $z^n=x^n+y^n>y^n$, άρα   z > y, και άρα , $z\geq y+1$.

Έτσι ,

$x^n=z^n-y^n\geq\left(y+1\right)^n-y^n=\left(y^n+ny^{n-1}+\underbrace{...}_{\text{+}}\right)-y^n>ny^{n-1}$.

Αφού  $x\leq n$, έχουμε  $ny^{n-1}\geq xy^{n-1}$.  Έτσι , $x^n>ny^{n-1}\geq xy^{n-1}$, και άρα  $x^{n-1}>y^{n-1}$, και άρα  x > y. Αυτό αντιτείθεται με το  $x\leq y$.Άρα άτοπο και δεν υπάρχουν λύσεις.

Έχω λύσει όλες τις ασκήσεις εκτός αυτής με τα S, P με τα αθροίσματα ψηφίων όμως ειδικά στις ακολουθίες οι λύσεις είναι λίγο μεγάλες......
Αν θέλετε κανένα hint.....

Για συμφιλίωση κάνω ποστ μια λύση:

Αρχική μου σκέψη , που παίζει το αn , παρατήρησα ότι κινείται μεταξύ 2(πάνω φράγμα) 0 κάτω χωρίς να είναι
καλό το κάτω φράγμα το αρχικό που βρίσκω. Μετά είδα κάτω από ρίζα το x^2 + 1 με χ=an/(2^n) και αμέσως
χωρίς σκέψη θέτω αn/(2^n) = tan(xn) , xn μία ακολουθία με χ1=π/4. (n εχω πάντα δείκτη, π/2>xn>0 για κάθε n). μετά από πράξεις γίνεται.

  2^(n+1) * tan(x(n+1)) = 2^(n+1) * sin(xn) / ( 1 + cos(xn) ) =>

   sin(x(n+1)) + sin(x(n+1))cos(xn) - cos(x(n+1))sin(xn) = 0 =>
   sin(x(n+1)) = sin(xn - x(n+1)) => xn = 2x(n+1) => x(n+1) = x1/ (2^n) = π / 2^(n+2) . Όμως
   tax>x για κάθε π/2>χ>0 και άρα tan(x(n+1))>x(n+1)= π / 2^(n+2) και άρα a(n+1) > π/2

Προσοχή , τα γράμματα n , π μοιάζουν!

Για την 1:

 Όλο το ζουμί είναι να γίνει η παρατήρηση ότι an= (4*sqrt(a(n-1)) - sqrt(a(n-2)))^2 .
  Αυτό το αποδεικνύουμε επαγωγικά

[quote="jj-the-king"]Για συμφιλίωση κάνω ποστ μια λύση:

Αρχική μου σκέψη , που παίζει το αn , παρατήρησα ότι κινείται μεταξύ 2(πάνω φράγμα) 0 κάτω χωρίς να είναι
καλό το κάτω φράγμα το αρχικό που βρίσκω. Μετά είδα κάτω από ρίζα το $x^2 + 1$ με $x= \frac{a_{n}}{2^n}$ και αμέσως
χωρίς σκέψη θέτω $\frac{a_{n}}{2^n} = tan(x_{n})$ ,$x_{n}$ μία ακολουθία με $x_{1} = \frac{\pi}{4}$. ($n$ εχω πάντα δείκτη, $\frac{\pi}{2}>x_{n} >0$ για κάθε $n$).

 μετά από πράξεις γίνεται.

  $\\ 2^{n+1} * tan(x_{n+1}) = 2^{n+1} \cdot \frac{ sin(x_{n})} { 1 + cos(x_{n}) } \Rightarrow \\ \\ sin(x_{n+1}) + sin(x_{n+1})cos(x_{n}) - cos(x_{n+1})sin(x_{n}) = 0 \Rightarrow \\ \\ sin(x_{n+1}) = sin(x_{n} - x_{n+1}) \Rightarrow x_{n} = 2 x_{n+1} \Rightarrow x_{n+1} = \frac{x_{1}}{2^n} = \frac{\pi} { 2^{n+2}}$ .

 Όμως $tan(x)>x$ για κάθε $\frac{\pi}{2}>x >0$ και άρα $tan(x_{n+1})>x_{n+1}=\frac{\pi} { 2^{n+2}}$ και άρα $a_{n+1} > \frac{\pi}{2}$

Προσοχή , τα γράμματα n , π μοιάζουν![/quote]

Πολύ όμορφη λύση  :lol:   :thumbup:

Μπραβο παιδια,ωραιες λυσεις!
<jj-the-king>,ιδια λυση με ην δικια μου.Επιασες το κλειδι της ασκησης!!!!
<nicolas>,πολυ κομψη λυση!!!Η δικια μου λυση ειναι 1,5 σελιδα...(την ελυσα τριγωνομετρικα και καταλαβαινεις τι εγινε με τις πραξεις.....Την εβαλα μηπως βρεθει καμια καλυτερη λυση και να τελικα που...)
<silouan>,σωστος!
Ε,εχω βρει μια πολυ ωραια λυση για το 1ο απο το καψιμο.Γραψτε και την δικια σας....
Και κατι δικο μου,που εβγαλα η αν θελετε προσεξα χθες το βραδυ:
Νδο ο αριθμος 37 διαιρει τον εξαψηφιο αριθμο αβγδεζ,αν και μονο αν,ο 37 διαιρει το αθροισμα των τριψηφιων (αβγ+δεζ) οπου αβγ=100α+10β+γ

ΡΑΛΛΗΣ :D

Για την πρώτη από το κάψιμο. Μια λύση στην $x^y+y^x=x-y$   είναι  η  x=y .

Για χ diaforo του y  θέτουμε  $x=y+z$  και αποδεικνύουμε ότι για $y>2$   aυτή δεν έχει λύση .(με Bernoulli)

Me τά είναι εύκολο για y=1  ή  y=2   .....

Σιλουάν για δείξε πως βγαίνει με Bernulli γιατί εγώ χρησιμοποίησα ότι
 η ln(x)/x είναι γν.φθίνουσα για χ>e και άρα αν είναι x,y>e και x>y τότε το
πρώτο μέλος είναι αρνητικό ενώ το δεύτερο θετικό.  Για x=y ισχύει η ισότητα.
Και μένουν 2 περιπτώσεις , y=1 => ισχύει για κάθε x ανήκει στο Ν και
y=2 όπου έχω x^2 - 2^x = x-2  όμως επαγωγικά προκύπτει ότι
x^2 - 2^x <= 0 για x>=4 , οπότε ελέγχω για x=3 και βρίσκω μια λύση.

Άρα οι λυσεις είναι οι (3,2) , (2,3) , (x,x) , (x,1) , (1,x)

Φωτιά στα μπατζάκια  1: Γραμμική μη ομογενής εξίσωση διαφορών δευτέρας τάξεως. Άρα

 a(n)=(1/6)[(26 - 15(3)^(1/2))(7 + 4(3)^(1/2))^n + (26 + 15(3)^(1/2))(7 - 4(3)^(1/2))^n] + (1/3) = [(1/6)[(9 - 5(3)^(1/2))(2 + 3^(1/2))^n +

(9 + 5(3)^(1/2))(2 - 3^(1/2))^n]]^2,

και η παράσταση μέσα στις εξωτερικές αγκύλες που είναι μέσα στο τετράγωνο είναι η λύση του προβλήματος αρχικών τιμών

 b(n + 2) = 4b(n + 1) - b(n),  b(1) = b(2) = 1.

ΟΕΔ.

Για το 1ο καψιμο:Οι λυσεις (χ,ψ)ειναι προφανως συμμετρικες.Θεωρουμε λοιπον χ>ψ<=>χ>=ψ+1 αφου χ,ψ ειναι ακεραιοι.
Με εφαρμογη της Bernoulli προκυπτει  ψ+1>ψ(ψ^(1/ψ))<=>ψ<(1+1/ψ)^Ψ    Ομως γνωριζουμε οτι οταν ν->απειρο, τοτε
lim(1+1/ν)^ν=e=2,7....<3 οποτε ψ<(1+1/ψ)^ψ<e<3 και τελικα βγαινουν οι λυσεις (2,3),(3,2),(1,κ),(κ,1),(κ,κ) με κ εΝ.
Και αβγδεζ=1000αβγ+δεζ=(37)(27)αβγ+(αβγ+δεζ) και προκυπτει το ζητουμενο.

ΡΑΛΛΗΣ :D

Και κατι ασχετο!
 Το 3ο θεμα της Μεσογειαδας ειναι σιγουρα λαθος οπως μας δωθηκε...
Εκτος απο το τριγωνο (3,4,5) και το τριγωνο (4,5,6) οπου η μια γωνια ειναι διπλασια της αλλης(αποδεικνυεται ευκολα)
και ολα τα τριγωνα (κ^2-2,κ^2-1,κ^2) με κ εΝ ,δεν πληρουν το αντιστροφο.
ΑΙΣΧΟΣ ΚΑΙ ΝΤΡΟΠΗ ΣΤΟΥΣ ΥΠΕΥΘΥΝΟΥΣ!!!!!!

ΡΑΛΛΗΣ :D

Dulac ή κ. Dulac (δεν ξέρω την ηλικία σας ) το φτιάχνω λίγο σε Latex για να φαίνεται καλό .Ελπίζω σιγά σιγά όλοι να μάθουμε Latex  ....Είναι θέμα συνήθειας....

[quote="DuLac"]Φωτιά στα μπατζάκια  1: Γραμμική μη ομογενής εξίσωση διαφορών δευτέρας τάξεως. Άρα

 $a_n=\frac{(26-15\sqrt{3})(7+4\sqrt{3})^n+(26+15\sqrt{3})(7-4\sqrt{3})^n}{6}+\frac{1}{3}=(\frac{(9-5\sqrt{3})(2+3\sqrt{3})^n+(9+5\sqrt{3})(2-\sqrt{3})^n}{6})^2$

και η παράσταση μέσα στις εξωτερικές αγκύλες που είναι μέσα στο τετράγωνο είναι η λύση του προβλήματος αρχικών τιμών

 b(n + 2) = 4b(n + 1) - b(n),  b(1) = b(2) = 1.

ΟΕΔ.[/quote]

[quote="DuLac"]Φωτιά στα μπατζάκια  1: Γραμμική μη ομογενής εξίσωση διαφορών δευτέρας τάξεως. Άρα

 $a_n=\frac{(26-15\sqrt{3})(7+4\sqrt{3})^n+(26+15\sqrt{3})(7-4\sqrt{3})^n}{6}+\frac{1}{3}=(\frac{(9-5\sqrt{3})(2+3\sqrt{3})^n+(9+5\sqrt{3})(2-\sqrt{3})^n}{6})^2$

και η παράσταση μέσα στις εξωτερικές αγκύλες που είναι μέσα στο τετράγωνο είναι η λύση του προβλήματος αρχικών τιμών

 b(n + 2) = 4b(n + 1) - b(n),  b(1) = b(2) = 1.

ΟΕΔ.[/quote]

Nomizw oti mia mikroula dikaiologisi gia to oti o arithmos mesa stin parenthesi einai akeraios xreiazetai, giati den einai teleiws profanes. Episis poli swstes oi eksiswseis diaforwn pou epikaleitai o k. Dulac alla den kserw kata poson einai didaktikos tropos gia kapoio mathiti olimpiadwn kathws apotelei ili tis grammikis algevras. Mipws iparxei kapoia kompsoteri lisi pou na min mplekei toso ta pragmata?

Alexandros

[quote="RAKAR"]Επειδη το χθεσινο ζεσταμα δεν φανικε να συγκινησε αρκετους εχουμε και λεμε....

     $2)$Νδο $\frac{20}{60} < sin20< \frac{21}{60}$  (Yποδ:Θ.Bolzano για την $f(x)=8x^{3}-6x+\sqrt{3}$)
[/quote]

Loipon exoume kai leme: $f(x)=8x^{3}-6x+\sqrt{3}$, ara ean thesw opou $x$ to $sinx$ exw $f(sinx)=8(sinx)^3-6sinx+\sqrt{3} = 2(4(sinx)^3-3sinx)+\sqrt{3}=-2sin3x +\sqrt{3}$  (Kathws opws gnwrizoume $sin3x=3sinx-4(sinx)^3$).
Ara exw  $f(sinx) = -2sin3x +\sqrt{3}$ ap'opou gia $x=20$ exoume $f(sin20)=-2sin60+\sqrt{3} = 0$, ara to $sin20$ einai riza tis $f(x)$.

Omws kathws $f(x)=8x^{3}-6x+\sqrt{3}$ exoume $f(\frac{20}{60})=$(prakseis)$=\frac{20^3(-46+27\sqrt{3})}{60^3}$ kai kathws $-46 +27\sqrt{3} < 0$, exoume $f(\frac{20}{60})<0$ kai omoia $f(\frac{21}{60}) >0$ sinepws apo to thewrima Bolzano, kathws i sinartisi einai sinexis sto diastima $[\frac{20}{60}, \frac{21}{60}]$, exoume mia riza tis $f(x)$. Omws i sinartisi $f(x)$ einai gnisiws auksousa gia $x\geq 1/2$(koitakste to p.x. me paragwgous), ara exei monadiki riza sto diastima auto. Koitakste twra oti to $sin20$ einai telika i monadiki riza tou diastimatos.

Alexandros

Για την 2)

καταρχήν είναι  $\frac{1+2+3+....+u}{u}\geq\sqrt[n]{u!}$  ή ισοδύναμα
$(\frac{u+1}{2})^u\geq u!$
Άρα το ένα σκέλος αποδείχτηκε.

Για το άλλο σκέλος : Παρατηρούμε ότι  to όριο της  $(\frac{u+1}{u})^u$  είναι $e$  όταν το u τείνει στο άπειρο....
Άρα $(\frac{u+1}{u})^u<3$ (1)   .

Θέτοντας u=1,2,....u  στην   (1)  και  πολλαπλασιάζοντας αυτές που θα βρούμε
παίρνουμε $\frac{1}{u!}<(\frac{3}{u+1})^u$   που είναι ισοδύναμη με τη ζητούμενη.

Σιλουανε πολυ καλα!Το β' μελος λυνεται με μια ανισοτητα απο το βιβλιο του κ.Στεργιου που λεγεται αν.του Gewichtete.

[quote="RAKAR"]      $2)$Για καθε φυσικο αριθμο $x$,εστω $S(x)$ το αθροισμα των ψηφιων του και $P(x)$ το γινομενο των ψηφιων του.
        Νδο για καθε φυσικο αριθμο $n$,υπαρχουν απειροι φυσικοι αριθμοι $x$ ωστε $S(S(x))+P(S(x))+S(P(x))+P(P(x))=n$[/quote][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=371&t=263118]εδώ[/url]


algebra
polynomial
Putnam
algebra proposed

[b]Suppose that the roots of  $ \displaystyle 4x^2+Bx+C=0   $   are distinct and located in $ (0,1) . $ Prove that at least one of coefficients $ B $ or $ C $ is'nt integer.[/b]

ok, so...
let's give an answer to this poor problem :)
we can translate the conditions into:
-8<B<0
0<C<4
0<B-16C<4
so... let's suppose they are both integers. we have that B-16C has to be equal to 1, and B == 1 (mod16).
but this implies that B=-1 or B=-7. if B=1, we have C=0, which is not possible.
if B=-7, we have C=3.
but the 4x-7x+3 = 0 has solution x=1, which is not possible. so they can't both be integers.

I remember that the original problem, was (approximatively):
[b],, Suppose $  f(x)=Ax^2+Bx+C  $ has integer coefficients. If $x_1,x_2 $ are the roots and $ 0<x_1<x_2<1 $ prove that  $ |A| \ge 5 .$[/b]

Unfortunately I not remember the exact reference.


The cards of a standard 52-card deck are dealt out in a circle. What is the expected number of pairs of adjacent cards which are both black? Express your answer as a common fraction.

For each of the $ 26$ black cards, there is a $ \frac{2\cdot26\cdot25}{51\cdot50}\equal{}\frac{26}{51}$ chance that $ 1$ of their neighbors is black, and a $ \frac{25\cdot24}{51\cdot50}\equal{}\frac4{17}$ chance that there are $ 2$ adjacent, black neighbors, so the expected value for the number of black neighbors for each of the $ 26$ cards is $ 1\cdot\frac{26}{51}\plus{}2\cdot\frac4{17}\equal{}\frac{50}{51}$.

Finally, the total expected value is $ \frac12\cdot26\cdot\frac{50}{51}\equal{}\boxed{\frac{650}{51}}$ (we had to divide by $ 2$ because each potential pair is counted twice).

The number of pairs of adjacent cards which are both black is equal to the number of black cards which have another black card to their right. For each black card, there is a 25/51 chance that the card to its right is also black, giving 1 pair, and a 26/51 chance that the card to its right is red, giving 0 pairs. There are 26 black cards, so the expected value of the number of pairs of adjacent black cards is \[ 26\left(\frac{25}{51}(1)\plus{}\frac{26}{51}(0)\right)\equal{}\frac{650}{51}\].

[quote="math154"] $ \frac{25\cdot24}{51\cdot50}$[quote]

please explain this part for me!

Just from my understanding, the part can be separated into 25/51 times 24/50. Because you are counting the number of cards after a black card is used to the numerator and denominator become less. And you multiply them because they are part of the probability (both have to work)

Also, I think the problem itself is a little ill-prepared. It could and should have stated the number in each color so that someone would know any be able to solve it instead of blindly guessing.

[hide="Or think about it this way"]
Consider any two adjacent cards in the circle. The probability both are black is given by the probability $\frac{26}{52}*\frac{25}{51}=\frac{25}{102}$. Since there're $52$ adjacent pairs, expectedly $52*\frac{25}{102}=\frac{650}{51}$ of them should be both black.

This is because 25/51 chance for a black card, and 26/51 is for the red card prob.

[hide=Solution]For any card $C$, define $X_C$ as:
$$p_C=\begin{cases}1&\text{if card C and the card to the right of C are both black}\\0&\text{otherwise}\end{cases}$$
Then, by Linearity of Expectation, the expected number of consecutive pairs of black cards would be:
$$E\left(\sum_Cp_C\right)=\sum_CE(p_C)$$
It suffices to solve for $E(p_C)$. Note that $E(p_C)$ is just the probability that card $C$ and the card to the right of $C$ are both black. Which means,
$$E(p_C)=\frac{26}{52}\cdot\frac{25}{51}=\frac{25}{102}$$
Therefore, the answer is
$$\sum_CE(p_C)=52\cdot\frac{25}{102}=\boxed{\frac{650}{51}}$$

[hide=Solution]The number of pairs of adjacent cards which are both black is equal to the number of black cards which have another black card to their right. For each black card, there is a $\dfrac{25}{51}$ chance that the card to its right is also black, giving 1 pair, and a $\dfrac{26}{51}$ chance that the card to its right is red, giving 0 pairs. There are 26 black cards, so the expected value of the number of pairs of adjacent black cards is $$26\left(\frac{25}{51}(1)+\frac{26}{51}(0)\right)=\boxed{\frac{650}{51}}$$

this is like the complicated version of that mathcounts nats question which luke won on. if I hadnt watched that I wouldve failed here lol

Sorry to bump, but this feels like each black card is independent. However, I don't think they should be independent, because after the first pair, I think that the number $\frac{25}{51}$ should change to $\frac{24}{50}$ or something.

[quote=Hilariouszeus]$$26\left(\frac{25}{51}(1)+\frac{26}{51}(0)\right)=\boxed{\frac{650}{51}}$$[/quote]

Hey I just noticed that if you'd posted 11 minutes later it would have been exactly three years later than the last post!

My thoughts: there are twenty-six black cards, divide that by two to get about thirteen pairs of cards. The actual problem stumped me until I thought of the probability of cards on each side of black cards being black. Good discussion!


inequalities
geometry
perimeter
inequalities unsolved

Let $ P$ be a mobile point, then

$ (s \minus{} a)R_1 \plus{} (s \minus{} b)R_2 \plus{} (s \minus{} c)R_3\geq 2S$

where $ s$ is the semi-perimeter, $ S$ is the area, $ a,b,c$ are the side-lengths of $ \triangle{ABC}$, $ R_1,R_2,R_3$ are  the distances from $ P$ to the vertices $ A,B,C$, respectively.

[color=purple] Here is my solution:
 It's not hard to check that $ x = \sqrt {a(s - a)},y = \sqrt {b(s - b)},z = \sqrt {c(s - c)}$ are three sides length of a triangle $ XYZ$ and $ S_{XYZ} = \frac {1}{2}S_{ABC}$ (check it normally by Heron formular). We have the generalized inequality :$ mx^2 + ny^2 + pz^2\ge4\sqrt {mn + np + pm}S_{XYZ}$ now let $ m = \frac {R_1}{a},n = \frac {R_2}{b},p = \frac {R_3}{c}$ we have
  $ R_1(s - a) + R_2(s - b) + R_3(s - c)\ge4\sqrt {\frac {R_1.R_2}{ab} + \frac {R_2.R_3}{bc} + \frac {R_3R_1}{ca}}S_{XYZ}$
 But on the other hand Hayashi'inequality give${ \frac {R_1.R_2}{ab} + \frac {R_2.R_3}{bc} + \frac {R_3R_1}{ca}}\ge 1$  and $ S_{XYZ} = \frac {1}{2}S_{ABC}$ then $ R_1(s - a) + R_2(s - b) + R_3(s - c)\ge2S$ which is the desired
@ Yu-Dong Wu: please check the message I've sent to you  :) [/color]

[color=purple] In China Inequality Research Group some one posted the following problem :
 $ R_1(s\minus{}a)\plus{}R_2(s\minus{}b)\plus{}R_3(s\minus{}c)\ge\frac{1}{2}\sqrt{a^2b^2\plus{}b^2c^2\plus{}c^2a^2}$
  I can not solve this problem and Yu-Dong Wu can you help me ? Thank you very much.[/color]

Jian liu has pointed out that the inequality is false.

[quote="evarist"][color=purple] Dear ityaa, could you please give me a counter-example ? In China do you have any stronger version of the first one inequality in this topic ?  :) [/color][/quote]

some topics are discused at here:[url]http://www.irgoc.org/bbs/dispbbs.asp?boardid=5&Id=3333[/url]
but the language is Chinese.

[color=purple] Ok, I can not read the Chinese but I can see the mathematical formulars. Anyway thank you very much,ityaa.
  And the last question of me : Could you please download this article for me, here is the link :
[url]http://www.3adisk.com/down_64208.shtml[/url]
  Thanks alot [/color]


conics
ellipse
AMC
AIME
Pythagorean Theorem
geometry

Find the locus of circle centers P of the circles tangent to $x^2+y^2=16$ and $(x-4)^2+y^2=100$

If you draw them, you see that the first circle lies entirely inside the second.  Therefore there is no line tangent to both.  That is, if you meant line.  Did you mean the set of lines tangent to $x^2+y^2=16$ and the set of lines tangent to $(x-4)^2+y^2=100$?  I'm not totally sure...maybe I missed something.

Oh...that makes it a nice problem!
You know that for any such center/circle
The radius of that circle to the larger circle is collinear/overlaps with the radius of the larger circle
The radius of that circle to the smaller circle is collinear to the radius of the smaller circle

Past that, I really don't know how to do this problem..it'll keep me busy for a while.  I guess there's some elegant solution.  I'm stuck atm, i'll come back to this thread if I get it.

Is it just not the locus of points who are $d$  from $(0,0)$ and $d+6$ from $(4,0)$
You could get an equation out of that...

[quote="seamusoboyle"]Is it just not the locus of points who are $d$  from $(0,0)$ and $d+96$ from $(4,0)$
You could get an equation out of that...[/quote]

?
$(0,0)$ and $(4,0)$ are only 4 apart.  So you can't have a point whose distance from one is 96 more than its distance from the other...
Something I'm missing?

[quote="K81o7"]Yeah, I think it's a circle...
It's the set of points equidistant from two circles.  The circle also has radius which is the arithmetic mean of the radii of the two other circles, and intersects the x-axis at midpoints of the intersections for the other two circles.  So I believe the equation is $(x-2)^2+y^2=49$[/quote]
How would you know the distance from point to circle?

Draw the tangent.  You could also find the circle with center at that point which is tangent to the circle.

You are correct.  I vaguely remember seeing this on the AIME II Math Jam.

I'm assuming this only includes circles which are externally tangent to the smaller circle, so the small circle is not allowed inside this circle...but yeah, you're right.
I probably did something incorrectly...but yeah, your method makes sense.
Suppose you wanted to find the locus of the points tangent to both circles, inside the larger circle and outside the smaller one?

This seems really familiar to me but I can't remember the exact thing I need.

[quote="Wumbate"]I think it's an ellipse with foci at the centers of the 2 circles and the sum of the distances from the foci=constant=14 (sum of 2 radii).

My reasoning:
Draw a token circle tangent to both of the other 2 circles anywhere, then draw the lines from the small circle to its center and from the big circle to its center.  Call r=radius of small circle, R=radius of big circle, and q=radius of the token circle.
Then the sum of the distacnes is:
(r+q)+(R-q)=r+R=constant, so it is an ellipse.

I think I've seen a problem like this here but I couldn't find it to link to.[/quote]
Yes, I agree. And since we have q in two lines (r+q and R-q), can we find what it's equal to for each and then create the ellipse equation?

To find the length of the axes:
You know one of them has length 8.
The other one can be found by drawing the isosceles triangle with the two foci and the point straight down from the center of the ellipse as parts of a triangle.
Calling this axis h, and using the Pythagorean theorem:
$\sqrt{4+h^2}+\sqrt{4+h^2}=14$
$4+h^2=49$
$h^2=45$
$h=\sqrt{45}$
So the equation of the ellipse will be:
$\frac{(x-2)^2}{64}+\frac{y^2}{45}=1$


One second, I had no rating and one post later I had a 2.1 rating. Is that real bad? Wouldn't I need a bunch of posts rated to get a 2.1? You can't get an average of 2.1 in one post,two posts, three posts or 4 posts, you need 10 posts rated. So someone must have a done a bunch of em at one time.

"Not rated yet" shows if you  have less than 20 posts rated.

What you said is not true. I have posted 26 posts and I am still be marked as not rated yet. I don't know why. So what does "NOT RATED YET" mean?

It means that you haven't had enough people rate your posts for an accurate rating to be represented.

Perhaps you didn't catch the word at the end; I said less than 20 posts [i]rated[/i].

sorry, my bad. I didn't read it clearly.


inequalities
geometry
trigonometry
area of a triangle
Heron\u0027s formula

For positive real numbers $a,b,c$, prove that
$(a^{2}+b^{2})^{2}\geq (a+b+c)(a+b-c)(a+c-b)(b+c-a)$

[hide]
At most one of $a+b-c$, $a+c-b$, or $b+c-a$ can be nonpositive, because if two were, say $x+y-z$ and $x+z-y$ where $(x,y,z)$ is some arrangement of $(a,b,c)$, then their sum, $2x$, would be nonpositive, which is BAD!!!! But if exactly one of them is nonpositive, the inequality is trivial; we have \[(a^{2}+b^{2})^{2}\geq 0\geq (a+b+c)(a+b-c)(a+c-b)(b+c-a)\]
Now we consider the case when $a+b-c$, $a+c-b$, and $b+c-a$ are all positive. Then $a,b,c$ are the sides of a triangle. Let $K$ be the area of the triangle and $\theta$ be the angle opposite the side of length $c$. Area formulas give us
\[(a+b+c)(a+b-c)(a+c-b)(b+c-a)=16K^{2}=4a^{2}b^{2}\sin^{2}\theta \leq 4a^{2}b^{2}\]
So we only need to prove that $(a^{2}+b^{2})^{2}\geq 4a^{2}b^{2}$, which follows from
\[(a^{2}+b^{2})^{2}-4a^{2}b^{2}=(a^{2}-b^{2})^{2}\geq 0\]
(Equality holds if and only if $a=b=\frac{c}{\sqrt{2}}$.)
[/hide]

The geometric way is the nice way to do it. The ugly way is the following one: expand everything, and reduce the inequality to $\left(2a^{2}-c^{2}\right)^{2}+\left(2b^{2}-c^{2}\right)^{2}\geq 0$. Done.

[hide]
We know that $(a-b)^{2}\geq 0$, so $a^{2}+b^{2}\geq 2ab$ and $\frac{a^{2}+b^{2}}{4}\geq \frac{ab}{2}$

We also know that the maximum area of a triangle that has two sides $a$ and $b$ is $\frac{ab}{2}$.
Now note that Heron's Formula can be rearranged to $\frac{1}{4}\sqrt{(a+b+c)(a+b-c)(a+c-b)(b+c-a)}$.

Therefore, $\frac{a^{2}+b^{2}}{4}\geq \frac{ab}{2}\geq \frac{1}{4}\sqrt{(a+b+c)(a+b-c)(a+c-b)(b+c-a)}$
$a^{2}+b^{2}\geq \sqrt{(a+b+c)(a+b-c)(a+c-b)(b+c-a)}$

$\boxed{(a^{2}+b^{2})^{2}\geq (a+b+c)(a+b-c)(a+c-b)(b+c-a)}$

Cases in which $c=a+b$ and in which $c>a+b$ are trivial - the first produces $LHS \geq 0$ and the second produces $LHS > n$ where $n$ is some negative number
[/hide]


conics
ellipse
rotation
geometry
geometric transformation
geometry proposed

Hi!

Let ${\cal E}$ be an ellipse, and consider an arbitrary point $A$ on ${\cal E}$. Prove that there  exist points $B$ and $C$ such that $\triangle ABC$ is equilateral and is inscribed in ${\cal E}$.

Michael

Rotate the ellipse around $A$ with an angle of $\frac \pi3$. You get another ellipse $\mathcal E'$, which intersects $\mathcal E$ in $A$. If there's another intersection point, then we're done. Otherwise, it must be that $\mathcal E,\mathcal E'$ are tangent in $A$. This means that the tangent in $A$ to $\mathcal E$ is the same as the tangent in $A$ to $\mathcal E'$. However, the latter is the image of the former through a rotation of angle $\frac \pi3$, so they cannot possibly coincide.


Prove that $ n\in\mathbb{N}$ could be written in the form $ 3x^2\plus{}y^2$ iff could be written in the form $ a^2\plus{}ab\plus{}b^2$.
 :D

[quote="pelao_malo"]Prove that $ n\in\mathbb{N}$ could be written in the form $ 3x^2 \plus{} y^2$ iff could be written in the form $ a^2 \plus{} ab \plus{} b^2$.
 :D[/quote]
Not too difficult to prove this problem  :oops:
If there is one of numbers a and b even ,we assume that a is even with a=2m
Then we set y=$ m\plus{}b$ and $ x\equal{}m$
if both a and b are odd
Represent:
$ a\equal{}2p\plus{}1$;$ b\equal{}2q\plus{}1$ then we set :
$ x\equal{}P\plus{}q\plus{}1$ and $ y\equal{}p\minus{}q$


which place do you think will host the 2016 olympics. right now the 4 places they are deciding on are Chicago  :lol: , Tokyo :P Madrid :mad: and Rio De janeiro( i think thats how you spell it)
 and don't just pick chicago because we live in america

Tokyo: ~14% chance of winning = they have hosted it before
Chicago: ~50% chance of winning = in 1904 they won but win was transferred to St. Louis
Madrid: ~34% chance of winning = they have bidded twice without success
Rio De Janeiro: ~2% chance of winning = Is this a serious bid?

wahh. chicago cuz i live like 40 min away!! umm... yeah i agree with james though on the percentage thing...

I'm going to say Madrid. Although, I hope it's Chicago -- then I can go watch what they do!

I dont give the last place much of a shot. However the final 3 split the chances pretty evenly. Here's my percentages.

Chicago-40%
Tokyo-35%
Madrid-25%

Comments:

Tokyo has hosted this before. Since the others haven't had a chance, less chance.
Madrid... actually has quite a bit of chance. 2 unsuccessful bids, 1976 and 2012. An unsuccessful bid in 2012 gives it lower chances. Chicago actually won the bid in 1904, but due to the World Fair it was moved to St. Louis.

(Slightly) Updated Probability

Tokyo: 20%
Chicago: 43%
Madrid: 37%

PEOPLE you're forgetting Rio de janeiro. it still ahs a chance. also fyi in 2010 china gets to hold the WORLD EXPEDITION THING. maybe its the same as teh world fair. w/e.I think its in shanghai

Err, I think we have kinda ruled out Rio De Janeiro.

1) One of the cities has to be removed in Round 1, likelier to be Rio De Janeiro than any other city.

2) Besides: its starting score: which determines "feasibility" of the Olympics was even lower than Doha, Qatar [Score was 6.1 compared to Doha's 6.7] and is lower than any of the other 3 cities.

Just a thought.

um this isn't a competition its just a prediction.who do you think will be hosting the 2016 olympics. not a competition.thers no city getting vote out

Actually this is the process used by the IOC to determine Host Cities.

chicago!! because i live in the city. but to say the truth, chicago's chances dont seem to be that high right now


Let given a positive real number $\alpha$ and a sequence of positive integers $c_{n}$ such that $c_{n}\geq\alpha\ .n$ for all $n$. Find all $k\geq\ 2$ an integer and a natural number sequence $a_{1},...,a_{k}$ such that there exists another natural number sequence $b_{n}$ : for all $n$: $\sum_{i=1}^n\ a_{i}^k=b_{n}^{c_{n}}$

The original question was : find all $(a,b)$ non-negative integers such that for any positive integer $n$: there exists an positive integer $c_{n}$ :$a^n+b^n=c_{n}^{n+1}$.


$ \frac{s^2}{h_a}\equal{}a$, $ \frac{s^2}{2r}\equal{}p$, $ \frac{s^2}{2(p\minus{}a)}\equal{}r_a$, then costruct the incircle and the A-excircle and the sideline BC that is one of the  internal tangent of these.


algorithm
vector
number theory
greatest common divisor
superior algebra
superior algebra unsolved

Take $ v_1,v_2$ and $ v_3$ in $ \mathbb{Z}^2$, and for simplicity, suppose they span a lattice of rank 2. Find an algorithm that produces a basis of the lattice they span.

Find a vector in the span whose first component is the gcd of the first components of the other vectors, then reduce the other vectors. Rinse and repeat for the other components (i.e. normal gaussian elimination, but since you can't divide you have to do it slightly different)


algebra
polynomial
quadratics
quadratic formula

[hide="Answer"]$ x \equal{} 5 \minus{} \sqrt{5}, y \equal{} 5 \plus{} \sqrt{5}$[/hide]
[hide="Solution"]Add 3 times the second equation to the first. This gives $ x^3 \plus{} 3x^2y \plus{} 3xy^2 \plus{} y^3 \equal{} 1000$, or $ (x\plus{}y)^3 \equal{} 1000$. Thus $ x\plus{}y \equal{} 10$. The second equation is $ xy(x\plus{}y) \equal{} 200$, so $ xy \equal{} 20$. Thus, $ x$ and $ y$ are the roots of the polynomial $ a^2 \minus{} 10a \plus{} 20 \equal{} 0$. Using the quadratic formula, we get $ 5 \pm \sqrt{5}$. Let $ x$ be the smaller root and $ y$ be the larger root, and we have our answer.[/hide]

[quote="Lulze"]If $ x < y$ , determine the exact ordered pair of real numbers $ (x,y)$ which satisfies the following:
$ x^3 \plus{} y^3 \equal{} 400$
$ x^2y \plus{} xy^2 \equal{} 200$[/quote][hide]
$ x^3\plus{}y^3\plus{}3(x^2y\plus{}xy^2)\equal{}(x\plus{}y)^3$
[/hide]

Hmm, I am confused as to how you obtained the polynomial $ a^2 \minus{} 10a \plus{} 20 \equal{} 0$.

See http://www.artofproblemsolving.com/Wiki/index.php/Vieta%27s_Formulas

This is a specific case: if you know the sum $ s$ and product $ p$ of two numbers, then the two numbers will be the roots of the polynomial $ x^2 \minus{} sx \plus{} p \equal{} 0$. This works even if your two numbers are complex.

[hide]$ (x \plus{} y)^3 \equal{} x^3 \plus{} y^3 \plus{} 3xy(x \plus{} y) \equal{} 400 \plus{} 3\times200 \equal{} 1000$
$ x \plus{} y \equal{} 10$
$ xy \equal{} 20$

Then, x,y are roots of $ t^2 \minus{} 10t \plus{} 20 \equal{} 0$, or $ (x,y) \equal{} (5 \minus{} \sqrt {5},5 \plus{} \sqrt {5})$[/hide]


Diophantine equation
number theory solved
number theory

It's a negative Pell equation: A^2-dB^2=-1. Some of them have infinitely many solns and some of them have none. This is probably one of the latter but I don't know why. Maybe I can find something in some books (then again, maybe I can't :D).

Pretty useless post, huh?  :D  :D  :D

[quote="bugzpodder"]show (k^2-4)x^2-y^2=1 has no integer solutions for k>3[/quote]

=====================
This can be solved using technics of Markov's equation:

If  eqn  (k^2-4)x^2-y^2=1 have integer sln then eqn  (k^2-4)x^2-y^2=4 (2) also have. Consider eqn
     z^2 - kxz + x^2 + 1 = 0 (3)
This eqn has integer sln in z iff D = (kx)^2 - 4x^2 - 4 is square, or
(k^2-4)x^2 - 4 = y^2. It mean (2) has integer sln. Vice versa, if (2) has integer sln (x,y) then (3) has integer sln (x,z). We proved that our initial problem is equivalent to so that if k> 3, (3) has no solution.

This fact can be done by following way: Let (3) has integer sln (x,z). Then it has minimal positive interger sln (x,z) with x \leq z. But if (x, z) positive interger sln of (3) then (x, kx-z) is also positive interger sln of (3). By minimalness of (x,z), we have  kx - z  \geq z. By Viet's theorem we have
x^2+1 = (kx-z)z \geq z^2. From that we have x = z. Put in (3) => (2-k)x^2 + 1 = 0. If k>3 it is impossible.

Technics of Markov eqn is very usefull. You can use it to solve the following:
1. Find all natural n s.t eqn  (x+y+z)^2 = nxyz has positive integer sln.
2. Let A - subset of N. Prove that we can find the set B such that B contains A and sum of square of all elements in B is equal to their product.

Try it.

Namdung


30.)"Right triangle ABC has one leg of length 6 cm, one leg of length 8 cm and a right angle at A..."

I know that legs typically refers to the perpendicular sides of a right triangle, but couldn't someone argue that 8 could be the hypotenuse and 6 one of the other sides, giving a third side of approximately 5 instead of 10, which would yield a different answer.

um... no?  legs are always the smaller sides :wink:

Yeah, leg always refers to the sides that are not the hypotenuse.

It is by universal definition. No arguing allowed. =D

[quote="magixter"]Why? Did anyone actually argue that?[/quote]
what do you mean by "argue". Do you mean challenge during competition, or just universally say "we're redefining the definition of 'leg'." The latter seems rather impossible  :) .

yeah, it would have to be "sides" in order to have a fair argument

anyways...

who wants to figure out 6^2+x^2=8^2????????????????(2sqrt7)

jorian

[quote="kyyuanmathcount"]At nationals (and supposedly all other levels), answers are posted after the round and you are given 30 minutes to write up a protest.[/quote]

Not at my Chapter or State. Not even at my school.

[quote="kyyuanmathcount"]Hence, "supposedly."  You can request though because it is written into the rule book.[/quote]

Who would want to do that though?  If one does really badly, they would just feel even worse.

Just saying you can.  It might actually be worth it if you can truly get the points.

Uh, anything more about the fact that "Leg" $\neq$ "Hypotenuse?"


Karen's three test grades to date in her science class are 82, 87, and 93. The final exam will count as two test grades. What grade must Karen earn on the final exam to have an overall mean of 90?

[quote=isabella2296]If the final exam is worth two tests, then we'll say she will have taken the equivalent of 5 tests in that whole marking period. She wants an average of  90, so the sum of the tests must be equal to $ 5 \times 90 \equal{} 450$. The three tests she'd already taken have a sum of $ 262$, so she needs 188 more points. The final exam grade is worth two tests, so she needs a $ 188 \div 2 \equal{} \boxed{94}$.[/quote]

Great Solution!


Let $x,y,z>0$ such that $xy+yz+zx= \sqrt{xyz}$. Prove that:
 \[  x+y+z+(\max\{x,y,z\}-\min\{x,y,z\})^2  \leq \frac13. \]

This is so EASY!!! But don't kill yourself harazi, we all love ya. ;)

You must be joking!!! Please, let me live a little bit!

dude,how do u come up with these wicked beautiful ones (was that an oxymoron or a paradox :D  :D ,if this is easy,i kill myself :D  :D

Maybe nobody will solve that one just to keep Harazi safe... :D

Pierre.

:) If someone finds a smart solution for this one, I promise I won't kill myself! Is it right now? Please, help harazi   :D .

We suppose : a \geq b \geq c>0
We denote :
x= \sqrt (ab)/c
y= \sqrt (ac)/b
z= \sqrt (ab)/c
=>x+y+z=1
The inequality becomes:
xy+yz+zx+(yz-xy) 2  \leq 1/3 (*)
Denote:
p=x+z \leq 1
q=xz
=> 0<q \leq p 2 /4
(*) <=>3p(1-p)+3q+3(1-p) 2 (p 2 -4q)-1 \leq 0
<=>3p 4 -6p 3 +3p-12p 2 q+24pq-9q-1 \leq 0
Let : f(q)=q(-12p 2 +24p-9)+3p 4 -6p 3 +3p-1
f(q) is a linear function.
So : f(q) \leq Max{f(0),f(p 2 /4)}
f(0)=3p 4 -6p 3 +3p-1 =g(p)
g'(p)=12p 3 -18p 2 +3=(2p-1)(6p 2 -6p-3) ( p in (0,1) )
=> g(p) \leq g(1/2)=-1/16
f(p 2 /4)=-9/4p 2 +3p-1=-(3/2p-1) 2  \leq 0
=>q.e.d

Maybe Harazi's promise make me feel confident to solve this problem  :D

Very nice solution, Tung Lam. I'm speaking to you now from heaven, since you killed me  :) . My solution was the following: with the same substitutions, let's make some observations:
  1/3-xy-yz-zx=1/3*(  \sum x 2 - \sum xy)=1/3( \sum x^3-3xyz) \geq 1/3*( \sum x(y-z) 2 ) (Use Schur). Now use Cauchy to find that  \sum x(y-z) 2  \geq ( \sum |xy-yz|) 2 and the fact that  \sum |xy-yz|=2(max{xy,yz,zx}-min{xy,yz,zx}).

2 "proofs from THE BOOK" :D  :D (wish erdos could have a look at these :D )


Let S be the set of all rational numbers which can be written as 0.abcabcabcabc... (where the integers a, b, c are not necessarily distinct). If the members of S are all written in the form r/s in lowest terms, how many different numerators r are required?

[hide]
all $0.\overline{abc}$ can be written as $\frac{abc}{999}=\frac{abc}{37\times 3\times 3}$

Since $001\leq abc\leq 998$ there are $998$ possible numerator before canceling with denominator.

we know that the denominator has factor of $3$ and $37$ . So there are $26$ number with factor $37$ ; $332$ number with factor $3$ ; $8$ number with factor $111$ .

hence the number of numerator is $998-(26+332-8)=\boxed{648}$
 :)
[/hide]

shyong,

[hide]$999=37 \cdot 3^3$[/hide]

but it doesn't make a difference. :P The answer is wrong however :(

[hide="hint"]multiples of 81[/hide]

oh ya , i thought that after the factor of three is cancelled then we dont need to compute $3^2,3^3,3^4$ :oops:  .... So for $81$ , there are $12$ . So add $12$ to make it
$\boxed{660}$ . :P  Is that correct ?

[quote="Wumbate"]Should you also include 0.000 = 0/1 ?[/quote]

It's in lowest terms, but $0/1$ isn't in lowest terms: $0/1=0$ so it doesn't work ;)


abstract algebra
linear algebra
linear algebra unsolved

Let a=(1,0,3), b=(0,2,0), c=(1,2,0), c*=(1,2,3) $\in$ $R^{3}$ AND LET x=(1,3), y=(0,2), z=(0,1), z*=(1,5) $\in$ $R^{2}$.
Giving reason, decide whether there are linear maps $R^{3}\to R^{2}$ which satisfy

a) $\varphi$:a $\to$ x , b$\to$ y and c $\to$z?

b) $\varphi$: a $\to$ x , b$\to$ y and c* $\to$z?

c) $\varphi$: a $\to$ x , b$\to$ y and c* $\to$z*?

In each case , state if such a linear map $\varphi$ is unique, and if so find a basis for its kernel and its image.

Try to construct $\varphi$ step by step and see if it works. First, define $\varphi(a)=x$. This determines the values of $\varphi$ on the linear span of $\{a\}$. Next, you want to set $\varphi(b)=y$ -- and you can, because $b\notin\mathrm{span}(\{a\})$. By now the values of $\varphi$ on the two-dimensional subspace spanned by $a$ and $b$ have been determined. $c$ does not belong to this span, which allows you to complete the construction for (a). But $c^{*}$ does, which means you are not free to change $\varphi(c^{*})$ -- it is already defined. The only thing you can do is to check if $\varphi(c^{*})$ happens to be the point you wanted.


[quote="Wickedestjr"]Okay I'm beginning a new game, and it is kind of random, but I didn't know where to put it, so I put it here. Basically it is a marathon of math problems, like the MC Marathon, and the MOEMS marathon, except you must put a problem that is in your homework, or you have recently solved for fun. The rules are simple. Solve the previous problem, and post the answer, and then post your problem after that. Here's the first problem;

This one was on my homework a week ago.

New Problem;

Out of 3 sports, rugby, association football, and hockey, 240 young people were asked to state their preference. 120 preferred association football, 172 preferred rugby, 128 preferred hockey. 64 liked rugby and association football, 76 liked rugby and hockey, while 68 liked association football and hockey. How many liked all three sports?[/quote]
@Mod: Please move to MATHCOUNTS Middle School Tournament forum.
Thanks.

Sorry, by concerning homework problems, this might lead to users posting their homework problems that are due, and we can't have this.


function
calculus
integration
number theory
number theory theorems

Hi, could anybody help mi with this problem:

Let [tex]x, y[/tex] be some positive irational numbers that satisfy equation [tex]x + y = xy[/tex]. Prove that for each natural number [tex]n[/tex] there exist some natural number [tex]m[/tex] such that [tex]n = [mx][/tex] or [tex]n = [my][/tex]. Where [tex][i][/tex] denotes the integral part of number [tex]i[/tex], examples: [5,1] = [5,4] = [5,7] = 5.

Thanks.

What you are looking for is Beatty's theorem (about Beatty's sequences) which states that :
For $x > 0$, let $S(x) = \{ [nx ] / n=1,2,3,... \}$.
Let $\alpha, \beta$ be two positive numbers. Then :

The sets $S(\alpha ) $ and $S(\beta)$ form a partition of $\mathbb {N}^*$ if and only if $\alpha, \beta$ are irrational and $\frac 1 {\alpha} + \frac 1 {\beta} = 1$.

The proof is not difficult but quite long and boring to write.
I think a googlization will leads to a already written proof... :D

Pierre.

Bingo :
http://www.cut-the-knot.com/proofs/Beatty.shtml

I didn't read it but it looks like a proof.
In another hand, you also could some days that our course in elementary number theory vol. 1 (for olympiad training in France) becomes in line  :D  A proof is in it.

Note that it can be proved (and it is done in the same book...) that there is no three (or more) real numbers $\alpha, \beta, \gamma$ such that $S(\alpha) , S(\beta), S(\gamma)$ form a partition of $\mathbb {N}^*$.

Pierre.

It was a Pulnam problem ,1993 or 1994, i don't remember. But the solution i read seems has no relation with the Beatty's sequences.  :?

[quote="pbornsztein"]Bingo :
http://www.cut-the-knot.com/proofs/Beatty.shtml

Note that it can be proved (and it is done in the same book...) that there is no three (or more) real numbers $ \alpha, \beta, \gamma$ such that $ S(\alpha) , S(\beta), S(\gamma)$ form a partition of $ \mathbb {N}^*$.

Pierre.[/quote]
sr , can someone prove this ???????????????????


complex numbers
algebra proposed
algebra

Let $A,B \subset \mathbb{C}^*$ be finite sets of non-zero complex numbers such that, for infinitely many integers $k$, we have
\[
\sum_{a \in A} a^k = \sum_{b \in B} b^k.
\]
Show that there exists an integer $k$ and two disjoint partitions $A = \cup_{i=1}^k A_i; B=\cup_{i=1}^k B_i$ with the following property. For each $i=1,2,\ldots,k$,
   [i](i)[/i] $\sum_{a \in A_i} a^k = \sum_{b \in B_i} b^k$ for infinitely many $k$; and
   [i](ii)[/i] for any $x,y \in A_i \cup B_i$, $x/y$ is a root of unity.

[Edited. ]

Excuse, maybe I am missing something... :blush:
Take $a_1=1$, $a_2=\exp(i\pi / A)$, $b_1=1$, $b_2=1$, $b_3=\exp(i\pi / B)$, $b_4=\exp(i\pi / C)$, $A, B, C$ are odd. It satisfies all conditions, doesn't it? Consider $k = (2j+1)ABC$.

Oh, excuse me, once again ...  :blush:  :blush:  :blush:   The problem I intended was to describe under which circumstances do we have $a_1^k + \cdots + a_m^k = b_1^k  + \cdots + b_n^k$ for infinitely many $k$; I overlooked the trivial cases involving roots of unity, mistakenly assuming the sufficiency of this condition:

[quote="vess"]
none of the sets  and  contains a subset of the form  with  a primitive th root of unity
[/quote]

Sorry about this  :blush:. Here is the revised version of the problem (I'll also edit the initial post):

[b]Problem. [/b] Let $A,B \subset \mathbb{C}^*$ be finite sets of non-zero complex numbers such that, for infinitely many integers $k$, we have
\[
\sum_{a \in A} a^k = \sum_{b \in B} b^k.
\]
Show that there exists an integer $k$ and two disjoint partitions $A = \cup_{i=1}^k A_i; B=\cup_{i=1}^k B_i$ with the following property. For each $i=1,2,\ldots,k$,
   [i](i)[/i] $\sum_{a \in A_i} a^k = \sum_{b \in B_i} b^k$ for infinitely many $k$; and
   [i](ii)[/i] for any $x,y \in A_i \cup B_i$, $x/y$ is a root of unity.

This was motivated by the following (absolutely  :) ) trivial problem: if $A,B \subset \mathbb{C} \setminus \{0\}$ are such that $\sum a^k = \sum b^k$ for all positive integers $k$, then $A \equiv B$.

--Vesselin


prove dat 20^(15) -1 is divisible by 11X31X61.

[hide]Prove individually for 11,31,61 on the expression[/hide]


Let $a,b,c>0$ such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$.

 Prove that $\frac{a}{bc(b+c)}+\frac{b}{ca(c+a)}+\frac{c}{ab(a+b)}\geq\frac{3}{2}$ ;)

$\sum{\frac{1}{a}}\ge \sum{\frac{a}{b}}|\cdot abc \Rightarrow \sum{ab}\ge \sum{a^2b}\Rightarrow \sum{a}\sum{ab}\ge \sum{a}\sum{a^2b}\ge (\sum{\sqrt{a\cdot ac^2}})^2=(\sum{ab})^2\Rightarrow \sum{a}\ge \sum{ab}\Rightarrow (\sum{a})^2\ge (\sum{ab})^2\ge 3abc(a+b+c)\Rightarrow \sum{a}\ge 3abc.$
I used Cauchy's inequality and the fact that $\sum{a^2}\ge \sum{ab}$.
Then, ${\sum{\frac{a}{bc(b+c)}}}=\frac{\sum{\frac{a^2}{b+c}}}{abc}\ge \frac{\frac{(\sum{a})^2}{2\sum{a}}}{abc}=\frac{a+b+c}{2abc}\ge \frac{3}{2}$.


Given triangle $ ABC$ (non right) with altitude $ AA',BB',CC'$. Let $ D,E,F$be excenters of $ \angle B'AC',\angle C'A'B,\angle B'A'C$ resp.$ A$-exradius of $ ABC$ touch $ BC,CA,AB$ at $ M,N,P$. Prove that circumcenter of $ DEF$ is orthocenter of $ MNP$.
Compare with my problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=864941#864941]Triangle problem 7[/url].

My solution:

Let $ I_a $ be the $ A- $ excenter of $ \triangle ABC $ .
Let $ G, T $ be the orthocenter, 9-point center of $ \triangle MNP $, respectively .

Since $ \triangle ABC \cup I_a \sim \triangle AB'C' \cup D \sim \triangle A'B'C \cup F $ ,
so $ \triangle I_aBC \sim \triangle DB'C' \sim \triangle FB'C \Longrightarrow \triangle B'C'C \sim \triangle B'DF $ ,

hence we get $ DF=C'C \cdot \frac{B'D}{B'C'}=C'C \cdot \frac{BI_a}{BC}=BI_a \cdot \sin \angle CBA =MP $ .

From $ \angle (C'C,DF)=\angle C'B'D=\angle I_aBC=PBI_a=\angle (C'C,MP) \Longrightarrow DF \parallel MP $ .
Similarly we can prove $ DE=MN $ and $ DE \parallel MN \Longrightarrow \triangle DEF $ and $ \triangle MNP $ are homothetic and congruent .

Since $ D $ is the reflection of $ I_a $ in $ NP $ (well-known) ,
so $ T $ is the midpoint of $ DM \Longrightarrow \triangle DEF $ is the image of $ \triangle MNP $ under homothety $ \mathbf{H} (T,-1) $ ,
hence we get $ G $ is the circumcenter of $ \triangle DEF $ ( $ \because G $ is the image of $ I_a $ under homothety $ \mathbf{H} (T,-1) $ ) .

Q.E.D

Let $I_a$ be the A-excenter of $\triangle ABC.$ Since $\triangle AB'C' \cup D \sim \triangle ABC \cup I_a$ $\Longrightarrow$ $\tfrac{AD}{AI_a}=\tfrac{B'C'}{BC}=\cos \widehat{A}=\cos \widehat{PAN}$ $\Longrightarrow$ $D$ is orthocenter of A-isosceles $\triangle ANP$ $\Longrightarrow$ $D$ is reflection of $I_a$ on $NP$ and by similar reasoning, $E,F$ are the reflections of $I_a$ on $PM,MN$ $\Longrightarrow$ $\triangle DEF$ is image of the medial triangle of $\triangle MNP$ under dilatation with center $I_a$ and coefficient 2. The conclusion follows.


[quote="nguoivn"]Given $ a, b, c \geq\ 0.$ Prove that:
$ \frac {a}{\sqrt []{a \plus{} b}} \plus{} \frac {b}{\sqrt []{b \plus{} c}} \plus{} \frac {c}{\sqrt []{c \plus{} a}} \geq\ \frac {1}{2}(\sqrt []{a \plus{} b} \plus{} \sqrt []{b \plus{} c} \plus{} \sqrt []{c \plus{} a})$
If it trues, equality occurs when $ (a \minus{} b)(b \minus{} c)(c \minus{} a) \equal{} 0$
And it's stronger than one old result:
$ \frac {a}{\sqrt []{a \plus{} b}} \plus{} \frac {b}{\sqrt []{b \plus{} c}} \plus{} \frac {c}{\sqrt []{c \plus{} a}} \geq\ \frac {1}{\sqrt []{2}}(\sqrt []{a} \plus{} \sqrt []{b} \plus{} \sqrt []{c})$
But I think it's not trues  :wink:
Can somebody give me some ideas ?[/quote]
Take a look at the case $ c\equal{}0,a\equal{}1,b\equal{}2$ ;)

[quote="arqady"][quote="nguoivn"]Given $ a, b, c \geq\ 0.$ Prove that:
$ \frac {a}{\sqrt []{a \plus{} b}} \plus{} \frac {b}{\sqrt []{b \plus{} c}} \plus{} \frac {c}{\sqrt []{c \plus{} a}} \geq\ \frac {1}{\sqrt []{2}}(\sqrt []{a} \plus{} \sqrt []{b} \plus{} \sqrt []{c})$
Can somebody give me some ideas ?[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=56071[/quote]
Oh,thank arqady,but ,i remember,Can_hang mentioned this problem in toanthpt site,and I think his proof is still nice.Post it Can_hang,please(I guess he used CSand pqr,but I cant find the full proof)

[quote="nguoivn"]Given $ a, b, c \geq\ 0.$ Prove that:
$ \frac {a}{\sqrt []{a \plus{} b}} \plus{} \frac {b}{\sqrt []{b \plus{} c}} \plus{} \frac {c}{\sqrt []{c \plus{} a}} \geq\ \frac {1}{\sqrt []{2}}(\sqrt []{a} \plus{} \sqrt []{b} \plus{} \sqrt []{c})$
Can somebody give me some ideas ?[/quote]
It's a problem which is in hungkhtn 's book .
 He let $ x\equal{}a^2;y\equal{}b^2;z\equal{}c^2$ then Square two HS
And apply Cyclic Ineq  :)


geometry
circumcircle
trigonometry
geometry solved

Let $O$ be the circumcenter of acutangle triangle $ABC$ and let $A_1$ be some point in the smallest arc $BC$ of the circumcircle of $ABC$. Let $A_2$ and $A_3$ points on sides $AB$ and $AC$, respectively, such that $\angle BA_1A_2 = \angle OAC$ and $\angle CA_1A_3 = \angle OAB$.

Prove that the line $A_2A_3$ passes through the orthocenter of $ABC$.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=25876 post #3.

  darij

I think I have it.

Since $\angle BA_1A_2 = const$  line $A_1A_2$ goes through constant point $M$ on circle.
Same holds for $A_1A_3$ for some point $N$. So we have two projective  map $A_2\mapsto A_1$ and $A_1\mapsto A_3$ wich give us another projective map $A_2\mapsto A_3$ fro line $AB$ to line $AC$. Since $A\mapsto A$ this map is perspective, so line $A_2A_3$ goes throught some constant point. Now it is not difficult to see that is orthocenter of triangle $ABC$. ;)

I suppose that $AA_1\cap (BC)\ne \emptyset$. I note $u=m(\widehat {BAA_1}),\ v=m(\widehat {CAA_1})\ (u+v=A),\ D\in AH\cap BC$.

$1.\ \dfrac {A_2B}{A_2A}=\dfrac {\sin u}{\sin (v+B)}\cdot \dfrac {\cos B}{\cos A},\ \dfrac {A_3C}{A_3A}=\dfrac {\sin v}{\sin (u+C)}\cdot \dfrac {\cos C}{\cos A}$

$2.\ \sin u\cdot \sin B +\sin v\cdot \sin C=\sin A\cdot \sin (u+C)$

$3.\ H\in A_2A_3\Longleftrightarrow \dfrac {A_2B}{A_2A}\cdot \tan B +\dfrac {A_3C}{A_3A}\cdot \tan C =\tan A$

\[ \overline {\underline {\left| \ \blacktriangle\ A.S.O.\ \blacktriangle\ \right| }} \]\[(1^{\circ}\wedge \ 2^{\circ}\Longrightarrow \ 3^{\circ})\]

Let the altitudes from $B$ respectively $C$ intersect the circumcircle at B' respectively C'. Then points $(C',A_2,A_1)$ and
$(A_1,A_3,B')$ are colliniar. Hence by [b]Pascal's [/b]theorem apllied to $ABB'A_1C'C$ we conclude that points $A_1,H,A_3$ are collinar.

My solution started like Saylors one but ended with an angle chaise, Anyway Saylors solution was done by like 5 of the contyestants.


i have a question about one of these reaction predictions....

Excess concentrated sulfuric acid is added to solid calcium phosphate.

the answer is:

$H_2SO_4 + Ca_3(PO_4)_2 -----> H_3PO_4 + CaSO_4$

why doesn't the sulfuric acid on the reactant side dissociate into H+ and HSO4- or H+ and SO42-??

I wouldn't worry too much about that.  I suppose that it might have something to do with "Excess concentrated," but I think that's just to ensure that it can replace (to use the term loosely) the Ca2+ in Ca3(PO4)2.  I would have broken it up into H+ and SO42-

I would still write with HSO4-.   Maybe it's because this test is old.  All the recent tests I've seen accept HSO4- or SO4 2- but not H2SO4.

It might have to do with the concentrated part.  Concentrated sulfuric acid is 98.something %, so there are more moles of H2S04 than H2O.  Then not a lot of ionization would happen.

pKa1 = -3, pKa2 = 2. Second dissication step should be treated as if it was a weak acid, not strong one. Especially in concentrated solutions.

Best,
Borek

[quote="pkothari13"]so is that just an incorrect answer or what? what would i do on the real AP test??[/quote]
The answer is correct, and remember, whatever test you are taking, it is always:

[b]Wrong to write: $\text{H}^{+}$ and $\text{SO}_4 ^{2-}$ if it is Concentrated acid.[/b]

[quote="shobber"][quote="pkothari13"]but then on the 1994 reaction predictions (concentrated hydrochloric acid added to solid potassium permanganate), they write the answer as H+ + Cl-[/quote]
You can do it with $HCl$ if it is concentrated, but not the $H_2SO_4$.[/quote]

so then which ones do you dissociate and which ones do you keep intact?? is there some sort of rule to remember this by?

I believe that you may dissociate HCl, HBr, HI, HNO3, HClO3, HClO4, and HMnO4 safely in most reasonable concentrations and that H2SO4 and H2SeO4 ionize only one proton in [i]concentrated[/i] solution.

As noted above, concentrated sulfuric acid (about 17.6M) does not dissociate very much at all because of the relatively inaqueous nature of the solution.  Even reagent grade H2SO4 is around 94% pure.  I am almost certain that this is the only acid which you might encounter on the AP test for which this consideration must be taken into mind.


function
induction
algebra unsolved
algebra

Prove that

 $ d! \equal{} \displaystyle\sum^d_{r \equal{} 0} (\minus{}1)^{r} {\binom{d}{r}}(x \plus{} d \minus{} r) ^ d \: \forall \: x$


how can I simplify this $ \frac{3^{2003}+2^{2003}}{3^{2001}+2^{2001}}$

they ask me for the biggest integer that isnt bigger than that fraction


algebra
polynomial
search
number theory
relatively prime
algebra proposed

let $A,B,C$ be polynoms in ${C}$

prove fermats last theorem for polynoms in ${C}$
hint: use mason's theorem

What is mason's theorem? And are you sure that the polynomial are in C? Not in Z? Can you formulate better the statement pls :) ?

Let $a(t), b(t), c(t)$ be relatively prime polynomials s.t. one of them has degree $\geq 1$, and s.t. $a(t)^{n}+b(t)^{n}=c(t)^{n}$. We need to prove that $n\leq 2$. By Mason-Stothers theorem, we get
\[n\text{ deg }a=\text{ deg }a(t)^{n}\leq \text{ deg }a(t)+\text{ deg }b(t)+\text{ deg }c(t)-1 \]
and similarly replacing $x$ by $y$ and $z$ on the left hand side. Add them up to get
\[n(\text{ deg }a+\text{ deg }b+\text{ deg }c)\leq 3(\text{ deg }a+\text{ deg }b+\text{ deg }c)-3. \]
This is a contradiction if $n\geq 3$.

[quote="ciprian"]Sry but I have to ask again: what is Mason Theorem?[/quote]
Search, again

http://mathworld.wolfram.com/MasonsTheorem.html


And what about the winner of the previous IMO - Bulgaria...!?! They have almost competely new team so..!?!? Any scores..?!?!

Nobody really knows those ones. I think that they finished in the top 5 countries, with 4 golds and 2 silvers, but I do not know any names ...


Is there a positive integer $ n$ such that, every digit, different from $ 0$, appears the same number of times in the decimal form of each of the numbers $ n, 2n, 3n, ..., 2008n$?

I am not sure what you mean with this problem  :(
  I'm not sure  what you did try to say  :(

   Can you clear it ?

When you write out $ n,2n,...,2008n$ in base $ 10$, you find the same number of $ 1$s in each of the $ 2008$ numbers, the same number of $ 2$s in each of the $ 2008$ numbers, ..., the same number of $ 9$s in each of the $ 2008$ numbers


I just can't seem to get this problem on my Algebra two homework!

Solve for x

$2x-6x=14$

edit:[size=150][color=red]ADD GENEROUS SUPPLY OF SARCASM[/color][/size] :wallbash_red:

This apparently wasn't getting noticed :mad:

[quote="mustafa"]I just can't seem to get this problem on my Algebra two homework!

Solve for x

$2x-6x=14$[/quote]
[hide="I think you made a mistake..."]$2x-6x=14$

$x=-\frac{7}{2}$[/hide]

[quote="Dezcaix"][quote="i_like_pie"][quote="mustafa"]I just can't seem to get this problem on my Algebra two homework!

Solve for x

$2x-6x=14$[/quote]
[hide="I think you made a mistake..."]$2x-6x=14$

$x=-\frac{14}{4}$[/hide][/quote]

yes i think so too...you sure this is alg. 2?

[hide="answer"]$2x-6x=14$

$-4x=14$

$x=-\frac{7}{2}$[/hide][/quote]
I forgot to simplify. :blush:  :ninja:

Mustafa, could you clarify the question? This clearly isn't hard.

[quote="mustafa"]I just can't seem to get this problem on my Algebra two homework!

Solve for x

$2x-6x=14$[/quote]

-4x=14
-2x=7
x=-7/2

you mean pre algebra?

[hide]$2(x-3x)=14\Rightarrow (-2)x=7\implies x=\frac{7}{-2}$[/hide]

 :maybe:  I think you made a mistake  :maybe:

EDIT:I added no sarcasm so i am wrong  :) !

[hide="solution"]We can simplify the LH to get the equation

-4x=14
x=-7/2[/hide]

Yeah this is rather easy...
so if you really had trouble let me explain what those guys did....
[hide] first they subtracted the variables to get
-4x=14
then they divided -4 out of x with the reciprocal -1/4
to get x=-14/4 which when simplified =-7/2

this does look like pre. algebra...[/hide]

Why are you guys posting the solution after 50 people had already solved it and its not even a real problem? Trying to get post count?

[quote="mustafa"]Sorry, I'll fix it[/quote]

ok it's still esay right now though... :wink:  :huh:  :huh:

This needs to be locked. Apparently no one realizes this whole topic was a joke.


Let n be a natural number. Let d(n) be the sum of the squares of the positive divisors of n. Find minimum n such that d(n)=(n+3)^2
Note: for example d(6)=1^2+2^2+3^2+6^2=50

Let $ d_0\equal{}1<d_1<...<d_k\equal{}n$ and $ d(n)\equal{}\sum_i d_i^2 \equal{}n^2\plus{}6n\plus{}9$, therefore $ \sum_{i\equal{}1}^{k\minus{}1}(d_i^2\plus{}\frac{n^2}{d_i^2})\equal{}12n\plus{}16$.
Obviosly $ n$ is not prime and $ n\not \equal{}p^2$. If n had 3 distinct prime divisors, then $ 6n\plus{}8\equal{}d(n)\minus{}n^2\minus{}1\ge p_1^2\plus{}(\frac{n}{p_1})^2\plus{}p_2^2\plus{}(\frac{n}{p_2})^2\plus{}p_3^2\plus{}(\frac{n}{p_3})^2>6n\plus{}8$.
If $ n\equal{}p_1p_2$, then $ d(n)\equal{}n^2\plus{}1\plus{}p_1^2\plus{}p_2^2\equal{}n^2\plus{}6n\plus{}9\to p_1^2\plus{}p_2^2\equal{}6p_1p_2\plus{}8.$
It give $ |p_1\minus{}3p_2|\plus{}2\sqrt 2 p_2 \equal{}2(1\plus{}\sqrt 2 )(3\plus{}2\sqrt 2 )^m\equal{}2(1\plus{}\sqrt 2 )^{2m\plus{}1}$.
Or $ |\frac{p_1\minus{}3p_2}{2}|\plus{}\sqrt 2 p_2\equal{}(1\plus{}\sqrt 2)^{2m\plus{}1}$ or
$ |p_1\minus{}3p_2|\equal{}\frac{(1\plus{}\sqrt 2 )^{2m\plus{}1}\plus{}\sqrt (1\minus{}\sqrt 2)^{2m\plus{}1}}{2},$
$ p_2\equal{}\frac{(1\plus{}\sqrt 2)^{2m\plus{}1}\minus{}(1\minus{}\sqrt 2 )^{2m\plus{}1}}{2\sqrt 2}.$
Because $ p_2$ prime $ 2m\plus{}1$ must be prime.


inequalities
search
inequalities unsolved

$ a,b,c\geq\minus{}\frac{3}{4}$ and $ a\plus{}b\plus{}c\equal{}1$
Proof
$ \frac{a}{a^2\plus{}1}\plus{}\frac{b}{b^2\plus{}1}\plus{}\frac{c}{c^2\plus{}1}\leq\frac{9}{10}$

[quote="colorfuldreams"]$ a,b,c\geq \minus{} \frac {3}{4}$ and $ a \plus{} b \plus{} c \equal{} 1$
Proof
$ \frac {a}{a^2 \plus{} 1} \plus{} \frac {b}{b^2 \plus{} 1} \plus{} \frac {c}{c^2 \plus{} 1}\leq\frac {9}{10}$[/quote]
Because $ \sum_{cyc}\frac {a}{a^2 \plus{} 1}\leq\frac {9}{10}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\left(\frac{3}{10}\minus{}\frac{a}{a^2\plus{}1}\plus{}\frac{18}{25}\left(a\minus{}\frac{1}{3}\right)\right)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac{(3a\minus{}1)^2(4a\plus{}3)}{a^2\plus{}1}\geq0.$

[quote="bokinhvan3"][quote="arqady"][quote="colorfuldreams"]$ a,b,c\geq \minus{} \frac {3}{4}$ and $ a \plus{} b \plus{} c \equal{} 1$
Proof
$ \frac {a}{a^2 \plus{} 1} \plus{} \frac {b}{b^2 \plus{} 1} \plus{} \frac {c}{c^2 \plus{} 1}\leq\frac {9}{10}$[/quote]
Because $ \sum_{cyc}\frac {a}{a^2 \plus{} 1}\leq\frac {9}{10}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\left(\frac {3}{10} \minus{} \frac {a}{a^2 \plus{} 1} \plus{} \frac {18}{25}\left(a \minus{} \frac {1}{3}\right)\right)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac {(3a \minus{} 1)^2(4a \plus{} 3)}{a^2 \plus{} 1}\geq0.$[/quote]

Because $ a,b,c\geq \minus{} \frac {3}{4}$ , if you use AM-GM inequality, you are wrong.
If $ a,b,c\geq 0$ , I have 3 solution.
Can you use holder inequality.[/quote]

Arqady's proof is perfectly true. Moreover this inequality holds for $ a,b,c\in \mathbb{R}$

@colorfuldreams: search for inequalities like this because they are famous and have been posted many times.

[quote="bokinhvan3"][quote="arqady"][quote="colorfuldreams"]$ a,b,c\geq \minus{} \frac {3}{4}$ and $ a \plus{} b \plus{} c \equal{} 1$
Proof
$ \frac {a}{a^2 \plus{} 1} \plus{} \frac {b}{b^2 \plus{} 1} \plus{} \frac {c}{c^2 \plus{} 1}\leq\frac {9}{10}$[/quote]
Because $ \sum_{cyc}\frac {a}{a^2 \plus{} 1}\leq\frac {9}{10}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\left(\frac {3}{10} \minus{} \frac {a}{a^2 \plus{} 1} \plus{} \frac {18}{25}\left(a \minus{} \frac {1}{3}\right)\right)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac {(3a \minus{} 1)^2(4a \plus{} 3)}{a^2 \plus{} 1}\geq0.$[/quote]

Because $ a,b,c\geq \minus{} \frac {3}{4}$ , if you use AM-GM inequality, you are wrong.
If $ a,b,c\geq 0$ , I have 3 solution.
Can you use holder inequality.[/quote]
Yes, I've made a mistake
Thank you for the comment

[b]The different solution:
[/b]Using the AM-GM ine, we have $ a^2 \plus{} 1 \ge 2\sqrt {a^2 .\frac {1}{9}} \plus{} \frac {8}{9} \equal{} \frac {{6a \plus{} 8}}{9}$
so $ \frac {a}{{a^2 \plus{} 1}} \le \frac {{9a}}{{6a \plus{} 8}} \equal{} \frac {3}{2} \minus{} \frac {6}{{3a \plus{} 4}}$
Similarly, we have
$ \frac {b}{{b^2 \plus{} 1}} \le \frac {3}{2} \minus{} \frac {6}{{3b \plus{} 4}}$
and $ \frac {c}{{c^2 \plus{} 1}} \le \frac {3}{2} \minus{} \frac {6}{{3c \plus{} 4}}$
Adding these three ine, we deduce
\[ \frac {a}{{a^2 \plus{} 1}} \plus{} \frac {b}{{b^2 \plus{} 1}} \plus{} \frac {c}{{c^2 \plus{} 1}} \le \frac {9}{2} \minus{} 6\left( {\frac {1}{{3a \plus{} 4}} \plus{} \frac {1}{{3b \plus{} 4}} \plus{} \frac {1}{{3b \plus{} 4}}} \right)\]
Because of $ a,b,c\geq\minus{}\frac{3}{4}$
so
\[ \frac {1}{{3a \plus{} 4}} \plus{} \frac {1}{{3b \plus{} 4}} \plus{} \frac {1}{{3c \plus{} 4}} \ge \frac {9}{{3(a \plus{} b \plus{} c) \plus{} 12}} \equal{} \frac {3}{5}\]
hence
\[ \frac {a}{{a^2 \plus{} 1}} \plus{} \frac {b}{{b^2 \plus{} 1}} \plus{} \frac {c}{{c^2 \plus{} 1}} \le \frac {9}{2} \minus{} 6.\frac {3}{5} \equal{} \frac {9}{{10}}\]
The ine is proved

[quote="Pain rinnegan"][quote="daigiaga1994"]
$ \frac {a}{{a^2 \plus{} 1}} \le \frac {{9a}}{{6a \plus{} 8}}$
[/quote]

Try $ a \equal{} \minus{} \frac {1}{2}$.[/quote]
Sorry, I've again made a mistake

[quote="bokinhvan3"]
Can you use holder inequality.[/quote]

How to solve this by holder inequality ?

[quote="daigiaga1994"][b]The different solution:
[/b]Using the AM-GM ine, we have $ a^2 \plus{} 1 \ge 2\sqrt {a^2 .\frac {1}{9}} \plus{} \frac {8}{9} \equal{} \frac {{6a \plus{} 8}}{9}$
so $ \frac {a}{{a^2 \plus{} 1}} \le \frac {{9a}}{{6a \plus{} 8}} \equal{} \frac {3}{2} \minus{} \frac {6}{{3a \plus{} 4}}$
Similarly, we have
$ \frac {b}{{b^2 \plus{} 1}} \le \frac {3}{2} \minus{} \frac {6}{{3b \plus{} 4}}$
and $ \frac {c}{{c^2 \plus{} 1}} \le \frac {3}{2} \minus{} \frac {6}{{3c \plus{} 4}}$
Adding these three ine, we deduce
\[ \frac {a}{{a^2 \plus{} 1}} \plus{} \frac {b}{{b^2 \plus{} 1}} \plus{} \frac {c}{{c^2 \plus{} 1}} \le \frac {9}{2} \minus{} 6\left( {\frac {1}{{3a \plus{} 4}} \plus{} \frac {1}{{3b \plus{} 4}} \plus{} \frac {1}{{3b \plus{} 4}}} \right)\]
Because of $ a,b,c\geq \minus{} \frac {3}{4}$
so
\[ \frac {1}{{3a \plus{} 4}} \plus{} \frac {1}{{3b \plus{} 4}} \plus{} \frac {1}{{3c \plus{} 4}} \ge \frac {9}{{3(a \plus{} b \plus{} c) \plus{} 12}} \equal{} \frac {3}{5}\]
hence
\[ \frac {a}{{a^2 \plus{} 1}} \plus{} \frac {b}{{b^2 \plus{} 1}} \plus{} \frac {c}{{c^2 \plus{} 1}} \le \frac {9}{2} \minus{} 6.\frac {3}{5} \equal{} \frac {9}{{10}}\]
The ine is proved[/quote]
The above may be incorrect. But I have a question.
We can assume $ a,b,c$ to be positive and prove the inequality as above.
Noting that $ \frac{a}{{a^{2}\plus{}1}}\plus{}\frac{b}{{b^{2}\plus{}1}}\plus{}\frac{c}{{c^{2}\plus{}1}}\le\frac{{\left| a\right|}}{{a^{2}\plus{}1}}\plus{}\frac{{\left| b\right|}}{{b^{2}\plus{}1}}\plus{}\frac{{\left| c\right|}}{{c^{2}\plus{}1}}\le\frac{9}{10}$, can we not say that it is valid for all real $ a,b,c$ ?
Please somebody answer.

[quote="Shanku"]
The above may be incorrect. But I have a question.
We can assume $ a,b,c$ to be positive and prove the inequality as above.
Noting that $ \frac {a}{{a^{2} \plus{} 1}} \plus{} \frac {b}{{b^{2} \plus{} 1}} \plus{} \frac {c}{{c^{2} \plus{} 1}}\le\frac {{\left| a\right|}}{{a^{2} \plus{} 1}} \plus{} \frac {{\left| b\right|}}{{b^{2} \plus{} 1}} \plus{} \frac {{\left| c\right|}}{{c^{2} \plus{} 1}}\le\frac {9}{10}$, can we not say that it is valid for all real $ a,b,c$ ?
Please somebody answer.[/quote]
But why $ |a|\plus{}|b|\plus{}|c|\equal{}1$ ? :wink:

[quote="arqady"][quote="Shanku"]
The above may be incorrect. But I have a question.
We can assume $ a,b,c$ to be positive and prove the inequality as above.
Noting that $ \frac {a}{{a^{2} \plus{} 1}} \plus{} \frac {b}{{b^{2} \plus{} 1}} \plus{} \frac {c}{{c^{2} \plus{} 1}}\le\frac {{\left| a\right|}}{{a^{2} \plus{} 1}} \plus{} \frac {{\left| b\right|}}{{b^{2} \plus{} 1}} \plus{} \frac {{\left| c\right|}}{{c^{2} \plus{} 1}}\le\frac {9}{10}$, can we not say that it is valid for all real $ a,b,c$ ?
Please somebody answer.[/quote]
But why $ |a| \plus{} |b| \plus{} |c| \equal{} 1$ ? :wink:[/quote]
Thanks arqady.

[quote="colorfuldreams"]like the  same  question :lol:  :P[/quote]
From $ \frac{1}{{a^{2}\plus{}1}}\plus{}\frac{1}{{b^{2}\plus{}1}}\plus{}\frac{1}{{c^{2}\plus{}1}}\le\frac{27}{10}$ it is simple by Chebyshev,

$ \frac{a}{{a^{2}\plus{}1}}\plus{}\frac{b}{{b^{2}\plus{}1}}\plus{}\frac{c}{{c^{2}\plus{}1}}\le \frac{1}{3}(a\plus{}b\plus{}c)\left(\frac{1}{{a^{2}\plus{}1}}\plus{}\frac{1}{{b^{2}\plus{}1}}\plus{}\frac{1}{{c^{2}\plus{}1}}\right)\le\frac{1}{3}\times\frac{27}{10}$ $ \equal{}\frac{9}{10}$
as $ \frac{1}{{a^{2}\plus{}1}}, \frac{1}{{b^{2}\plus{}1}}, \frac{1}{{c^{2}\plus{}1}}$ and $ a,b,c$ are oppositely sorted.

[quote="Shanku"]
as $ \frac {1}{{a^{2} \plus{} 1}}, \frac {1}{{b^{2} \plus{} 1}}, \frac {1}{{c^{2} \plus{} 1}}$ and $ a,b,c$ are oppositely sorted.[/quote]

Try $ a\equal{}\minus{}\frac{1}{2}, b\equal{}0, c\equal{}1$

[quote="Pain rinnegan"][quote="Shanku"]
as $ \frac {1}{{a^{2} \plus{} 1}}, \frac {1}{{b^{2} \plus{} 1}}, \frac {1}{{c^{2} \plus{} 1}}$ and $ a,b,c$ are oppositely sorted.[/quote]

Try $ a \equal{} \minus{} \frac {1}{2}, b \equal{} 0, c \equal{} 1$[/quote]

Sorry Pain. Forgot $ a,b,c$ are any real.


trigonometry
inequalities
geometry solved
geometry

$x+y+z=\frac{\pi}2$, $x,y,z>0$
$(1-\sin x)(1-\sin y)(1-\sin z)\geq \sin x\sin y\sin z$

Since the angles x, y, z are positive and add up to $\frac{\pi }{2}$, all of them are $<\frac{\pi }{2}$.

Let $A=\frac{\pi }{2}-x$, $B=\frac{\pi }{2}-y$ and $C=\frac{\pi }{2}-z$. Then, all of the angles A, B, C are $<\frac{\pi }{2}$ (since the angles x, y, z are positive) and > 0 (since the angles x, y, z are $<\frac{\pi }{2}$). Further,

$A+B+C=\left( \frac{\pi }{2}-x\right) +\left( \frac{\pi }{2}-y\right) +\left( \frac{\pi }{2}-z\right) =\frac{3\pi }{2}-\left( x+y+z\right) $
$=\frac{3\pi }{2}-\frac{\pi }{2}=\pi $.

Thus, A, B, C are the angles of a triangle, and this triangle is acute (since all of the angles A, B, C are $<\frac{\pi }{2}$). Now, since $A=\frac{\pi }{2}-x$, $B=\frac{\pi }{2}-y$, $C=\frac{\pi }{2}-z$, we have sin A = cos x, sin B = cos y and sin C = cos z. Thus, your inequality rewrites as

$\left( 1-\cos A\right) \left( 1-\cos B\right) \left( 1-\cos C\right) \geq \cos A\cos B\cos C$.

And for this one, see http://www.mathlinks.ro/viewtopic.php?t=4972 .

  Darij

Yes, I've just solved it today. I posted my proof on the page with the link you have provided. But what about trigonometry1 Darij ??


What's everybody's favorite type of candy bar? mine is 100 Grand, but i also like twix bars a lot.

a poll would be nice, but there's just way too many candy bars out there. no matter what options i put up, i think other would just run away with it

Haha...Luna bars...For women. :-D  (haha, MY SISTER).

And Vegan pie.

shrek m&ms. these arent candy bars either, but they are closer to candy bars than slurpies, so im putting them!

Twix, Hundred Grand, and Crunch.

None of which, unfortunately, are orthodontic work-compatible.  Confound it all.

[quote="JSRosen3"][quote="Yasha"]Definitely any kind of Russian chocolate.[/quote]

I concur!!!  :D  Wonder why...[/quote]
gah! can't believe i forgot!  :-x
if there were a drooling smilie, i'd use it here

i've never had russian chocolate.  I like 3 musketeers

Milky way is just a 3 musketeers with caramel.  I also like Snickers

Anyone here tried the European Milky Way bars? Totally different from the American ones. They are so good....... drool........

[quote="GoBraves"]Yay for 3 musketeers but Milky Way is the best.[/quote]

[quote="math92"]Milky way is just a 3 musketeers with caramel.[/quote]

 I agree with GoBraves and math92 cause the caramel makes it taste better.

Butterfinger, Milky Way, Three Musketeers, Crunch, Kit Kat...

Oh brother now I'm hungry  :P

[quote="mcalderbank"][quote="math92"]Anyone had Aero or Maltesers?[/quote]

Yes....Aero is cool, because it is so airy that the chocolate melts easier and it is nice and crumblier. When I went to England, I had all these Cadbury bars that were very nice, but I cannot remember their nice. All I remember is that they were better than Hershey's!  :D[/quote]

Yeah Cadbury is great stuff. I wish they sold them here

I like : Crunch, Twix, Reese's, and Baby Ruth


The sentence game is a game that stretches creativity.
First, you have a sentence. You have to think up a sentence that has at least one word in common to it.
Example: The bird can fly.
                The fish can swim.

Now, there are levels in the game. We start on level one. On the first level, you have to think up a sentence that has at least one word in common to the last sentence. On level x, you need to think up a sentence that has at least x words in common. Every week I'll check back. If someone has posted since Friday, I'll advance the level by one. If not, the level changes back to one. :diablo:
I'll start.

The boulder blocking the cave entrence was moved.

Meh, that didn't come out write.
My sentence is "He was".

he was what???? You need a full sentence, not two words. Like "She was a bad girl."

"He was" is a sentence.
It has a subject and a verb.

ok.

The dog was mean at first, but he calmed down.

The thunderstorm calmed after a fortnight, after igniting the plains with flashes of fire and sparkles;  my iPod broke.

i ran to the door to open it ,only to find count dracula waiting for  me there! :(

I might believe that it's not butter.  :)

Believe.

[color=cyan][size=9]this is because the message was too small[/size][/color]

Social Studies are the Weapons of Mass Destruction.

A teacher was arrested last night for boarding a plane with weapons of math instruction.

The very same teacher was released today for some unknown reasons.

my teacher was retarded and taught me to find this bird that was unknown but then he disappeared...

I hereby advance this topic to level three!!!!!!!!

My teacher taught to the class that he loves sushi.

Believe me, he really loves sushi with the love of a brother!

[quote="SplashD"]He loves sushi.[/quote] :P

He loves sushi so much that he has it for breakfast, lunch, dinner, and dessert every day.

He has breakfast, lunch, dinner, and dessert every day.

He has breakfast, lunch, dinner, and dessert with his wife every day.


symmetry
inequalities
ratio
vector
analytic geometry
triangle inequality

demonstrate that a triangle of points A(X1, y1), B(X2,Y2) and C(X3,Y3) has its baricentre in the point ((x1+x2+x3)/3, (y1+y2+y3)/3)  :)

baricentre=centroid, i suppose that all you have to do is find the midpoints, and find the equation of the median, and show that the point is on that, and by symmetry, it will be on all the medians

[hide="just an observation, not about the problem"] is the 1º people from Spain that I saw in MathLinks[/hide]

Let $D$ be the midpoint of $BC$. Then we have $D(\frac{x_2+x_3}{2},\frac{y_2+y_3}{2}).$ Then $\overrightarrow{AD}=\langle \frac{x_2+x_3}{2}-x_1, \frac{y_2+y_3}{2}-y_1\rangle=\langle \frac{x_2+x_3-2x_1}{2},\frac{y_2+y_3-2y_1}{2}\rangle$. The centroid $G$ is two-thirds of the length of $AD$ from $A$, hence $\overrightarrow{AG}={2\over 3}\overrightarrow{AD}=\langle \frac{x_2+x_3-2x_1}{3},\frac{y_2+y_3-2y_1}{3}\rangle.$ Now $\overrightarrow{OG}=\overrightarrow{OA}+\overrightarrow{AG}=\langle \frac{x_2+x_3-2x_1}{3}+x_1,\frac{y_2+y_3-2y_1}{3}+y_1\rangle = \langle \frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\rangle$

just to clarify, fahrenheit, did you use the triangle inequality to show that G lied on the median?

As far as I know, altheman, the baricenter (a.k.a. the centroid) is the intersection point of the medians of the triangle, and it has the property of dividing the medians in the ratio 1:2. That's about the first "advanced" property of a triangle taught in the elementary school. (I must say I'm surprised, coz it's not like you to overlook such simple facts.)

lol...i did not really read the proof that carefully, but initially i thought the way of deriving that property would be done along these lines, but now i realize that the problem is not really asking you to prove various things like the 2:1 median ratio, and the fact that they are concurrent, etc, i have been thinking that you could use purely vectors to show collinearity by using the triangle inequality

:) The way I understood the problem, we only had to prove that the point with the given coordinates IS the baricenter. And since it's already named "baricenter", I thought that implied we were free to use all its known properties, such that it exists (in the sense of concurrency) and that it divides the medians in the ratio 2:1.

Of course, the baricenter can be defined differently, as the (unique - which is also provable) point $G$ for which it holds $\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=\overrightarrow{0}$ (that definition is almost indispensable when working with systems of points). There's a simple exercise in basic vector algebra... to prove that "classic" and vector definition of the centroid are equivalent.


function
complex analysis
real analysis
real analysis unsolved

Is the following true: for any harmonic function with real values defined on a neighborhood of 0 there exists  a continuous curve $h(t)$ starting from $0$, leading in a suitable point $x_{0}\ne 0$ and such that $f(h(t))$ is constant?

If $f$ is a harmonic function of two real variables, then you can find a function $F$ such that $\mathrm{Re}\, F=f$ and $F$ is holomorphic in a neighborhood of $0$. The level curves of $f$ are the preimages of vertical lines under $F$. Therefore, each level curve is piecewise analytic, with isolated branch points at which it has an even number of branches.

I suppose the level sets of harmonic functions in higher dimensions are piecewise analytic surfaces, but do not have a proof.


A recipe for crispy rice treats results in a mixture that fills a 9-inch by 13-inch pan to a depth of one inch. If a crispy rice treats mixture resulting from 1.5 times the original recipe is poured into a pan that is 10 inches by 15 inches, to what depth, in inches, will the pan be filled? Express your answer as a decimal to the nearest hundredth.

[hide="solution"] The original recipe has a volume of $ 9*13*1\equal{}117$.
1.5 times that would fill up a volume of $ 1.5*117\equal{}175.5$.
When poured into a pan measuring 10 by 15, the depth would be $ \frac{175.5}{10*15}\equal{}\boxed{1.17}$[/hide]

Well first find the volume of the pan:

$ 9\times13\times1\equal{}117$

So $ 1.5$ times that amount would be:

$ 117\times1.5\equal{}175.5$

Then divide the amount of mix be the area of the base of the tray which is:

$ 10\times15\equal{}150$

Divide:

$ 175.5\div150\equal{}\boxed{1.17}$


geometry
3D geometry
number theory unsolved
number theory

Find an infinite nonconstant arithmetic progression of positive integers such that each term is not a sum of two perfect cubes.

[quote="TomciO"]$ 7k \plus{} 4$.[/quote]

???? Where did that come from? Can you explain it more to me?

Take cubes modulo 7. They can only take values 0, 1 and -1. So their sums can be only -2, -1, 0, 1, 2 modulo 7. 4 is none of these.


Given two positive numbers $x,y$ such that $x^2+y^2=1$, prove that
\[ \frac1x+\frac1y\geq2\sqrt2+\Bigl(\sqrt{\frac xy}-\sqrt{\frac yx}\Bigl)^2. \]

easy and nice ineq, but tricky i think...

$\frac1x+\frac1y\geq2\sqrt2+\Bigl(\sqrt{\frac xy}-\sqrt{\frac yx}\Bigl)^2 \\\Leftrightarrow \frac{1-y}{x}+\frac{1-x}{y}+2\geq 2\sqrt2 \\\Leftrightarrow (\frac{1-y}{x}+\frac{1-x}{y}+2)^2\geq 8 \\\Leftrightarrow (\frac{1-y}{1+y}+\frac{1-x}{1+x})+(\frac{4(1-x)}{y}+\frac{4(1-y)}{x}+\frac{2(1-x-y+xy)}{xy})\geq 4 \\\Leftrightarrow \frac{1-xy}{(1+x)(1+y)}+\frac{x+y-1}{xy}\geq 1 \\\Leftrightarrow (xy-x^2y^2)+(x+y-1)(1+x+y+xy)\geq xy(1+x+y+xy) \\\Leftrightarrow xy-x^2y^2+(x+y)^2-1+xy(x+y-1)\geq xy+x^2y+xy^2+x^2y^2 \\\Leftrightarrow xy\geq 2x^2y^2 \\\Leftrightarrow 1\geq 2xy \\\Leftrightarrow x^2+y^2\geq 2xy$
and finally... we're done!

[quote="babylearnmath294"]campos,I think your solution isn't nice
        I think there is another solution[/quote]
Yes, I know that there are at least two other solutions.

solution from math and young
$\frac{1}{x}+\frac{1}{y}-(\frac{x}{y}+\frac{y}{x}-2)=\frac{1-x+y}{x}+\frac{1+x-y}{y}\geq 2\sqrt{2}$
so :$x^2+y^2 =1$ and $x,y>0$
$\to 0<x<1 , 0<y<1 \to 1-x+y>0 , 1+x-y>0$.
using cauchy :
$\frac{1-x+y}{y}+\frac{1+x-y}{x}\geq 2\sqrt{\frac{(1-x+y)(1+y-x)}{xy}}=2\sqrt{\frac{1-x^2-y^2-2xy}{xy}} =2\sqrt{\frac{2xy}{xy}}=2\sqrt{2}$

@babylearnmath : đừng post như vậy nữa bạn nhé

[quote="black_puma"]solution from math and young
$\frac{1}{x}+\frac{1}{y}-(\frac{x}{y}+\frac{y}{x}-2)=\frac{1-x+y}{x}+{1+x-y}{x}\geq 2\sqrt{2}$
so :$x^2+y^2 =1$ and $x,y>0$
$\to 0<x<1 , 0<y<1 \to 1-x+y>0 , 1+x-y>0$.
using cauchy :
$\frac{1-x+y}{y}+\frac{1+x-y}{x}\geq 2\sqrt{\frac{(1-x+y)(1+y-x)}{xy}}=2\sqrt{\frac{1-x^2-y^2-2xy}{xy}} =2\sqrt{\frac{2xy}{xy}}=2\sqrt{2}$

@babylearnmath : đừng post như vậy nữa bạn nhé[/quote]

In Vietnam, students have been taught to call Cauchy instead of AM-GM. However, in this forum, we should follow the international conventions. Nho nhe!

I correct some typos. Now it is better
[quote="black_puma"]$\frac{1-x+y}{y}+\frac{1+x-y}{x}\geq 2\sqrt{\frac{(1-x+y)(1-y+x)}{xy}}=2\sqrt{\frac{1-x^2-y^2+2xy}{xy}} =2\sqrt{\frac{2xy}{xy}}=2\sqrt{2}$

[/quote]

This problem is easy, :D
Since $1<x+y\leq \sqrt{2},$
$\frac{1}{x}+\frac{1}{y}-\left(\sqrt{\frac{x}{y}}-\sqrt{\frac{y}{x}}\right)^2=\frac{2}{x+y+1}+2\geq \frac{2}{\sqrt{2}+1}+2=2\sqrt{2}$


Hi! hope anybody can help me to solve this problem:

Eliminate the recursion of the following function, based on ADT BinaryTree(T), and find an iterative algorithm:

replace: BinaryTree(T) x T x T -> BinaryTree(T)
replace (Nil, x, y) = Nil
replace (Bin (lt, r, rt), x, y) = [b]if[/b] (r==x) [b]then[/b] Bin (replace (lt, x, y), y, replace (rt, x, y)) [b]else[/b] Bin (replace (lt, x, y), r, replace (rt, x, y)) [b]endif[/b]

Note: BinaryTree has this constructors:
Nil: -> BinaryTree(T)
Bin: BinaryTree(T) x T x BinaryTree(T) -> BinaryTree(T)

Ask me anything you don't understand (or is missing) about this problem.. thanks

oh.. sorry.. it's in functional language.. Haskell

Hm... I thought so, kubus keeps talking about it quite a lot since they're learning Haskell in school :). Unfortunately, I still don't quite understand it. Perhaps you should try sending a PM to [url=http://www.mathlinks.ro/Forum/profile.php?mode=viewprofile&u=6828]kubus[/url], I'm sure he'll be able to help. (I haven't seen him around here for quite a while but he should read his PMs.)


modular arithmetic
number theory
least common multiple
relatively prime

I've made a conjecture. Here it is.

If \[x \equiv y \pmod{a}\] and \[x \equiv y \pmod{b}\] then does \[x \equiv y \pmod{[a,b]}\] hold?

I've proved it when a and b are relatively prime.

P.S. [a,b] = LCM(a,b)

P.S.S. Do you want me to post my proof if a and b are relatively prime?

[hide]Consider first $x-y=p\in\mathbb{P}$.  Then you have $x\equiv y\bmod p$ so $\textrm{lcm}\left(p,p\right)=p$ and the result follows.  Now consider $x-y=mn$ for $m,n\in\mathbb{Z}$.  Then we have $x\equiv y\bmod m$ and $x\equiv y\bmod n$.  This implies that $x\equiv y\bmod mn$ which clearly means that $x\equiv y\bmod\textrm{lcm}\left(m,n\right)$.[/hide]

Let $u = x-y$. We have that $x\equiv y\pmod{a}$ and $x\equiv y\pmod{b}$, i.e. - according to the definition of modulo - $a | u$ and $b | u$. We want to prove that $\text{lcm}(a,b) | u$. Now let \[a = \prod_{p_{i}\in\mathbb{P}}p_{i}^{\alpha_{i}}\ \text{and}\ b = \prod_{p_{i}\in\mathbb{P}}p_{i}^{\beta_{i}}\ \text{and}\ u = \prod_{p_{i}\in\mathbb{P}}p_{i}^{\gamma_{i}}\] for some non-negative integers $\alpha_{i}, \beta_{i}, \gamma_{i}$. Because $a | u$, we have that $\alpha_{i}\leq \gamma_{i}$ for all $i$. Because $b | u$, we have that $\beta_{i}\leq \gamma_{i}$ for all $i$. Therefore, $\alpha_{i},\beta_{i}\leq \gamma_{i}$ and so $\max(\alpha_{i},\beta_{i}) \leq \gamma_{i}$ for all $i$. But \[\text{lcm}(a,b) = \prod_{p_{i}\in\mathbb{P}}p_{i}^{\max(\alpha_{i},\beta_{i})}.\] Therefore, $\text{lcm}(a,b) | u$, which proves the theorem.

thanks a lot.. I couldn't figure it out!


can someone explain how to prove that

| a + b |^r \leq  | a |^r  + | b |^r

for 0 < r \leq 1 (ie no holder like)

thanks

Here's (I think) a nice trick.

put $x=a/b$, so the inequality becomes (via dividing through) $|1+x|^r\leq 1+|x|^r$.  It is clear that we may assume $x>0$.  Hence it's enough to show that $(1+x)^r\leq 1+x^r$.  This is not hard, use calculus or whatever.


The hypotenuse of a right triangle measures 10 inches and one angle is $ 45^{\circ}$. What is the number of square inches in the area of the triangle?

In a 45-45-90 triangle, the hypotenuse is the $ leg*\sqrt2$ this makes the legs $ \frac{10}{\sqrt{2}}\equal{}5\sqrt{2}$
$ \frac{(5\sqrt{2})^2}{2} \equal{} \boxed{25}$


geometry
parallelogram
geometry proposed

A circle contacts BC and AC edge of ABCD quadrangle at C and D .Straight line in here cut AB at K and L .AC cut BD at P.M is centre CD.proving if CL = DL, K,P,M are straight
[img]http://i209.photobucket.com/albums/bb172/Ineq/hay44.jpg[/img]
 :D

$ LD \equal{} LC$ mean $ AB$ is external bisector of $ \angle DKC$. ie, $ \angle DKA \equal{} \angle CKB$. Let $ X$ be a point opposite side of $ K$ wrt $ DC$ such $ \angle XDC \equal{} \angle KDA,\angle XCD \equal{} \angle KCB$. Then obviously, $ XDKC$ is parallelogram. By jacobi theorem, $ KX,BD,AC$ are concurrent.  Together with fact that $ XDKC$ is parallelogram, we obtain $ K,P,M$ are collinear.

[quote="vodanh_tieutot"][quote="Leonhard Euler"]$ LD \equal{} LC$ mean $ AB$ is external bisector of $ \angle DKC$. ie, $ \angle DKA \equal{} \angle CKB$.[/quote]
I don't think so  :(
$ \angle DKA \neq \angle CKB$
and $ AB$ isn't external bisector of $ \angle DKC$[/quote]
You are wrong,
$ \angle DKA\equal{}\angle DCL\equal{}\angle LDC\equal{}\angle CKL\equal{}\angle CKB$.

Orther proof
[img]http://i209.photobucket.com/albums/bb172/Ineq/ant.jpg[/img]
We have
$ \hat{ADC} \equal{} \hat{DCB}$

 $ \Rightarrow\frac {sin{\hat{O}}}{sin{\hat{ODA}}} \equal{} \frac {sin{\hat{O}}}{sin {\hat{DCB}}}$

$ \Rightarrow \frac {AD^2}{BC^2} \equal{} \frac {OA^2}{OB^2} \equal{} \frac {P_{A|(O')}}{P_{B|(O')}} \equal{} \frac {LA.KA}{LB.KB}$

$ \Rightarrow \frac {LA}{LB}: \frac {OA}{OB} \equal{} \frac {KB}{KA}: \frac {OB}{OA}$ so $ (ABLO) \equal{} (BAKO)$

We have $ (ABLO) \equal{} (DCMO)$

$ \Rightarrow (DCMO) \equal{} (BAKO)\Rightarrow DB,AC,KM$ are concurrent
$ \Rightarrow K,P,M$ are collinear


May I know what is "notion of closure", I don't need comprehensive explanation, I just want to understand what is meant by a set is closed under another set in order to proceed.

For example, Set B is closed under Set A,
does it mean that Set A is a strict superset of Set B?

jack1234, you could look at http://mathworld.wolfram.com/SetClosure.html


Why are vivid changes in color so common when transition metals' hydrates are heated?

I tried to find some information about this on the internet, but could not find many resources.

Perhaps because some of the molecules of water that act as ligands are passing to vapour phase, thereby leaving the transition metal with a different coordination number. This means different separation of d orbitals and so electronic transitions with different energies. One well known example are the aqua complexes of cobalt, which were (are?) used as "weather signals".


trigonometry
inequalities
inequalities unsolved

the direction of inequality should be reversed.

[quote="hxtung"]Let $0<x<\frac {\pi}{2}$. Prove that: $\displaystyle \sin(\tan(x))>\tan(\sin(x))$.[/quote]I do believe thats not true(for example in degrees, sin(tan(89.999999)) is negative so 89.999999 converted to radians should be negative too (unless your working in degrees when you say pi/2). perhaps this would work when 0<x<arctan( :pi: /2)

hello, for [tex]x=\frac{\pi}{4}[/tex]we get:
[tex]\frac{\sin(1) \cdot \cos\left(\frac{\sqrt 2}{2}\right)-\sin\left(\frac{\sqrt 2}{2}\right)}{\cos\left(\frac{\sqrt2}{2}\right)}[/tex]
and this is negative.Maple says -0,01303944720..
Regards,Sonnhard.

I conjecture:

Let f(x)=tan(sin(x)) and g(x)=sin(tan(x)). Define f(n)(x)=f(f(f(...f(x))...) (n times), where n is a positive integer.

Then f(n)(x)>g(n)(x) for all 0<x<:pi:/2 and n=2k, where k is a non-negative integer.


Find positive integer n, such that $ \displaystyle\sqrt{\frac{9n\minus{}1}{n\plus{}7}}$ is rational.

[hide]
$ (n \plus{} 7,9n \minus{} 1) \equal{} (n \plus{} 7,64) \equal{} 2^s$ ($ s \equal{} 0,1,2,3,4,5,6$),so we can let
$ n \plus{} 7 \equal{} 2^sa^2,9n \minus{} 1 \equal{} 2^sb^2$ with $ a,b\in \mathbb{N}$ and $ (a,b) \equal{} 1$,then we have

$ 9a^2 \minus{} b^2 \equal{} 2^{6 \minus{} s}$.

and $ b^2\equiv ( \minus{} 1)^{7 \minus{} s}\pmod{3}\Longrightarrow 7 \minus{} s$ is even because $ \minus{} 1$ is not quadratic residue of $ \text{mod 3}\Longrightarrow s$ is odd,that is $ s \equal{} 1,3,5$.

1)$ s \equal{} 1$
we have $ (3a \minus{} b)(3a \plus{} b) \equal{} 2^5$ so $ (3a \minus{} b,3a \plus{} b) \equal{} (2,16)(4,8)$ hence $ (a,b) \equal{} (3,7)$ because $ (a,b) \equal{} 1$ so we get $ n \equal{} 2^sa^2 \minus{} 7 \equal{} 11$.

2)$ s \equal{} 3$
we have $ (3a \minus{} b)(3a \plus{} b) \equal{} 2^3$ then similarly we have $ (a,b) \equal{} (1,1)$ hence $ n \equal{} 1$.

3)$ s \equal{} 5$
obviously there are no solutions.

therefore we have the solutions $ n \equal{} \boxed{1,11}$[/hide]


$ P(x)$ is a polynomial of even degree with integer coefficients. If $ P(x)$ is a perfect square for infinitely many integers $ x$, prove that $ P(x)$ is a perfect square for all integers $ x$.

[hide="Small hint"]
Prove the slightly stronger statement that $ P(x)\equal{}[Q(x)]^2$ for some polynomial $ Q(x)$ with integer coefficients.
[/hide]

You need the condition that $ P(x)$ is monic; otherwise, we have counterexamples like $ P(x) \equal{} 2x^2 \plus{} 2x$.  :)

See http://www.artofproblemsolving.com/Forum/viewtopic.php?t=148696 .


This also seems to work:
[code][asy]
size(300);

xaxis(-4, 20, linewidth(1), Ticks("%", 0, 0, 2, 0, extend=true));
xaxis(YEquals(20), 0, 20, invisible, Ticks("%", 0, 0, 2, 0, extend=true, linewidth(0.5)));
xaxis(YEquals(-20), 0, 20, invisible, Ticks("%", 0, 0, 2, 0, extend=true, linewidth(0.5)));

yaxis(-20, 20, linewidth(1), Ticks("%", 0, 0, 2, 0, extend=true));
yaxis(XEquals(20), 1, 20, invisible, Ticks("%", 0, 0, 2, 0, extend=true, linewidth(0.5)));
yaxis(XEquals(-5), 1, 20, invisible, Ticks("%", 0, 0, 2, 0, extend=true, linewidth(0.5)));
yaxis(XEquals(20), -20, -1, invisible, Ticks("%", 0, 0, 2, 0, extend=true, linewidth(0.5)));
yaxis(XEquals(-5), -20, -1, invisible, Ticks("%", 0, 0, 2, 0, extend=true, linewidth(0.5)));
[/asy]
[/code]
[asy]size(300);

xaxis(-4, 20, linewidth(1), Ticks("%", 0, 0, 2, 0, extend=true));
xaxis(YEquals(20), 0, 20, invisible, Ticks("%", 0, 0, 2, 0, extend=true, linewidth(0.5)));
xaxis(YEquals(-20), 0, 20, invisible, Ticks("%", 0, 0, 2, 0, extend=true, linewidth(0.5)));

yaxis(-20, 20, linewidth(1), Ticks("%", 0, 0, 2, 0, extend=true));
yaxis(XEquals(20), 1, 20, invisible, Ticks("%", 0, 0, 2, 0, extend=true, linewidth(0.5)));
yaxis(XEquals(-5), 1, 20, invisible, Ticks("%", 0, 0, 2, 0, extend=true, linewidth(0.5)));
yaxis(XEquals(20), -20, -1, invisible, Ticks("%", 0, 0, 2, 0, extend=true, linewidth(0.5)));
yaxis(XEquals(-5), -20, -1, invisible, Ticks("%", 0, 0, 2, 0, extend=true, linewidth(0.5)));[/asy]


abstract algebra
group theory
Ring Theory
induction
superior algebra
superior algebra unsolved

Let $ G$ be an abelian group, and let $ H$ be a subgroup of finite index. Show that any character $ \chi : H \to \mathbb{C}^1$ can be extended to a character of $ G$. Is this still true if $ G$ is non-abelian, but $ H$ is abelian?

I'm unclear on if you meant $ \mathbb{C}^1$, the additive group, or $ \mathbb{C}^*$, the multiplicative group.  Either way, both groups are divisible, hence injective objects in the category of abelian groups, so the character extends, whether $ H$ is finite index or not.

For the non-abelian case, again I don't know if you meant $ \mathbb{C}^1$ or $ \mathbb{C}^*$, but either way, $ G\equal{}D_{\infty}$, the infinite dihedral group, and $ H\equal{}\mathbb{Z}$, with $ [G: H]\equal{}2$, should give a counterexample.

Steve

I mean $ \mathbb{C}^*$, the multiplicative group. I'm sorry, I don't really see your proof, or your counterexample, could you please take some time to elaborate it for me?

$ \mathbb{C}^*$ is an injective module because it's divisible. check out any introduction on homological algebra.

nevertheless, it is interesting if there is a proof which is constructive (injective = divisible over PID needs AC).

Here is something for your proof:

Let $ g\in G\setminus H$ and take a minimal natural number $ n>1$ with $ g^n\in H$ (Why does $ n$ exist?). Then every element $ h'$ of the subgroup $ H'$ generated by $ g$ and $ H$ is uniquely represented by $ g^mh$ with $ (m,h)\in \mathbb Z/n\times H$. Pick $ \alpha\in\mathbb C^*$ with $ \chi(g^n)\equal{}\alpha^n$ and extend $ \chi$ to $ H'$ by setting $ \chi'(g^mh)\equal{}\alpha^m\chi(h)$. (Why is $ \chi': H'\to\mathbb C^*$ a homomorphism?)

Since $ [G: H']<[G: H]$ you can finish your proof by induction on $ [G: H]$.

BTW: This is one part of the proof that a divisible Abelian group is an injective $ \mathbb Z$-module.


geometry
rectangle
rotation
combinatorics unsolved
combinatorics

hi,
i'll be happy if anyone help me to solve this problem :
[u]The problem :[/u]

suppose that ABCD is a square .let's divide it into $8\times 8$

cases such that each case is a square .how many rectangles

are there in ABCD ?

thanks in advance.

thank you very much for the result.i will be happy if you show

me the demonstration,please .

[quote="kevinatcausa"]I think you're leaving out some rectangles.  For example, the grid has plenty of squares with side $\sqrt{2}$.[/quote]
The problem just states to find the number of rectangles inside ABCD, and that can mean anything: rectangles that have their sides along the grid lines; rectangles that have vertices in the knots of the gird, or maybe even just any rectangles inside ABCD(then the answer is infinity).
The problem statement should be clearer.

[quote="rem"]Notice that for every rectangle $mXn$ where $m\ge n$, there are $2(9-m)(9-n)$ such rectangles in the $8X8$ square. We add a $2$ because the rectangle can be rotated by $90^{o}$ to get a different image unless it's a square( We will later subsrtact off that case), and if we have the given $mXn$ rectangle with its longer side horizontal, there are $(9-m)(9-n)$ places for its left top vertice to go.
Because $1\le m,n\le 8$, the answer is:
$\sum_{i=1 j=1}^{8}2(9-i)(9-j) = 36^{2}= 1296$ (the case for the squares got substracted off automatically because each rectangle with $i=j$ would have $(9-i)(9-j)$ counted only once, which is what we need.)[/quote]

Hmmm... you got the right answer, but with the wrong derivation.  Actually

$\sum_{i=1 j=1}^{8}2(9-i)(9-j) = 2 * 1296 = 2592$

You don't want the two in the sum;  without it, you count each square once, and the $mXn$ rectangles are counted both at (m, n) and (n, m), which is exactly what you want.

A shorter proof is to notice that each rectangle is uniquely defined by a pair of rows and a pair of columns;  there are $C(9,2) = 36$ pairs of each, so there are $36^{2}= 1296$ rectangles.


Prove that for any positive real numbers $ a_1,...,a_n$ with product 1 we have
$ n^n (a_1^n+1)...(a_n^n+1)\geq (a_1+...+a_n+1/a_1+...+1/a_n)^n $


let $ x,y,z,a,b,c,r>0$, prove that:

$ \frac{x\plus{}y\plus{}a\plus{}b}{x\plus{}y\plus{}a\plus{}b\plus{}c\plus{}r}\plus{}\frac{y\plus{}z\plus{}b\plus{}c}{y\plus{}z\plus{}a\plus{}b\plus{}c\plus{}r}>\frac{z\plus{}x\plus{}a\plus{}b}{z\plus{}x\plus{}a\plus{}b\plus{}c\plus{}r}$

[quote="tobeno_1"]let $ x,y,z,a,b,c,r > 0$, prove that:

$ \frac {x \plus{} y \plus{} a \plus{} b}{x \plus{} y \plus{} a \plus{} b \plus{} c \plus{} r} \plus{} \frac {y \plus{} z \plus{} b \plus{} c}{y \plus{} z \plus{} a \plus{} b \plus{} c \plus{} r} > \frac {z \plus{} x \plus{} a \plus{} b}{z \plus{} x \plus{} a \plus{} b \plus{} c \plus{} r}$[/quote]
Maybe we need to prove that
\[ \frac {x \plus{} y \plus{} a \plus{} b}{x \plus{} y \plus{} a \plus{} b \plus{} c \plus{} r} \plus{} \frac {y \plus{} z \plus{} b \plus{} c}{y \plus{} z \plus{} a \plus{} b \plus{} c \plus{} r} > \frac {z \plus{} x \plus{} a \plus{} c}{z \plus{} x \plus{} a \plus{} b \plus{} c \plus{} r}
\]
:?:
If so, assume $ a \plus{} b \plus{} c \plus{} x \plus{} y \plus{} z \equal{} p,$ rewrite your inequality in the form $ Ar^2 \plus{} Br \plus{} C > 0$
and prove that $ A > 0,$ $ B > 0$ and $ C > 0.$ :wink:


Ok, I checked websites, but I really do not know. I am applying for several summer programs, if they accept me, do I need to go. Can I just say, thanks but no thanks. Any similar experiences would be nice.

MathCamp, PROMYS, Ross. I hope that I can get into at least one  :D

"I am applying for several summer programs, if they accept me, do I need to go. Can I just say, thanks but no thanks."

As far as I know (and someone please correct me if I'm wrong), none of the national-level summer programs require that you attend if you've applied and are accepted.  I can say with certainty that Mathcamp and HCSSiM do not view an application as an promise to attend.  Of course, any program that invites you to attend wants you to go, but we all understand that applicants may be interested in multiple programs.

Yeah, the application is not binding if you get in. I applied to MathCamp and PROMYS, and am hoping that I get into at least one, but if I get into both, I plan on going to MathCamp...