Εύχομαι ότι ακολουθεί είναι σωστό . Η άσκηση για όσους την ξέρουν είναι μεγάλος μπελάς ...
Θα βάλω τη λύση σε hint αφήνοντας όσους δεν την προσπάθησαν να την προσπαθήσουν .

Να Βρείτε τις λύσεις στους θετικούς ακεραίους της εξίσωσης

$ (2^n\minus{}1)(3^n\minus{}1)\equal{}m^2$

[hide]Η λυση μου
Με mod 4 εύκολα βρίσκουμε ότι n άρτιος και έστω $ n\equal{}2N$ . Θα έχουμε τότε ότι  $ 2^n\minus{}1\equal{}aA^2$  και $ 3^n\minus{}1\equal{}aB^2$
όπου ο a είναι περιττός και square free .
Οι παραπάνω μπορούν να γραφούν όμως και ως $ (2^N)^2\minus{}aA^2\equal{}1$ και $ (3^N)^2\minus{}aB^2\equal{}1$

Γνωρίζουμε ότι οι λύσεις τις εξίσωσης $ x^2\minus{}ay^2\equal{}1$ δίνονται από μία αρχική  $ (u,v)$  σε μία ακολουθία τέτοια ώστε
$ u_n\plus{}\sqrt{a}v_n\equal{}(u\plus{}\sqrt{a}v)^n$  όπου οι $ u_n, v_n$ είναι εναλλάξ άρτιος και περιττός και επιπλέον $ u_{2n}\equal{}2u_n^2\minus{}1$

Επομένως θα έχουμε ότι $ 2^N\plus{}\sqrt{a}A\equal{}u_k\plus{}\sqrt{a} v_k$  και $ 3^N\plus{}\sqrt{a}B\equal{}u_l\plus{}\sqrt{a}v_l$ όπου είτε ο k είναι άρτιος είτε ο $ l$

Aν κ άρτιος τότε $ k\equal{}2K$ και $ u_k\equal{}2u_K^2\minus{}1$ άτοπο προφανώς
Αν $ l$ άρτιος τότε $ l\equal{}2L$ και $ u_l\equal{}2u_L^2\minus{}1\equal{}3^N$ το οποίο δίνει άτοπο με mod 3 άρα η μόνη λύση είναι $ n\equal{}0$ qed  :D [/hide]

$ (2^n \minus{} 1)(3^n \minus{} 1) \equal{} m^2$
Με $ mod4$ βρισκουμε οτι $ n \equal{} 2k$ και στην συνεχεια με mod16 βρισκουμε οτι $ k \equal{} 2x$
Εχουμε:$ (4^{2x} \minus{} 1)(9^{2x} \minus{} 1)$=m^2 επομενως $ (6^{2x} \plus{} 1)^2 \equal{} (2^{2x} \plus{} 3^{2x})^2 \plus{} m^2$
Αρα $ 6^{2x} \plus{} 1 \equal{} d(z^2 \plus{} w^2)$  και $ 2^{2x} \plus{} 3^{2x} \equal{} d(z^2 \minus{} w^2)$ ($ d$ περιττος,$ (z,w) \equal{} 1$ και $ z$ αρτιος και $ w$ περιττος)
Προσθετωντας τις 2 εξισωσεις εχουμε $ 6^{2x} \plus{} 3^{2x} \plus{} 2^{2x} \plus{} 1 \equal{} 2dz^2$ Oμως αφου $ z$ αρτιος θα πρεπει $ 6^{2x} \plus{} 3^{2x} \plus{} 2^{2x} \plus{} 1 \equal{} 0mod8 \equal{} > 3^{2x} \plus{} 1 \equal{} 0mod8 \equal{} > 9^x \equal{} 7mod8$ ΑΔΥΝΑΤΟ...Μενει βεβαια να ελεγξουμε για $ x \equal{} 0$ αρα $ n \equal{} 1$ μοναδικη λυση...

για x=0 ειναι n=0


calculus
derivative
analytic geometry
graphing lines
slope
search
function

Let $ f$,$ f: \mathbb{R}\rightarrow\mathbb{R}$, have a continuous derivative and let $ A\equal{}(a,b)$ be a point in the plane, which does not lie on the graph of $ f$.

[list]$ (i)$Show, that there always exists a point $ B$ on the graph of $ f$, which is the closest to $ A$.[/list]
[list]$ (ii)$Show, that the normal to the graph in the point $ B$ goes through $ a$.[/list]
[list]$ (iii)$Find the points on $ g(x)\equal{}\sqrt{x}$, which are the closest to $ (3,0)$.[/list]

Thank you, Rajiv.

$ (ii)$ Let $ d(x)$ be a distance from the point $ (a,b)$ to some point on the curve. Then $ d^{2}(x)\equal{}(x\minus{}a)^{2}\plus{}(y\minus{}b)^{2}$, where $ x\in \mathbb{R}$. We search a minimum of $ d(x)$. Hence it is among points $ (x,y)$, for which $ (d^{2})^{\prime}\equal{}0 \rightarrow y^{\prime}\equal{}\minus{}\frac{x\minus{}a}{y\minus{}b}$ or the points where the derivative does not exist. But the slope of the line through $ (a,b)$ and $ (x,y)$ is $ k\equal{}\frac{y\minus{}b}{x\minus{}a} \rightarrow k\equal{}\minus{}\frac{1}{y^{\prime}}.$ Of course we must observe also $ lim_{x \to \infty}d(x)$ and $ lim_{x \to \minus{}\infty}d(x)$, but we know, that it goes toward $ \infty$.

[quote="Rajiv"][quote="tmrfea"]$ (iii)$Find the points on $ g(x) = \sqrt {x}$, which are the closest to $ (3,0)$[/quote]

Let $ (h, k)$ be the point to which $ (3,0)$ is closest. This means that the distance of $ (3, 0)$ from the tangent to $ (h, k)$ is the least amongst its distance from the tangent to any other point on the curve. Evidently, $ (3,0)$ lies on the normal to the curve at $ (h,k)$.

The slope of the normal is $ - 2\sqrt {h}$.
The equation of the normal is: $ y + 2x\sqrt {h} = c$. Since this line satisfies $ (3,0)$, we get $ c = 6\sqrt h$. Now, you can easily find slope of normal since you know three points on it, $ (0, c), (3, 0), (h, \sqrt h)$.
The required point on the curve is: ${ (\frac {5}{2}, \sqrt {\frac {5}{2}}})$

we can verify this by drawing a circle with $ (3,0)$ as center and $ \sqrt {2.75}$ as radius. It should touch $ g(x)$ only in one point, and it does![/quote]

This also explains $ (ii)$ to some extent.

Easy!

iii) Let $ P(t^2,\ t)\in {y\equal{}\sqrt{x}} (t\geq 0)$, for the point $ A(3,\ 0)$ we have $ \overline{AP}^2\equal{}(t^2\minus{}3)^2\plus{}t^2\equal{}t^4\minus{}5t^2\plus{}9$

$ \equal{}\left(t^2\minus{}\frac{5}{2}\right)^2\plus{}\frac{11}{4}\geq \frac{11}{4}$, the equality holds when $ t\equal{}\sqrt{\frac{5}{2}}$, or $ P\left(\frac{5}{2},\ \sqrt{\frac{5}{2}}\right)$.

[quote="kunny"]Easy!

iii) Let $ P(t^2,\ t)\in {y \equal{} \sqrt {x}} (t\geq 0)$, for the point $ A(3,\ 0)$ we have $ \overline{AP}^2 \equal{} (t^2 \minus{} 3)^2 \plus{} t^2 \equal{} t^4 \minus{} 5t^2 \plus{} 9$

$ \equal{} \left(t^2 \minus{} \frac {5}{2}\right)^2 \plus{} \frac {11}{4}\geq \frac {11}{4}$, the equality holds when $ t \equal{} \sqrt {\frac {5}{2}}$, or $ P\left(\frac {5}{2},\ \sqrt {\frac {5}{2}}\right)$.[/quote]

Pretty clean...atleast faster than what I did...!!!

Great!

Now I have also the first one.

[list]$ (i)$ We have to show that previously defined function $ d^{2}(x)\equal{}(x\minus{}a)^{2}\plus{}(y\minus{}b)^{2}$, which is continuos (since it is a sum of two continuous functions), has a global minimum on $ \mathbb{R}$. It is obvious that $ \displaystyle lim_{x \to \infty}\equal{}lim_{x \to \minus{}\infty}\equal{}\infty$. Therefore we can observe the function $ d^{2}$ on some closed interval $ [\minus{}t,t]$ (since for each $ A \in \mathbb{R}^{\plus{}}$ we can find $ t$, such that $ d^{2}(x)>A$ for all $ |x|>t$. Now since $ d^{2}(x)$ is continuos, it has a minimum on $ [\minus{}t,t]$. For $ t$ great enough it is also a global minimum.[/list]

[list][b]Geometric solution of [/b] $ (i)$: We draw a graph of the function $ f(x)$ and the point $ (a,b)$. Then draw a circle with the center in $ (a,b)$. We are increasing the radius of this circle until it touches the curve. Sooner or later this will happen. (this is in fact true for any function, not just continuos). We can also see that the tangent to the circle in the touching point is also the tangent to $ f(x)$. If it wasn`t, then there would exist a point of the graph lying within the circle. Contradiction. [/list]


vector
induction
real analysis
real analysis unsolved

Prove that if we have $n$ vectors of norm at most 1 in $R^k$  we can order them in some way, let's say $ v_1,...,v_n$ such that all partial sums have norms at most $\frac{5^{k-1}}{2}$. In some references I saw that actually we can reduce this to $k$ and for $k=2$ to $\frac{\sqrt{5}}{2}$. It seems that proving it for $k$ it's a very difficult job, but I don't get it with  $\frac{\sqrt{5}}{2}$ in the case $k=2$.

ouch, thats like impossible to do.  It looks like college stuff!

[quote="harazi"]Prove that if we have $n$ vectors of norm at most 1 in $R^k$  we can order them in some way, let's say $ v_1,...,v_n$ such that all partial sums have norms at most $\frac{5^{k-1}}{2}$. [/quote]
Shouldn't there be some additional assumption (that the sum of all vectors is $0$ or something like that)?

I think I found a fairly simple proof of the estimate $2^k-1$. It goes by induction on $k$. When $k=1$, the construction is easy: take $v_1$ to be any vector and then take $v_j$ to be a vector whose direction is opposite to that of $v_1+\dots+v_{j-1}$. Then all partial sums are bounded by $1$. Now suppose that the statement is proved for $k-1$. Take any $n$ vectors in $\mathbb R^k$ and consider the longest vector $w$ that can be obtained by summing several of our vectors. Without loss of generality, $w=v_1+\dots+v_\ell$. It is not hard to see that the projections of $v_1,\dots,v_\ell$ to the direction of the vector $w$ are non-negative and the projections of $v_{\ell+1},\dots,v_n$ to the direction of $w$ are non-positive (otherwise we could drop or add the corresponding vector to make $w$ longer). Let now $v'_j$ be the projection of $v_j$ to the hyperplane $H$ orthogonal to $w$. Note that $v'_1+\dots+v'_\ell=v'_{\ell+1}+\dots+v'_n=0$. According to the induction assumption, we can enumerate the vectors $v_1,\dots,v_\ell$ in such a way that the partial sums of the vectors $v'_1,\dots,v'_\ell$ are bounded by $2^{k-1}-1$ and similarly for $v_{\ell+1},\dots,v_n$. Now take $v_1$ to be the first vector; then take several of the vectors $v_{\ell+1},\dots$ until the projection of the sum to the direction of $w$ becomes negative; then take several of the vectors $v_2,\dots$ until it becomes positive again; and so forth. Note that the projection of any partial sum to the direction of $w$ is bounded by $1$ while its projection to $H$ is bounded by $2(2^{k-1}-1)$ because each partial sum obtained in this way is a sum of two partial sums of the vectors $v_1,\dots,v_\ell$ and $v_{\ell+1},\dots,v_n$. Thus the length of each partial sum does not exceed $2(2^{k-1}-1)+1=2^k-1$.

The end  :)


:lol: Λοιπόν  αν  και  νέος  θα  ήθελα   να  προτείνω  μια  άσκηση  οχι  τόσο  δύσκολη  αλλα  ισα  ίσα  για  να  ξανα  δώσω  ζωή  σε  αυτό  το  Φορουμ  !!!!!!!!!!!!!!!!!!!!!!

  Θα  ηθελα  την  συμβολη όλων  των  λινκερς .  Είναι  πραγματικά   κρίμα  να  υπάρχει  αυτό  το  φόρουμ  και  να  μην  συμμετεχει  κανείς  για  πολύ  καιρό .

Δινεται  τρίγωνο  ΑΒΓ  να κατασκευασθεί  ευθεία  ΔΕ  τετοια  ώστε  να  τεμνει  τις  πλευρές  ΑΒ  και ΑΓ στα Δ  και Ε  αντίστοιχα  και να ισχύει  οτι  ΔΒ =ΔΕ =ΕΓ .
  Αν  και  το  σχήμα  δεν  ανταποκρίνεται  στην  πραγματικότητα  μπορεί   να  βοηθήσει

Θα μ'ενδιέφερε η λύση με την αναλυτική γεωμετρία

[hide="Η δική μου λύση"]
Έστω $\widehat{ADE}=2\theta, \widehat{AED}=2\phi$.

Τότε $\theta$ είναι η γωνία βάσης του ισοσκελούς τριγώνου $DBE$ και $\ \phi$ η γωνία βάσης του $\triangle EDC$
Από το τρίγωνο $ADE$ έχουμε

$2\phi+2\theta+\widehat{A} = 180^o \Rightarrow 2(\theta+\phi)=180-\widehat{A} \Rightarrow \theta+\phi = 90^o-\frac{\widehat{A}}{2}$

Φέρουμε τις μεσοκαθέτους των $BE,DC$. Αυτές τέμνονται σε σημείο $K$.

Έχουμε $KB=KE$ και $KD=KC$.
Επίσης, οι μεσοκάθετοι $DK,EK$ είναι ταυτόχρονα και διχοτόμοι των αντίστοιχων γωνιών $\widehat{BDE},\widehat{DEC}$.

Θα δείξουμε η $AK$ είναι διχοτόμος της γωνίας $A$.

Θα υπολογίσουμε πρώτα τη γωνία $KDC$

$\theta+\phi = 90^o-\frac{\widehat{A}}{2} \Rightarrow$

$(90-\phi)-\theta=\frac{\widehat{A}}{2} \Rightarrow$

$\widehat{EDK}-\widehat{EDC} = \frac{\widehat{A}}{2}\Rightarrow$

$\widehat{CDK}=\frac{\widehat{A}}{2}$

(αποδείχθηκε ταυτόχρονα, οτι η γωνία $\widehat{EDK}$ είναι μεγαλύτερη από την $\widehat{EDC}$, δηλαδή η ημιευθεία $DC$ περιέχεται στη γωνία $\widehat{EDK}$)

Εξάλλου, το  τρίγωνο $KDC$ είναι  ισοσκελές, οπότε η γωνία της κορυφής είναι $\widehat{DKC} = 180-\widehat{A}$.

Το τετράπλευρο $ADKC$ έχει τις δύο απέναντι γωνίες $A$ και $K$ παραπληρωματικές, επομένως είναι εγγράψιμο.
Και επειδή $KD=KC$ συμπεραίνουμε ότι $\widehat{DAK}=\widehat{KAC}$, δηλαδή το $K$ ανήκει στη διχοτόμο της γωνίας $A$ του τριγώνου $ABC$.

Πάμε να βρούμε τώρα σε ποιο ακριβώς σημείο της διχοτόμου βρίσκεται το $K$.
Δηλαδή θέλουμε να βρούμε ακόμη μία ευθεία ή κύκλο στον οποίο ανήκει το $K$.

$\triangle KDE:$

$\widehat{DKE}+ \widehat{KDE} + \widehat{KED}= 180^o \Rightarrow$
$\widehat{DKE} + (90-\phi)+(90-\theta) = 180 \Rightarrow$
$\widehat{DKE} = \phi+\theta \Rightarrow$
$\widehat{DKE} = 90^o-\frac{\widehat{A}}{2}$

Τα τρίγωνα $BDK,EDK,ECK$ είναι ίσα αφού έχουν τις πλευρές τους μία προς μία ίσες.

Επομένως είναι $\widehat{BKD}=\widehat{DKE}=\widehat{EKC}$.

Το άθροισμα των τριών γωνιών είναι ίσο με $3\widehat{DKE} = 3(90^o-\frac{\widehat{A}}{2}) = 270^o - 3\frac{\widehat{A}}{2}$

$3\widehat{DKE} \geq 180^o \Leftrightarrow$
$\widehat{A}\leq 60^o$

επομένως αν είναι $\widehat{A} < 60^o$, τότε το $K$ είναι εσωτερικό του τριγώνου και βλέπει την πλευρά $BC$ υπό γωνία $360-3\widehat{DKE} = 3\frac{\widehat{A}}{2}+90$
Αν $\widehat{A}=60^o$ το $K$ ανήκει στη $BC$ και

Αν $\widehat{A}>60^o$, το $K$ είναι εξωτερικό του τριγώνου και βλέπει τη $BC$ υπό γωνία $3\widehat{DKE} = 270^o - 3\frac{\widehat{A}}{2}$

(Για να δούμε την αναλογία που υπάρχει μεταξύ των δύο περιπτώσεων, κατασκευάζουμε τον περιγεγραμμένο κύκλο του τριγώνου $KBC$, τότε τα σημεία του κύκλου που βρίσκονται στο τόξο $BC$ που δεν περιέχει το $K$  θα βλέπουν τη $BC$ υπό γωνία $180-(270 - 3\frac{\widehat{A}}{2}) = 3\frac{\widehat{A}}{2}-90$)

[size=125][b]Συγκεντρωτικά:[/b][/size]

Έστω $AH$ η διχοτόμος του τριγώνου $ABC$
Έχουμε τις περιπτώσεις:

$\blacksquare \ \ \widehat{A}=60^o$ Το $K$ ταυτίζεται με  το σημείο $H$

$\blacksquare \ \ \widehat{A} < 60^o$ Τα $K,A$ βρίσκονται στο ίδιο μέρος της $BC$ και το $K$ βλέπει τη $BC$ υπό γωνία $3\frac{\widehat{A}}{2}+90$

$\blacksquare \widehat{A} > 60^o$ Τα $K,A$ βρίσκονται εκατέρωθεν της $BC$ και το $K$ βλέπει τη $BC$ υπό γωνία $270^o - 3\frac{\widehat{A}}{2}$

[size=150][b][color=blue]Κατασκευή[/color][/b][/size]

Βρίσκουμε το σημείο $K$.

Αν $AB=AC$, απλώς κατασκευάζουμε τον κύκλο $(K,KC)$, που θα περνάει και από το $B$ και θα τέμνει τις πλευρές στα ζητούμενα σημεία.

Αν $AB>AC$, τότε υπάρχει σημείο $D'$ της πλευράς $AB$, τέτοιο ώστε $AD'=AC$, και προφανώς το $D'$ θα ανήκει στον κύκλο $(K,KC)$. Εμείς τότε διαλέγουμε σαν $D$ το άλλο σημείο τομής του $(K,KC)$ με την $AB$

Επίσης, υπάρχει ένα $E'$ στην προέκταση της $AC$, τέτοιο ώστε $AB=AE'$.
Ο κύκλος $(K,KC)$ θα τέμνει την πλευρά $AC$ στο $E'$ και σε ένα εσωτερικό της σημείο $E$.

Τα σημεία $D,E$ είναι τα ζητούμενα.

[color=red][b]ΥΓ[/b]
Παρατήρησα ότι προκύπτουν και άλλες ιδιότητες, οπότε ίσως να υπάρχει και πιο απλή κατασκευή.

Ίσως αν χρησιμοποιήσουμε το γεγονός ότι τα $D',B,C,I$ είναι ομοκυκλικά ($I$ το έγκεντρο)
[/color][/hide]

Λουκά τι να πω , εκπληκτική λύση.

Άλλος τρόπος.
Αν ΒΔ=ΔΕ=ΕΓ=κ και εφαρμόσουμε το νόμο των συνημιτόνων στο ΑΔΕ τρίγωνο
καταλήγουμε σε μια δευτεροβάθμια ως προς κ που έχει διακρίνουσα

$D=4(cosA)^2(a-b)^2-8ab(cosA-1)>0$ .Άρα αφού γωνίες πλευρές γνωστές το κ προσδιορίζεται

Ελπίζω σωστό

Λοιπόν  pontios   ωραία  η  λύση  σου  αλλα  αρκετά  μεγάλη .Τελος  πάντων  θα  γραψω  την  δικιά  μου  λύση .
 Μου  πήρε  σχεδόν  δύο  εβδομάδες  να  το  σκεφτώ  αλλα  πιστεύω  οτι  έκανα  μια  κομψή  προσεγγιση .

 Ανάλυση....

  Εστω  ότι  η  ζητούμενη  ευθεία  εχει  κατασκευασθεί .  Φέρουμε από  την κορυφή  Α  ευθεία  παράλληλη  στην DΕ
και πέρνουμε AZ = AB  φέρουμε  ακόμη  την  ΒΖ . Μετον  τρόπο  αυτό τα  τρίγωνα ΑΒΖ  και  DΒΖ  είναι  όμοια  καθώς  $\angle DEB =\angle AZB$  ωσ  εντό  εκτός  και  επι  τα  αυτά .Ετσι  έχουμε $\frac{AB}{DB}=\frac{BZ}{BE}=\frac{AZ}{DE}$  (1)έτσι  AB  = AZ . Ακόμη  φέρουμε   από  το  Z  παράλληλη  στην BC
έτσι τα  τρίγωνα  ΚΕΖ  είναι  όμοιο με  το ΕΒC  . Oπότε  $\frac{KE}{EC}=\frac{EZ}{BE}$  άρα $\frac{KC}{EC} =\frac{BZ}{BE}$ (2) . Aπό (1) και (2)    προκύπτει  ότι ΚC = AB  .

Σύνθεση .........

Στο  δεδομένο  τρίγωνο  κατασκευάζουμε  κύκλο  με  ακτίνα ΑΒ  και  κέντρο Α . Παιρνουμε ΚC =ΑΒ  και φέρουμε  από  το  Κ παράλληλη  στην  ΒC  που  τέμνει  τον  κύκλο  στό  Z . Ενώνουμε  το  ΒΖ το  οποίο  τέμνει  την ΑC  στό Ε .  Από  το  Ε  φέρουμε  Παράλληλη  στην ΑΖ  η οπόια τέμνει  την ΑΒ  στο  D . AND THATS  ALL,.................................

 Η   Ζητούμενη  ευθεία έχει  κατασκευασθεί......... Γιώργος
 :lol:  :lol:  :lol:  :lol:  :lol:

Γιώργο, πολύ ωραία η λύση σου. Και πολύ απλή!

Μόνο που δυσκολεύτηκα αρκετά μέχρι να καταλάβω αυτά που γράφεις

[quote="gdeskos"] Ανάλυση....

  Εστω  ότι  η  ζητούμενη  ευθεία  εχει  κατασκευασθεί .  Φέρουμε από  την κορυφή  Α  ευθεία  παράλληλη  στην DΕ
και παίρνουμε AZ = AB [color=red] -> Ενώ το κατασκευάζεις έτσι, παρακάτω αποδεικνύεις ότι είναι ΑΖ=ΑΒ. Φαντάζομαι ότι εννοείς "η παράλληλη από το Α τέμνει την ευθεία ΒΕ στο Ζ" [/color]
..φέρουμε  ακόμη  την  ΒΖ . Με τον  τρόπο  αυτό τα  τρίγωνα ΑΒΖ  και  DΒ[color=red]Ε[/color]  είναι  όμοια  καθώς  $\angle DEB =\angle AZB$  ως  εντός  εκτός  και  επι  τα  αυτά .Ετσι  έχουμε $\frac{AB}{DB}=\frac{BZ}{BE}=\frac{AZ}{DE}$  (1) έτσι  AB  = AZ . Ακόμη  φέρουμε   από  το  Z  παράλληλη  στην BC
έτσι το  τρίγωνο  ΚΕΖ  είναι  όμοιο με  το ΕΒC. [color=red]Ποιο είναι το σημείο Κ;;; Έπρεπε να παρατηρήσω τις αναλογίες που ακολουθούν για να καταλάβω ότι ΚΖ\\BC και Κ ανήκει στην AC[/color]  Oπότε  $\frac{KE}{EC}=\frac{EZ}{BE}$  άρα $\frac{KC}{EC} =\frac{BZ}{BE}$ (2) .
Aπό (1) και (2)    προκύπτει  ότι ΚC = AB  .

Σύνθεση .........

Στο  δεδομένο  τρίγωνο  κατασκευάζουμε  κύκλο  με  ακτίνα ΑΒ  και  κέντρο Α . Παιρνουμε ΚC =ΑΒ [color=red]..εννοείς ότι το Κ θα βρίσκεται πάνω στην πλευρά AC[/color] και φέρουμε  από  το  Κ παράλληλη  στην  ΒC  που  τέμνει  τον  κύκλο  στό  Z . Ενώνουμε  το  ΒΖ το  οποίο  τέμνει  την ΑC  στό Ε .  Από  το  Ε  φέρουμε  Παράλληλη  στην ΑΖ  η οπόια τέμνει  την ΑΒ  στο  D . AND THATS  ALL,.................................

[/quote]

Μετά λοιπόν από την .. αποκωδικοποίηση είδα ότι ήταν μία πολύ ωραία λύση, αφού χρησιμοποιείς μόνο παράλληλες ευθείες (ειδικά σε σύγκριση με τη δική μου που έφτασα σε εκείνο το "τραγικό" $90+3\frac{A}{2}$). Ήξερα βέβαια ότι η λύση μου ήταν πολύπλοκη, αλλά αυτή βρήκα, αυτήν έγραψα.

Αν και είμαι βέβαιος ότι δεν μπορεί να υπάρχει πιο απλή λύση από τη δική σου, προτείνω να ψάξουμε και για άλλες λύσεις γιατί φαίνεται να έχει πολλές ακόμη...

Α! πες τη λύση και στον άλλον που περιμένει ακόμα [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=71717]εδώ[/url] (αλλά γραψ'τα κανονικά γιατί θα τον τρελάνεις τον τύπο  ;) )

Σιλουανέ,

Δοκίμασα τη δεύτερη λύση που προτείνεις και δε βρήκα την ίδια διακρίνουσα

Οκ  Λουκά  με  συγχωρείς  έχεις  απόλυτο  δίκιο  . Την  άσκηση  την έγγραψα  στο  σχολείο  και  μαλιστα    βιαστικά . Λοιπόν  ψάχνω  είδη  να  βρώ  αυτό  που  λές  με  τα  ομοκυκλικά  σημεία . Ψαχνω  όμως παράλληλα  την  λύση στο  νέο  πρόβλημα   του  Σιλουανού ..... :lol: Τέλος πάντων
Φαίνεται  ότι η μέθοδος  που  χρησιμοποίησα  γενικεύεται  και  για  άλλες  ιδιότητες  των  ευθειών  που  ζητούνται να  κατασκευασθούν π.χ

       Να  κατασκευασθεί  ευθεία  που  να  διέρχεται  από  τις  ΑΒ  και  ΑΓ  και  να  ισχύει ότι ΔΒ =ΔΕ+ΕΓ
Φέροντας  πάλι  τις  παράλληλες  αυτες  και  μετατρέποντας  τους  λόγους  στους  ζητόυμενους , θα  καταφέρουμε  να  επιλύσουμε  τέτοιου  είδους  προβλήματα .. :lol:  ;)

Αυτό με τα ομοκυκλικά το έψαξα λίγο και φαίνεται να μεταφράζεται σε μία πολύ χρήσιμη ιδιότητα:
Αν $BH \parallel DE$ και $H$ σημείο του περιγεγραμμένου κύκλου, τότε $CH=AC-AB$
(θεωρώντας ότι $AC>AB$)

Εδώ τα ομοκυκλικά είναι τα $B,C,I$ και $H$

[quote="gdeskos"]
Φαίνεται  ότι η μέθοδος  που  χρησιμοποίησα  γενικεύεται  και  για  άλλες  ιδιότητες  των  ευθειών  που  ζητούνται να  κατασκευασθούν π.χ

       Να  κατασκευασθεί  ευθεία  που  να  διέρχεται  από  τις  ΑΒ  και  ΑΓ  και  να  ισχύει ότι ΔΒ =ΔΕ+ΕΓ
[/quote]Καλή η ιδέα σου. Ίσως να γίνεται.

Θα τα προσπαθήσω και τα δύο , όταν βρω χρόνο ...

[quote="pontios"]
Σιλουανέ,

Δοκίμασα τη δεύτερη λύση που προτείνεις και δε βρήκα την ίδια διακρίνουσα[/quote]

Τι διακρίνουσα βρήκες ? Αυτή με την αναλυτική την κοίταξε κάποιος ? Είναι σωστή?

Το τριώνυμο που βρήκα είναι

$(1-2cosA)x^2+2(b+c)(cosA-1)x+a^2=0$

Ναι πράγματι έτσι είναι , εγώ είχα κάνει λάθος κάπου. Αλλά και έστι αν δεν κάνω πάλι λάθος η διακρίνουσα μετά από πράξεις γράφεται

$D=4(b-c)^2(cosA)^2+8bc(1-cosA)>0$

Άρα στην ουσία είναι το ίδιο. έχω κάνει πάλι λάθος ?Αν ναι πες μου.


[quote="Phelpedo"]A parking lot for compact cars has $12$ adjacent space, $8$ of which are occupied. A large sport-utility vehicle arrives, needing 2 adjacent open spaces. What is the probability that it will be able to park?[/quote]

[hide]Our goal is to choose four empty spaces so that two of them are next to each other.

Let us find the number of ways chosing empty spaces such that none of them are next to eachother.

This is just $\binom{9}{4}=126$.

The total number of chosing four spaces is $\binom{12}{4}=495$.

So the number of these that works is $\binom{12}{4}-\binom{9}{4}=369$.

So the probability that it will be able to park is

$\frac{369}{495}=\frac{41}{55}$[/hide]

[quote="Phelpedo"][quote="mathgeniuse^ln(x)"]
[hide]Our goal is to choose four empty spaces so that two of them are next to each other.

Let us find the number of ways chosing empty spaces such that none of them are next to eachother.

This is just $\binom{9}{4}=126$.
[/hide][/quote]

How did you come up with this? I got the same thing but I had an intermediate step, so I'm wondering if you have a shortcut.[/quote]

[hide]Wait so are you saying my answer is right?

Well, I put the twelve slots in a line.

I took three slots out to have 9 remaining.

Then I choose four spaces.

now, I can put the three slots anywhere to guarantee that no spaces will be next to eachother, and it goes from there.[/hide]

[quote="Phelpedo"]Followup- a related problem: A gardener plants three maple trees, four oak trees, and five birch trees in a row. He plants them in random order, each arrangement being equally liekly. Find the probability that no two birch trees are next to one another.

More generally: given $g$ girls and $b$ boys, how many different ways can you seat them in a row of $g+b$ seats so that no two boys sit together? Consider two cases: boys/girls distinguishable and boys/girls not distinguishable.[/quote]

[hide]a) $\frac{\binom{11}{5}*\frac{7!}{3! 4!}}{\frac{12!}{3! 4! 5!}=\frac{7}{12}$[/hide]


Let $ABCDEF$ be a hexagon (which may not be regular), which circumscribes a circle $S$. (That is, $S$ is tangent to each of the six sides of the hexagon.) The circle $S$ touches $AB, CD, EF$ at their midpoints $P,Q,R$ respectively. Let $X,Y,Z$ be the points of contact of $S$ with $BC, DE, FA$ respectively. Prove that $PY, QZ, RX$ are concurrent.


Let $ A_1A_2...A_n$,$ n\geq3$ a regular poligon,A,B,C three consecutive tops and M a point on the circumscriptible circle so that AC separes the points B and M.Prove that : \[ MA\plus{}MC\equal{}2MBcos\frac{\pi}{n}\].

I will post a proof but in roumain...I hope you will understand...if knot...ask:D
Fie $ z_A \equal{} 1$, $ z_B \equal{} \epsilon$ , $ z_C \equal{} \epsilon^2$, unde $ \epsilon \equal{} cos \frac {2\pi}{n} \plus{} isin \frac {2\pi}{n}$ iar afixul lui M il vom nota $ z_M \equal{} cost \plus{} isint$.
\[ MA \equal{} |z_M \minus{} 1| \equal{} \sqrt {(cost \minus{} 1)^2 \equal{} sin^2t} \equal{} \sqrt {2 \minus{} 2cost} \equal{} 2sin\frac {t}{2}
\]

\[ MB \equal{} 2sin(\frac {t}{2} \minus{} \frac {\pi}{n})
\]

\[ MC \equal{} 2sin(\frac {t}{2} \minus{} \frac {2 \pi}{n})
\]
utilizand aceste rezultate egalitatea din enunt devine
\[ 2sin \frac {t}{2} \plus{} 2sin( \frac {t}{2} \minus{} \frac {2\pi}{n}) \equal{} 4sin( \frac {t}{2} \minus{} \frac {\pi}{n})cos \frac {\pi}{n}
\]
Se transforma in produs membrul stang si inegalitatea este evidenta.


topology
algebra
polynomial
conics
real analysis
real analysis solved

Consider the following two topologies on $\mathbb{R}^2$:
$1)$ the Zariski topology ${\cal{T} }_1$.
$2)$ the topology ${\cal{T}}_2$,the weakest topology where are straight lines are closed set.
show that$(\mathbb{R}^2,{\cal{T}}_1)$ and $(\mathbb{R}^2,{\cal{T}}_2)$ are not homeomorphic.

I'm beginner in topology and i found it really exciting but it seems that you're not  :D

What is Zariski topology.I think I have heard about it before but I cant remember it right now :blush: .Can you tell me about it.

The problem is I've heard of more than one construction being called the Zariski topology.

First of all, there's [url=http://planetmath.org/encyclopedia/ZariskiTopology.html]this[/url]. Another one is number $4$ on [url=http://www-history.mcs.st-and.ac.uk/~john/MT4522/Lectures/L11.html]this page[/url]. They are not the same in $\mathbb R^2$ (although they are the same in $\mathbb R$).  So, sam-n, could you please give the exact definition of what you're referring to as being the Zariski topology?

i mean by Zariski topology where a base open set is complement of zero sets of polynomials in two variables

Assume they are homeomorphic.

$\mathcal T_2$ has a subbasis of closed sets formed by sets (the straight lines) with the property that each two intersect in at most one point (this means that each closed set can be written as a finite union of arbitrary intersections of lines). The same must then be true for $\mathcal T_1$.

Consider the curve determined by $xy=1$ (a rectangular hyperbola). This must be a finite union of intersections of sets in ur subbasis for $\mathcal T_1$. Since it's a finite union, it means that one of these intersections contains infinitely many points $(x,y)$ with the property that $xy=1$. Since this set is the zero set of a polynomial $P(x,y)=\sum_{i,j=0}^{m,n}a_{ij}x^iy^j$, it means that $\sum_{i,j=0}a_{ij}x^{i-j}$ is identically null (we get a polynomial with infinitely many roots out of that if we multiply it with a certain power of $x$), which, in turn, means that $P(x,y)=0,\ \forall xy=1$. This means that our set, which is an intersection of elements in the subbasis, contains the set $S=\{(x,y)|xy=1\}$, so there is at least one set in our subbasis which has $S$ as a subset.

We can do the same for another conic whose intersection with $S$ has at least two points, and we have a contradiction: we have found two elements of the subbase with intersection consisting of at least two points.

Is it correct?


Given a regular $n$-gon inscribed in a circle of radius 1, where $n > 3$.
Define $G(n)$ as the average length of the diagonals of this $n$-gon.

Prove that if $n \rightarrow \infty,  G(n) \rightarrow \frac{4}{\pi}$.

The elementary angle is a = 2*pi/n. For the angle k*a , the diagonal lenght is 2*sin(k*a/2) = 2*sin(k*pi/n) , with k=2,3,...,[n/2]
For any k < [n/2] there are n diagonals . For k=[n/2] there are n diagonals(if n is odd) or [n/2] diagonals (if n is even).
Finally G(n) looks as a Riemann (or Darboux) sum for 2*sin(x*pi/2), and its limit is Sum[0 =< x =< 1){2*sin(x*pi/2).dx} = 4/pi

Riemann ? Darboux ?
In Flanders students know only elementary stuff (such as if x -> 0, (sin x)/x -> 1  :D).
The official (ELEMENTARY) solution is about two pages long.

But do they know things like Sum(a =< x =< b){sin(c*x)} = (cos(c*a) - cos(c*b))/c , and the definition of this kind of things ?

No, they don't (I didn't know this neither).

The official solution involves finding and proving a formula for the sum

sin(a) + sin(2a) + sin(3a) + ... + sin(na)

but that's only a small part of the solution.

A way to compute it is:

sum (sin (kx)) (k=1,n)= sin[(n+1)/2*x]*sin(nx/2)/sin(x/2) which is done with induction.

Or it can also be done using complex numbers. computing both sums sum (cos(kx) and also sum (sin (kx)) k=1,n.

cheers! :D

I guess the complex numbers - proof is as follows :

Denote cis x = cos x + i sin x. Then consider the sum

cis x + cis 2x + ... + cis nx.

This sum equals \sum (cis x)^k for 1 \leq k \leq n (de Moivre's formula).
This is the sum of a geometric progression. Summing this progression gives

\sum (cis x)^k = (cis nx - 1)/(cis x - 1).

I think the result should follow by comparing the real and imaginary parts, isn't it ?
I like complex numbers :D But in general Flemish students are not used to use them.
So instead of proving the formula like this they should guess it and prove it by induction.
Another very elegant way to prove it is using a telescoping sum ! (Try it !) :)


[quote="hungkhtn"]Let $ a,b,c$ be non-negative with sum $ 1$. Prove that
\[ \frac{a(b+c)}{c+a}+\frac{b(c+a)}{a+b}+\frac{c(a+b)}{b+c}\ge 1-\max\{(a-b)^{2},(b-c)^{2},(c-a)^{2}\}. \]
[/quote]
Very nice, Hung! And it can be solved in exactly the same way as your solution here [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=110449[/url]  :lol:

Yes, Ductrung. You are totally right. This is an other application of Rearrangement Inequality. I also really like this kind of Solution. Actually, Song Yoon Kim has this form of solution before me.


Putnam
invariant
symmetry
algebra
polynomial
complex numbers
college contests

Fixed the typo:

[quote="justdudxit"]A3. Prove that if $11z^{10}+10iz^{9}+10iz-11 = 0$, then $|z|$ = 1[/quote]

I think I found it. :)

The equation $11z^{10}+10iz^{9}+10iz-11 = 0$ is equivalent to both
$\begin{cases}10(z+i)(z^{9}+i)=1-z^{10}\\11(\bar{z}+i)(\bar{z}^{9}+i)= i(z+z^{9})\end{cases}$.

So multiplying the first equation with the conjugate of the second, we get \[110(|z|^{2}-i(z+\bar{z})+1) (|z^{9}|^{2}-i(z^{9}+\bar{z}^{9})+1) = (1-\bar{z}^{10})(z+z^{9})I.\] Now an interesting observation is that the LHS is invariant under conjugation, so must be real, and so must the RHS.

RHS real gives $z-\bar{z}^{1}0z+z^{9}-\bar{z}^{1}0z^{9}$ imaginary, so adding its conjugate gives $0$: \[(z^{9}+\bar{z}^{9}+z+\bar{z})(|z|^{2}-1)=0.\] So either $|z|=1$ or we assume $z^{9}+\bar{z}^{9}=-(z+\bar{z})$. Now the imaginary part of the LHS must also be zero, yielding \[(z+\bar{z})(|z^{9}|^{2}+1)+(z^{9}+\bar{z}^{9})(|z|^{2}+1)=0,\] under $z^{9}+\bar{z}^{9}=-(z+\bar{z})$ this would give $|z|=|z|^{9}$, so $|z|=0$ (trivially impossible) or $|z|=1$ (previously excluded).

So we conclude that $|z|=1$. :)

[quote]Now an interesting observation is that the LHS is invariant under conjugation, so must be real, and so must the RHS. [/quote]

There is a symmetry changing z into its conjugate but the LHS isn't invariant under conjugation.

conjugate of -i is +i

Use properties of the following transformation:  $ z \mapsto \frac {1 \minus{} a z}{1 \minus{} \overline{a}z}$.   For instance,
you can look at the unit disk ($ \{ |z| \equal{} 1 \}$.

Solve for $ z^9$ and see what $ a$ is and then argue that $ |z|$ cannot be greater than 1 and cannot be
less than 1.

Writing $ w \equal{} iz$, we get $ 11(w^{10} \plus{} 1) \minus{} 10(w^9 \plus{} w) \equal{} 0.$  Obviously if this polynomial has all roots on the unit disk then all roots $ z$ will be on the unit disk.  We note that when $ w \equal{} 0$ this polynomial is nonzero, hence $ w \neq 0$.  We divide through by $ w^5$ and get a quintic polynomial in $ u \equal{} w \plus{} w^{ \minus{} 1}$ - namely $ q(u) \equal{} 11(u^5 \minus{} 5u^3 \plus{} 5u) \minus{} 10(u^4 \minus{} 4u^2 \plus{} 2)$.  If this polynomial has all real roots in the interval $ [ \minus{} 2,2]$ then we will be done.  Then it is just a matter of substituting appropriate roots of $ u^5 \minus{} 5u^3 \plus{} 5u$ and $ u^4 \minus{} 4u^2 \plus{} 2$ and we find five real roots.


trigonometry
real analysis
real analysis unsolved

Discuss the convergence or divergence of

$ \sum_{n\equal{}1}^{\infty}\frac{(\minus{}1)^{n} |\sin n|}{n}$

It seems, it was discuss on the forum, it converge.


[hide="My Solution"]
Combining like terms works best here.

$ x^2\cdot 2^x \plus{} 8 \equal{} 2x^2 \plus{} 2^{x \plus{} 2}\implies (2x^2 \minus{} 8) \plus{} (4\cdot 2^x \minus{} x^2\cdot 2^x) \equal{} 2(x^2 \minus{} 4) \minus{} 2^{x}(x^2 \minus{} 4) \equal{} (x^2 \minus{} 4)(2 \minus{} 2^{x}) \equal{} 0.$

Therefore, $ \boxed{x \in \{\minus{}2,1,2\}}$.
[/hide]

The answer is that $ x$ can be $ 1$, $ 2$, or $ \minus{}2$.

The answer is the sum of the numbers  :wink:

So yep, 1 is right.


abstract algebra
superior algebra
superior algebra theorems

To prove that $\ds M\otimes_R\frac{R}{I}\cong\frac{M}{IM}$ where $M$ is an $R$-module and $I$ is an ideal of $R$
is the following correct?

Consider $g: M\rightarrow M\otimes R/I$ defined by $m\mapsto m\otimes\overline{1}.$  This is clearly well-defined.  Notice that $\ker(g)=IM$ since
$m\otimes \overline{1}=m\otimes\overline{i}=im\otimes\overline{1}$ for any $i\in I$
and
$g(im)=im\otimes \overline{1}=m\otimes \overline{i}=m\otimes \overline{1}$ implying that $g(i_1m_1+\cdots+i_sm_s)=m_1+\cdots+m_s\otimes\overline{1}.$
Thus, the First Isomorphism Theorem (or if you call it the Fundamental Isomorphism Theorem or whatever) yields the result.

Or, what about showing that
$\frac{R}{I}\otimes\frac{R}{J}\cong\frac{R}{I+J}?$
My thoughts were to define $g: R\rightarrow\frac{R}{I}\otimes\frac{R}{J}$ such that $r\mapsto \overline{r}\otimes\hat{r}$ where $\overline{r}$ is the conjugacy class of $r$ modulo $I$ and $\hat{r}$ is the same but modulo $J$.  I wanted to show that the $\ker g= I+J$ in this case.  But, I am stuck.

[quote="baz"]To prove that $\ds M\otimes_R\frac{R}{I}\cong\frac{M}{IM}$ where $M$ is an $R$-module and $I$ is an ideal of $R$
is the following correct?

Consider $g: M\rightarrow M\otimes R/I$ defined by $m\mapsto m\otimes\overline{1}.$  This is clearly well-defined.  Notice that $\ker(g)=IM$ since
$m\otimes \overline{1}=m\otimes\overline{i}=im\otimes\overline{1}$ for any $i\in I$
and
$g(im)=im\otimes \overline{1}=m\otimes \overline{i}=m\otimes \overline{1}$ implying that $g(i_1m_1+\cdots+i_sm_s)=m_1+\cdots+m_s\otimes\overline{1}.$
Thus, the First Isomorphism Theorem (or if you call it the Fundamental Isomorphism Theorem or whatever) yields the result.[/quote]

It is not true that $m\otimes\bar 1=m\otimes\bar i$. It is true that $0=m\otimes\bar 0=m\otimes\bar i$.
If $im\in IM$, then $g(im)=im\otimes\bar 1=m\otimes i\cdot\bar 1=m\otimes\bar i=0$, so $im\in\ker g$. Since elements of the form $im$ generate $IM$, we get $IM\subseteq\ker g$.
You have not proved the reverse inclusion. So you have just proved that there exists a surjection $M/IM\to M\otimes_RR/I: m+IM\stackrel{\varphi}{\to }m\otimes(1+I)$. $m+IM$ denotes the class on $m$ modulo $IM$ in $M$ and $1+I$ denotes the class of $1$ modulo $I$ in $R$.
I don't know if it can be proved directly by following your idea that $\ker g\subseteq IM$.
Let $\bar f: M\times R/I\to M/IM$ be defined by $\bar f(m,r+I)=rm+IM$. $\bar f$ is clearly $R$-bilinear, hence by the universality property of the tensor product, it induces an $R$-modules homomorphism  $f: M\otimes_RR/I\to M/IM$ which on the generators of $M\otimes_RR/I$ acts by $f(m\otimes (r+I))=rm+IM$
$f\circ\varphi(m+IM)=f(m\otimes(1+I))=m+IM$. Therefore $f\circ\varphi=1_{M/IM}$ i.e. the identity map of $M/IM$. This implies that $\varphi$ is injective. Since we knew that $\varphi$ was a surjection, we get that $\varphi$ is an isomorphism.

[quote="baz"]Or, what about showing that
$\frac{R}{I}\otimes\frac{R}{J}\cong\frac{R}{I+J}?$
My thoughts were to define $g: R\rightarrow\frac{R}{I}\otimes\frac{R}{J}$ such that $r\mapsto \overline{r}\otimes\hat{r}$ where $\overline{r}$ is the conjugacy class of $r$ modulo $I$ and $\hat{r}$ is the same but modulo $J$.  I wanted to show that the $\ker g= I+J$ in this case.  But, I am stuck.[/quote]

Why not use the previous problem?
Take $M=R/I$. Then $R/I\otimes_RR/J\simeq\frac{R/I}{JR/I}=\frac{R/I}{(I+J)/I}\simeq\frac{R}{I+J}$ by another theorem of isomorphism.


I don't want to cross any boundaries as far as rules go and sharing information with others, but I need clarification for the definition of sequence in problem 5 of round 4.

Does it refer to a sequence where no element of the sequence may appear twice, or is it a sequence that can have elements multiple times in itself, as long as it satifies the restrictions stated?

We cannot answer this question while the round is in progress.


inequalities
algebra
polynomial
inequalities unsolved

prove or disprove the following inequality($a,b,c>0$):
$a^4b^2+b^4c^2+c^4a^2\geq abc(a^3+b^3+c^3)$

while trying to solve another inequality, that one here was sufficient but i wasn't able to prove or disprove it.

Peter

ok, it's wrong(sorry for posting it): set $a=t^2,b=1,c=t$. on the left, you get a polynomial of degree 8 in t, on the right a polynomial of degree 9.


AMC
AIME
geometry
3D geometry
octahedron
probability
number theory

The numbers 1, 2, 3, 4, 5, 6, 7, and 8 are randomly written on the faces of a regular octahedron so that each face contains a different number. The probability that no two consecutive numbers, where 8 and 1 are considered to be consecutive, are written on faces that share an edge is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

[quote="PenguinIntegral"]Use the resources section, don't spam up the boards...[/quote]

The resources section doesn't have a solution so he's not spamming the boards, he's trying to get help. Don't discourage people and especially when you haven't checked the topic first.

[quote="PenguinIntegral"]Yes it does. Click the thread linky and scroll down.[/quote]

One solution posted is not complete and/or correct, the other has something messed up about quadrants or something

[quote="PenguinIntegral"][quote="Ignite168"]and the other has something messed up about quadrants or something[/quote]

He got the right answer. So what if he messed up the notation, it's not hard to figure out what he means.

In any case, he should probably be reviving that thread.[/quote]

Reviving old threads is about the worst possible thing you could do. You shouldn't ever do this. Besides, Phelpedo wasn't doing this purposely.

[quote="Karth"]
Reviving old threads is about the worst possible thing you could do. You shouldn't ever do this. [/quote]
Why not? There's a place linked to in the resources section to talk about this problem, hence, we should talk about it there, instead of breaking it all up.

This is mostly my personal opinion, and other moderators, feel free to correct me here.  If you don't really understand the solutions posted by people to a problem at all, then feel free to start a new thread, provided you ask a specific question about the problem.  Like "How do we go about counting the number of distinct ways to arrange these people, I'm afraid I'm just not quite understanding the easiest way to do it. "

However, if you have a question about a specific solution... use moderation in bringing back a really old thread.  If its really pertinent... sure.. go ahead.  Its generally frowned upon.  I would honestly try and see if there is anything the solution has to offer you, try it again, and if you still have no ideas... then say something about it.  Hope this helps

Also, most if not all of the posts in the thread should be kept to PM. (Most of the information presented is pretty much forum courtesy/rules)

[quote]I thought you were allowed to revive threads if you have a question about a solution?[/quote]

Yes, I think we're allowed to revive threads in these occasions. But don't revive old threads if you just feel like it. In other words, do it if it's very important. (But then I can't really say anything, because I'm not a mod.)

Anyway, PenguinIntegral, you've caused a lot more spam than you've solved, so let's return to the problem in question. In case you didn't pick it up, I don't find the solutions in the other thread very satisfying.

[quote="Phelpedo"]Anyway, PenguinIntegral, you've caused a lot more spam than you've solved,[/quote]
???

[quote="Phelpedo"]In case you didn't pick it up, I don't find the solutions in the other thread very satisfying.[/quote]

[quote="Pakman2012"] If you don't really understand the solutions posted by people to a problem at all, then feel free to start a new thread, provided you ask a specific question about the problem.  Like "How do we go about counting the number of distinct ways to arrange these people, I'm afraid I'm just not quite understanding the easiest way to do it. "

[/quote]

Mods:

This thread should probably be....er...moderated.  :wink:  Perhaps a split...

[quote="Elemennop"][quote="PenguinIntegral"][quote="Phelpedo"]Anyway, PenguinIntegral, you've caused a lot more spam than you've solved,[/quote]
???
[/quote]

Instead of keeping your mouth shut and letting the thread go it's course, you made several posts detailing how the thread [i]shouldn't[/i] be up, consequently resulting in [i]more[/i] posts (including this one) in reply to your original posts, none of which contain any help towards the original poster's question.[/quote]
It's an intelligent discussion, just in the wrong place...

[quote="PenguinIntegral"][/quote][quote="Elemennop"][quote="PenguinIntegral"][quote="Phelpedo"]Anyway, PenguinIntegral, you've caused a lot more spam than you've solved,[/quote]
???
[/quote]

Instead of keeping your mouth shut and letting the thread go it's course, you made several posts detailing how the thread [i]shouldn't[/i] be up, consequently resulting in [i]more[/i] posts (including this one) in reply to your original posts, none of which contain any help towards the original poster's question.[/quote][quote="PenguinIntegral"]
It's an intelligent discussion, just in the wrong place...[/quote]

You just proved his point. It's in the WRONG place, the original poster still hasn't got an answer to his question and instead, has to sit through a debate about something completely irrelevant to what he wants to know.

[quote="Ignite168"][/quote][quote="PenguinIntegral"][/quote][quote="Elemennop"][quote="PenguinIntegral"][quote="Phelpedo"]Anyway, PenguinIntegral, you've caused a lot more spam than you've solved,[/quote]
???
[/quote]

Instead of keeping your mouth shut and letting the thread go it's course, you made several posts detailing how the thread [i]shouldn't[/i] be up, consequently resulting in [i]more[/i] posts (including this one) in reply to your original posts, none of which contain any help towards the original poster's question.[/quote][quote="PenguinIntegral"]
It's an intelligent discussion, just in the wrong place...[/quote][quote="Ignite168"]

You just proved his point. It's in the WRONG place, the original poster still hasn't got an answer to his question and instead, has to sit through a debate about something completely irrelevant to what he wants to know.[/quote]
Yeah, I know. Sarcasm doesn't translate right without facial expressions :(

This really needs to be split. However, my orginal point still holds. When the resources section has the answer, use it. When you need clarification on an ambiguous answer, revive the thread. When you have a specific question about the problem, go post it in the correct forum. In any case, don't just repost it...

[quote="Karth"]Um, somebody should lock this thread... it's getting sort of out of control. Insults are eventually going to start flying...

EDIT: Which year's AIME was this?[/quote]
2001. :wink:

I'll see if I can get a solution to this problem and I'll try to post it.  :)

[quote="mysmartmouth"][quote="Ignite168"]
You just proved his point. It's in the WRONG place, the original poster still hasn't got an answer to his question and instead, has to sit through a debate about something completely irrelevant to what he wants to know.[/quote]

[size=117]
[b]SPLIT THE THREAD INTO ROUND TABLE[/b][/size][/quote]

I DON'T HAVE MODERATION POWERS SO STFU.

[quote="Ignite168"]I DON'T HAVE MODERATION POWERS SO STFU.[/quote]Well I do, and I'm just going to lock this thread because it's out of control.

Jeez.... :noo:


Phil loves socks. He has 6 red socks, 6 blue socks, and 6 green socks. Phil pulls out socks at random without looking, and he does not replace them. What is the probability of Phil pulling out a red sock, then a green sock, then a red sock again?

Explain your answer, please.

[hide="this is 10% skill, 20% luck, and 15% the power of pure concentrated will"]Probablity of 1st sock=$\frac{6}{18}$=$\frac{1}{3}$
Probablity of 2nd sock=$\frac{6}{18-1}$=$\frac{6}{17}$
Probablity of 3rd sock=$\frac{6-1}{17-1}$=$\frac{5}{16}$
$\frac{1}{3}*\frac{6}{17}*\frac{5}{16}=\boxed{\frac{5}{136}}$[/hide]

is it about ?
Firstly,the probability of Phils pulling out a red sock is 1/3
Secondly,there are 6green socks in 17 socks, so the probability of Phils pulling out a green sock is 6/17
Finally, there are 5 red socks in 16 socks, so the probability of Phils pulling out a red sock is 5/16.
So the probability of Phils pulling out a red sock, then a green sock and a red sock again is (1/3)(6/17)(5/12).100%=3.68%
I'm not sure of this result and my English is so bad that i can't explain clearlier.

The real answer:

[hide]tz is right, listen to him [/hide] :D


For how many values of b does the equation  \[ \frac{2}{2\minus{}\frac{2}{2\minus{}\frac{2}{2\plus{}x}}}\equal{}b\] have no solution?

(what are the values, and how do you do this?)

I simplified it to (1+x)/(2+x) = b, but I can't get the answer after that. Any ideas from anyone?

-rts2007

hello, i have simplifyed the given equation to $ b\equal{}2\plus{}\frac{2}{x}$ with the solution $ x\equal{}\frac{2}{b\minus{}2}$ if $ b\ne2$.
Sonnhard.

so, for how many values is it no solution?
The answer is 3. But i have no clue how.

Obviously, there is no solution when $ b \equal{} 0$.

Looking for last one...

EDIT: Got it.

Value of $ b$ that makes $ x\equal{}\minus{}2$ will be the third one.

We have $ \minus{}2\equal{}\frac{2}{b\minus{}2} \implies 2\equal{}\minus{}2b\plus{}4 \implies b\equal{}1$

Thus, three values of $ b$ that make the expression solution-less is $ 0, 1, 2$

hello, no, i think my answer is correct, because $ 2\minus{}\frac{2}{2\plus{}x}\equal{}2\left(1\minus{}\frac{1}{2\plus{}x}\right)\equal{}2\left(\frac{1\plus{}x}{2\plus{}x}\right)$ and further $ 2\minus{}\frac{2}{\frac{2(1\plus{}x)}{2\plus{}x}}\equal{}2\minus{}\frac{2\plus{}x}{1\plus{}x}\equal{}\frac{x}{1\plus{}x}$ and finally $ \frac{2}{\frac{x}{1\plus{}x}}\equal{}\frac{2\plus{}2x}{x}\equal{}\frac{2}{x}\plus{}2$ and from here we get my equation
$ \frac{2}{x}\plus{}2\equal{}b$ with the solution $ x\equal{}\frac{2}{b\minus{}2}$ if $ b\ne2$.
Sonnhard.

The problem asked, 'For what values of b does the equation have no solution?'

You said that if $ b\ne2$, then the equation will have solution, which means that in order for the equation to not have solution, $ b\equal{}2$, which is one of three values of $ b$ that I mentioned.

hello, no, your answer is wrong, if $ b\equal{}0$ we have the solution $ x\equal{}\minus{}1$ and if $ b\equal{}1$ you will have $ x\equal{}\minus{}2$ .
Sonnhard.

If $ x\equal{}\minus{}1$, the denominator of $ \frac{2}{2\minus{}\frac{2}{2\plus{}x}}$ is 0.

If $ x\equal{}\minus{}2$, the denominator of $ \frac{2}{2\plus{}x}$ is 0.

hello, the only case where the equation has no solution is $ b\equal{}2$.
Sonnhard.

hello, you have right, i was not carfully enough, because of the range of definition it must be $ x\ne\minus{}2$,$ x\ne\minus{}1$ and $ x\ne 0$.
Sorry,Sonnhard.

Fantasy Lover is correct. For the the equation to be without solution, the value of $ b$ must be undefined or infinite ($ \infty$) For that to happen $ x$ is equal to either $ \minus{}1$ or $ \minus{}2$. Could not come up with another solution.


modular arithmetic
number theory
relatively prime

Let $ b_1<b_2<...<b_{\phi(m)}$ be the integers between $ 1$ and $ m$ that are relatively prime to $ m$(including $ 1$), and let $ B\equal{}b_1b_2b_3\cdots b_{\phi(m)}$ be their product.

(i) Show that $ B\equiv 1\pmod{m}$ or $ B\equiv \minus{}1\pmod{m}$

(ii) Find a patter for when $ B$ is equal to $ 1\pmod{m}$ and when it is equal to $ \minus{}1\pmod{m}$

[quote="countdownmath"]Let $ b_1 < b_2 < ... < b_{\phi(m)}$ be the integers between $ 1$ and $ m$ that are relatively prime to $ m$(including $ 1$), and let $ B \equal{} b_1b_2b_3\cdots b_{\phi(m)}$ be their product.

(i) Show that $ B\equiv 1\pmod{m}$ or $ B\equiv \minus{} 1\pmod{m}$

(ii) Find a patter for when $ B$ is equal to $ 1\pmod{m}$ and when it is equal to $ \minus{} 1\pmod{m}$[/quote]
[hide="1"]
Notice that each number coprime to $ m$, say $ b_i$, in $ \bmod m$ has an inverse, say $ b_j$, such that $ b_i\ne b_j$ other than $ b_i \equal{} 1$ or $ b_i \equal{} m \minus{} 1$.  Thus, we can pair each $ b_i$ with its $ b_j$ and their product in $ \bmod m$ will be $ 1$.  Now, we are left with $ 1$ and $ m \minus{} 1$, whose product in $ \bmod m$ is $ \minus{} 1$.  Thus, the product of all of the numbers coprime to $ m$ under $ m$ is $ 1\cdot 1\cdot 1\cdots 1\cdot \minus{} 1\equiv \minus{} 1\bmod m$.

I think I'm missing something, though, because I'm not getting any possibility of $ 1\bmod m$ except when $ m \equal{} 2$, in which case $ \minus{} 1\equiv 1\bmod 2$... [/hide]
EDIT: Oh, sorry about that!  Thanks, [b]t0rajir0u [/b]and [b]scorpius119[/b].

[quote="The QuattoMaster 6000"][quote="countdownmath"]Let $ b_1 < b_2 < ... < b_{\phi(m)}$ be the integers between $ 1$ and $ m$ that are relatively prime to $ m$(including $ 1$), and let $ B \equal{} b_1b_2b_3\cdots b_{\phi(m)}$ be their product.

(i) Show that $ B\equiv 1\pmod{m}$ or $ B\equiv \minus{} 1\pmod{m}$

(ii) Find a patter for when $ B$ is equal to $ 1\pmod{m}$ and when it is equal to $ \minus{} 1\pmod{m}$[/quote]
[hide="1"]
Notice that each number coprime to $ m$, say $ b_i$, in $ \bmod m$ has an inverse, say $ b_j$, such that $ b_i\ne b_j$ other than $ b_i \equal{} 1$ or $ b_i \equal{} m \minus{} 1$.  Thus, we can pair each $ b_i$ with its $ b_j$ and their product in $ \bmod m$ will be $ 1$.  Now, we are left with $ 1$ and $ m \minus{} 1$, whose product in $ \bmod m$ is $ \minus{} 1$.  Thus, the product of all of the numbers coprime to $ m$ under $ m$ is $ 1\cdot 1\cdot 1\cdots 1\cdot \minus{} 1\equiv \minus{} 1\bmod m$.

I think I'm missing something, though, because I'm not getting any possibility of $ 1\bmod m$ except when $ m \equal{} 2$, in which case $ \minus{} 1\equiv 1\bmod 2$... [/hide][/quote]
Yes, you're missing something. You might be left with more than $ 1$ and $ m \minus{} 1$: in fact, if you take $ m$ to be a divisor of 24, you'll be left with [i]everything[/i] (For example, 5's inverse modulo 24 is 5.)

[hide="The issue"] As scorpius points out, the difficulty here is that the equation $ x^2 \equiv 1 \bmod m$ has more than two solutions in general; in other words, more than two elements are their own inverses.  You can even show that $ B^2 \equiv 1 \bmod m$ without much difficulty, and you can't conclude that $ B \equiv 1, \minus{} 1 \bmod m$.  :) [/hide]


Hi everyone,

I quite get the answer out for this question and i just feel like i have been beating my head into a wall for 4 hours! aghh....
[img]http://img245.imageshack.us/img245/5388/picture18ye1.png[/img]
but then if i sub in the non-homogenous BC, [img]http://img147.imageshack.us/img147/8363/picture17zg4.png[/img]

i get 0 = x

lol, what have i done this time! :P

any help would be so very much appreciated! :)

Cheers
Sarah

Replace $\sinh(n\pi y)$ with $\sinh(n\pi (1-y))$.


How do you set up a loop so that your computer runs your mouse through a process (for example, rigging online voting by doing it repeatedly)?

Thanks,
-diophantient

It depends on your operating system but if you use Windows, there are windows API functions for moving mouse, sending mouse clicks and keyboard strokes. I made a program once that could all by itself play a simple space shooter game by taking screenshots, analyzing them and sending back keyboard strokes. I don't remember the names of the functions to move your mouse but it shouldn't be too hard to google up. (I assume that you have a knowledge of some programming laguage and you're able to use windows API in that language. There [i]might[/i] be an already existing program that can move your mouse and do with it what you tell it to do, but I'm not aware of this.)


A copper bar with a mass of 12.340 g is dipped into 255 mL of 0.125 M AgNO3 solution.  When the reaction that occurs has finally ceased, what will be the mass of unreacted copper in the bar?  If all the silver that forms adheres to the copper bar, what will be the total mass of the bar after the reaction?

its simple.

just follow the basic equation for the rxn:
so you have:

$Cu+2Ag^+\rightarrow Cu^{2+}+Ag$

find out which reactant is limiting and then all you have to do is some basic stocheometry to find the answer.

its just a simple Redox (Reduction~Oxidation) reaction.

one ion gets reduced while the other atom gets oxidized.

also dont forget that you are dealing with Ag(NO3), not Ag^+, when finding the mol of Ag,

the rxn, that i wrote was the Net Ionic one.

oh i understand now, i didn't realize it was the net ionic and was confused at what happened to the NO3.  thanks a lot.

actually...just in case, what would the molecular equation be?  Cu + AgNO3 --> ?

$Cu + 2Ag(NO_3) \rightarrow 2Ag + Cu(NO_3)_2$

but i'm not 100% sure... we're just now learning this in my ap class so don't quote me on it lol

that is correct.
but keep in mind that NO3 never percipitates!

Oh yeah my bad. It should be:

$Cu_{(s)} + 2Ag(NO_3)_{(aq)} \rightarrow 2Ag_{(s)} + Cu(NO_3)_{2(aq)}$


I was wondering if there was a easy formula to use for thi

if have two points

and i need to solve for X (ie x???, y0)

does anyone have any advice i could use?

example if i had point1(x1119,y8129) point2(3623,2730)

I will now attempt to understand the problem.

So you have 2 points:
$(x_{1},y_{1}) (x_{2},y_{2})$
You know another point on the line is $(x_{0},y_{0})$
If you know $y_{0}$ and you're solving for $x_{0}$:
The formula is $\frac{(x_{2}-x_{1})(y_{0}-y_{2})}{y_{2}-y_{1}}+x_{2}$
If you know $x_{0}$ and you're solving for $y_{0}$:
The formula is $\frac{(y_{2}-y_{1})(x_{0}-x_{2})}{x_{2}-x_{1}}+y_{2}$

..I think.

actually i dont know if a thrid point is x0 y0

actually that seems to work

answer = (((X2 - X1)*(0 - Y2)) / (Y2 - Y1)) + X2;

is what i am useing to find x

anyone have seen anything wrong?


function
number theory unsolved
number theory

$ (1)$.Prove that if $ lcm(a,a\plus{}5)\equal{}lcm(b,b\plus{}5)$ then $ a\equal{}b$.
$ (2)$.Prove that the following equality doesn't take place ever:
$ lcm(a,b)\equal{}lcm(a\plus{}c,b\plus{}c)$,for $ a,b,c$ positive integers.

for the first one,note that $ \text{lcm} (x,y) = \frac {xy}{\gcd (x,y)}$ hence our equiation is equivalent to:

$ \frac {a(a + 5)}{A} = \frac {b(b + 5)}{B}$

where $ A = \gcd (a,a + 5)$ and $ B = \gcd (b,b + 5)$

now note that $ A\mid a$ and $ A\mid a + 5$ hence $ A\mid 5$ so we must have either $ A = 1$ or $ A = 5$ the same thing holds for $ B$,i.e. $ B = 1$ or $ 5$.

now if for example $ A = 5$ then we must have $ a = 5k$ for some $ k\in\mathbb{N}$ hence:

$ \frac {5k(5k + 5)}5 = \frac {b(b + 5)}B$

$ \Rightarrow 5\mid\frac {b(b + 5)}B$

$ \Rightarrow 5\mid b(b + 5)$

$ \Rightarrow 5\mid b$ and thus $ 5\mid b + 5$ so we must have $ 5\mid B$ but as we said before $ B = 1$ or $ 5$ thus we must have $ B = 5$,therefore $ A = 5$ implies $ B = 5$.

so we are up to 2 cases,$ A = B = 1$ or $ A = B = 5$,in either cases our equation is equivalent to:

$ a(a + 5) = b(b + 5)$

$ \Rightarrow a^2 + 5a = b^2 + 5b$

but this equality can't hold,furthermore let $ f(x) = x^2 + 5x$ then $ f'(x) = 2x + 5$ which is positive for $ x\geq 0$ hence this function is stricly increasing on the non-negative integers...

so $ a(a + 5) = b(b + 5)$ implies $ a = b$.


[quote="fishln"]Suppose $f(x) \in C^1[0,1]$, proof that
\[ \lim_{n \rightarrow \infty}n[\int_0^1f(t)dt-\frac{1}{n}\sum_{k=0}^{n-1}f(\frac{k}{n})] = \frac{f(1)-f(0)}{2}  \][/quote]
I remember  somebody posting this, try to search it also.
I'll try to give a sketch for the proof:
Denote $F(x)=\int_{0}^x f(t)dt$, then use the fact that $\int_0^1f(t)dt =\sum_{k=0}^{n-1}(F(\frac{k+1}{n})-F(\frac{k}{n}))$ and write Taylor expansions ${F(\frac{k+1}{n})-F(\frac{k}{n}) =\frac{1}{n}f(\frac{k}{n})+\frac{1}{2n^2}f'(u_k})$ and ${F(\frac{k}{n})-F(\frac{k+1}{n}) =\frac{1}{n}f(\frac{k+1}{n})+\frac{1}{2n^2}f'(u_{1k}})$, substract the second expansion from the first you get $2(F(\frac{k+1}{n})-F(\frac{k}{n}))$, then to this end you just need to sum this representation.
If it doesn't work i'll write more exactly.

It has been discussed before. Try to search the forum.

[quote="eugene"]
If it doesn't work i'll write more exactly.[/quote]
Sorry,eugene,I can't figure out with what you wrote,can you explain more?Thanks


function
number theory proposed
number theory

Provr for all $a \gneqq b$ and $n$ positive whole numbers that:
$n|$$\varphi (a^{n}-b^{n})$

Read the whole thread : http://www.mathlinks.ro/Forum/viewtopic.php?t=135914


Let $S$ be a set of $n$ distinct real numbers. Let $A_{s}$ be the set of numbers that occur as averages of two distinct elements of $S$. For a given $n \ge 2$, what is the smallest possible number of elements in $A_{s}$?

Source: 1992 Putnam

Is it $2n-3$?

I'm guessing that values 1 to n in a linear function f(x) = ax + b would create the smallest set of averages, not quite sure though.

[quote="mna851"]Do you mean distinct elements of $\ A_{s}$ ?[/quote]

The elements of $A_{s}$ are defined to be distinct.

Treething: Yes, can you prove it?


For positive integers $ x$ and $ y$, $ \frac{1}{x}\plus{}\frac{1}{y}\equal{}\frac{5}{12}$. What is the least possible value of $ x\plus{}y$?

From the expression, we get $ x\plus{}y\equal{}\frac{5xy}{12}$, so $ xy$ is divisible by 12. Then the least value could be 5, but cannot be attained. 10 maybe (for xy=24)? Yes, for x=6 and y=4, we get x+y=10


[hide="solution"]
This can be solved using the properties of logarithms.

From Equation 1
$ \log_{10} (2000xy) \minus{}(\log_{10} x)(\log_{10} y) \rightarrow \log_{10} 1000 \plus{} \log_{10} 2 \plus{} \log_{10} x \plus{} \log_{10} y   \minus{} (\log_{10} x)(\log_{10} y)\equal{} 4 \rightarrow \log_{10}2 \plus{} \log_{10}x \plus{} \log_{10}y \minus{} (\log_{10}x)(\log_{10}y) \equal{} 1$.

From Equation 2
$ \log_{10} 2xy \minus{}(\log_{10} y)(\log_{10} z) \rightarrow \log_{10} 2 \plus{}\log_{10} x \plus{}\log_{10} y \minus{}(\log_{10} y )(\log_{10} z)\equal{} 1$.

Set Equation 1 equal to Equation 2.
$ \minus{}(\log_{10} x)(\log_{10} y) \equal{} \minus{}(\log_{10} y)(\log_{10}z) \rightarrow x\equal{}z$ or $ \log_{10}y\equal{}0$, which makes $ y\equal{}1$.

Case 1 $ (x\equal{}z)$
$ \log_{10}zx \minus{}(\log_{10}x)(\log_{10}z) \rightarrow 2\log_{10}x \minus{}(\log_{10}x)^2\equal{}0$
$ (\log_{10}x)(2\minus{}\log_{10}x)\equal{}0$  $ x\equal{}z\equal{} 1$ or $ 100$

Case A: $ x\equal{}z\equal{}1$.  Substitute to get $ y\equal{}5$
Case B: $ x\equal{}z\equal{} 100$.  Substitute to get $ y\equal{}20$

Case 2 $ (y\equal{}1)$
Substitute into equation one to get that $ x\equal{}5$.  Then, substitute into equation 3 to get $ z\equal{}10^{\minus{}\log_{2} 5}$

Solution set at $ (1, 5, 1), (100, 20, 100), (5, 1 , 10^{\minus{}\log_{2} 5})$.
[/hide]

[quote="azjps"]AoPSWiki/resources: [[2000 AIME I Problems/Problem 9]].[/quote]

Except equation two is $ \log_{10} 2xy$ instead of $ \log_{10} 2yz$.
I was curious about the asymmetry of the equations.


ratio
geometry
rectangle
geometry proposed

It's problem 3 from the first Romanian TST in 2002. I have a computational proof and I won't post it because it's really boring: we use Menelaus to find the ratio in which side BC is divided by AA1, then we write Ceva's condition for concurrence with the ratios we have just calculated and we get something like (tanA-tanB)(tanB-tanC)(tanC-tanA)=0, so 2 of them must be equal, so the triangle is isosceles.

I remember slaving on this one for 45 minutes or so (during th contest):).

[quote="Valiowk"]Let M and N be the midpoints of the respective sides AB and AC of an acute-angled triangle ABC.  Let P be the foot of the perpendicular from N onto BC and let A1 be the midpoint of MP.  Points B1 and C1 are obtained similarly.[/quote]

Well, this looks a bit strange since the definition of $A_1$ is not symmetric in $B$ and $C$. However, we can easily see that if we take the foot $Q$ of the perpendicular from $M$ onto $BC$, and define $A_1$ as the midpoint of $NQ$, then we get the same point $A_1$. In fact, the quadrilateral $MNQP$ is a rectangle.

[quote="Valiowk"]If $AA_1$, $BB_1$ and $CC_1$ are concurrent, show that triangle $ABC$ is isosceles.[/quote]

Well, I have reduced this to

$(b^2+3c^2-a^2)(c^2+3a^2-b^2)(a^2+3b^2-c^2)$ = $(c^2+3b^2-a^2)(a^2+3c^2-b^2)(b^2+3a^2-c^2)$  ==> triangle is isosceles.

But this is algebra, again. Any nice proof?

  Darij

I think I got here as well during that contest. We replace expressions of the type a2+b2-c2 with 2ab*cosC, we divide both sides by 8 and by abc, and we get  \pi{cyl}(b*cosA+c)=\pi{cyl}(c*cosA+b)=>\pi{cyl}(b*cosA+c)/(a*cosB+c)=1.

b*cosA+a*cosB=c and b*cosA/a*cosB=tanB/tanA=>(b*cosA+c)/(a*cosB+c)=(2tanB+tanA)/(2tanA+tanB)=>\pi{cyl}(2tanB+tanA)/(2tanA+tanB)=1, and from here it's easy to derive \pi{cyl}(tanA-tanB)=0.

P.S.
{cyl} comes from cyclic.

See the attached drawing, the triangles $ MNA_1$ and $ BCA_2$ are centrally similar, centered at $ A$. $ A_2$ is the center of the rectangle erected outwardly on $ BC$, of dimensions $ a, h_a$. If $ A' \equiv BC \cap AA_2$, we have then:
$ \frac {BA'}{CA'} \equal{} \frac {S_{[ABA_2]}}{S_{[ACA_2]}}\equal{}\frac{AB \cdot BA_2 \cdot sin (\widehat {ABC}\plus{}\widehat {CBA_2})}{AC \cdot CA_2 \cdot sin (\widehat {ACB}\plus{}\widehat {BCA_2})}\equal{}\frac{c \cdot sin (\angle B \plus{} \angle \alpha)}{b \cdot sin (\angle C \plus{} \angle \alpha )}$ ( 1 ), where $ a, b, c$ and $ h_a, h_b, h_c$ are the sides and altitudes of the triangle, $ S_{[XYZ]}$ is the area of the $ \triangle XYZ$, $ \alpha$ being the half-angle of the diagonals of the rectangle constructed as explained and other two similar relations. If $ AA', BB', CC'$ are concurrent ($ B', C'$ are obtained similarly), then, by Ceva we get as the product of fractions (1) the unit and from this we get the same results as [b]Darij[/b].

Best regards,
sunken rock

Computational solution, although I'll post it since it seems sort of instructive on ``how to bash'' :D Not that there's anything terribly new here...

Employ barycentric coordinates.  Let $S_A = \frac{b^2+c^2-a^2}{2}$, et cetera.  By taking the midpoint of the foot of the perpendicular $(0: S_C : S_B)$ with $C = (0 : 0 : a^2)$ we find that $P=(0 : S_C : S_B + a^2)$.  With $M = (a^2 : a^2 : 0)$ we get $A_1 = \left( a^2 : S_C+a^2 : S_B+a^2 \right)$.  Hence $AA_1$ divides $BC$ in a $S_C+a^2$ to $S_B+a^2$ ratio.

Now by Ceva's Theorem, the condition is equivalent to \begin{align*}
		0 &= (S_B+a^2)(S_C+b^2)(S_A+c^2)-(S_B+c^2)(S_C+a^2)(S_A+b^2) \\
		&= \sum_{\text{cyc}} a^2(b^2-c^2)S_a + \sum_{\text{cyc}} (c^2-b^2)S_{BC} \\
		&= \sum_{\text{cyc}} (b^2-c^2)(a^2S_A-S_{BC}) \\
		&= \frac{1}{4} \sum_{\text{cyc}} (b^2-c^2)(2a^2(b^2+c^2-a^2)-(b^2+a^2-c^2)(c^2+a^2-b^2)) \\
		&= \frac{1}{4} \sum_{\text{cyc}} (b^2-c^2)(2a^2b^2+2b^2c^2+2c^2a^2+b^4+c^4-3a^4) \\
		&= \frac{1}{4} \sum_{\text{cyc}} (b^2-c^2)(b^4+c^4-3a^4) \\
		&= \frac{1}{4} \sum_{\text{cyc}} (b^2-c^2)(a^4+b^4+c^4-4a^4) \\
		&= - \sum_{\text{cyc}} a^4(b^2-c^2) \\
		&= (a^2-b^2)(b^2-c^2)(c^2-a^2)
	\end{align*} which holds if and only if $ABC$ is isosceles.

[quote=Valiowk]Let $M$ and $N$ be the midpoints of the respective sides $AB$ and $AC$ of an acute-angled triangle $ABC$.  Let $P$ be the foot of the perpendicular from $N$ onto $BC$ and let $A_1$ be the midpoint of $MP$.  Points $B_1$ and $C_1$ are obtained similarly.  If $AA_1$, $BB_1$ and $CC_1$ are concurrent, show that the triangle $ABC$ is isosceles.

[i]Mircea Becheanu[/i][/quote]

Is there a full proof

[quote=Valiowk]...Points $B_1$ and $C_1$ are obtained similarly...

[i]Mircea Becheanu[/i][/quote]

This is not a clear definition for $B_1, C_1.$

Solved from LiOG 3.8

Let the foot from $M$ to $BC$ be $T$, so $A_1$ is the center of rectangle $MNPT$. Let $AA_1$ meet $MN$ at $R$ and $BC$ at $S$. Let $MR=d$. Due to symmetry, since $AA_1$ passes through the center of the rectangle, we have $SP=d$ as well. Furthermore, by homothety at $A$, $BS=2d$, so $BP=3d$. However, we also have $$BP=BT+TP=\frac{c}{2}\cos\beta+\frac{a}{2},$$ so therefore, $$3d=\frac{c}{2}\cos\beta+\frac{a}{2}.$$ Thus, $$d=\frac{c\cos\beta+a}{6},$$ so $$\frac{BS}{CS}=\frac{\frac{c\cos\beta+a}{3}}{a-\frac{c\cos\beta+a}{3}}=\frac{a+c\cos\beta}{2a-c\cos\beta}=\frac{2a^2+2ac\cos\beta}{4a^2-2ac\cos\beta}.$$ However, by Law of Cosines, this is equal to $$\frac{3a^2+c^2-b^2}{3a^2+b^2-c^2}.$$ Thus, by Ceva's theorem, it suffices to show that $$\prod_{cyc}(3a^2+b^2-c^2)=\prod_{cyc}(3a^2-b^2+c^2)$$ implies that $\triangle ABC$ is isosceles. Let $p=a^2,q=b^2,c=r^2$ so that this is $$\prod_{cyc}(3p+q-r)=\prod_{cyc}(3p-q+r).$$ Let $x=q-r,y=r-p,z=p-q$ so that this is $$(3p+x)(3q+y)(3r+z)=(3p-x)(3q-y)(3r-z)$$ Since terms with an even number of factors in $\{x,y,z\}$ cancel, this is just $$9(pqz+pry+qrx)+xyz=0.$$ Substituting $x,y,z$ back in, this is $$9(\sum_{cyc}pq(p-q))+(p-q)(q-r)(r-p)=\sum_{cyc}8p^2q+p^2r-9pq^2,$$ and evaluating the cyclic sum and setting it to 0, this is simply just $$8(p^2q+q^2r+r^2p)-8(p^2r+q^2p+r^2q),$$ so the condition is just $$p^2q+q^2r+r^2p-(p^2r+q^2p+r^2q)=0.$$ This factors as $$(p-q)(q-r)(r-p)=0,$$ so two of $p,q,r$ are equal so $\triangle ABC$ is isosceles and we are finally done as we step back from this crazy bash.


In the figure, the visible gray area within the larger circle is equal to three times the area of the white circular region. What is the ratio of the radius of the small circle to the radius of the large circle? Express your answer as a common fraction.

[asy]size(101);
filldraw(Circle((0,0),2)^^Circle((.8,-.3),1),gray(.6)+fillrule(1),black);[/asy]

The ratio of the area of the larger circle to the smaller circle is (3+1):1=4:1, so the ratio of the radii is 2:1. Thus the ratio is $ \frac{1}{2}$


Last year, we had 2 qualifiers from South Korea and 1 from Taiwan.

To note, there are no international qualifiers on the 2009 List.....

I wonder why  :(

↓

I qualified for 2008 USAMO, and participated in it.

this year [b]I got 144.0 AMC 10A, 11 AIME I. [/b]

I'm from South Korea, and I'm an American citizen, which makes me eligible.

However, although I was on last year's list (with an idex of 230)

I'm not on this year's list (with an index of 254)

[b]Can anybody tell me what happened??

I took the AIME I this year, so I don't think there is any reason my name should go on late....

Quite worried[/b]  :(

The list is "preliminary" for a reason - just wait until April 13. I realize it's difficult, but just be patient. Then ask the AMC. Note that AIMEs from South Korea may take much longer to reach the AMC in Nebraska.

i'm sure your name will be put on it soon since i only have 216.5 indices and i'm still on the list

Daewon-Math, could you please repeat in English. I am also an international 12th grader and I have definitely scored above the 201.0 index. However,  I did not see my name on the list.

Call the telephone number above, and send an email to the address just in case. But you must have an US citizenship, I think.

We had about 2300 students or more outside the US and Canada take the AIME.
Based on the last two years' experience, there are perhaps 1,2, or 3 who fall under our qualification rules by virtue of US citizenship.  We have no way of figuring out who those 1,2 or 3 might be since we do not collect citizenship status on the answer forms.  So you will have to contact us now that you know what the qualification values are.  You can contact us by any of the means noted on our website, and printed on all of our materials.  Email will probably be most efficient.  We will reply, and verify your status, and tell you how to proceed.

physisnu:  I contacted you yesterday, April 13, approximately 5:00 pm CST in the US. Of course, it may have been April 14 in Asia by then, but you will need to respond to us to proceed further.

Steve Dunbar
MAA Director, American Mathematics Competitions

[quote="Daewon-Math"]음....나도 목록에 안올라가서 일단
미국 800-527-3690에 전화하고, amcinfo@maa.org에 이메일을 보냈어
난 미국 시민권이 없어서.....ㅠㅠ
-대원외고[/quote]

A Korean responding to a post written in Korean (in Engilsh): You need be a U.S. citizen to qualify if you are enrolled in a school outside U.S. or Canada.

[quote="physisnu"]I have received the e-mail, and I'll send the appropriate documents ASAP.

Thank you for the correspondence!!  :lol:[/quote]

Looks like the list has been updated...

[quote="Ipparchos"]Unfortunately I am not a U.S. citizen. I wish physisnu luck. I believe there must have been a mistake. It is impossible not to qualify with an 11 on the AIME.[/quote]

Well, unless you did atrociously on the AMC and didn't make the index, but then you wouldn't be writing the AIME anyway.

Anyway, I'd better get to sending those documents in.

[quote="kohjhsd"][quote="Daewon-Math"]
Quote removed for privacy reasons, as requested
[/quote]

A Korean responding to a post written in Korean (in Engilsh): You need be a U.S. citizen to qualify if you are enrolled in a school outside U.S. or Canada.

[/quote]

Uh...yeah. Duh. Obviously I realized that. I said I didnt have a US citizenship.

[quote]
Uh...yeah. Duh. Obviously I realized that.[/quote]
And I was pointing out that your name cannot, in principle, be on the list for precisely that reason. While I apologize for misunderstanding, keep in mind that elaborating your special situation (such as... transfer) would have easily prevented it in the first place. And please, let our public conversation stop here.

EDIT: Thank you.


[quote="Char_Aznable"]a,b,c be positive reals prove that (a-1+1/b)(b-1+1/c)+(b-1+1/c)(c-1+1/a)+(c-1+1/a)(a-1+1/b) <= 3[/quote]

http://www.mathlinks.ro/Forum/viewtopic.php?t=14567. Post #7 Darij Grinberg :)
Sign reverse!


conics
parabola
trigonometry
geometry
angle bisector
calculus
calculus computations

Let we have the parabola $ y^{2}\equal{}4px, p>0$ with the focus $ F(p,0)$. Let $ T$ be any point on the parabola. Show that the angle between the normal to the curve in $ T$ and $ x$-axis is the same as the angle between the line $ FT$ and the normal.

$ T(x_{0},y_{0}), k_{n}\equal{}\minus{}\frac{1}{k_{t}}$

$ 2y\frac{dy}{dx}\equal{}4p \rightarrow \frac{dy}{dx}\equal{}\frac{2p}{y}$

$ k_{n}(x_{0})\equal{}\minus{}\frac{y_{0}}{2p}$

$ \tan{\alpha}\equal{}k_{n},$ where $ \alpha$ is the angle between the normal and $ x$-axis.

$ \tan{\beta}\equal{}FT\equal{}\frac{y_{0}}{x_{0}\minus{}p}$, where $ \beta$ is the angle between the normal and $ FT$.

$ \alpha\minus{}\beta$ is one of the angles between the normal and $ FT$.

$ \left|\tan{(\alpha\minus{}\beta)}\right|\equal{}\left|\frac{\tan{\alpha}\minus{}\tan{\beta}}{1\plus{}\tan{\alpha}\tan{\beta}}\right|\equal{}
\left|\frac{\minus{}\frac{y_{0}}{2p}\minus{}\frac{y_{0}}{x_{0}\minus{}p}}{1\plus{}(\minus{}\frac{y_{0}}{2p})\frac{y_{0}}{x_{0}\minus{}p}}\right|\equal{}\left|\minus{}\frac{y_{0}}{2p}\right|$

$ \left|\tan{(\alpha\minus{}\beta)}\right|\equal{}\left|\tan{\alpha}\right| \rightarrow \alpha\minus{}\beta$ and $ \alpha$ are suplementary. Hence one of the angles between the normal and $ FT$ is the same to $ \alpha$. (drawing the sketch shows us, that we get isosceles triangle $ FTN$, where $ N$ is the intersection between the normal and $ x$-axis.

Now, my question is, whether I forgot on something?

[b]Another option:[/b] We have to show that $ TFN$ ($ N$ the intersection of the normal with $ x$-axis) is an isosceles triangle. So $ TF \equal{} FN$. We know, that the angle bisector at $ F$ should be perpendicular to the base (in our case normal), for the triangle to be isosceles. But if the angle bisector is perpendicular to the normal, then it is paralel to the tangent in the point $ T$.

Therefore we must just show that $ \tan{\frac {\beta}{2}} \equal{} \frac {dy}{dx}(x_{0})$. $ \left(\tan{\beta} \equal{} FT \equal{} \frac {y_{0}}{x_{0} \minus{} p}, \frac {dy}{dx}(x_{0}) \equal{} \frac {2p}{y_ {0}}\right)$

Since $ \beta \in (0,\pi)$ and $ \tan{\frac {\beta}{2}} \equal{} \frac { \minus{} 1\pm \sqrt {\tan^2{\beta} \plus{} 1}}{\tan{\beta}},$ we must study two options.

[b]1. option:[/b] $ \beta \in (0,\frac {\pi}{2})$   and   $ \tan{\frac {\beta}{2}} \equal{} \frac { \minus{} 1 \plus{} \sqrt {\tan^2{\beta} \plus{} 1}}{\tan{\beta}},$

Since $ \beta \in (0,\frac {\pi}{2}) \Rightarrow x_{0} > p \Rightarrow x_{0} \minus{} p > 0$.
\[ \tan{\frac {\beta}{2}} \equal{} \frac { \minus{} 1 \plus{} \sqrt {\left(\frac {y_{0}}{x_{0} \minus{} p}\right)^{2} \plus{} 1}}{\frac {y_{0}}{x_{0} \minus{} p}} \equal{} \frac { \minus{} 1 \plus{} \frac {\sqrt {y^{2}_{0} \plus{} (x_{0} \minus{} p)^{2}}}{\sqrt {(x_{0} \minus{} p)^{2}}}}{\frac {y_{0}}{x_{0} \minus{} p}} \equal{} \frac { \minus{} 1 \plus{} \frac {\sqrt {y^{2}_{0} \plus{} (x_{0} \minus{} p)^{2}}}{x_{0} \minus{} p}}{\frac {y_{0}}{x_{0} \minus{} p}}
\]
[b]2. option:[/b] $ \beta \in (\frac {\pi}{2},\pi)$ and $ \tan{\frac {\beta}{2}} \equal{} \frac { \minus{} 1 \minus{} \sqrt {\tan^2{\beta} \plus{} 1}}{\tan{\beta}},$

Since $ \beta \in (\frac {\pi}{2},\pi) \Rightarrow x_{0} < p \Rightarrow x_{0} \minus{} p < 0$.
\[ \tan{\frac {\beta}{2}} \equal{} \frac { \minus{} 1 \minus{} \sqrt {\left(\frac {y_{0}}{x_{0} \minus{} p}\right)^{2} \plus{} 1}}{\frac {y_{0}}{x_{0} \minus{} p}} \equal{} \frac { \minus{} 1 \minus{} \frac {\sqrt {y^{2}_{0} \plus{} (x_{0} \minus{} p)^{2}}}{\sqrt {(x_{0} \minus{} p)^{2}}}}{\frac {y_{0}}{x_{0} \minus{} p}} \equal{} \frac { \minus{} 1 \minus{} \frac {\sqrt {y^{2}_{0} \plus{} (x_{0} \minus{} p)^{2}}}{|x_{0} \minus{} p|}}{\frac {y_{0}}{x_{0} \minus{} p}} \equal{}
\]

\[ \frac { \minus{} 1 \plus{} \frac {\sqrt {y^{2}_{0} \plus{} (x_{0} \minus{} p)^{2}}}{x_{0} \minus{} p}}{\frac {y_{0}}{x_{0} \minus{} p}}
\]
[b]Both options:[/b]

For both options we get $ \tan{\frac {\beta}{2}} \equal{} \frac { \minus{} 1 \plus{} \frac {\sqrt {y^{2}_{0} \plus{} (x_{0} \minus{} p)^{2}}}{x_{0} \minus{} p}}{\frac {y_{0}}{x_{0} \minus{} p}}$. And now we just confirm
\[ \tan{\frac {\beta}{2}} \equal{} \frac { \minus{} 1 \plus{} \frac {\sqrt {y^{2}_{0} \plus{} (x_{0} \minus{} p)^{2}}}{x_{0} \minus{} p}}{\frac {y_{0}}{x_{0} \minus{} p}} \equal{} \frac {2p}{y_ {0}} \equal{} \frac {dy}{dx}(x_{0})
\]
.

[b]3. solution:[/b]We just find the intersection of the normal and the $ x$-axis, which is denoted by $ N$ and show that $ FT=FN$, for the triangle $ TFN$ to be isosceles.

We already know, that $ k_{n}=-\frac{y}{2p}$. We know that $ N$ is of the form $ N(x_{n},0)$. So
\begin{eqnarray*}
0-y_{0}&=&k_{n}(x_{0})(x-x_{0})\\
-y_{0}&=&-\frac{y_{0}}{2p}(x-x_{0})\\
2p&=&x-x_{0}\\
x&=&x_{0}+2p
\end{eqnarray*}
Hence $ N=(x_{0}+2p,0)$. Therefore

$ FN^{2}=(x_{0}+p)^{2},\\
 FT^{2}=(x_{0}-p)^{2}+y^{2}_{0}=x^{2}_{0}-2px_{0}+p^{2}+4x_{0}p=(x_{0}+p)^{2}$.

So we get $ FN=FT,$ which was needed.


function
limit
integration
calculus
real analysis
real analysis unsolved

1)Let $ f : (o, \plus{} \infty) \to [0, \plus{} \infty)$ be a bounded function
If : $ \lim_{x \to 0}{(f(x) \minus{} \frac {1}{2}\sqrt {f(\frac {x}{2})})} \equal{} 0$ , and $ \lim_{x \to 0}{(f(x) \minus{} 2f^2(2x))} \equal{} 0$
Prove that : $ \lim_{x \to 0}{f(x)} \equal{} 0$
2) Let $ f : [0,1] \to \mathbb{R}$ be an integrable function such that :
     $ \lim_{n \to \plus{}\infty}{n \int_{x}^{x\plus{}\frac{1}{n}}{f(t)dt}}\equal{}0 ,\forall x \in [0,1)$
 Prove that : $ \int_{a}^{b}{f(t)dt}\equal{}0$ , $ \forall a,b \in (0,1)$

1)
$ \lim_{x\to 0}{(f(x)-\frac{1}{2}\sqrt{f(\frac{x}{2})})}= 0$
Since f(x) is bounded, the product equals zero aswell:
$ \lim_{x\to 0}{(f(x)-\frac{1}{2}\sqrt{f(\frac{x}{2})})}\lim_{x\to 0}{(f(x)+\frac{1}{2}\sqrt{f(\frac{x}{2})})}= \lim_{x\to 0}({f^2(x)-\frac{1}{4}f(\frac{x}{2})} )=0$
And of course we can substitute: $ x=2x$ to get $ \lim_{x\to 0}({f^2(2x)-\frac{1}{4}f(x)} )=0$
Now, since both the limits exist and are finite:
$ 0=0+0= \lim_{x\to 0}({4f^2(2x)-f(x)} ) +  \lim_{x\to 0}{(f(x)-2f^{2}(2x))}= \lim_{x\to 0}({2f^2(2x)} )$
and obviously it means that $ \lim_{x\to 0}{f(x)}= 0$

2)
If we use the basic theorem of Calculus, $ F(x) ={\int_{0}^{x}{f(t)dt}}$ is differentiable for all $ x \in (0,1)$
So if we take $ h=\frac{1}{n}$,
$ 0= \lim_{n\to+\infty}{n\int_{x}^{x+\frac{1}{n}}{f(t)dt}}$ = $ \lim_{h\to 0}{\frac{\int_{x}^{x+h}{f(t)dt}}{h}}= \lim_{h \to 0}{\frac{F(x+h)-F(x)}{h}=F'(x) = f(x)}$
Especialy, $ \int_{a}^{b}{f(t)dt}=0 , \forall a,b \in (0,1).$

[quote="Yotam"]
2)
If we use the basic theorem of Calculus, $ F(x) \equal{} {\int_{0}^{x}{f(t)dt}}$ is differentiable for all $ x \in (0,1)$
[/quote]
we need continuity of $ f$ in theorem


geometry
geometric transformation
vector
parallelogram
geometry proposed

On the sides $BC$, $CA$, $AB$ of an acute-angled triangle $ABC$, we erect (outwardly) the squares $BB_aC_aC$, $CC_bA_bA$, $AA_cB_cB$, respectively. On the sides $B_cB_a$ and $C_aC_b$ of the triangles $BB_cB_a$ and $CC_aC_b$, we erect (outwardly) the squares $B_cB_vB_uB_a$ and $C_aC_uC_vC_b$.

Prove that $B_uC_u\parallel BC$.

[i]Comment.[/i] This problem originates in the 68th Moscow MO 2005, and a solution was posted in http://www.mathlinks.ro/Forum/viewtopic.php?t=30184 . However ingenious this solution is, there is a different one which shows a bit more: $B_uC_u=4\cdot BC$.

  Darij

Let $B'_a$ be the simmetrical point of $B_a$ wrt $C_a$ and similarly $C'_a$ of $C_a$ wrt $B_a$.

Let $A_1$, $A_2$ and $A_3$ be the translations of point $C_u$  of the vectors $C_aB_a$, 2$C_aB_a$ and 3$C_aB_a$ respectively.

Looking at the rigth triangles and parallelograms, it is easy to verify the claim.


i just leanred about a robotics competition for highschoolers
i dont know if its recomended for middle schoolers because i go to  a secondary school so i can learn from the high schoolers
the competition is called bot balls and more info can be found at
http://www.botballs.org
i just want to point this out because a lot of intelectuals are on aops and this program (get it :rotfl: ) is a great way to start learning programming
the language u use for this competition is interactive C

Yes, botball is also for middle schoolers.

My middle school had a botball team the last time I checked. I believe the middle schools have a separate division, but I cannot be too sure.


If $ a^2\plus{}b^2\plus{}c^2\equal{}1$ for non negative real $ a,b,c$

Prove that

$ 1\leq \frac{a}{1\plus{}bc}\plus{}\frac{b}{1\plus{}ac}\plus{}\frac{c}{1\plus{}ab}\leq\sqrt{2}$

[quote="shyong"]If $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$ for non negative real $ a,b,c$

Prove that

$ 1\leq \frac {a}{1 \plus{} bc} \plus{} \frac {b}{1 \plus{} ac} \plus{} \frac {c}{1 \plus{} ab}\leq\sqrt {2}$[/quote]

This is Faruk Zejnulahi ' Inequalitty

We have

      $ (a \plus{} b \plus{} c)^2 \leq 2(1 \plus{} bc)^2 \Leftrightarrow 2(ab \plus{} bc \plus{} ca) \leq 1 \plus{} 4bc \plus{} 2b^2c^2 \Leftrightarrow (b \plus{} c \minus{} a)^2 \plus{} 2b^2c^2 \geq 0$  ( True !)

   Whence $ \frac {a}{1 \plus{} bc} \plus{} \frac {b}{1 \plus{} ac} \plus{} \frac {c}{1 \plus{} ab} \leq \frac {\sqrt2 a}{a \plus{} b \plus{} c} \plus{} \frac {\sqrt2 b}{a \plus{} b \plus{} c} \plus{} \frac {\sqrt2 c}{a \plus{} b \plus{} c} \equal{} \sqrt2$
Equality holds if and only if $ a \equal{} b \equal{} 1/ \sqrt2,c \equal{} 0$ and its cyclic permutations

On the otherhand
$ a \plus{} abc \leq a \plus{} \frac {a(1 \minus{} a^2)}{2} \equal{} 1 \minus{} \frac {(a \minus{} 1)^2 (a \plus{} 2)}{2} \leq 1$
Consequently
$ \frac {a}{1 \plus{} bc} \plus{} \frac {b}{1 \plus{} ac} \plus{} \frac {c}{1 \plus{} ab} \equal{} \frac {a^2}{a \plus{} abc} \plus{} \frac {b^2}{b \plus{} abc} \plus{} \frac {c^2}{c \plus{} abc} \geq a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$
Equality holds if and only if $ a \equal{} 1 ,b \equal{} c \equal{} 0$ and its cyclic permutions

 :lol:  :lol:


geometry
trigonometry
complex numbers
geometry unsolved

Let triangle ABC. D,E,F on BC,CA,AB satisying $ \angle BAD\equal{}\angle ACF\equal{}\angle CBE$ and AD=BE=CF. Prove that  triangle ABC is equilateral.

[quote="Agr_94_Math"]A hint: Working with complex numbers along with Stewatr's Theorem makes the problem very simple. COmplex numbers help here because of the given angle equalties.[/quote]
You can explain it, please.

[quote="borislav_mirchev"]i've been posted similar problems ... and there is a trigonometry solutions in bulgarian ... similar or exactly that problem was proposed in romanian tst 1993[/quote]
I can borrow it ( link ),please.


modular arithmetic
number theory
number theory proposed

Determine all integers $ a$ and $ b$ such that:

$ (19a\plus{}b)^{18}\plus{}(a\plus{}b)^{18}\plus{}(a\plus{}19b)^{18}$ is a perfect square.

[hide]Let $ f(a,b) \equal{} (19a\plus{}b)^{18}\plus{}(a\plus{}b)^{18}\plus{}(a\plus{}19b)^{18}$

By Fermat's Little Theorem, if $ n \not \equiv 0 \pmod{19}$, then $ n^{18} \equiv 0 \pmod{19}$.

If $ n \equiv 0 \pmod{19}$, then $ n^{18} \equiv 0 \pmod{19}$.

Therefore, $ f(a,b) \equiv 0,1,2,3 \pmod{19}$.

The residues $ \pmod{19}$ that are perfect squares are: $ 0,1,4,9,16,6,17,11,7,5 \pmod{19}$.

Thus, $ f(a,b) \equiv 0,1 \pmod{19}$.

So, at least two of $ 19a\plus{}b \equiv b$, $ a\plus{}b$, $ a\plus{}19b \equiv a$ must be equivalent to $ 0 \pmod{19}$.

If any two of $ a,b,a\plus{}b$ are equivalent to $ 0 \pmod{19}$ then so is the third one.

Thus, $ a,b \equiv 0\pmod{19}$

From here, we can use proof by infinite descent to prove that the only $ a,b \in \mathbb{Z}$ such that $ f(a,b)$ is a perfect square is $ (a,b)\equal{}(0,0)$. [/hide]


The number 4 can be written as a sum of one or more natural numbers in exactly five ways: $ 4,\ 3\plus{}1,\ 2\plus{}1\plus{}1,\ 2\plus{}2,$ and $ 1\plus{}1\plus{}1\plus{}1$; and so 4 is said to have five \textit{partitions}.  What is the number of partitions for the number 7?

We  have to list these out:

$ 7$
$ 1 \plus{} 6$
$ 2 \plus{} 5$
$ 3 \plus{} 4$
$ 1 \plus{} 3 \plus{} 3$
$ 1 \plus{} 1 \plus{} 5$
$ 2 \plus{} 1 \plus{} 4$
$ 2 \plus{} 2 \plus{} 3$
$ 1 \plus{} 1 \plus{} 2 \plus{} 3$
$ 1 \plus{} 1 \plus{} 1 \plus{} 1 \plus{} 3$
$ 1 \plus{} 1 \plus{} 1 \plus{} 1 \plus{} 1 \plus{} 1 \plus{} 1$
$ 1 \plus{} 1 \plus{} 1 \plus{} 1 \plus{} 2$
$ 1 \plus{} 2 \plus{} 2 \plus{} 2$
$ 1 \plus{} 1 \plus{} 1 \plus{} 4$
$ 3 \plus{} 2 \plus{} 1 \plus{} 1$

Counting these up, we have $ \boxed{15}$ ways.

EDIT: Beaten to it.

When you write these out, I recommend starting with the largest possible number, and then go down.

answer is 18, not fifteen, says in answer book

there are 5 ways to make 4 and 3 ways to make 3 so it is just 5*3 ways to make 7


Let there be a cone of any dimension. Slice the cone along a plane parellel to its base. Then, the cone is now cut into two parts. Compare the bottom of the top part to the top of the bottom part. Are the two circles congruent or different?

If they are congruent, then wouldn't the cone be a cylander?

If they are different, wouldn't the cone be a "tiered figure"?

This paradox first arose in the ancient times.  :)

huh??? i don't see how that makes it a cylinder, all it does is show that one has the same bottom as the other's top, it gives no relation between other parts of the solids...

Well, I think that's the solution. I got this from a book that didn't explain the answer to this "paradox". But I think you found the glitch.  :)


linear algebra
matrix
algebra
polynomial
function
linear algebra unsolved

Let $ M\in\mathcal{M}_n(\mathbb{C})$, s.t. no eigenvalue of $ M$ belongs to $ \mathbb{R}_\minus{}$.
Show that there is a unique $ R\in\mathcal{M}_n(\mathbb{C})$, s.t. $ M\equal{}R^2$ and each eigenvalue of $ R$ has a positive real part.

1) There is:
We may assume $ M\equal{}diag(\lambda_i(I\plus{}N_i))$ where $ N_i$ is a nilpotent matrix and $ \lambda_i\notin{\mathbb{R}_\minus{}}$. There exists a unique $ \mu_i$ s.t. $ \mu_i^2\equal{}\lambda_i,Re(\mu_i)>0$. Take $ R\equal{}diag(\mu_i(I\plus{}1/2N_i\minus{}1/8N_i^2\plus{}\cdots)$.
Note that $ R$ is a polynomial in $ M$. (Use the Hermite's interpolation).
2) a unique:
$ R$ is in the form $ R\equal{}diag(\mu_i(I\plus{}N_i))$ where $ Re(\mu_i)>0$, if $ i\not\equal{}j$ then $ \mu_i\not\equal{}\mu_j$ and $ N_i$ is a nilpotent matrix.
Let $ S$ be another solution ($ R^2\equal{}S^2,\cdots$).
We know that $ R$ is a polynomial in $ R^2$. $ R^2,S$ commute then $ R,S$ commute and $ S\equal{}diag(S_i)$ where $ S_i^2\equal{}\mu_i^2(I\plus{}N_i)^2$. Thus $ S_i$ admits only the eigenvalue $ \mu_i$ and $ S_i\equal{}\mu_i(I\plus{}P_i)$ where $ P_i$ is a nilpotent matrix s.t. $ (I\plus{}P_i)^2\equal{}(I\plus{}N_i)^2$.
$ P_i,P_i\plus{}1/2P_i^2$ are similar because it's true for a Jordan nilpotent matrix. $ P_i\plus{}1/2P_i^2\equal{}N_i\plus{}1/2N_i^2$ then $ P_i,N_i$ are similar : $ P_i\equal{}QN_iQ^{\minus{}1}$ and $ Q,N_i\plus{}1/2N_i^2$ commute. It remains to show that $ Q,N_i$ commute what is a consequence of the following result:
Lemma: if $ N$ is a nilpotent matrix then $ N$ is a polynomial in $ N\plus{}1/2N^2$.
Proof: Let $ U\equal{}N\plus{}1/2N^2$ ; $ N^k\equal{}0$; if $ p<k$ then $ U^p\equal{}N^p\plus{}\cdots$, $ U^{k\minus{}1}\equal{}N^{k\minus{}1}$. Conversely we can calculate $ N$ as a linear function of $ U,\cdots,U^{k\minus{}1}$.


inequalities
trigonometry
geometry
incenter
geometry solved

Let $M$ be point inside triangle $ABC$.
Prove that $MA \cos{{ \measuredangle A \over 2}}+MB \cos{{ \measuredangle B \over 2}}+MC \cos{{ \measuredangle C \over 2}}\geq{a+b+c\over 2}$.

[quote="dule_00"]Let $M$ be point inside triangle $ABC$.
Prove that $MA \cos{{ \measuredangle A \over 2}}+MB \cos{{ \measuredangle B \over 2}}+MC \cos{{ \measuredangle C \over 2}}\geq {a+b+c\over 2}$.[/quote]

Let Ma, Mb, Mc be the feet of perpendiculars from M to BC, AC, AB respectively.
Then $AC=AM\cos\angle CAM+CM\cos\angle ACM$  (even if triangle ACM isn't acute)
We have similar:
$AB=AM\cos \angle BAM+BM\cos\angle ABM$
$BC=BM\cos\angle CBM+CM\cos\angle BCM \Rightarrow$
$a+b+c=AM(\cos\angle BAM +\cos\angle CAM)+...=$
$AM*2*\cos\frac{\angle BAM+\angle CAM}{2}\cos\frac{\angle BAM-\angle CAM}{2}+...\leq$
$2AM\cos\frac{\angle A}{2}+... \Rightarrow$
$MA \cos{{ \measuredangle A \over 2}}+MB \cos{{ \measuredangle B \over 2}}+MC \cos{{ \measuredangle C \over 2}}\geq {a+b+c\over 2}$
V:)

Slight extension of the problem, and a different solution:

[color=blue][b]Theorem 1.[/b] If M is a point in the plane of a triangle ABC, then $MA\cos\frac{A}{2}+MB\cos\frac{B}{2}+MC\cos\frac{C}{2}\geq\frac{a+b+c}{2}$.[/color]

[i]Proof.[/i] Let I be the incenter of triangle ABC. Then, of course, I is a point inside triangle ABC. Hence, according to http://www.mathlinks.ro/Forum/viewtopic.php?t=30976 , we have

[color=green][b](1)[/b][/color] $MA\cdot\sin x+MB\cdot\sin y+MC\cdot\sin z\geq IA\cdot\sin x+IB\cdot\sin y+IC\cdot\sin z$,

where x = < BIC, y = < CIA and z = < AIB.

Now, since I is the incenter of triangle ABC, we have $\measuredangle IBC=\frac{B}{2}$ and $\measuredangle ICB=\frac{C}{2}$. Thus, by the sum of angles in triangle BIC, we have

$x=\measuredangle BIC=180^{\circ}-\measuredangle IBC-\measuredangle ICB=180^{\circ}-\frac{B}{2}-\frac{C}{2}=180^{\circ}-\frac{B+C}{2}$
$=180^{\circ}-\frac{180^{\circ}-A}{2}$           (since B + C = 180° - A by the sum of angles in triangle ABC)
$=90^{\circ}+\frac{A}{2}$.

Hence, $\sin x=\cos\frac{A}{2}$. Similarly, $\sin y=\cos\frac{B}{2}$ and $\sin z=\cos\frac{C}{2}$. Hence, the inequality [color=green][b](1)[/b][/color] becomes

$MA\cdot\cos\frac{A}{2}+MB\cdot\cos\frac{B}{2}+MC\cdot\cos\frac{C}{2}\geq IA\cdot\cos\frac{A}{2}+IB\cdot\cos\frac{B}{2}+IC\cdot\cos\frac{C}{2}$.

Now, let X, Y, Z be the orthogonal projections of the point I on the sides BC, CA, AB of triangle ABC. Then, in the right-angled triangle IYA, we have $AY=IA\cdot\cos\measuredangle IAY$. But since I is the incenter of triangle ABC, we have $\measuredangle IAY=\frac{A}{2}$, so that $AY=IA\cdot\cos\frac{A}{2}$. Similarly, $AZ=IA\cdot\cos\frac{A}{2}$. Hence, $IA\cdot\cos\frac{A}{2}=\frac{IA\cdot\cos\frac{A}{2}+IA\cdot\cos\frac{A}{2}}{2}=\frac{AY+AZ}{2}$. Similarly, $IB\cdot\cos\frac{B}{2}=\frac{BZ+BX}{2}$ and $IC\cdot\cos\frac{C}{2}=\frac{CX+CY}{2}$. Hence,

$MA\cdot\cos\frac{A}{2}+MB\cdot\cos\frac{B}{2}+MC\cdot\cos\frac{C}{2}\geq IA\cdot\cos\frac{A}{2}+IB\cdot\cos\frac{B}{2}+IC\cdot\cos\frac{C}{2}$
$=\frac{AY+AZ}{2}+\frac{BZ+BX}{2}+\frac{CX+CY}{2}=\frac{\left(BX+CX\right)+\left(CY+AY\right)+\left(AZ+BZ\right)}{2}$
$=\frac{BC+CA+AB}{2}=\frac{a+b+c}{2}$,

and Theorem 1 is proven.

  darij


Let ABC be a triangle  such each interior angle is $\leq \frac {\pi}{2}$.

The points of the triangle sides are colored with $red$ and $blue$ .

Prove that there are 3 points   M,N,P colored with the same color, such that the triangle MNP is right .
.
 :cool:  What kind of color do you prefer ? :cool:

:cool:  I dont know ! Try to solve it !  :cool:


That is correct, but I did it slightly differently.

[hide]$6^{3}\cdot6^{6}=6\cdot6^{x}$

$6^{9}=6^{x+1}$

$x+1=9$

$x=8$

I didn't divide by 6.[/hide]


The isosceles triangle and the square shown here have the
same area in square units. What is the height of the triangle,
$ h$, in terms of the side length of the square, $ s$?

[asy]draw((0,0)--(0,10)--(10,10)--(10,0)--cycle);
fill((0,0)--(17,5)--(0,10)--cycle,white);
draw((0,0)--(17,5)--(0,10)--cycle);
label("$s$",(5,10),N);
label("$h$",(6,5),N);
draw((0,5)--(17,5),dashed);
draw((0,7)--(2,7)--(2,5));[/asy]

area of triangle = $ h\times s\times \frac{1}{2}\equal{}\frac{hs}{2}$

area of square = $ s^2$

$ s^2\equal{}\frac{hs}{2} \implies s\equal{}\frac{h}{2} \implies h\equal{}2s$

answer : $ 2s$

$\text{Area of Triangle}=\text{Area of Square}$
$\frac{s\cdot h}{2}=s^2$
$s\cdot h=2s^2$
$h=2s$


[color=blue][size=150]RUSIA 1997[/size][/color]

 Mostrar que los numeros desde 1 a 16,
pueden ser escritos en una linea, pero no en un circulo,
de tal manera que la suma de dos numeros adjuntos cualesquiera ...
sea un cuadrado perfecto.

(Sugerencia)
[hide=" Hint"]Tomar la posibilidad de escribirlos sobre el circulo y llegar a una contradiccion[/hide]
[hide="English Version"]
Show that the numbers from 1 to 16 can be written in a line, but not in a circle, so that the sum of any two adjacent numbers is a perfect square.
[/hide]

Carlos Bravo :)
Lima - PERU

mi solución:
[hide]Sup. que esta en un circulo, entonces vamos a ver que pasa con el 16, se tienen que formar dos cuadrados mayores que 16, como no pueden ser iguales y la máxima suma de dos seria 31 no se puede.

Los numeros sobre una linea se pueden poner de esta forma:
16-9-7-2-14-11-5-4-12-13-3-6-10-15-1-8[/hide]


Find the number of ordered palindromic partitions of an integer $n$.

Can you answer to this question : Find the number of ordered comic partitions of an integer $n$ ?

I think, it's impossible, for you to answer, even more you have to be crazy to try to answer.  :)

[hide=I'm obviously oversimplifying. How?]$P_n = 1 + P_{n-2} + P_{n-4} + ...$ ,  with $P_0 = 1$ and $P_1 = 1$

$1=1$
$2=2$
$2=1+1$
$3=3$
$3=1+1+1$
$4=4$
$4=2+2$
$4=1+2+1$
$4=2+2$

which are all the palindromic partitions for $n=1, 2, 3, 4$. This agrees with my expression for $P_n$. It can't be this easy![/hide]


This may sound like an odd question..

A mongrel dog has 3 equal proportions of breeds in it..

Let's call them A ,B, and C ..in other words , a long,long time ago the breed started from Dogs A,B and C.

What would the matings be down through the generations to result in a dog containing  3 equal proportions of the original breeds..?

I've run out of paper trying to work this family tree out !

Bullseye..

Making the normal assumptions about heredity, this is in fact impossible.  The proportion of a given breed in a given dog can be proven to always be expressible as a rational number with denominator $ 2^n$, where $ n$ is the generation number of the dog (prove by induction!).


MATHCOUNTS
geometry
geometric transformation
reflection

Thanks so much to solafidefarms for proofreading and editing these questions. He corrected so many errors and made the problems much more understandable. Be sure to send over your thanks to him!

Anyway, here are the PDF's. PM me your answers when you're done. Please put all the answers for one round in a single PM. I do not care whether Sprint or Target answers are together or not.

Sprint Round Instructions:
This round of the competition consists of 30 problems. You will have 40 minutes to complete the problems. You are not allowed to use calculators, books, or any other aids during this period. Calculations may be done on scratch paper. All answers must be complete, legible and simplified to lowest terms. If you complete the problems before time is called, use the remaining time to check your answers.

Target Round Instructions:
This round of the competition consists of eight problems. They will be presented to you in pairs. Work on one pair of the problems will be completed and answers should be recorded before the next pair is seen. The time limit for each set of two problems is six minutes. All answers must be complete, legible and simplified to lowest terms. This round assumes the use of calculators, and calculations may also be done on scratch paper, but no other aids are allowed. If you complete the problems before time is called, use the remaining time to check your answers.

You guys know the drills.

Good luck everyone!

Lastly, some notes: The first number in #9 is $26\frac{2}{3}\%$. In #11, the V's should actually be triangle symbols. Just pretend they do not exist.

You said to pm you the answers in one pm, then said we could do them separately on together. which?

No, I meant PM answers for one round in one PM (don't give me sprint #1 in one message, and #29 in another), but you may separate sprint and target, especially if you cannot do the rounds together.

ok
About what lvel are they (chapter state national)

I intended the level of the test to be on par with those of the more difficult state level tests. solafidefarms said that sprint round was good, but target was a little easier than average.

wait for #16 with the information on the line and stuff, which line are you referring to?

Find the equation of the line which does not cross the $y$-axis at $y=3$.

If we just pm them to you like 5 minutes ago, how long will we have to wait

I'm REALLY anxious to know my score.

Is there a deadline to when the answers are due? (just wondering)  :P

Sorry Aznwithabrain, I'll get right to grading what I have. I'll post answers, results, and statistics no earlier than Monday evening.

Sorry, I made a mistake in my answer to Sprint Round #30. Thanks to ch1n353ch3s54a1l who pointed this out. The scores of 236factorial, ch1n353ch3s54a1l, and nonie have been increased to reflect this.

wait, in 16, does the line we're trying to find cross the y axis at y=3 or does the line its perpendicular with?

Find the equation of the line to which it is perpendicular.

I am actually pretty busy tonight, and since I'm up-to-date on the grading, I'll extend the deadline to late tomorrow evening. Preferrably, sometime before then:
Captain Sugar, please send me your Sprint answers
k8reindeer and tz, please send me your Target answers
For others, if you plan to participate, yes, please send me your answers...

I am going to send you my answers too (is that okay?)

Sure! They have been graded and your score has been returned.

I will also pm you my answers. I like these mock mathcounts, they will get me ready for the Florida State Competition (March 31st). BTW, Anyone who is participating in the Florida State Comp., please PM me and tell me since I want to know people before I go against them, lol!!!  :lol:  :lol:  :lol:

i finished my sprint, here are the answers.  :)

I thought overall the test was really great.

A couple of things.
Sometimes you didn't say to write as a common fraction or in terms of $\pi$. That threw me off because I thought I had the wrong answer.

I difficulty, I think it is about the level of an easy state for the sprint, and hard target for the chapter.

Very rarely is it that I get 43 on a state compeition, and perfect on target.

i thought the target was a little harder than normal, and the sprint was a little easier than normal. of course i'm better at sprint  :D i thought it was really helpful practice since my state is this saturday.

Sorry, there was another error. My answer to #11 on Sprint was incorrect, and so the scores of 236factorial, ch1n353ch3s54a1l, and goldendomer have been increased by one to accommodate for this mistake.

Yay! Thanks. My score is still pretty pitiful though...
Can we discuss these problems yet?  :)

Nope! Please wait until tomorrow evening before discussing the problems or lodging complaints. At that time, a new thread with results, statistics, and some notes will be opened. Your cooperation is appreciated.

I still haven't received Target answers from tz and k8reindeer. I will close this mock Mathcounts three hours from now. Please submit your answers soon.

Umm...do we discuss here? I don't want to post in the wrong area... :?

No, please discuss problems here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=77168


probability
integration
calculus
calculus computations

The joint distribution of X and Y is given by:
\[f(x,y)=\frac{e^{-y}}y\qquad 0<x<y<+\infty\]
Compute:
\[E\left(X^3\;\lvert\; Y=y\right)\]

If you are given that $Y=y$, then $X$ is uniformly distributed on the interval $(0,y).$ Therefore

$E(X^3\,|\,Y=y)=\frac1y\int_0^yx^3\,dx=\frac{y^3}4.$

Going further: the marginal density of $Y$ is $\int_0^y\frac{e^{-y}}y\,dy=e^{-y}.$

Thus,

$E(X^3)=E(E(X^3\,|\,Y=y))=\int_0^{\infty}\frac{y^3}4e^{-y}\,dy=\frac32.$


function
integration
limit
inequalities
real analysis
real analysis unsolved

Hi guys,

Could you provide me some hints on solving this one?

Let $ \left < f_n\right >$ be a sequence of functions in $ L^p,1 < p < \infty$, which converge almost everywhere to a function $ f$ in $ L^p$, and suppose that there is a constant $ M$ such that $ \left\parallel{}f_n\right\parallel{} \le M$ for all $ n$. Then for each function $ g$ in $ L^q$ we have
$ \int fg \equal{} \lim\int f_n g$.
Is the result true for $ p \equal{} 1$?

Thanks in advance,
ftr

Let's rephrase it.  We have $ f_n,f\in L^p(\Omega)$, $ f_n\to f$ a.e., $ f_n,f$ uniformly bdd by $ M > 0$, and we want to show that $ f_n\rightharpoonup f$.

First of all, this is not that hard if $ m(\Omega) < \infty$; we can use Egorov's theorem to find a set $ A_\delta$ such that $ m(A_\delta)\leq\delta$ and $ f_n\to f$ [i]uniformly[/i] on $ \Omega\setminus A_\delta$.  Now by Holder,
\[ \|(f_n \minus{} f)g\|_{L^p(A_\delta)} \leq 2M\|g\|_{L^q(A_\delta)}
\]
If $ \delta$ is small enough, we can make this as small as we like.  For the other half,
\[ \|(f_n \minus{} f)g\|_{L^p(\Omega\setminus A_\delta)} \leq \|f_n \minus{} f\|_{L^\infty(\Omega\setminus A_\delta)}\|g\|_{L^1(\Omega\setminus A_\delta)}
\]
by Holder again, and this is small by uniform convergence if $ n$ is large enough.

For the case where $ m(\Omega) \equal{} \infty$, take a set of finite measure $ \Omega'$ such that $ \|g\|_{L^q(\Omega\setminus\Omega')}$ is small and apply the previous result.

Finally, if $ p \equal{} 1$, this doesn't hold (take $ g\equiv 1$, $ f_n \equal{} \tau_n\chi_{(0,1)}$ for example, or any of the usual examples of functions which converge pointwise to $ 0$ a.e. but not in $ L^1$).

I think there is probably a quicker proof with the Banach Alaoglu theorem but I wanted to keep this somewhat elementary.

Hi Blahblahblah,

Could you explain to me how you apply Holder's Inequality to get
$ \parallel{}(f_n-f)g\parallel{}_{L^p(\Omega-A_\delta)}\le\parallel{}f_n-f \parallel{}_{L^\infty(\Omega-A_\delta)}\parallel{}g\parallel{}_{L^1(\Omega-A_\delta)}$

please?

Thank you very much,
ftr

sorry, it should be $ \parallel{}(f_n - f)g\parallel{}_{L^1(\Omega - A_\delta)}\le\parallel{}f_n - f \parallel{}_{L^\infty(\Omega - A_\delta)}\parallel{}g\parallel{}_{L^1(\Omega - A_\delta)}$

the $ L^1$ norm of $ g$ is finite since the $ q$ norm dominates the $ 1$ norm on a set of finite measure.


given triangle ABC prove
$\sum a^{2}\geq 36r^{2}$
(or, in other words, $a^{2}+b^{2}+c^{2}\geq 36r^{2}$)

At least brute force works !

r = (s-a)(s-b)(s-c)/s from Heron

Inequality is equivalent to :

$5\cdot \sum a^{3}+9abc\geq 4\cdot \sum \left( ab^{2}+ac^{2}\right)$ which is true from Schur and rearrangement.

This is another way:
We have $64S^{3}\leq \left( abc\right) ^{2}$  and   $S=\frac{\left( a+b+c\right) r}{2}$.
So just by Cauchy, we have what we need. :)