Βάζω εδώ μία άσκηση που αν την έχω λύση σωστά πιστεύω ότι έχει κάποιο ενδιαφέρον...

Έστω $ f: [1,\infty)\rightarrow R$ η οποία είναι συνεχής και θετική συνάρτηση.Αν ισχύει

$ \int_{1}^{x}f(t)dt\leq f^{2}(x)$ ,για $ x\geq1$

Να δειχθεί ότι $ f(x) \geq \frac {1}{2}(x - 1)$

Υ.Γ:Επειδή φεύγω για διακοπές αύριο θα βάλω την λύση που έχω σκεφτεί ή το βράδυ για να μου πείτε αν είναι σωστή ή αύριο το μεσημέρι....

Aν $ g(x)\equal{} \int_{1}^{x}f(t)dt$ τότε η $ 2\sqrt{g(x)}\minus{}x$ βγαίνει αυξουσα και $ 1\leq x , g(1)\equal{}0$ άρα $ \sqrt{g(x)}\geq \frac{x\minus{}1}{2}$ ακόμη $ f(x)\equal{}g'(x)\geq \sqrt{g(x)}\geq \frac{x\minus{}1}{2}$

[quote="giannis"]Μήπως εννοείς $ \int_{1}^{x}f(t)dt$?[/quote]

Ήταν λάθος ταχύτητας.... :blush:
Το άλλαξα τώρα...

Η λύση που έχω σκεφτεί είναι η παρακάτω...

Αν $ F(x)\equal{}\int_{1}^{x}f(t)dt$ τότε έχουμε ότι η συνάρτηση $ G(x)\equal{}\int_{1}^{x}\sqrt{F(t)}dt$ έχει πρώτη και δεύτερη πράγωγο τις εξής:

$ G'(x)\equal{}\sqrt{F(x)}$ για $ x\geq1$ και
$ G''(x)\equal{}\frac{F'(x)}{2\sqrt{F(x)}}\equal{}\frac{f(x)}{2\sqrt{F(x)}}$ για $ x>1$
Ισχύει ότι $ G(1)\equal{}G'(1)\equal{}0$ οπότε από θεώρημα Taylor έχουμε ότι

$ G(x)\equal{}\frac{f(c)(x\minus{}1)^{2}}{4\sqrt{F(c)}}$ με $ c\in(1,x)$

Εύκολα δείχνουμε ότι $ G(x)\leq f^{2}(x)$ και ισχύει από την σχέση που μας έχουν δώσει ότι $ \frac{f(c)}{\sqrt{F(c)}} \geq 1$.
Από τις δύο αυτές σχέσεις βγαίνει το ζητούμενο...

Καλό καλοκαίρι σε όλους και θα τα πούμε ξανά από Σεπτέμβρη με περισσότερες δυνάμεις...
Ελπίζω ότι η λύση είναι σωστή.Στις παρατηρήσεις θα απαντήσω σε ένα μήνα που θα έχω γυρίσει από διακοπές...

Καλά να περάσετε όλοι.....

Λοιπόν ρίχνω και εγώ τη λύση μου, η οποία υποπτεύομαι ότι μοιάζει πολύ με του κ Ροδόλφου. Ο λόγος που τη γράφω είναι ότι προσπαθώ να περιγράψω αναλυτικά τον τρόπο που σκέφτηκα μέχρι να φτάσω στη λύση. Τελικά προκύπτει ότι μπορούμε να λύσουμε την άσκηση με κλασσικό τρόπο χωρίς κάτι περίεργο.

Λοιπόν πρώτο μέλημα να διώξουμε  το ολοκλήρωμα. Άρα ορίζουμε
$ g(x)\equal{}\int_1^x f(t)dt$
Οπότε αμέσως προκύπτει $ g(1)\equal{}0$
και $ g'(x)\equal{}f(x)$

Πάμε να διώξουμε την f λόγω του ολοκληρώματος και να διατυπώσουμε την άσκηση συναρτήσει της g.
Παίρνοντας τη ρίζα στη σχέση της εκφώνησης η άσκηση διατυπώνεται ισοδύναμα:
Αν ισχύει ότι $ g'(x)\ge \sqrt{g(x)}$ τότε να δειχτεί ότι ισχύει $ g'(x)\ge\frac{x\minus{}1}{2}$

Κάνω την αρκετά λογική σκέψη να παρεμβάλω την $ \sqrt{g(x)}$ ανάμεσα στην $ g'(x)$ και την $ \frac{x\minus{}1}{2}$ και μετά να πάω με μεταβατική ιδιότητα.

Οπότε έχω να δείξω ότι $ \sqrt{g(x)}\ge \frac{x\minus{}1}{2}$.

Επίσης κλασσικό κόλπο είναι το να τα τραβήξω όλα στο πρώτο μέλος και να ορίσω τη συνάρτηση
$ h(x)\equal{}\sqrt{g(x)} \minus{}\frac{x\minus{}1}{2}$
Αυτή θέλω να δείξω ότι είναι αύξουσα οπότε μετά θα έχω $ h(x)\ge h(1)$ για $ x\ge 1$
Παραγωγίζοντας την $ h$ και με τη σχέση της νέας εκφώνησης (συναρτήσει της $ g$) βγάζω ότι η $ h$ είναι αύξουσα. Επίσης $ h(1)\equal{}0$. Άρα $ h(x)>h(1) \Longleftrightarrow \sqrt{g(x)} \minus{}\frac{x\minus{}1}{2} \ge 0 \Longleftrightarrow  \sqrt{g(x)} \ge \frac{x\minus{}1}{2}$

Άρα τελικά έχω:

$ f(x)\equal{}g'(x) \ge \sqrt{g(x)} \ge \frac{x\minus{}1}{2}$

Είναι βασικό κατά τη γνώμη μου ότι η άσκηση λύνεται χωρίς κάποια μεγάλη ουρανοκατέβατη σκέψη και μπορούμε να φέρουμε το μέσο μαθητή στο επίπεδο να τη λύνει.

Βρήκα μια λύση γεωμετρική/διαισθητική που αν είναι σωστή θα κάνει μπαμ. Αν όχι… το πολύ πολύ να με πάρουν με τις ντομάτες. Αλλά νομίζω ότι αξίζει τον κόπο να το ρισκάρω  :D  .
Θα τη στείλω Δευτέρα για σχολιασμό. Αν κάποιος θέλει να την παλέψει… δίνω hint… ας δει τι σημαίνει γεωμετρικά το ολοκλήρωμα και το $ f^2(x)$. Μετά ας δει τι σημαίνει η σχέση που πρέπει να αποδειχτεί.

Γράφω τη γεωμετρική/διαισθητική λύση. Αν είναι σωστή την αφιερώνω στον r_boris επειδή στο παρθενικό μου μήνυμα στο forum απάντησε μια απορία γρήγορα και σωστά και κατάλαβα ότι έχω να μάθω πολλά εδώ μέσα. Ξέρω ότι του αρέσουν και οι διαισθητικές λύσεις...

Θεωρώ σύστημα συντεταγμένων με οριζόντιο άξονα το t και κατακόρυφο το f(t).

Σχεδιάζω στο σχήμα 1 (που είναι το κάτω από τα 2) την ευθεία $ y\equal{}\frac{t\minus{}1}{2}$ και παρατηρώ με απλή εμβαδομέτρηση ότι στο τυχαίο σημείο της (x,f(x)) το ολοκλήρωμά της (εμβαδό τριγώνου) είναι ίσο με το εμβαδό του τετραγώνου με πλευρά $ f(x)\equal{}\frac{x\minus{}1}{2}$

Πάμε τώρα στην απόδειξη της άσκησης. Αντί να δείξω την πρόταση που ζητάει η εκφώνηση θα δείξω την ισοδύναμη αντιθετοαντίστροφή της η οποία είναι η εξής:

Αν υπάρχει ένα σημείο $ x_1$ στο οποίο ισχύει $ f(x_1) < \frac{x_1\minus{}1}{2}$ δηλαδή η f σε εκείνο τη σημείο είναι κάτω από την ευθεία $ y\equal{}\frac{t\minus{}1}{2}$ τότε υπάρχει τουλάχιστο ένα σημείο $ x_2$ στο οποίο ισχύει $ \int_1^{x_2} f(t)dt > f^2(x_2)$

Στο σχήμα 2 (που είναι το πάνω από τα 2) βλέπω καθαρά τι συμβαίνει. Αφού υπάρχει σημείο $ x_1$ στο οποίο η f πέφτει κάτω από την ευθεία $ y\equal{}\frac{t\minus{}1}{2}$ και ξέρουμε ότι στο $ x\equal{}1$ ξεκινάει από πάνω της (λόγω του ότι η f είναι θετική) σημαίνει αναγκαστικά ότι κάπου θα πρέπει να συναντήσει την ευθεία. Αν κάποιου δεν του αρκεί αυτό μπορεί να χρησιμοποιήσει το θεώρημα Bolzano στην $ h(t)\equal{}f(t)\minus{}\frac{t\minus{}1}{2}$ στο $ [1,x_1]$για να αποδείξει ότι θα γίνει η συνάντηση σε κάποιο $ x_2$

Από το σχήμα λοιπόν είναι προφανές ότι το εμβαδό κάτω από την f δηλαδή το $ \int_1^{x_2} f(t) dt$ είναι μεγαλύτερο από το εμβαδό κάτω από την ευθεία το οποίο είναι ίσο με το εμβαδό του τετραγώνου. Άρα υπάρχει το $ x_2$ το οποίο απαιτεί η ισοδύναμη εκφώνηση.


Show that for $ a,b,c,d>0$,we have
$ \frac{1}{a(a\plus{}b)}\plus{}\frac{1}{b(b\plus{}c)}\plus{}\frac{1}{c(c\plus{}d)}\plus{}\frac{1}{d(d\plus{}a)}$≥$ \frac{2}{\sqrt{abcd}}$

This inequality is homogeneous, then we can assume that $ abcd\equal{}1$
Then we have to prove that:

$ \sum_{cyc}\frac{bcd}{a\plus{}b} \geq 2$

Because it is cyclic we can assume WLOG $ a\geq{b,c,d}$

If $ b\geq d$:
Then the inequality is equivalent to:

$ \frac{bd(b\minus{}d)(a\minus{}c)(ab\plus{}ac\plus{}ad\plus{}bc\plus{}bd\plus{}cd\plus{}a^2\plus{}c^2)}{(a\plus{}b)(b\plus{}c)(c\plus{}d)(d\plus{}a)} \plus{}bc\plus{}da \geq 2$

Which is obvious by $ MA\geq MG$
$ bc\plus{}da \geq 2$

If $ d\geq b$:
Then the inequality is equivalent to:

$ \frac{bd(d\minus{}b)(a\minus{}c)(ab\plus{}ac\plus{}ad\plus{}bc\plus{}bd\plus{}cd\plus{}b^2\plus{}d^2)}{(a\plus{}b)(b\plus{}c)(c\plus{}d)(d\plus{}a)} \plus{}ab\plus{}cd \geq 2$

Which is obvious again, and we finish the proof.

Why $ \sum{{1\over{a(a\plus{}b)}}}\ge\sum{{1\over{b(a\plus{}b)}}}$,
you cannot assume any order between $ a,b,c,d$.

why not , $ WLOG a\le{b}\le{c}\le{d}$  but , it doesn't make any clue.....

Of course you can't assume $ a\le b\le c\le d$
The inequality is cyclic, but not symmetric in 4 variables..

Oh, that does not matter. $ \sum \frac{1}{a(a\plus{}b)}\geq \frac{1}{b(a\plus{}b)}$ will not hold for each and every pair of $ a, b, c, d$ so $ \sum\frac{1}{a(a\plus{}b)}\ge \sum\frac{1}{b(a\plus{}b)}$ is not true at all.


Wrong section. :)
Just basic calculations. :roll:

This is basically a restatement of Adam Hesterberg's factoring trick. I suppose if you really wanted to, you could expand the right side and see that you get the left side.

[quote="mdk"]Show that a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)[/quote]

Yeah, no offense, but problems that involve pure mechanical calculation probably shouldn't be in this forum.

[quote="Lazarus"]Adam Hesterberg's factoring trick[/quote]

huh? i've never heard it called that...i've seen it in books published before the time of Adam Hesterberg...at least I think so

It's new name since when? Since that post at the beginning of the topic?

More like the later part of MOP. We didn't get to hear your nice inequality until the second team contest.

[quote="Elemennop"][quote="Lazarus"]Uhh since the middle of MOP 2007.[/quote]

so pretty much never?[/quote]

Pretty much.

In any case, this thread's getting pretty off-topic...


ratio
advanced fields
advanced fields theorems

Hi,

I have two questions :

1) the first is a bit dumb, what do you call this in english : "the set of lines in a plane through a fixed point",
I can't find it anywhere
I'll just denote that set with $A_p$

2) this is my real question
apparently you can define quadrics with cross ratio's
consider two points p and q, and $A_p$ and $A_q$
now let $\phi$ be a map from $A_p$ to $A_q$, such that the crossratio of lines is preserved [url]http://www.cut-the-knot.org/pythagoras/Cross-Ratio.shtml[/url]
consider all intersections of lines L in $A_p$ with $\phi(L)$

all those points should be a quadric, and all quadrics are formed in this way

I would like to prove this, but I would like to refrain from heavy calculation, therefore I would like to find the polarity of such a quadric, can it be seen geometrically what it is?

I can help you at 1: It's called a pencil of lines. I've seen it in many projective solutions.

What do you mean by [color=blue]polarity[/color]?

when just talking about a projective plane : a polarity is a map $\alpha$from points to lines, and lines to points, that is bijective and involutive

if p lies on M, then $M^{\alpha}$ lies on $p^{\alpha}$

a point p is absolute  if and only if it lies on its own image

in a projective plane over the reals : the quadrics (nondegenerate) are exactly the sets of absolute points of polarities


Find the number os all Integer Sided Obtuse Isosceles Triangles wth Perimeter 2008
( RMO 2008  , Delhi)
Please Help me out. I just want to confirm my answer . Please Solve it and tell me what answer did you get??


function
calculus
integration
calculus computations

try to calculate sum:
$ \sum_{k \equal{} 0}^{\infty}\frac {1}{((2k\plus{}1)!)^{2}}$

It's a sum or difference of two values of one of the Bessel functions ... for example, one can use formula (76) in http://mathworld.wolfram.com/BesselFunctionoftheFirstKind.html to express it in this way.  (This also allows you to replace it with a definite integral, if you prefer.)


Sa se ilustreze printr-o figura geometrica identitatea $x^{2}-y^{2}=(x-y)(x+y)$.

Simpatica problema...Dat fiind ca sunt la ora de info momentan nu pot atasa un desen. Am sa il descriu doar. Luam patratul ABCD de latura x si luam E pe AB si F pe AD asa incat se formeaza patratul AEGF de latura y.
Prelungim EG pana intersecteaza CD in H.
$x^{2}-y^{2}$ corespunde ariei poligonului BCDFGE
Aria lui DFGH este $(x-y)y$, iar a lui EHCB este $(x-y)x$. Prin insumarea acestor arii obtinem aria lui BCDFGE.

Buna ziua !

Variatiune pe aceeasi tema : Ilustrati geometriceste egalitatea : \[(a-b)^{2}=a^{2}-2ab+b^{2}\] .
("mon" manuel 3ème) :wink:
P.S. Asa faceau arabii (si chinezii ?) acum vreo mie de ani... :maybe:

Desenul e cam acelasi ca si cel de mai sus. Patratul de latura $a$ este format din:
-un patrat de latura $b$
-doua dreptunghiuri de laturi $a-b$ si $b$
-un patrat de latura $a-b$


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $ G$ be a finite group that acts transitively on a set $ X$, and let $ x\in X$ be a fixed element of $ X$. Let $ G_x\equal{}\left\{g\in G\mid g\cdot x\equal{}x\right\}$ be the isotropy subgroup of $ x\in X$, and let $ H\subseteq G_x$ be a subgroup of $ G_x$. Let $ N_G\left(H\right)$ be the normalizer of $ H$ in the (whole) group $ G$. Prove that:
[b](a)[/b] The group $ N_G\left(H\right)$ acts on the set $ \text{Fix }H$, where $ \text{Fix }H\equal{}\left\{y\in X\mid g\cdot y\equal{}y\text{ for all }g\in H\right\}$.
[b](b)[/b] If $ H$ is a $ p$-Sylow subgroup of $ G_x$, then this action is transitive.

The proof of [b](a)[/b] is straightforward. The proof of [b](b)[/b] is very similar to the proof of Burnside's theorem.

[hide="Proof:"]Let $ y \in \text{Fix }H$. It will suffice to show that there exists $ u \in N_G(H)$ such that $ ux \equal{} y$. By assumption, $ G$ acts transitively on $ X$ and so there exists $ g \in G$ such that $ gx \equal{} y$. Thus, $ gHg^{\minus{}1} \leq gG_xg^{\minus{}1} \equal{} G_{gx} \equal{} G_y$. Moreover, as $ y \in \text{Fix }H$, we have $ H \leq G_y$. Therefore, $ H, gHg^{\minus{}1} \in Syl_p(G_y)$ and so there exists $ z \in G_y$ such that $ H \equal{} zgHg^{\minus{}1}z^{\minus{}1}$, i.e., $ zg \in N_G(H)$. Set $ u \equal{} zg$. Then $ ux \equal{} zgx \equal{} zy \equal{} y$.[/hide]


algebra
polynomial
linear algebra
matrix
linear algebra unsolved

A,B matrices in $M(n,C)$ such that AB=BA
prove that $deg(\mu_{AB})\leq deg(\mu_{A})deg(\mu_{B})$

where $\mu_{M}$ is minimal polynomial

Probably I'm stupid, but this one is obvious: the algebra generated by $A$ and $B^$ is actually generated by $A^{i}B^{j}$ for $i$ between 0 and $deg (f_{A})-1$ and $j$ between $0$ and $deg(f_{B})-1$. Now, just use that $AB$ is in that subalgebra, so the subalgebra generated by this matrix has dimension at most the dimension of the first algebra, which is the product of degrees;


trigonometry
geometry
geometric transformation
rotation
analytic geometry
function

If you have a point (x, y) and you want to rotate it around the origin by theta degrees, you can use these formulas :

new_x = x*cos(theta) - y*sin(theta)
new_y = x*sin(theta) + y*cos(theta)

I can not understand how these formulas work (other than you can derive them from the rotation matrix). Can someone please explain these formulas conceptually to me?

Why does x*cos(theta) - y*sin(theta) give you the x component of the new line? This is probably so simple but I'm just not thinking of it the right way.

Any help/ideas are appreciated.

rewrite the point (x,y) as (rcos(a),rsin(a))

then we get
New x = rcos(a)cos(theta)-rsin(a)sin(theta) = r cos(a+theta)
new y = r cos(a)sin(theta) + rcos(theta)sin(a) = r sin(a+theta)

which means that the original point was rotated by an angle of theta.

maybe this will help:
think of polar coordinates, rotating the point $ (r, \alpha)$ to $ (r, \alpha \plus{}\theta)$
then we have the old coordinates: $ x\equal{}r\cos{\alpha}$, $ y\equal{}r\sin{\alpha}$
and the new coordinates: $ x'\equal{}r\cos{(\alpha \plus{}\theta)}$, $ y'\equal{}r\sin{(\alpha \plus{}\theta)}$.
we can expand these to get: $ x'\equal{}r\cos{\alpha}\cos{\theta}\minus{}r\sin{\alpha}\sin{\theta}\equal{}x\cos{\theta}\minus{}y\sin{\theta}$
and do the same thing for $ y'$

Thanks for replying.

I was trying to avoid the identity :
[code]cos(a + b) = cos(a)cos(b) - sin(a)sin(b)[/code]
because it's so hard to prove.

But I think I can prove it now and I understand both of your explanations.

Thank you very much for the help. :)

[quote="the_innovator"]I was trying to avoid the identity :
[code]cos(a + b) = cos(a)cos(b) - sin(a)sin(b)[/code]
because it's so hard to prove.[/quote]

The proofs without words on the side include a nice proof of this, if I recall correctly.

Rotation is a [url=http://en.wikipedia.org/wiki/Linear_map]linear map[/url] from the Cartesian plane to itself. Think of it as a function that, given a point, returns another point.

Any rotation function satisfies two important properties of linear maps. Let $ f$ be a rotation function -- ie, it takes in a point, as an input, rotates it counterclockwise by an angle of $ \theta$, and returns the result. For example, if $ \theta \equal{} 90 ^{\circ}$, then $ f((1,2)) \equal{} (\minus{}2,1)$. For two points $ P$ and $ Q$ and for a real number $ c$:

$ f(P \plus{} Q) \equal{} f(P) \plus{} f(Q)$. Basically, adding the coordinates of two points and then rotating is the same as rotating both points and then adding.

$ f(cP) \equal{} c \cdot f(P)$. Basically, moving a point $ c$ times further from the origin and then rotating is the same as rotating and then moving the point $ c$ times further away.

Do you see why these properties make sense?

Using these properties, we find that:
$ f((x,y)) \equal{} f((x,0) \plus{} (0,y))$
$ \equal{} f((x,0)) \plus{} f((0,y))$
$ \equal{} f(x \cdot (1,0)) \plus{} f(y \cdot (0,1))$
$ \equal{} x \cdot f((1,0)) \plus{} y \cdot f((0,1))$

What does it do for us? We've simplified the problem of rotating any point to the problem of rotating the two specific points $ (1,0)$ and $ (0,1)$, which are conveniently on the unit circle!

Rotating $ (1,0)$ by an angle of $ \theta$ gives $ ( \cos \theta, \sin \theta )$, and rotating $ (0,1)$ gives $ ( \minus{} \sin \theta, \cos \theta )$

So, rotating $ (x,y)$ gives $ x(\cos \theta, \sin \theta ) \plus{} y( \minus{} \sin \theta, \cos \theta )$

Combining, we get $ (x \cos \theta \minus{} y \sin \theta, x \sin \theta \plus{} y \cos \theta )$

Thank you very much, that is a perfect explanation! So clear and easy to derive, and I'm sure I understand it.
... I knew I could avoid trigonometric identities :lol:

@t0rajir0u : Thank you, I will look for that.


algebra
polynomial
probability
analytic geometry
function
AMC

Whenever I get sometimes, I'll try to post some AMC practice sets in here. Each set will contains 10 questions, some are from AHSME and others from other contests that are in U.S.A. or in Europe. This is the first one. Enjoy!

1. If a polynomial $x^3-2$ is divided by $x^2-2$, then the remainder is:
A) 2
B) -2
C) -2x-2
D) 2x+2
E) 2x-2

2. How many base 10-four digit numbers, N = abcd, satisfy all three of the following conditions?
(I) $4000 \leq N<6000$
(II) N is a multiple of 5
(III) $3 \leq b <c \leq <6$
A) 10
B) 18
C) 24
D) 36
E) 48

3. If a,b and c are three (not necessarily different) numbers chosen randomly and with replacement from the set {1,2,3,4,5}, the probability that ab+c is even is:
A) 2/5
B) 59/125
C)
D) 64/125
E) 3/5

4. The sides of a triangle have lengths 11,15, and k, where k is an integer. For how many values of k is the triangle obtuse?
A) 5
B) 7
C) 12
D) 13
E) 14

5. In an isosceles triangle ABC, AC = BC. The inscribed circle of the triangle has radius 5 and touches AC in the point of D with CD = 12. Then AC equals:
A) 20
B) 39/2
C) 98/5
D) 200/11
E) 216/13

6. For what values, is this an integer?
$\frac {[(n-2)(n-4)(n-6)(n-8)-1]}{2}$
A) only 1
B) only 2
C) only 9
D) any odd integer
E) any even integer

7. Five candidates ran for the office of dog catcher. No two had the same number? of votes. The winner had 10 votes, What is the largest possible number of votes for the candidate who finished last?
A) 3
B) 4
C) 5
D) 6
E) 7

8. If ABCDE is a regular pentagon, then the angle ACE, measured in degrees, is in the interval
A) 0-15
B) 16-30
C) 31-45
D) 45-60
E) 61-90

9. If we add 329 to three digit number 2x4, we get 5y3. 5y3 is divisible by 3, then the greatest possible value of x is:
A) 1
B) 4
C) 7
D) 8
E) 9

10. Two rays with common endpoint O form a 30 degrees angle. Point A lies on one ray, point B on the other ray, and AB = 1. The maximum possible length of OB is:
A) 1
B) $\frac {1+\sqrt 3}{\sqrt 2}$
C) $\sqrt3$
D) 2
E) $\frac {4}{\sqrt 3}$

Nice problems!

I'll just post some solutions I guess.

[hide]1) Well, by long division, I got $2x-2$, so I guess (E), though I might be wrong.

2) Um... Organize this into cases. You have for the middle two digits: (5,6),(4,6),(3,6),(4,5),(3,5)(3,4), and for the first digit either 4 or 5. For the last, either 0 or 5. So, $2*2*6$ = $24$. So, (C)?

3)Cases again.

P(A)+P(B) = Total.

P(A) = (ab = odd, and c = odd)
P(B) = (ab = even, and c = even)

So, $\frac{27}{125} + \frac{32}{125} = \frac{59}{125}$.[/hide]

Yep. Those are correct.

Just note to #1:

[hide="hint"]Note that we can rewrite $x^3-2$ as $x^3+0x^2+0x-2$ and $x^2-2$ as $x^2+0x-2$. By doing this, this makes the question a lot neater in the long division.[/hide]

Anyone want to try more?

They're good questions, not too hard.

4.
[hide]
pythagorean:  sqrt(346) = 18.6 will yield right triangle, longer will yield obtuse triangle.
19<= k < 11+25
there are 7 integer k
11^2+ k^2 <225
6 int k
7+6 = 13
D
[/hide]

5.
[hide]
Let the center be E, by pythagorean CE = 13, altitude of the base is 13+5 = 18
let DA = x, so is A to the intersecting point of altitude of AB
(12+x)^2 = x^2 + 18^2
24x = 180
x = 15/2
AC = 15/2 + 24/2 = 39/2
B
[/hide]

6.
[hide]
(n-2)(n-4)(n-6)(n-8) have to be odd
n = all odds satisfies this condition
D
[/hide]

7.
[hide]
10,9,8,7,6
6
D
[/hide]

9.
[hide]
5y3 is div by 9, possible candidate for y is 1,4,7
y = x+3 x = 4
B
[/hide]

10.
[hide]
not sure if it can be longer than 2
D?
[/hide]

You know, it's not custom to quadruple post. There is that handy dandy button called the edit button.  :D

10.

[hide]

Yes, you're right, isaacchao; it's 2.

think about it on a coordinate grid ( O is the origin and A along the x-axis).

B is maximized when AB is perpendicular to A.  We then have a 30-60-90 triangle, B being the hypotenuse.

Since AB is 1, ray B is 2

[/hide]

Kent Merryfield's Generilization of  10:

[hide]

same idea as above, but now we have a variable angle.  Simply use the law of sines.
[tex]
\frac{1}{sin\left(\alpha\right)} = \frac{OA}{sin\left(90-\alpha\right)}\\
[/tex]
[tex]
\frac{1}{sin\left(\alpha\right)} = \frac{OA}{cos\left(\alpha\right)}\\
[/tex]
[tex]
\frac{cos\left(\alpha\right)}{sin\left(\alpha\right)} = OA
[/tex]
[tex]
cot\left(\alpha\right) = OA
[/tex]

[/hide]

Isaacchao: you can take the question mark off of your answer in #10. The correct answer is one of the five choices, and you have the correct one.

Generalization of #10: If the angle between the rays is $\alpha$ (assume $\alpha$ is acute) and $AB=1,$ what is the maximum possible length of $OA$? Express your answer as a function of $\alpha$. (This will presumably involve trigonometric functions of $\alpha.$)


I'm in AP Chem right now and we're working on coordination compounds. The only thing we're doing is naming complex ions and the compounds. I have a test coming up and I'm a little bit nervous. I don't really understand how you can tell the coordination number just by looking at the formula. I just have a few questions of my own that came up while I was studying.

1) When naming coordination compounds, my how come sometimes brackets are used to enclose the complex ions and sometimes they aren't? Like in $ \text{K}_{3} \text{Fe(CN)}_{6}$, it would be named Potassium hexacyanoferrate (III) if I am correct. That would make the complex ion $ \text{Fe(CN)}_{6}^{3 \minus{} }$. In all the other exmaples in the textbook, they use brackets to enclose the complex ion (ie $ \text{K}_{4}[\text{PtCl}_{6}]$).

2) When I'm trying to figure out the coordination number of a complex ion, do I just count up the number of ligands? Does that method always work? It seemed to in all of my examples but I'm not sure if it will [i]always[/i] work.

3) My textbook gives me a chart for coordination numbers of "some common metal ions." Is there anyway to predict coordination numbers that isn't overly complicated? I say that because I know the AP doesn't go too in-depth with some explanations.

Any replies are greatly appreciated - the test is on Tuesday.

2. Yes, given that all of the ligands are present in the complex ion.
3. If in doubt, coordination number = 2 x cation charge. Now, there are a few exceptions to the rule, aka $ \mathrm{Al(OH)_4^\minus{}}$ and hexacyanoferrous ion.

1. It is common practice to use [] to enclose the complex ion in a coordination compound.

2.

[quote="Blkmage"]2) When I'm trying to figure out the coordination number of a complex ion, do I just count up the number of ligands? Does that method always work? It seemed to in all of my examples but I'm not sure if it will always work.[/quote]

That will only work if all ligands are monodentate. For polydentate ligands (i.e., chelating agents) you have to figure out (by counting the number of donor atoms capable of making a bond with a metal ion) the number of bonds. For example, in the complex ion Ca-EDTA there is only one ligand, but since EDTA makes six bonds with metals, the coordination number is 6.

3.

[quote="Jdlkem"]there are a few exceptions to the rule[/quote]

I have never seen that ridiculous rule. In fact, there are *a lot* of exceptions - just look any EDTA complex.

lol i don't think its a valid "rule" at all. Its just that in AP Chem, the guideline to figuring out coordination numbers is charge x 2.

at the ap chem level, u dont need to know about chelating agents...


topology
advanced fields
advanced fields unsolved

most books just point out that spin(n), ($ n>\equal{}3$) is a non-trivial double cover of SO(n).
Then "since the fundamental group of SO(n) is $ \{1,\minus{}1\}$,  spin(n) is simply connected."
Why? did I miss something obvious?

[quote="ZetaX"]Normally, you define $ Spin(n)$ to be the universal covering space of $ SO(n)$, thus it's simply connected by defintion.
The mentioned property (being a double covering of a space with $ \pi_1 \equal{} \mathbb Z/ 2\mathbb Z$) suffices if that covering is already connected. Take a look at Deck-transformations (e.g. in Hatcher).[/quote]

thanks for the reminder! basicly there is following thm:
If Y is a covering of connected X, then Y is simply connected iff $ \pi(X)$ acts transitively and freely on each fiber in Y.


Find all functions $ f: \mathbb N_0 \to \mathbb R$ that satisfy the following condition for all non-negative integers $x$ and $y$:
\[f(x\plus{}y)\plus{}f(x\minus{}y)\equal{}f(3x).\]

[quote="NikolayKaz"]Find all function $ f: N_0 \Rightarrow R$ that satisfy the following condition

$ f(x \plus{} y) \plus{} f(x \minus{} y) \equal{} f(3x)$

[hide]
After few easy steps I got that $ f(a) \plus{} f(b) \equal{} f(a \plus{} b)$ , it's cauchy's equation so i get the only family of solutions is f(x)=cx. Substituting this gives me f(x)=0 but i'm not sure if i did it right

[/hide][/quote]

$ f(3) \plus{} f(1) \equal{} f(2) \plus{} f(2)\qquad \equal{} f(6)$
$ f(2) \plus{} f(0) \equal{} f(3)$
$ f(4) \plus{} f(2) \equal{} f(5) \plus{} f(1)\qquad \equal{} f(9)$
$ .\qquad\ \ \ f(6) \equal{} f(4) \plus{} f(0)$

Adding it all up yields $ f(6) \equal{} f(5)$.  Further,

$ f(6) \plus{} f(2)\ \equal{} f(5) \plus{} f(3) \equal{} f(12) \Longrightarrow$ $ f(2) \equal{} f(3)\Longrightarrow f(0) \equal{} 0$
$ f(6) \plus{} f(0)\ \equal{} f(5) \plus{} f(1) \equal{} f(9) \Longrightarrow$ $ f(0) \equal{} f(1) \equal{} 0$

Now, from the first two above we get $ f(1) \equal{} f(2)$. So we have $ f(1) \plus{} f(2k \plus{} 1) \equal{} f(2) \plus{} f(2k) \equal{} f(3) \plus{} f(2k \minus{} 1)$ for $ k\ge1$, i.e.$ f(x) \equal{} 0\forall x\in\mathbb N_0$  is the only solution :(

[quote="NikolayKaz"]Find all function $ f: N_0 \Rightarrow R$ that satisfy the following condition

$ f(x \plus{} y) \plus{} f(x \minus{} y) \equal{} f(3x)$

[hide]
After few easy steps I got that $ f(a) \plus{} f(b) \equal{} f(a \plus{} b)$ , it's cauchy's equation so i get the only family of solutions is f(x)=cx. Substituting this gives me f(x)=0 but i'm not sure if i did it right

[/hide][/quote]

$ y\equal{}0$ $ \implies$ $ f(3x)\equal{}2f(x)$ and so $ f(0)\equal{}0$
$ y\equal{}x$ $ \implies$ $ f(3x)\equal{}f(2x)$ and so $ f(2x)\equal{}2f(x)$
So $ f(6x)\equal{}4f(x)$
Using then $ x\equal{}2y$ in the original equation, we get $ f(3y)\plus{}f(y)\equal{}f(6y)$ and so $ 2f(y)\plus{}f(y)\equal{}4f(y)$ and so $ f(y)\equal{}0$

And the unique solution is $ f(x)\equal{}0$ $ \forall x$


geometry
trigonometry
rotation
area of a triangle
Heron\u0027s formula

Hello. I need help with this problem. It seems easy but it isn't so! (for me).

"I live in the countryside, in an area that forms an equilateral triangle ABC. My house is very near, $a$ km., $b$ km. and $c$ km from three different villages A, B and C, respectively.
Find the distance between the villages."

That is, exists a point M inside ABC such that MA=a, MB=b and MC=c. I must find AB=BC=CA.

I tried Geometry and Trigonometry but it didn't work.

Thanks.

I will reformulate the problem...

$\triangle ABC$ is an equilateral triangle with side $L$.
$M$ is a point inside the triangle such that $MA=a$, $MB=b$, $MC=c$.
Find $L$ in terms of $a$, $b$ and $c$.

Rotate ABC 60 degrees about a vertex and then use law of cosines.

[hide]One way to do this  is to consider point N ('outside the triangle and  'other side'  of line BC)  such that  MCN is an equilateral triangle (That is $MC=CN=NM=c$)
Now $\angle MCN$  is $60^o$  so $\angle BCN = \angle MCA$  and thus the triangle BCN ( conguant to triangle  ACM)  has sides  $(a,c,L)$  and of course, $\angle BNC = 60^o+ \angle MNC$ ) Easy to find the $\angle MNC$ (Its a triangle with sides ($a,b,c$) - so it is $\cos^{-1} (\frac{c^2+a^2-b^2}{2ac})$   Now you know the two sides ($a$ and $c$) and the angle between them ($60^o + above \ angle$) so can find $L$. [/hide]

Added later : Sorry, did not see the above post, it is essentially the same idea.

Hello.

Gyan I think you got confused when typing but, working with your suggestion and WarpedKlown1335 I found that

$L^2=\frac{1}{2}(a^2+b^2+c^2)+2\sqrt{3}\cdot\mathrm{area}(T)\qquad(*)$

where T is the triangle with sides $a, b, c$.
I had to use the Law of Cosine and Heron's Formula of area of a triangle!

Now I wonder if there is a synthetic proof.

Also, now I see that this problem is closely related to Fermat-Napolen configuration. I extracted the following from Forum Geometricorum, Volume 2 (2002) 39–46, Gilles Boute.

Let ABC a triangle with sidea $a, b, c$. Let $F_a^+$ be the point such that $CBF_a^+$ is equilateral with the same orientation as $ABC$; similarly for $F_b^+$ and $F_c^+$. Then, the author states, without proof that the segments $AF_a^+$, $BF_b^+$, $CF_c^+$ have the same lenght $L$ and is pricesly (*) !!!

Thanks.

[quote="hpoincare"]Hello.

Gyan I think you got confused when typing but, working with your suggestion and WarpedKlown1335 I found that

$L^2=\frac{1}{2}(a^2+b^2+c^2)+2\sqrt{3}\cdot\mathrm{area}(T)\qquad(*)$

where T is the triangle with sides $a, b, c$.
I.[/quote]

This is EXACTLY the same (or equivalent to)  as what I typed  above. (Just use the formula $\cos(60^o+\theta)=(1/2) \cos \theta -(\sqrt 3 /2) sin\theta$  and you get the same result.


I am trying to create an equation to use in a computer program. The aim is to find the value of Z, if the values of X and Y are known.

The Z value is going to be used to order some data, so the range does not matter.

I would like Z to increase as X and the result of the Y EQUATION increase.

The Y EQUATION needs to be able to be able to fit the following:
 - When Y is between 30 and 40, the result of the Y EQUATION needs to be at its highest.
 - When Y is at 1, the result of the Y EQUATION needs to be roughly a quater of its highest possible value.
 - When Y is at 200, the result of the Y EQUATION needs to be roughly a quater of its highest possible value.
 - No result of the Y EQUATION should be less than or equal to 0 when Y is greater than 0

Any help would be a godsend.
Paul


floor function
number theory unsolved
number theory

Prove that: $ (k!)^{k^2\plus{}k\plus{}1}$ divides $ (k^3)!$
he has quoted from the exam Poland.But I think it is incorrect.
you please check it.

Example: for $ k\equal{}2$  it is wrong   :)

actually for $ k\equal{}2$ it is correct, because $ 2^7 | 8!$ :)

Adriana N wrote:"Example: for $ k\equal{}2$ it is wrong"
you can prove or negative it??

[quote="Adriana N."]I learned that a bigger number can't divides a smaller number...    :|[/quote]

Sure, but $ (2!)^{2^2\plus{}2\plus{}1}\equal{}2^7\equal{}128$ and $ (2^3)!\equal{}8!\equal{}40320\equal{}128\times 315$ :P

[quote="math10"]Prove that: $ (k!)^{k^2 \plus{} k \plus{} 1}$ divides $ (k^3)!$
he has quoted from the exam Poland.But I think it is incorrect.
you please check it.[/quote]

This statement is correct.

It's easy to see that we only need to prove that:

For every prime $ p$, $ (k^2\plus{}k\plus{}1)[\lfloor \frac{k}{p} \rfloor]\plus{}\lfloor \frac{k}{p^2} \rfloor]\plus{}....\plus{}\lfloor \frac{k}{p^n} \rfloor] \le \lfloor \frac{k^3}{p} \rfloor]\plus{}\lfloor \frac{k^3}{p^2} \rfloor]\plus{}....\plus{}\lfloor \frac{k^3}{p^{3n}} \rfloor$,

where $ \lfloor \frac{k}{p^n} \rfloor$ is the last non-zero term.

[b]Proof:[/b]

We only need to prove that $ (k^2\plus{}k\plus{}1)(\lfloor \frac{k}{p^m} \rfloor) \le \lfloor \frac{k^3}{p^{m}} \rfloor \plus{} \lfloor \frac{k^3}{p^{n\plus{}m}} \rfloor\plus{}\lfloor \frac{k^3}{p^{2n\plus{}m}} \rfloor$.

As $ \lfloor \frac{k}{p^n} \rfloor$ is non-zero, we have $ k \ge p^n$.

So $ (k^2\plus{}k\plus{}1)(\lfloor \frac{k}{p^m} \rfloor)$
$ \le \lfloor \frac{k^3}{p^{m}} \rfloor \plus{} \lfloor \frac{k^3}{p^{m}k} \rfloor\plus{}\lfloor \frac{k^3}{p^{m}k^2} \rfloor$
$ \le \lfloor \frac{k^3}{p^{m}} \rfloor \plus{} \lfloor \frac{k^3}{p^{m}(p^n)} \rfloor\plus{}\lfloor \frac{k^3}{p^{m}(p^n)^2} \rfloor$
$ \equal{}\lfloor \frac{k^3}{p^{m}} \rfloor \plus{} \lfloor \frac{k^3}{p^{n\plus{}m}} \rfloor\plus{}\lfloor \frac{k^3}{p^{2n\plus{}m}} \rfloor$, done.

may i know this is from which book of titu ? :?:

perhaps, it is the book:"problem from a book"

[quote="stephencheng"][quote="math10"]Prove that: $ (k!)^{k^2 \plus{} k \plus{} 1}$ divides $ (k^3)!$
he has quoted from the exam Poland.But I think it is incorrect.
you please check it.[/quote]

This statement is correct.

It's easy to see that we only need to prove that:

For every prime $ p$, $ (k^2 \plus{} k \plus{} 1)[\lfloor \frac {k}{p} \rfloor] \plus{} \lfloor \frac {k}{p^2} \rfloor] \plus{} .... \plus{} \lfloor \frac {k}{p^n} \rfloor] \le \lfloor \frac {k^3}{p} \rfloor] \plus{} \lfloor \frac {k^3}{p^2} \rfloor] \plus{} .... \plus{} \lfloor \frac {k^3}{p^{3n}} \rfloor$,

where $ \lfloor \frac {k}{p^n} \rfloor$ is the last non-zero term.

[b]Proof:[/b]

We only need to prove that $ (k^2 \plus{} k \plus{} 1)(\lfloor \frac {k}{p^m} \rfloor) \le \lfloor \frac {k^3}{p^{m}} \rfloor \plus{} \lfloor \frac {k^3}{p^{n \plus{} m}} \rfloor \plus{} \lfloor \frac {k^3}{p^{2n \plus{} m}} \rfloor$.

As $ \lfloor \frac {k}{p^n} \rfloor$ is non-zero, we have $ k \ge p^n$.

So $ (k^2 \plus{} k \plus{} 1)(\lfloor \frac {k}{p^m} \rfloor)$
$ \le \lfloor \frac {k^3}{p^{m}} \rfloor \plus{} \lfloor \frac {k^3}{p^{m}k} \rfloor \plus{} \lfloor \frac {k^3}{p^{m}k^2} \rfloor$
$ \le \lfloor \frac {k^3}{p^{m}} \rfloor \plus{} \lfloor \frac {k^3}{p^{m}(p^n)} \rfloor \plus{} \lfloor \frac {k^3}{p^{m}(p^n)^2} \rfloor$
$ \equal{} \lfloor \frac {k^3}{p^{m}} \rfloor \plus{} \lfloor \frac {k^3}{p^{n \plus{} m}} \rfloor \plus{} \lfloor \frac {k^3}{p^{2n \plus{} m}} \rfloor$, done.[/quote]
thank a lot


Prove it the following way:

Let $\mathcal{P}_n$ be the statement $A_n\ge{G}_n$ for any nonnegative numbers $a_1,\ldots,a_n$, where $A_n=\frac1n(a_1+\ldots+a_n)$ and $G_n=\sqrt[n]{a_1\cdots{a}_n}$.

First prove $\mathcal{P}_{2^m}$ for all $m\ge0$, and then prove $\mathcal{P}_n\Longrightarrow\mathcal{P}_{n-1}$ for $n>1$.

Then we hope that all this implies $\mathcal{P}_n$ for all $n\ge1$. I think it's called something like transfinite induction, but I have to look that up.

I think this is Cauchy's (?) proof of AM-GM (I might be mistaking him for someone else.)

I've heard "reverse" induction, but I don't know the common name. It's certainly a cool proof.

Indeed, this type of induction proof is often called "reverse induction".

Alright, reverse induction then.

I guess everyone likes using the AM-GM, but no one wants to prove it. :(

edit: [hide="hint for the second part"]Let $a_n=\frac1{n-1}(a_1+\ldots+a_{n-1})$. I think that's right.[/hide]


probability
integration
probability and stats

Suppose we consider a particle on a grid $ \{(rh,sh): r,s \in \mathbb{Z}\}$. Let the particle move according to the following rule:

If the position $ Y_{n}$ of the particle at the end of the n-th step is given by $ Y_{n} \equal{} (rh,sh)$ , then $ Y_{n\plus{}1} \equal{} ((r\plus{}u)h, (s\plus{}v)h)$ with the probability $ \frac{1}{4}$ if $ |u| \equal{} |v| \equal{} 1$ and probability $ 0$ otherwise.
For this process, we need to prove that

$ Pr\left\{ \frac{(Y_{m\plus{}n} \minus{} Y_{n})}{h \sqrt{m}} \in [a,b] \times [c,d]\right\} \to \frac{1}{2\pi} \int_{a}^{b} \int_{c}^{d} \exp\left(\minus{}\frac{x^2 \plus{} y^2}{2}\right) dxdy$ as $ m \to \infty$

the first thing to notice is that, since the movements along the x-axis and y-axis are independent, it suffices to prove the $ 1$-dimensional analogue of your problem.

isn't it exactly the central limit theorem: take $ S_n\equal{}\sum_1^n B_k$ a symmetric random walk, with $ P(B_k\equal{}\minus{}1)\equal{}P(B_k\equal{}\plus{}1)\equal{}\frac{1}{2}$. Then $ \frac{Y_{n\plus{}m}\minus{}Y_n}{h \sqrt{m}} \equal{} \frac{S_m}{\sqrt{m}}$. You should be able to conclude, isn't it ?


If you think you have a cool avatar, plz post here

Edit: Such as mine!

i think i'll actually nominate myself on this one. my av rocks.

I consider mine pretty cool because it was designed personally by one of the users of this site.

Thanks Treething. :)

Mine is cool because I have the smartest person who lived on this planet.

haha, I think my avator is the best.  It's my name spelled using math symbols.

I like mine (of course) but I've been meaning to get a new one...I haven't been able to find something I like to use as an avatar though...

My avatar is cool because.......

[b]It's U2. Enough said.[/b] :)

I think my avatar is cool. It's from Warcraft III: The Frozen THrone. One of my favorite games.

Mine is some anti Microsoft icon,  Too bad the words are too small to read
It says "Where do you want to go today?"
Took me a while to resize it so it's small enough for AoPS.

Yeah, I had that same problem. I think you can just barely read mine. Can you?

i can only read "Grim reaper, parking only"

It says:

"Grim Reaper

Parking only

Violators are next in line"

 :ninja:

lol, your avatar would be great if more people could see what it says.

Yah, it would... I have another for  LOTR fans, not sure thar many people would get it though... Let me try and make it wider to see if you can read it.

i can see it a little better now.  But it is still hard to read.

You know what's funny about this whole thing? Is that I didn't change it...

 :rotfl:

oh, well i guess since i knew what it read i can see it better

i hate my avatar and i loathe slim shady( rap sucks ,anyhow ,ROCK rules especially Guns n roses, metallica, dire straits and deep purple ) However i couldnt find nething else so i kept it..........

If you hate your avatar, why don't you delete it?

i think he has it becuase it goes with his username

This topic is [b][i]8[/i][/b] years old!!

How about mine? Play Sushi Cat 2 if you don't know what it is :3

[quote="dragonrowe"]You people don't know what a avatar is!!!!!!!!!!!! math92 is a camera lense. IT IS UNDER THE STARS![/quote]
Maybe they changed it.


[hide]So, we need to find the multiples of three.  Because there are eight sides to each die, the smallest sum is 2 and the largest is 16.  The number of possibilities is 64.

The multiples of three possible are 3,6,9,12, and 15.  There are 2 ways of getting three, 5 ways of getting six, 8 ways of getting nine, 5 ways of getting twelve, and 2 ways of getting fifteen.

So the probability is 22/64 = 11/32

I think that that's right.[/hide]

[quote="ckck"][quote="mr. math"]two eight sided dice are rolled. what is the probability that the dice will add up to a number divisible by 3?

how would you do this problem?

ive tried, but i didn't get the right answer.[/quote]
I have a long way involving casework.
[hide]So we have a 1/8 probability chance of choosing a 1 on the first dice. Then the numbers that would add up to a multiple of three are 2,5,8. So a 3/8 chance. So for 1 are 1/8*3/8=3/64. But there are three numbers so multiple this by 3 to get 9/64.

Then we have a 1/8 chance of chossing a 2. The numbers that work are 1,4, and 7. That's another 3/8. So thats another 9/64.

Then theres three. And the numbers that work with three are 3, and 6. So thats 2/64. But theres two of them is 4/64.

Then 4 with 2,5, and 8. So thats another 9/64.

Then 5 with 1,4, and 7. So another 9/64.

Then 6 with 3, and 6. So thats another 4/64.

Then 7 with 2,5, and 8. So thats another 9/64.

Finally 8 with 1,4,a dn 7. So that another 9/64.

Adding all these up the final answer is 79/64.[/hide]

I think I did this wrong.[/quote]

Chris, you [i][u]think[/u][/i] you did this wrong?  Or is it possible that its more that 100%.   :D

[im not being mean, ckck is my friend so im just messin' with him]

[hide]There's a 3/8 probability of rolling a number equal to 1 mod 3, 3/8 probability of rolling a number equivalent to 2 mod 3, and 1/4 probability of rolling a multiple of 3.  Then if we roll a number equivalent to 1 mod 3, we need a number equal to 2 mod 3 on the other die, etc.  Then the total probability is:
$\frac{3}{8}\times \frac{3}{8}+\frac{3}{8}\times \frac{3}{8}+\frac{1}{4}\times \frac{1}{4}=\frac{9+9+4}{64}=\frac{22}{64}=\frac{11}{32}$[/hide]

the answer book says 13/32. thats why im a little confused


superior algebra
superior algebra solved

[quote="Stewie"]If $ f$ is an automorphism of a finte group such that $ f(x) \equal{} x \iff x \equal{} e$ and $ f(f(x)) \equal{} x$ for all elements $ x$ of the group, prove that it is the unique $ f(x) \equal{} x^{ \minus{} 1}$.[/quote]

[b]Lemma.[/b] If $ f$ is an automorphism of a finte group G s.t. $ f(x) \equal{} x \iff x \equal{} e$, then $ (\forall g\in G)(\exists x\in G)(g \equal{} x^{ \minus{} 1}f(x))$.
[i]Proof.[/i] The map $ \hat f: G\to G: x\mapsto x^{ \minus{} 1}f(x)$ is one-one, since if $ \hat f(x) \equal{} \hat f(y)$ then $ yx^{ \minus{} 1} \equal{} f(y)(f(x))^{ \minus{} 1} \equal{} f(yx^{ \minus{} 1})$.
As G is [i]finite[/i], we have $ \hat f$ onto as well.$ \Box$

Now come to the main point. If $ f^2 \equal{} I$, given any $ g$, we have by Lemma some x with $ g \equal{} x^{ \minus{} 1}f(x)$. So,
$ f(g) \equal{} f(x^{ \minus{} 1})f^2(x) \equal{} (f(x))^{ \minus{} 1}x \equal{} g^{ \minus{} 1}$. $ \;\Box$


inequalities
function
logarithms
real analysis
real analysis unsolved

solve the following equation:
[img]http://www.mathramz.com/xyz/latexrender/pictures/62823bccfe7bb0fde083b1d055cbfd94.png[/img]

hello, your solution is $ x\equal{}2$.
Sonnhard.

Prove that the left hand side is never less than the right hand side with an equality achieved only in one case (clearly, the left hand side is larger for $ x\to 1$ and $ x\to\infty$, so if it could be less, the root wouldn't be unique, and since there is no other nice number solving the equation, but hte problem should be solvable, the inequality is the only logical possibility left). The monstrous expressions on the left make one suspect that the exact formulae are almost irrelevant, so let us denote $ a\equal{}(x\minus{}1)^2$ and $ b\equal{}x^2\minus{}3$. The right side is then just $ a\plus{}2b$, so we get the inequality
\[ \frac{a^2}{b^2}\plus{}b^4\plus{}\frac 1a\ge a\plus{}2b\]
to prove. Since it is hard to relate $ a$ and $ b$ explicitly, it is reasonable to believe that the exact relation does not really matter and the inequality should be true for any $ a$ and $ b$. At last, since it isn't an IMO question or Vasc's invention, it shouldn't be much more than a straight application of AM-GM, which for 3 numbers is $ \alpha x\plus{}\beta y\plus{}\gamma z\ge x^\alpha y^\beta z^\gamma$ ($ x,y,z>0$, $ \alpha,\beta,\gamma>0; \alpha\plus{}\beta\plus{}\gamma\equal{}1$). Now there is not much choice as to what to do next, so I'll leave it to you to finish.

first of all thank you.
but i didnt understand how did you get the last inequality,and how should i use it to solve the first inequality you proposed?

The last inequality I mentioned is just very standard. You will learn it sooner or later anyway (it is just the [[Jensen's inequality]] for the function $ \log x$). As to the application, you want to get $ a$ dominated by a part of the sum $ \frac {a^2}{b^2}\plus{}b^4\plus{}\frac 1a$ so, what $ \alpha,\beta,\gamma$ should you choose to get $ \alpha\frac {a^2}{b^2}\plus{}\beta b^4\plus{}\gamma \frac 1a\ge \left(\frac {a^2}{b^2}\right)^\alpha(b^4)^\beta\left(\frac 1a\right)^\gamma\equal{}a$? Then look at what is left.

I think I'm just double-posting because [b]fedja[/b] has given solution to the problem already... Anyway, the equation is
\[ \frac {(x \minus{} 1)^4}{(x^2 \minus{} 3)^2} \plus{} (x^2 \minus{} 3)^4 \plus{} \frac {1}{(x \minus{} 1)^2} \equal{} 3x^2 \minus{} 2x \minus{} 5.
\]
As [b]fedja[/b] suggested, take $ a \equal{} (x \minus{} 1)^2$, $ b \equal{} x^2 \minus{} 3$ (note that $ a\geq 0$). So the equation becomes
\[ \frac {a^2}{b^2} \plus{} b^4 \plus{} \frac {1}{a} \equal{} a \plus{} 2b.
\]
But by AM-GM we actually have that
\[ \frac {4}{7}\cdot \frac {a^2}{b^2} \plus{} \frac {2}{7}b^4 \plus{} \frac {1}{7}\cdot\frac {1}{a}\geq (a^2b^{ \minus{} 2})^{4/7}(b^4)^{2/7}(a^{ \minus{} 1})^{1/7}\geq a
\]

\[ \frac {3}{7}\cdot \frac {a^2}{b^2} \plus{} \frac {5}{7}b^4 \plus{} \frac {6}{7}\cdot\frac {1}{a}\geq 2b.
\]
and adding up yields
\[ \frac {a^2}{b^2} \plus{} b^4 \plus{} \frac {1}{a}\geq a \plus{} 2b.
\]
Hence we must have equality in the inequality above; it's easy to realize it's the case when $ a \equal{} b \equal{} 1$, so that $ x \equal{} 2$ is the only solution.

I hope this is helpful to you. :)

the thing is that the equation is solvable with simpler "tools".
anyway,can you help me a lil bit to find apropriates alpha beta and gamma?
thank you.

to jutaro:

[img]http://alt2.mathlinks.ro/Forum/latexrender/pictures/f/8/c/f8c867018ad2f8a6b319b07ebbd117f0f7025ed6.gif[/img]
here on the right you mean (= a) right?
also i didnt understand how did you get this
 [img]http://alt1.mathlinks.ro/Forum/latexrender/pictures/4/2/8/428d99fd92bc10d414ef44f042b1d9086f7cd500.gif[/img]

[quote="lock345"]the thing is that the equation is solvable with simpler "tools".
[/quote]
Perhaps. I just showed you the way I would follow if confronted with such a question. Of course, there may exist plenty of other approaches. So, what tools do you want to restrict to? (I'll accept everything except the idea to multiply by the denominators and open the parentheses). Or, in other words, where did you get this problem from?

As to finding $ \alpha,\beta,\gamma$, Jutaro did it already. Of course, the way to find them is to compare the degrees of $ a$ and $ b$ on both sides and use the condition $ \alpha\plus{}\beta\plus{}\gamma\equal{}1$. You'll  get 3 linear equations with 3 unknowns.

okay i understood but i dont think my teacher will accept this way because it uses a way we did not learn yet.
the equation is solvable with very simple tools like sufisticated tricks and some algebra, and i think that it has a conection to [b]very smple[/b] calculus.
thank you anyway for your help.

Well, if you write the solution accurately, honestly state that you got some help with it, and learn all the related things like Jensen's inequality, etc., your teacher will hardly have any choice but to accept the solution. After all, mathematics is about being able to solve problems, not about just following the instructions, and one cannot be punished for knowing more of it than he is assumed to know. The solution I outlined is rather elementary (except, maybe, Jensen) and the problem is, clearly, not one of the standard exercises, so you are sort of expected to do something unusual. Thus, I wouldn't worry too much about whether this solution will be accepted or not. I'll try to think of whether there is some clever trick (and, if I find one, I'll give you a hint), but not right now :).

[quote="Jutaro"][quote="lock345"]
also i didnt understand how did you get this
 [img]http://alt1.mathlinks.ro/Forum/latexrender/pictures/4/2/8/428d99fd92bc10d414ef44f042b1d9086f7cd500.gif[/img][/quote]

Er... When trying to use weighted AM-GM and you think you are not clever enough to find suitable values for the coefficients (like me :D)  you can just name them, say $ \alpha$, $ \beta$, $ \gamma$, etc., and then establish what you want to get. For example, as for the second inequality, we want to prove something like $ xa^2b^{ \minus{} 2} \plus{} yb^4 \plus{} za^{ \minus{} 1} \geq b$, where $ x \plus{} y \plus{} z \equal{} 1$. Applying AM-GM to the left-hand side we get $ xa^2b^{ \minus{} 2} \plus{} yb^4 \plus{} za^{ \minus{} 1} \geq (a^2b^{ \minus{} 2})^x(b^4)^y(a^{ \minus{} 1})^z \equal{} a^{2x \minus{} z}b^{4y \minus{} 2x}$. So, if we want the right-hand side to be something like $ b$, we should have $ 2x \minus{} z \equal{} 0$ and $ 4y \minus{} 2x \equal{} 1$. Putting this together with $ x \plus{} y \plus{} z \equal{} 1$ we just have to solve a linear system of equations. In this case you get as solution $ (x,y,z) \equal{} \left(\dfrac{3}{14},\dfrac{5}{14},\dfrac{6}{14}\right)$.[/quote]
when i look on it again:
i understood how you picked up alpha beta and gamma.
but what i did not understand is how from this line
 [img]http://alt2.mathlinks.ro/Forum/latexrender/pictures/f/8/c/f8c867018ad2f8a6b319b07ebbd117f0f7025ed6.gif[/img]
you got this line : [img]http://alt1.mathlinks.ro/Forum/latexrender/pictures/4/2/8/428d99fd92bc10d414ef44f042b1d9086f7cd500.gif[/img]

and also i did not understant how from this line you got the last line.
thank you.

[quote="lock345"]
but what i did not understand is how from this line
 [img]http://alt2.mathlinks.ro/Forum/latexrender/pictures/f/8/c/f8c867018ad2f8a6b319b07ebbd117f0f7025ed6.gif[/img]
you got this line : [img]http://alt1.mathlinks.ro/Forum/latexrender/pictures/4/2/8/428d99fd92bc10d414ef44f042b1d9086f7cd500.gif[/img]

and also i did not understant how from this line you got the last line.
thank you.[/quote]

The two inequalities you say are two different applications of AM-GM. Keep in mind that what we want to prove is that   $ \frac {a^2}{b^2} \plus{} b^4 \plus{} \frac {1}{a}\geq a \plus{} 2b$. What we do is to establish two separate inequalities and add them up. First we have

 \[ \frac {4}{7}\cdot \frac {a^2}{b^2} \plus{} \frac {2}{7}b^4 \plus{} \frac {1}{7}\cdot\frac {1}{a}\geq (a^2b^{ \minus{} 2})^{4/7}(b^4)^{2/7}(a^{ \minus{} 1})^{1/7}\geq a,\] and on the other hand, \[ \frac {3}{14}\cdot \frac {a^2}{b^2} \plus{} \frac {5}{14}b^4 \plus{} \frac {6}{14}\cdot\frac {1}{a}\geq (a^2b^{ \minus{} 2})^{3/14}(b^4)^{5/14}(a^{ \minus{} 1})^{6/14}\geq b,\] or in other words, \[ \frac {3}{7}\cdot \frac {a^2}{b^2} \plus{} \frac {5}{7}b^4 \plus{} \frac {6}{7}\cdot\frac {1}{a}\geq 2b.\] Summing the two inequalities above we get just what we wanted.  :)

ok understood.
last question:
how did you relize that, and how will i'll write it in a 'math' way?:

[quote]Hence we must have equality in the inequality above; it's easy to realize it's the case when $ a \equal{} b \equal{} 1$, so that $ x \equal{} 2$ is the only solution.[/quote]
why only a=b=1?
maybe there are more solutions?

thank you.

[quote="lock345"]ok understood.
why only a=b=1?
maybe there are more solutions?
[/quote]

When applying AM-GM equality is reached if and only if all the numbers are equal, so in our case you must have $ \dfrac{a^2}{b^2}\equal{}b^4\equal{}\dfrac{1}{a}$. From this, and given that $ a \geq 0$, it should be straightforward that $ a\equal{}b\equal{}1$, right?  :)


Syntax Error wrote:write  :sqrt:[5-2:sqrt:6)] as a difference of the square roots of two integers

[hide]:sqrt:[5-2:sqrt:6)]=:sqrt:a - :sqrt:b

[5-2:sqrt:6)]=a+b-2:sqrt:ab (squared both sides)

a+b=5

:sqrt:ab=:sqrt:6  --> ab=6

Factors of 6: {1,6},{2,3}

Partitions of 5: {1,4},{2,3},{5}

The only one in common is {2,3}, so {a,b} :cong: {2,3}

So we have the possible solutions :sqrt:2-:sqrt:3 and :sqrt:3-:sqrt:2, but :sqrt:[5-2:sqrt:6)] implies a positive value so the sol'n is :sqrt:3-:sqrt:2 because the other is negative.

[/hide]

I made that a bit more complicated than necessary and was very verbose. I did it for those that didn't understand other answers.

[quote="Tare"][quote="Syntax Error"]yes, but what were your thoughts on it, like how you approached, though i think this was in aops v1[/quote]
Do you remember what chapter it was on? I can't find it anywhere...[/quote]

The one about quadratics, i.e. chapter 6.


calculus
function
LaTeX
calculus computations

can anyone solve this poisson formula to get the mean (m) as the subject?

i've had a go but just got stuck.

p = (m^x) (e^ (-m))
     __________________

                    x!

i want it rearranged to m =

thanks in advance.  :)

You can't (but what has this got to do with calculus?) write it in terms of elementary functions.

If you look up the[url=http://mathworld.wolfram.com/LambertW-Function.html]LambertW function[/url] it can be written as (thank you Maple)

$m=\exp\left(-\frac{x \mathrm{LambertW\left(-\frac{\exp\left(\frac{\ln(px!)}{x}\right)}{x}\right)}-\ln(px!)}{x}\right)$
or equivlantly (blah ugly Latex!)

$\left( p\Gamma  \left( x+1 \right)  \right) ^{{x}^{-1}} \left( {e^{{\mathrm{LambertW}} \left( -{\frac { \left( p\Gamma  \left( x+1 \right)
 \right) ^{{x}^{-1}}}{x}} \right) }} \right) ^{-1}
$

Sorry  - new to this. Saw something else about poisson in this forum, so put the question in here.

The answer is over my head, but perhaps I’m asking a difficult question that can’t have a simple answer.

Is there a way I can practically calculate the mean? Lambert is confusing.

Apologies for putting this in the wrong place.


Maria has 4 identical apples and 7 identical oranges. In how many ways can she distribute the fruits among her 4 friends?

We will consider the number of ways to distribute the apples and then the number of ways to distribute the oranges and then multiply them.

The number of ways to distribute the apples is in one-to-one correspondence with the number of nonnegative integer solutions to $ a\plus{}b\plus{}c\plus{}d\equal{}4$. To solve this problem, we will denote an ordered quadruple $ (a,b,c,d)$ using dots and dividers; for example, $ (1,1,1,1)$ is denoted as $ \cdot|\cdot|\cdot|\cdot$, and $ (2, 0, 2, 0)$ is denoted as $ \cdot\cdot\parallel{}\cdot\cdot|$. Notice how if $ a\plus{}b\plus{}c\plus{}d\equal{}4$, then there are $ 7$ symbols altogether, and we need to choose $ 4$ of them to be dots. So the number of ways to distribute the apples is $ \binom{7}{4}$.

We are not done! We also need to distribute the oranges: the number of ways to do this is $ \binom{10}{7}$ (confirm). So our final answer is $ \binom{10}{7}\binom{7}{4}$.

why is that not the same as

$ \dfrac{14!}{4!7!3!}$

Because it's overcounting by a lot--for example, for one divider, we might count both ...|ao|... and ...|oa|..., which are considered the same, but different by that method.

oh yeah of coarse  :blush:
I should have realized that


Let $ ABC $ be a triangle with $ \angle B \geq \frac{\pi}{2} $ . Prove that :
 $ \frac{1}{a}+\frac{2\sqrt2}{c} \geq \frac{3\sqrt3}{b} $

Here is my approach:
Acording cosin law we have :$b\geq \sqrt{a^2+c^2}$.We will prove that:
$(\dfrac{1}{a}+\dfrac{2\sqrt2}{c})^2(a^2+c^2)\geq 27$
But we always have :(by Cauchy's inequal)
${\dfrac{1}{a}+\dfrac{\sqrt2}{c}}+\dfrac{\sqrt2}{c}\geq 3\sqrt[3]{\dfrac{2}{ac^2}} \to (\dfrac{1}{a}+\dfrac{2\sqrt2}{c})^2 \geq 9 \sqrt[3]{\dfrac{4}{a^2c^4}}(1)$
On the other hand:$a^2+c^2=a^2+\dfrac{c^2}{2}+\dfrac{c^2}{2}\geq 3\sqrt[3]{\dfrac{a^2c^4}{4}}(2)$
Multiplying (1) and (2) we are done.
The equal ocurs when $b=\sqrt3,c=\sqrt2,a=1$
---------------------------------
Hey ,nttu.
What's Old and New inequalities??And where did you buy it???

This problem is not hard . Indeed , my solution is the same with yours . The equality holds for $%Error. "measure" is a bad command.
B =\dfrac{\pi}{2} , 2a^2 =c^2$
  You can just need to solve it in case $%Error. "measure" is a bad command.
B = \dfrac{\pi}{2}$  :)
  And my teacher said this problem can be solved by Holder  ;)

Yes,Ur teacher was right!By Holder's inequal:
$(\dfrac{1}{a}+\dfrac{2\sqrt2}{c})(\dfrac{1}{a}+\dfrac{2\sqrt2}{c})(a^2+c^2)\geq (1+2)^3=27$(desired ineq)
And plz ask my question,nttu.I want to know more detail about it. :D


Max lies on Fridays, saturdays, and sundays, be he is truthful on all other days.  Minnnie lies on Tues., Wednesday, and Thurs., but she is truthful on all other days.  On what day of the week could they both say "tommorow i will lie"?

i don't think so because Minnie tells the truth on Fridays and Satudays so if she said "tommorow i will lie"  on friday that wouldn't be true because on Saturday she tells the truth and she has to telll the truth on Friday.  Does that make sense? :)

yes it is Thursday because Max tells the truth on thursday and on friday he lies. minnie lies on thursdays but tells truths on friday. so she would lie and say she was going to lie on friday. therefore, Thursday they will both be able to say "tomorrow i will lie" truthfully. :P

woops!  I wasn't thinking here.  I would've posted an explanation if I was more patient but i thought about it and said that they both were different on friday than the day before.  And gr8sk8r said that they will say "i will lie" truthfully, but minnie lies about it.  See i tried some other examples and found that they say that if they both do different things the next day.

well she's telling the truth about lying. i know it's confusing.

That's even hard for the person who's saying it. It's sort of like saying the color of the word

[color=orange]Blue[/color]

[quote="236factorial"] It's sort of like saying the color of the word

[color=orange]Blue[/color][/quote]
i love those brainteasers

Yeah, I agree with everyone. It's Thursday. Kinda confusing though :oops:

where did you get that question?
i'm just wondering 'cause it's the same question as question 6 of the first round of the FMO :)

Minnie is written with 2 n's not 3  :D  :D

quite confusing yes. The solution is max tells the truth and Minnie (2 times n :P ) lies. So its Thursday :-)

This also proves women lie more even if they lie equaly as often as men  :P


This is a very nice problem, which can be solved with a very common method!
Think of infinite descent.....


trigonometry
function
geometry
trig identities
Law of Sines

Please explain why it is true.  Isn't SSA and exception?

An oblique triangle in which two sides and an angle are given always results in at least one triangle.

Yeah, if a triangle is defined by two sides and an angle there could be two possible solutions to the other side lengths and angles to the triangle.

Mathematically this can be demonstrated using the law of sines,

$\frac{a}{\sin{A}} = \frac{b}{\sin{B}}$

Suppose you know $a$, $b$ and $A$, and you want to find $B$.  Then, rearanging,

$\sin{B} = \frac{b*\sin{A}}a$
$B = \sin^{-1}{\frac{b*\sin{A}}a}$

When you take the inverse of the sine function make sure you take all of the solutions in the interval $[0,\pi]$ radians.  This way you can get two solutions to the triangle if it is not a right triangle.

[quote="pkothari13"]i think it's trying to say two sides and an INCLUDED angle[/quote]

That would be SAS, if I understand what you are saying correctly, which implies congruency in two triangles. I think he wants to know why SSA does not imply congruency.

oh oops. you only use SAS and ASA and stuff when youre trying to prove congruency.  in your case, youre not trying to say its congruent to anything.

when you have two sides and then an angle, you can use the Law of Sines to solve for the triangle.  when you have two sides and an included angle, you can use the Law of Cosines.

SSA is a not possible congruency because......... well I cant really explain to you why as I do not have the program to draw diagrams but I know why.

Consider this picture:

[img]http://img.photobucket.com/albums/v213/joml88/SSA.png[/img]

We have $DF=AC$, $DE=AB$ and $m\angle EFD=m\angle BCA$ and yet the two triangles aren't congruent.  Thus you can't use SSA to prove congruency.

I think I know one exception. If $AB=DE$ $AC=DF$ and $\angle ABC=\angle DEF$ but $\angle ABC$ and $\angle DEF$ are obtuse angles, SSA is OK. Can you proof?

The SSS, SAS, ASA, SsA "theorems" can be proved from each other, but I think at least one has to be postulated. Here is the proof of SsA using SAS:

[color=blue][b]SsA:[/b] If 2 triangles have congruent 2 sides and the angle against the larger one of them, then they are congruent.[/color]

Let the triangles $\triangle ABC, \triangle A'B'C'$ have the sides AB = A'B', BC = B'C' and the angles $\angle CAB = \angle C'A'B'$ congruent and let BC > AB. Assume that these 2 triangles are not congruent, i.e., $CA \neq C'A'$, for example, CA > C'A'. Then there exists a point D on the ray AC, different from the point C, such that DA = C'A'. Since the sides DA = C'A', AB = A'B' and the angles $\angle DAB \equiv \angle CAB = \angle C'A'B'$ are congruent, the triangles $\triangle ABD \cong \triangle A'B'C'$ are congruent by SAS. Therefore, their remaining sides BD = B'C' are also congruent. Since BD = B'C' and BC = B'C', it follows that BD = BC. This means that the triangle $\triangle BCD$ is isosceles and the angles $\angle BCD = \angle BDC < 90^o$ are congruent. Thus the angle $\angle BDA$, complementary to the angle $\angle BCD$ is greater than 90o, i.e., obtuse. This means that the angle $\angle BDA$ is the largest angle of the triangle $\triangle ABD$, which in turn implies, that the side AB against this angle is the largest side of this triangle, AB > BD = BC. This is a contradiction with the assumption AB < BC. Therefore, the isosceles triangle $\triangle BCD$ cannot exist, which means that the points $C \equiv D$ are identical, $CA \equiv CD = C'A'$ and the triangles $\triangle ABC \cong \triangle A'B'C'$ are congruent. The possibility of these 2 triangle not being congruent when CA < C'A' is disproved similarly.

It is also clear, why the assumption BC > AB is necessary in SsA. If BC < AB, nothing prevents the existence of the isosceles triangle $\triangle BCD$ and it actually does exist.


induction
number theory proposed
number theory

$n$ is a positive integer and $d_{1}(n)$ is the number of divisors of $n$ that are in the form $4k+1$ and $d_{3}(n)$ is the number of divisors of $n$ that are in the form $4k+3$. Prove that $d_{1}(n)\geq d_{3}(n)$. :ninja:

Let $n=2^{k}m$, were m is odd. If $m=3(mod 4)$, then $d_{1}(n)=d_{3}(n)$ because if k is divisor m form 1(mod 4), then m/k is divisor form 3(mod 4). If m=1(mod 4) and all prime divisors had form 1(mod 4), then $d_{1}(n)>d_{3}(n)=0$. Else exist prime divisor p=3(mod 4), suth that $ord_{p}(m)=2k.$ It give $d_{1}(n)=(k+1)N+kM, d_{3}(N)=kN+(k+1)M$, were N is number of divisors k  form 1(mod 4), suth that $p\not |k$, M is number of divisors k form 3(mod 4), suth that $p\not |k$. By induction $M\le N$. It give $d_{1}(n)\ge d_{3}(n).$

Let f(n)=d1(n)-d3(n).
Then we can show that if (m,n)=1 -> f(mn)=f(m)f(n)
and f(p^e) >= 0
qed

Similary, we can also prove that:
$d_{i}(n)$is the number of divisors of n that are in the form 8k+i.
Prove that  $Max(d_{1}(n), d_{3}(n), d_{5}(n), d_{7}(n) )=d_{1}(n)$
It is harder than d1(n)>=d3(n).

Take a look at http://www.mathlinks.ro/Forum/viewtopic.php?t=113024 .


a triangle has verticices (2,2),(5,6), and (6,2). What is the number of units in length of the longest side of the triangle

please explain step by step
thanks

[hide]
The vertices are (2,2), (5,6), and (6,2).
The distance from (2,2) to (5,6) is $\sqrt{(6-2)^2+(5-2)^2}=\sqrt{3^2+4^2}=5$ by the distance formula.
Similarly, the distance from (5,6) to (6,2) is $\sqrt{(2-6)^2+(6-5)^2}=\sqrt{4^2+1^2}=\sqrt{17}$.
The distance from (2,2) to (6,2) is obviously $4$.

Since $\sqrt{17}$ is a little bit more than 4, the longest side is $\boxed{5}$.
[/hide]

I am so dumb  cause i plugged it wrong into the equation  thank you so much!!!!! :)  :D

[color=blue][size=75]Kyyuanmathcount: Dont say that![/size][/color]


You know how they say after you take a test you should relax. Well i can't. After a test all I can do is think about it. I can't stop myself no matter how hard I try. The worst part is I remember all teh questions on teh test and then do it in my head again. and then I'm like oh no I did this wrong. or I did this stupid mistake. I then panick(sometimes though after I get teh test back i find out that I didn;'t do that problem wrong). Any advice or anything for this problem of mine.

[quote="(^_^)"]well, think of it this way

what is the worst thing that can ever happen?

you die

not so bad now is it?

...

and remember that watever happens, you die anyways

so chill[/quote]

There you have it.

There's nothing wrong with thinking about a test right after the test. Maybe it'll be easier to get it out and over if you talk to some other people about it.

After a point, you have to realize and accept that you don't know what's going to happen to the test, and convince yourself that it'll be most effective for you to wait until you get your test back to think about it more.

Try writing down your comments or questions to yourself, the grader, or the instructor after the test, if any, as well as any questions. Then, put it in a folder and put it away until you get the test back. Then, take it out and look at it again.

Try playing music. Or working out. Or eating.

I know EXACTLY how you feel, as I'm feeling that way right now. Whenever in anxious about a test result I always think in my head how stupid that teacher was for not teaching that lesson clearly enough. If that doesn't work because I know it was my fault, I'll just tell myself that the results are weeks away and to just chill and take my mind off it playing a game.


geometry
radius
ratio
incircle
IMO 1970
IMO

$M$ is any point on the side $AB$ of the triangle $ABC$. $r,r_1,r_2$ are the radii of the circles inscribed in $ABC,AMC,BMC$. $q$ is the radius of the circle on the opposite side of $AB$ to $C$, touching the three sides of $AB$ and the extensions of $CA$ and $CB$. Similarly, $q_1$ and $q_2$. Prove that $r_1r_2q=rq_1q_2$.

We have to show that $\frac r q = \frac{r_1}{q_1} \cdot \frac{r_2}{q_2}$. Let $s, s_1, s_2$ be the semiperimeters, $\triangle, \triangle_1, \triangle_2$ the areas and h the common C-altitude of the triangles $\triangle ABC, \triangle AMC, \triangle BMC$. The circle pairs $(r, q), (r_1, q_1), (r_2, q_2)$ are centrally similar with the common external homothety center C, the intersection of their common external tangents. Their pairwise homothety coefficients are equal to the distances of their tangency points on from the vertex C:

$\frac r q = \frac{s - AB}{s},\ \ \frac{r_1}{q_1} = \frac{s_1 - AM}{s_1},\ \ \frac{r_2}{q_2} = \frac{s_2 - BM}{s_2}$

and we have to show that

$(?)\ \ \ \frac{s - AB}{s} = \frac{s_1 - AM}{s_1} \cdot \frac{s_2 - BM}{s_2}$

$(?)\ \ \ 1 - \frac{AB}{s} = \left(1 - \frac{AM}{s_1}\right) \left(1 - \frac{BM}{s_2}\right)$

$(?)\ \ \ \frac{AB}{s} = \frac{AM}{s_1} + \frac{BM}{s_2} - \frac{AM}{s_1} \cdot \frac{BM}{s_2}$

Substituting $AB = \frac{2 \triangle}{h},\ AM = \frac{2 \triangle_1}{h},\ BM = \frac{2 \triangle_2}{h}$ and $s = \frac{\triangle}{r},\ s_1 = \frac{\triangle_1}{r_1},\ s_2 = \frac{\triangle_2}{r_2}$, this becomes

$(?)\ \ \ \frac{2r}{h} = \frac{2r_1}{h} + \frac{2r_2}{h} - \frac{4r_1 r_2}{h^2}$

Reducing by $\frac{2h}{r_1 r_2}$ and rearranging,

$(?)\ \ \ \frac{2}{h} = \frac{1}{r_1} + \frac{1}{r_2} - \frac{r}{r_1 r_2}$

This relation has been proved in the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=19468]Inradii problem[/url] and it keeps popping up - I already used it to solve the easy problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=60409]Congruent inscribed circles[/url] and not so easy problems [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=39268]A very difficult one[/url] and [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=18580]Finding ratio[/url].

Yetti

[quote="armpist"]It is maybe not so easy to see that all 6 radii in the expression q*r1*r2 = r*q1*q2

  are segments of the sides of  a triangle in which  concurrent cevians are

  approprietly cutting the  sides into. Ceva theorem works just fine in this

  particular case.  Some may have preferrence for Menelaos theorem. It will

  produce the desired result too.

  In the case of Ceva, any idea what the point of concurrency may be?

  Thank you.

  M.T.[/quote]

It is quite easy to see that a triangle does exist, the sides of which can be cut in any ratios whatsoever. On the other hand, it is also quite easy to find a triangle and a point M on one of its sides, such as q + r > q > q1 + r1 + q2 + r2, which means that triangles with the sides q + r, q1 + r1, q2 + r2 or q, q1, q2 cannot exit. So, it has to be a triangle with the sides r, r1, r2, which trivially exists, because r1 + r2 = r + 2*r1*r2/h > r. But the thingy is this: To use Ceva's theorem to deduce q*r1*r2 = r*q1*q2, you have to know the concurrency point to begin with, don't you?

For a change, here are my questions, instead of solution:

   Any way of easy construction of the triangle for a proof without words?

   How does this radii triangle gets transformed as point M moves along side AB?

   At some points the break-up of the sides changes: if a side was (q + q1)  then

   it changes to (q + q2) (and appropriate change for other sides)

   What is the path of the cevian concurrency point?  When does it change it's

   position from inside to the outside of this triangle with sides divided in proportions

   of  6 radii?  Any connection with Ph-An's favorite problem of equal length

   cevians intersecting in circumscribed quadrilateral (that you Yetti solved recently)?

   You must have noticed that the 4 centers  r1,r2,q1,q2  are concyclic when

   ABC is isosceles ! That's what we should have looked at then.

   Any 'after the fact' beneficial use of the present identity in the old problem?

   Any connection to axial inversion?   The identity  r / q = r1 / q1 * r2 / q2

  r / q - constant    and r1 /q1 * r2 / q2  look very much like  tan(X) * tan(Y).

   One more thing I forgot to tell you:  the configuration in Inradii problem works for a

   simple reason - if you cross-connect ex-centers with their ex-circles tangency points

   they will intersect on  altitude right in the middle of it,  and not because of

   complicated multiplication and/or  division. There are much stronger forces,

   stronger than arithmetic operators,  something like projectivity (or collinear UFOs) !

   Thank you.

    M.T.

[quote="armpist"]
   Any 'after the fact' beneficial use of the present identity in the old problem?

   Any connection to axial inversion?   The identity  r / q = r1 / q1 * r2 / q2

  r / q - constant    and r1 /q1 * r2 / q2  look very much like  tan(X) * tan(Y)[/quote]
As for the 1st question, I already gave 3 examples. How can you pose a question that has been answered shortly before you even asked it? As for the following expressions, you are welcome to use some multiplications, divisions, etc.

[quote="armpist"]
   One more thing I forgot to tell you:  the configuration in Inradii problem works for a

   simple reason - if you cross-connect ex-centers with their ex-circles tangency points

   they will intersect on  altitude right in the middle of it,  and not because of

   complicated multiplication and/or  division. There are much stronger forces,

   stronger than arithmetic operators,  something like projectivity (or collinear UFOs) !

   Thank you.

    M.T.[/quote]
That's interesting. Can you prove it? It could well be called Armpist's paradox, because it is patently false (just draw the figure). What makes simple and even not so simple mathematical apparatus useful is the fact that it is rigged by axioms to reflect reality. If the rigging is right, the "forces" are contained in it.

[quote="armpist"]Take a look at this seemingly related topic

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

     in particular  H. Stengel  contribution.

   Thank you.

   M.T.[/quote]
Just leave me alone, will you? I really do not want to talk to you, but it requires some effort on your side as well.


MIT
college
Stanford
Harvard
Princeton
University of Chicago
Columbia

I know that there are mostly math people here but I would be grateful if anyone can suggest universities (other than Caltech, MIT etc.) with a good, challenging undergrad physics programme.
I am an Indian student. My SAT I scores are math:800, CR:700, writ:570, SAT II scores yet to come and not too good school grades.

i don't think us news ranks undergrad programs, but these are some grad school rankings from a magazine i have from 2001:
tied for #1: caltech, stanford
tied for #3: harvard, mit, princeton, uc berkeley
tied for #7: cornell, uchicago
#9: u of illinois-urbana-champaign
tied for #10: columbia, uc santa barbara, yale

anyways, notice these:
3. uc berkeley
10. uc santa barbara
14. uc los angeles
18. uc san diego

these ranking probably didn't change much. also, uc irvine is on the rise and they have a great honors program with lots of undergrad research, especially in physics (i know this cause i live in irvine).

anyways, those uc's might be some schools that are not as hard to get into as "caltech, mit, etc." but are still ranked quite well. that's not to say that you should not apply to a few of those schools, you never know. also, if you do, keep in mind that harvard, princeton, and yale are financially much better to international students than other schools like stanford, mit, and caltech (in other words, you won't be rejected cause you can't pay).

uc santa barbara is probably easier to get into than berkeley, los angeles, or san diego, but still has a very good physics program.

of the privates up there in the top 10, cornell is probably the easiest to get into...

if you think you might like engineering, also consider:
berkeley (a repeat from above list, just because it is probably better than all ivy leagues at engineering)
carnegie mellon university
rose-hulman inst of tech
harvey mudd college
cooper union (free tuition)
olin school of engineering (free tuition)

[quote="shbmsc90"]
if you do, keep in mind that harvard, princeton, and yale are financially much better to international students than other schools like stanford, mit, and caltech (in other words, you won't be rejected cause you can't pay). [/quote]

Actually, MIT is need blind.  So your financial status isn't taken into account at all when you're applying.

True, but MIT has a quota - I know that for a fact since I was an international student applying there and they warned that being international puts you at  a distinct disadvantage. The admission rate for internationals is about 4% as opposed to around 10% overall.

But I agree that MIT is probably better (more fair) than Stanford on this issue, because at Stanford you have an advantage if you have money, while at MIT it's the same rules for all international students, though it's still really hard to get in.

Harvard, Princeton, and Yale do not have quotas because they are wealthy enough to pay for anyone who needs finaid. Being international does not disadvanage you at these three schools in any way except if you really can't speak any English.

[quote]But I agree that MIT is probably better (more fair) than Stanford on this issue, because at Stanford you have an advantage if you have money,[/quote]

What are you talking about? Is this advantage for internationals? As far as I know Stanford has actually pledged to begin financial aid for internationals, which is an innovation among peer institutions in itself.

For your info, please read the official policies from the respective institutions' websites:

Harvard:
http://www.admissions.college.harvard.edu/prospective/international/financial_aid/index.html

Princeton:
http://www.princeton.edu/main/prospective/affordability/

Yale:
http://www.yale.edu/admit/freshmen/financial_aid/international.html

Stanford:
http://www.stanford.edu/dept/uga/applying/1_4b_financialaid.html
Also,
http://www.stanford.edu/dept/finaid/pdf/07FAOhdbook.pdf
(in particlar, pay attention to page 2, paragraph 3 which hints at Stanford's policy for intrntnls)

MIT:
http://www.mitadmissions.org/topics/apply/international_applicants_helpful_tips/index.shtml
(look at paragraph 3)

Also, this is an interesting Stanford article:
http://daily.stanford.org/article/2007/3/14/internationalStudentAidLacks
(especially pay attention to end of paragraph 9)
A quote:
“In terms of being need-blind for international students, we’re still a long way off,” Hennessy said. “When I set that goal in 2005, we were on a 10-year horizon. I was hoping we could finish it sooner than that, but that’s going to depend a lot on fundraising.”

Now, I'm not saying MIT is the best, but 100 vs. 35, come on?

Thanks for your replies. I should have mentioned that financial aid is very important hence need blind/ need based does not make much difference. I know that UCLA does not but do UC Irvine, San Diego offer financial aid?
Also are Ivy Leagues such as Columbia, UPenn, Dartmouth that good? That tough to get in?

Sorry that we got off track a bit...

Anyways, the UC's will give some finaid but not nearly as good as the privates. The UC's are public schools and lots of their finaid money comes from the federal gov't. Lots of the federal grants are for US citizens or permanent residents only, so the finaid for internationals is very limited.

Brown, Columbia, Dartmouth, Duke, and UPenn are also great schools each of which is unique in its own way. They also offer finaid like Stanford.


Sorry if this has already been asked, but since i couldn't find anything after searching, how do you post images?  I tried adding an attachment but it didn't work...

[quote="probability1.01"] I tried adding an attachment but it didn't work...[/quote]

Well, thats how to do it. Make sure that the file is small enough in memory and in dimention, and make sure that it is in the right format.

JPG, GIF or BMP. BMP is strongly advised against, because it takes too much space. You should be able to convert BMP to GIF (preferred for Geometer's Sketchpad graphics) or JPG (otherwise) in MSPaint.


[quote="valeriummaximum"]Problem no 29

@ shruthi-tis is ur first relevant post...

I think the answer for ur question is A is CuCl2[/quote]

Paethal. CuCl2 shows Chocolate brown ppt with Ferrocyanide. Ranga don't answer blindly.

Thanks for letting me know that....But then you are wrong, on both counts..

Firstly, copper reacts with potassium ferrocyanide

Secondly, I had posted a problem with WOP before

@Shre: Not it aint ferrous ion

@Rohit: Elementary my dear BabyK :) Thats correct.. But then identify A first da :D

Edit: @Ranga: Close, but not correct.....

Maybe its Bi salt bcos of pearl white  :maybe:  But i'll give it a try later  :D  I'm doing pardesi's  :D

[quote="valeriummaximum"]@shreyas don blabber ferrous gives prussian blue with K4FeCN6

THe answer is--a is Bi2Cl3[/quote]
excuse this naive person but could anyone here tell where that relevant post is about the above discussion you are all doing ie in a nutshell what is the question???? :ninja:  :D

dont blabber its true for Bi,Sb,Sn  :D  :D  Fiitjee ques  :P

@Srinitrs see the iitjee thread ques no 29

Then the only other possibility is HgCl2

[quote="shruthibaskaran"]@Ranga: Aiyo, you were close.. But now I fear you've gone off tangent
@Shre: Ada pavi, its not anitmony :p

@Rohit: Weeell, it is a FIITJEE Q.. Bang on target, does this mean, BabyK already knows? :D

@Srini: Refer IIT JEE Thread da[/quote]

I knew that the white turbidity is a fiitjee ques  :D  (It was in our FT you know!). I think A is $ \ce{BiCl3}$
B is $ \ce{Bi2S3}$  C is $ \ce{H2S}$ D is $ \ce{Bi(NO3)3}$ E is $ \ce{Bi(OH)3}$ white turbidity is $ \ce{BiOCl}$

I am positively aure the answer is Bi Cl3 and not BiCl2

@madness- enda loose a da ne??
                 how is +2 state of Bi stable???

sorry  was thinking abt the math sum. You people dunno how many pm's ive been handling and how many sums ive been doing . The above post was a result of overwork :blush:

I was mad in answer inf=g it as Bi2Cl3
cos Cl never has -2 state?????? :blush:  :blush:  :blush:

@Rohit: Bang on target man! :D

@Ranga: Lol

Actually congratulate ranga for the start  :D   :)

PS: [size=59]otw he might cry[/size]  :P

E is not actually Bi(OH)3, it is a white  basic salt of variable composition. it is a mixture of Bi(OH)3 and Bi(OH)2NO3. :D  :D


Έστω {$ a_{1},a_{2},...,a_{50},b_{1},b_{2},...,b_{50}$} ένα σύνολο $ 1000$ πραγματικών αριθμών. Υποθέτουμε ότι η εξίσωση $ \sum^{50}_{k \equal{} 1}\left|x \minus{} a_{k}\right| \equal{} \sum^{50}_{k \equal{} 1}\left|x \minus{} b_{k}\right|$ έχει $ N$ λύσεις, όπου $ N$ διάφορου του απείρου. Βρείτε τη μέγιστη τιμή του $ N$.


A quadrilateral that has consecutive sides of lengths 70,90,130 and 110 is inscribed in a circle and also has a circle inscribed in it. The point of tangency of the incribed circle to the side of length 130 divides that side into segments of length $x$ and $y$. Find $|x-y|$.

$\text{(A)} \ 12 \qquad \text{(B)} \ 13 \qquad \text{(C)} \ 14 \qquad \text{(D)} \ 15 \qquad \text{(E)} \ 16$

cyclic quad. have alternate supplemetary angles.  Draw the radius of the inscribed angle to the tangent points, we see two pairs of similar right triangles. using ratio we get:

\begin{eqnarray*} \frac{r}{x} &=& \frac{x-40}{r}\\ \frac{r}{x-110} &=& \frac{x-130}{r}\\ \end{eqnarray*}
cross-multiplying and equating, we get $x^2 - 40x = -240x + 14300$, solving for $x$ get $x = \frac{143}{2}$.  calculating $|130 - 143|$ we find the answer is $13$. QED.


calculus
integration
trigonometry
calculus computations

$ \int_{\frac{1}{n}}^{\pi \minus{}\frac{1}{n}} \frac{1}{\sin ^{\frac{1}{3}}(x)} \, dx\leq \int_0^{\pi }
   \frac{1}{\sin ^{\frac{1}{3}}(x)} \, dx\equal{}\int_0^{\frac{\pi }{2}} \frac{2}{\sin ^{\frac{1}{3}}(x)} \,
   dx\equal{}\int_0^1 \frac{2}{\left(1\minus{}\theta ^2\right)^{2/3}} \, d\theta$

$ \int_0^1 \frac{2}{\left(1\minus{}\theta ^2\right)^{2/3}} \, d\theta \leq \int_0^1 \frac{\pi }{\sqrt[3]{1\minus{}\theta   }} \, d\theta \equal{}\frac{3 \pi }{2}$

I don`t know a good way to prove the last integral. My way is too confusing even by myself. I Will try to find a better integral latter.

I`m very beginner at integral  :(   so I integrate like this

$\int_{0}^{1}\frac{\pi }{\sqrt[3]{1-\theta }}\, d\theta  = \pi \int_{0}^{1}\frac{1 }{\sqrt[3]{1-\theta }}\, d\theta$

Now assume $\int \frac{1 }{\sqrt[3]{1-\theta }}\, d\theta = a(1- \theta)^b +C$
then we need $b= \frac{-1}{3} +1$ and $ab=-1$ so

$\int \frac{1 }{\sqrt[3]{1-\theta }}\, d\theta = \frac{-3}{2} (1- \theta)^{\frac{2}{3}}$

Thus $\int_{0}^{1}\frac{\pi }{\sqrt[3]{1-\theta }}\, d\theta= \pi \int_{0}^{1}\frac{1 }{\sqrt[3]{1-\theta }}\, d\theta = \pi ( 0 - \frac{-3}{2})= \frac{3 \pi}{2}$

Is it true ?


limit
integration
calculus
calculus computations

you can use the Cesaro-Stolz's lemma !

[b][u]C-S lemma :[/u][/b] $\left\{\begin{array}{c}u_{n}\rightarrow\infty\ ,\ v_{n}\nearrow\infty\\\\ \frac{u_{n+1}-u_{n}}{v_{n+1}-v_{n}}\rightarrow L\in\overline{\mathcal R}\end{array}\right\|$ $\implies$ $\frac{u_{n}}{v_{n}}\rightarrow L$.


Just out of curiousity...
I'm NOT racist, so relax..  :P
And I'm Asian here.(South Korea)

[quote="Nerd_of_the_Ages"]who ISNT asian?

anyway, i'm asian[/quote]

Then who voted Austrailian?

 :?:

[quote="davidyko"]Polls aren't accurate. [/quote]
Yeah, anybody here remember the poll that showed that maybe like 20 people lived in Antarctica? lol

I tried to edit it, but it won't allow me..

[quote="DonkeyKong"][quote="davidyko"]Polls aren't accurate. Like 20 people vote that they got 15 on the AIME every year.

Asian.[/quote]

Huh? Then how can I make this poll more accurate?
Any suggestions?

[quote="K81o7"]"Caucasian (including people born here, but different race)"

Okay...my parents are Indian, but I was born here. I would say Asian, but why have you defined Caucasian as above? Does that make me Caucasian?[/quote]
Well, I thought of Caucasian as mixed race, but if you are born here, you are officially a citizen of United States of America. So I thought I should put that in with it..
If you don't want it, I'll take it off.[/quote]
I thought of Caucasian as a synonym for European.

Anyways, I'm Asian (Chinese)

I'm Asian but born in America  :D So it's like the second choice or something.

is there any other!?!?!

my mom is  japanese, my dad is 90% brazilian, 8% armenian, 1% italian, and 1% austrian

so I am 50% Japanese, 45% Brazilian, 4% Armenian, 1/2 % Italian, and 1/2 % Austrian

Asian here too. This was a bigger surprise than the age one...

Asian
parents indian
I was born here in teh US though


AMC
AIME
calculus
number theory
greatest common divisor
quadratics
algebra

Ok heres the deal. I became enamored with math mainly due to the elegance and rigor (The usually opposite nature of these statements explain in entirety why its math i enjoy the most compared to something like chemistry).

I want to find more of it. Lets start an elegance chain on this post, similar to those begun on the aime in the past. The rules are as follows:

1. The problem must have an elegant solution, one that is reasonably short and doesnt use any advanced math (at least what is considered this such as linear algebra and calculus)
2. Post the problem, and then the problem stays until an elegant solution is found or it is agreed upon that the solution (the original poster's) should be presented. Then the solver posts a new problem.

Heres the first problem:
In a finite sequence of real numbers, the sum of any seven consecutive terms is negative, and the sum of any eleven consec terms is positive. Determine the maximum number of terms in the sequence.

we show that 16 is the maximum number of terms in the sequence:

suppose the sequence has at least 17 terms, and consider the following table:

[tex]a_1 + a_2 + ... + a_7 < 0 [/tex]
[tex]a_2 + a_3 + ... + a_8 < 0 [/tex]
[tex]...................................[/tex]
[tex]a_{11}+a_{12}+...+a_{17} < 0 [/tex]

summing the rows we get a negative number, while summing the columns gives us a positive number, contradiction.

Therefore the sequence can contain at most 16 terms, and with a little effort one can show that the following sequence with 16 terms satisfies the problem:

5, 5, -13, 5, 5, 5, -13, 5, 5, -13, 5, 5, 5, -13, 5, 5

[edited to add problem]

find all pairs of nonnegative integers (x,y) so that
[tex]x^3+8x^2-6x+8=y^3[/tex]

Yep, thats the elegant solution I was thinking of.... Ill get to work on your problem after work.

For [tex]x^3+8x^2-6x+8=y^3[/tex] we can subtract 8 from both sides and factor to get [tex]x(x^2+8x-6)=(y-2)(y^2+2y+4)[/tex].  Now we can set [tex]x=y-2[/tex] which forces [tex]x^2+8x-6=y^2+2y+4[/tex].  After substituting, [tex](y-2)^2+8(y-2)-6=y^2+2y+4[/tex] which yields [tex]y^2+4y-18=y^2+2y+4[/tex] which gives [tex]y=11[/tex] and thus [tex]x=11-2=9[/tex]

Now for my problem:

find one pair of positive integers [tex](x,y)[/tex] (whose gcd=1) such that y :le: 10000 and the fraction [tex]\frac{x}{y}[/tex] satisfies

[tex]\displastyle{\frac{2003}{2004}<\frac{x}{y}<\frac{2004}{2005}}[/tex]

You only found one solution!  You have to find (and prove that you've found) all pairs!

A try for the two problems :

x^3+ 8x-6x+8 = (x+2)^3 + 2x-18x = (x+3)^3 - (x+33x+19)

So we must have (but it's not enough) :
either
2x-18x <= 0 and (x+33x+19) >= 0 if x+2>= 0
or
2x-18x >= 0 and (x+33x+19) <= 0 if x+3<=0

This quadratic inegs are easy to solve. I find that  0<=x<= 9.
"Manual" check showed that (0, 2) and (9,11) are the only solutions.

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

For the second one we must have x < y or x = y-z with z >= 1.
The inegs become 2004z < y < 2005z. y <= 10000 implies z < 5 i.e. z=1,2,3,4 for which there are 0,1,2,3 values for y.

So there are 5 (2 numbers are equal) solutions to these inegs (4007/4009, 6010/6013, 6011/6014, 8013/8017, 4007/4009, 8015/8019).

A nice - a bit difficult but since we are in advanced - problem with an [b]amazing[/b] solution (not mine : I had been unable to solve it :-( ) :

Let a1, .., an and b1,.., bn two different groups of positive integers such that the numbers ai+aj (1<=i<j<=n) are the same as the numbers bi+bj (1<=i<j<=n).

Show that n is a power of 2.

Ok, I probably misunderstood something, but:

[tex]a_1+a_2 = b_1+b_2[/tex]
[tex]a_2+a_3 = b_2+b_3[/tex]
Subtracting the two equations
[tex]a_1-a_3 = b_1 - b_3[/tex]
Adding the equation [tex]a_1+a_3 = b_1 + b_3[/tex] to both sides,
[tex]2a_1 = 2b_1[/tex]
[tex]a_1 = b_1[/tex]
Contradiction

It doesn't say that they are the same numbers [i]in order[/i].  Just that every number represented in one such set is also represented in the other.

I'm not quite sure what you mean.  I had thought it followed that [tex]a_i = b_i[/tex], which seems to me like the same thing as saying the sequences are identical.  Could you please clarify?  Thanks.

If I'm interpreting the problem correctly, it means we are looking at sequences {a}  and {b} of n elements such that for every h,i there exist j,k such that a[size=75]h[/size] + a[size=75]i[/size] = b[size=75]j[/size] + b[size=75]k[/size]
For example, {-1,1,1,2} and {0,1,1,2} both have {0,1,2,3} as the possible results of pairwise sums. If this is the problem, you'd also have to specify that the set of unique elements in each sequence is different, for example {0,1,1} would not qualify as a different sequence than {0,0,1}

Right -- we're saying "{x | there exists i, j, x = a_i + a_j} = {y | there exists h, k, y = b_h + b_k}," [b][i]not[/i][/b] "for all i, j, a_i + a_j = b_i + b_j."

Hint :

[hide]Let :

f(x) =  :Sigma: x^ai

g(x) =  :Sigma: x^bi

Consider f(x)^2 - f(x^2) and the same for g[/hide].

I've seen this problem with slightly different statement:
Take n numbers a_1,...,a_n and write n(n-1)/2 numbers A_ij=a_i+a_j. For which n can we restore numbers a_i given numbers A_ij (multiset of them, i.e. set that can contain multiple copies of each element -only their order is unknown).So in JBL's definition {x | there exists i, j, x = a_i + a_j} and {y | there exists h, k, y = b_h + b_k} should be multisets (as well as "groups" a_i and b_i in original statement).
The solution I've seen is valid not only for positive integers, but for any reals. It used some advanced techniques like symmetric polynomials and isn't very elegant, so I won't post it here.

[quote="zabelman"]wait a second... In this problem ARE we dealing with multisets?[/quote]
I think we SHOULD deal with them. Else the statement won't hold: consider {0,0,1,1,1} and {0,0,0,1,1}, sets of pairwise sums are {0,1,2} for both, but multisets are different.


Let $a, b, c, d$ be real numbers such that
$a = \sqrt{4 - \sqrt{5 - a}}$, $b = \sqrt{4 + \sqrt{5 - b}}$,
$c = \sqrt{4 - \sqrt{5 + c}}$, $d = \sqrt{4 + \sqrt{5 + d}}$.
Calculate $abcd$

It is not very tricky indeed  :lol:
$a$ and $b$ are the solution of the equation $x^4-8x^2+x+11=0$ ; and $c$ and $d$ are the solution of the equation $y^4-8y^2-y+11=0$. which means that $a,b,-c,-d$ are the roots of the polynominal $p(x)=x^4-8x^2+x+11$.
By viet relations we deduce that $abcd=11$  :lol:


In case you've noticed me asking a lot of questions, a friend just sent me an old practice test for a contest and I've been trying to do it, pity the explanations for the answers are terrible.

Let $ A \equal{} ABCDE$, find all prime factors of $ A$.

By itself, it's nonsense.  Is this question part of a series of related questions?

I don't know, maybe I'm missing a page :(


please help me to  solve this problem
For a,b,c >0 and x,y,z>0 :
 such that x+y+z=1
(a+b/x)^4+(a+b/y)^4+(a+b/z)^4 :ge: 3(a+3b)^4
thanks

Dumb typo, but the answer follows from Power-Mean and AM-HM.

[quote="blahblahblah"]I think that there might be a typo, we should have instead that:

(a+b/x)^4 + (a+b/y)^4 + (a+b/z)^4 >= 3(3a + b)^4

Anyways, by Power-Mean, we have that:

LHS/3 >= (a+a+a+b(1/(x+y+z))^4 = (3a + b)^4

Thus, LHS >= 3(3a+b)^4 as desired.[/quote]

Hi, I think you are wrong.
I think you remember power mean inequality wrongly.

L.H.S./3 >= ((a+a+a+b(1/x+1/y+1/z))/3)^4
>= ((3a+9b)/3)^4 (As 1/x+1/y+1/z >= 9/(x+y+z) = 9)
=(a+3b)^4
So L.H.S. >= 3(a+3b)^4


trigonometry
geometry
trig identities
Law of Sines

Let be $ ABC$ a triangle . Prove that :

$ (b\plus{}c)\sin \ \frac{A}{2}\equal{}a\sin \ \left(\frac{A}{2}\plus{}C\right)$

[hide="Solution"]We start with the identities $ cos(A) \equal{} 2cos^2(\frac {A}{2}) \minus{} 1$ and $ sin(A) \equal{} 2sin(\frac {A}{2})$. From these, we can derive that $ cos(A)sin(\frac {A}{2}) \plus{} sin(\frac {A}{2}) \equal{} sin(A)cos(\frac {A}{2})$. Multiplying everything by $ sin(C)$ results in $ sin(C)cos(A)sin(\frac {A}{2}) \plus{} sin(\frac {A}{2}) \equal{} sin(C)sin(A)cos(\frac {A}{2})$. Adding $ sin(A)sin(\frac {A}{2})cos(C)$ makes $ \nobreak{sin(A)sin(\frac {A}{2})cos(C) \plus{} sin(C)cos(A)sin(\frac {A}{2}) \plus{} sin(\frac {A}{2}) \equal{} sin(A)sin(\frac {A}{2})cos(C)sin(C)sin(A)cos(\frac {A}{2})}$. On the right hand side, we can factor out $ sin(A)$, leaving $ sin(A)$ multiplied by the expansion of $ sin(\frac {A}{2} \plus{} C)$. On the left hand side, we can factor out $ sin(\frac {A}{2})$, making $ sin(\frac {A}{2})$ multiplied by $ sin(A \plus{} C) \plus{} sin(C)$. However, since ABC is a triangle, $ sin(A \plus{} C) \equal{} sin(B)$. The whole equation has become $ sin(A)sin(\frac {A}{2} \plus{} C) \equal{} sin(\frac {A}{2})(sin(B) \plus{} sin(C))$. Finally, by the law of Sines, we reduce to $ a*sin(\frac {A}{2} \plus{} C) \equal{} sin(\frac {A}{2})(b \plus{} c)$.[/hide]

Take a point $P$ on the ray $BA$ such that $AP = AC = c.$ $\triangle APC$ is isosceles with $\angle APC= \angle ACP=\frac {_1}{^2}\angle A$ and notice that $\angle PCB= \angle C + \frac {_1}{^2}\angle A.$  Using sine law in the triangle $\triangle PBC,$ we get:

$\frac {PB}{BC}=\frac {b + c}{a}=\frac {\sin ( C + \frac {A}{2})}{\sin (\frac {A}{2})} \Longrightarrow (b+c) \cdot \sin \frac {A}{2}=a \cdot \sin \left ( C + \frac {A}{2} \right).$


function
integration
limit
real analysis
real analysis unsolved

Let $ (\Omega, \Sigma, \mu)$ be a finite measure space. Give an example of complex-valued measurable functions on $ \Omega$: $ \phi$ and $ \{\phi_i\}_{i\in N}$ such that the conditions $ \int_{\Omega}|\phi_{i}|^2 d\mu < 1$ and $ \int_{\Omega}|\phi|^2 d\mu < \infty$ hold but $ \int_{\Omega}|\phi_{i} \minus{} \phi|^2 d\mu\nrightarrow 0$?

Thanks!

I bet that you are also assuming that $ \phi_i$ is converging towards $ \phi$ in some sense to be precise ...

Ah, I'm sorry I forgot to add the condition that $ \lim_{i\to\infty}\varphi_i(x) \equal{} \varphi(x)$ for almost every $ x\in\Omega$. :blush:

I am, sorry it was easy question, just I've solved:
Let's $ \Omega \equal{} [0, 1]$ and $ \varphi_j(x) \equal{} \frac {\sqrt {j}}{2}$, if $ x\in [0, \frac {1}{j}]$ and $ 0$ if $ x\in (\frac {1}{j}, 1]$. Then the statement holds.


whoa...he just took over the spot for #1 member. He joined Dec. 31, 1969?...how?

maybe that's removed accounts, because when you click on jobro_gold it says he doesn't exist
I think he changed to jophus

[quote="joml88"][quote="Scrambled"]whoah...
how do people notice these things?[/quote]

Go to the memberlist tab at the top of the screen and you can get a bunch of info like that.[/quote]

i know.  I meant y do people on a whim open the members list?

O ok.  Well I actually noticed it.  Forgot why I was looking at the memberlist though.  Normally I wouldn't notice something like that.  Well to answer your question I guess that people are very observant sometimes.

I noticed it as well. I accidentally clicked on the Memberlist tab, and it seemed a bit unusual, so I investigated a bit and came to the same conclusion as Dark_Shadow.

i'm still curious why that's stored as the first member though.

All fixed.  Little weird computer glitch.


请问大家那有中文的关于几何中 "inversion" 和 “和谐分割” 的材料？
好像一般中国的书都不大讲这方面的东西。

近代欧式几何学 和 几何变换与几何证题。

请问那里可以下载这些书？

http://math.elmo.net.cn/info.asp?rno=214  这是《近代欧式几何学》。

《几何变换和几何证题》这本书你去aoshoo找几何区的版主山雨老师邮购。


ceiling function
floor function
number theory unsolved
number theory

For a given positive integer $ a_0$ define the sequence $ (a_n)$ by

$ a_{i\plus{}1}\equal{}\frac{a_i}{2}$ if $ a_i$ is even, and $ a_{i\plus{}1}\equal{}3a_i\minus{}1$ if $ a_i$ is odd.

Prove that if $ n$ is a natural number such that $ a_n\equal{}a_0$, then $ 2^n>a_0$.

Notice that the closed-form definition of $ a_n$ is
\[ a_n =
\begin{cases}
\left( \frac{3}{2} \right) ^{\lceil \frac{n}{2} \rceil} a_0 - \sum_{i=0}^{\lfloor \frac{n}{2} \rfloor} \left( \frac{3}{2} \right) ^i & \text{if } n \text{ is odd} \\
\frac{a_{n-1}}{2} & \text{if } n \text{ is even} \\
\end{cases}
\]
which can be proven trivially by induction.

Assume the hypothesis: there exists $ n \in \mathbb{N}$ such that $ a_0 = a_n$. Then
\begin{align*}
a_0
&= \begin{cases}
\left( \frac{3}{2} \right) ^{\lceil \frac{n}{2} \rceil} a_0 - \sum_{i=0}^{\lfloor \frac{n}{2} \rfloor} \left( \frac{3}{2} \right) ^i & \text{if } n \text{ is odd} \\
\frac{a_{n-1}}{2} & \text{if } n \text{ is even} \\
\end{cases} \\
&< \left( \frac{3}{2} \right)^{\lceil \frac{n}{2} \rceil} a_0 \\
&\leq \left( \frac{3}{2} \right)^n \\
&< 2^n \\
\end{align*}
QED.


analytic geometry
number theory
relatively prime
number theory proposed

Let s be a positive real.

Consider a two-dimensional Cartesian coordinate system. A [i]lattice point[/i] is defined as a point whose coordinates in this system are both integers. At each lattice point of our coordinate system, there is a lamp.

Initially, only the lamp in the origin of the Cartesian coordinate system is turned on; all other lamps are turned off. Each minute, we additionally turn on every lamp L for which there exists another lamp M such that

- the lamp M is already turned on,
and
- the distance between the lamps L and M equals s.

Prove that each lamp will be turned on after some time ...

[b](a)[/b] ... if s = 13. [This was the problem for class 11.]
[b](b)[/b] ... if s = 2005. [This was the problem for classes 12/13.]
[b](c)[/b] ... if s is an integer of the form $s=p_1p_2...p_k$ if $p_1$, $p_2$, ..., $p_k$ are different primes which are all $\equiv 1\mod 4$. [This is my extension of the problem, generalizing both parts [b](a)[/b] and [b](b)[/b].]
[b](d)[/b] ... if s is an integer whose prime factors are all $\equiv 1\mod 4$. [This is ZetaX's extension of the problem, and it is stronger than [b](c)[/b].]

  Darij

I think this may work for (c):
[hide]
Let $s=p_1p_2...p_k$

To prove that all lamps are eventually turned on, it suffices to show that $(1,0)$ is eventually turned on, and to do that it suffices to show there is a path from $(0,0)$ to $(1,0)$ such that each step has length $s$.

Since each $p_i$ is 1 mod 4, we can write $p_i=x_i^2+y_i^2$.  Repeatedly using the formula $(q^2+r^2)(s^2+t^2)=(qs-rt)^2+(qt+rs)^2$, we can write $s=x^2+y^2$.  Since $s$ is square free, it follows that $(x,y)=(x,s)=(y,s)=1$ (any term dividing two must divide the third...but then its square divides $x^2+y^2$).  We now use the formula once more time to write $s^2=(x^2-y^2)^2+(2xy)^2=c^2+d^2$, where $c=x^2-y^2, d=2xy$  $d$ is relatively prime to $s$ since $2, x, $ and $y$ are.  Therefore $c$ is relatively prime to $s$ as well, so $2cd$ is relatively prime to $s$.  Let $m>0$ and $n>0$ be such that $2cd*m-s*n=1$

Making $d$ $(c,d)$ steps, $c$ $(d,c)$ steps, and  $s$ $(0,-s)$ steps puts us at $(2cd,0)$.   Repeating this process $m$ times at taking $n$ $(-s,0)$ steps puts us at $(1,0)$, and we are done.

This method should work for (d) as well, but the catch is showing that $s$ can still be written as the sum of two relatively prime squares.[/hide]


Represent 651 as a sum of several natural numbers, such that the product of those numbers would be also 651.

[hide]$ 651$ factors as $ 3 \cdot 7 \cdot 31$, so:

$ 3\plus{}7\plus{}31\plus{}1\plus{}1\plus{}1\plus{}1\plus{}1\plus{}...\equal{}651$ (there are $ 610$ $ 1$s)
and
$ 3 \cdot 7 \cdot 31 \cdot 1 \cdot 1 \cdot 1...\equal{}651$[/hide]