geometry
geometric transformation
conics
parabola
search
special factorizations

[quote="Patterns_34"][hide="Try Completing the Square, with Fractions"]
Convert it to the form $ y \equal{} a(x \minus{} b)^2 \plus{} h$, knowing that $ (b,h)$ is the vertex of the parabola. It's a little more difficult since this is not working with integers but with fractions.
An example for the first portion:
[list]$ y \equal{} x^2 \minus{} 3x \plus{} 2$
$ y \plus{} \left(\frac {9}{4}\right) \equal{} \left(x^2 \minus{} 3x \plus{} \frac {9}{4}\right) \plus{} 2$
[hide]$ \boxed{y \equal{} \left(x \minus{} \frac {3}{2}\right)^2 \minus{} \frac {1}{4}}$[/hide][/list]
If still having trouble with the second part, $ y \equal{} x^2 \plus{} x \plus{} 1$ [hide="Another Hint"]
Try adding both sides by $ \frac {1}{4}$ following a similar procedure illustrated above.[/hide]
As for the translation, focus on the vertex of the parabolas.[/hide]
Next time if you're having trouble with homework and want some help, ask for help in [url=http://www.artofproblemsolving.com/Forum/index.php?f=149]High School Basics[/url].[/quote]That was very helpful, I appreciate the help.

Also, I wasn't sure where to post so I did a search on "graphs" and saw another person had posted in here.  Will post in HS Basics next time though, thanks a lot!

[quote="rocomath"]By how much must the parabola $ y \equal{} x^2 \minus{} 3x \plus{} 2$ be translated along the x-axis and along the y-axis so that the parabola $ y \equal{} x^2 \plus{} x \plus{} 1$?

Completing the square gets me nowhere ... I'm not really sure what to do, help :-][/quote]

[hide="solution"]Find the vertex of both equations, which is where x is the average of the roots. Then it is easy from there.[/hide]


[quote="jhaussmann5"]I just wanted to thank the administrators for getting the site back up, and their hard work in general.  I do appreciate the service you provide.[/quote]Thanks, we appreciate that.


Find the sum of the first five terms in the geometric sequence $ \frac13,\frac19,\frac1{27},\dots$. Express your answer as a common fraction.


In a contest, 8 people are in a group with the top 3 advancing.
If everyone is to play each other once, what is the maximum # of losses you can have and still qualify? How many combinations possible are there?

1. 7-0, 6-1, 5-2.
2. 6-1, 6-1, 6-1
3. 6-1, 6-1, 5-2?

Are these correct? Is there anyway to prove it anayltically, instead of empircally.

One problem with the question: what if several people are tied for third place?  (In fact, in an 8 person tournament it's possible to have six people tie for third place, i.e. everyone except the people in first and second place.)

An heuristic method (i.e. one which will help you get the answer, but isn't a proof): if you're borderline qualifying, how do you want the losses to be distributed among the lower-ranked players?  Does it hurt you or help you if someone goes 0-7?

A method that will allow you to actually enumerate the possibilities: consider the total number of wins (there's one in each game).  These must be distributed among the 8 players.  This puts some limitations on what can happen, e.g. the first-ranked player can't have only 2 wins because then no player would have more than 2 wins, but this doesn't result in a large enough number of wins total.  So, analyze what happens for the various possible wins for a given player, subject to the constraint that five other players must do less well.

After you've gone through those possibilities, you should double-check that such a tournament is possible.  (One of the things that could go wrong is including more than one player with a 7-0 record.)


calculus
integration
logarithms
function
parameterization
real analysis
real analysis theorems

the textbook states as follows:

If the integral logarithm is defined as the Cauchy principal value intergral

   $ li x \equal{} \int^x_0 \frac 1{logt} dt \equal{} lim _ {e \to 0}(\int^{1 \minus{} e}_0 \plus{} \int^x_{1 \plus{} e}) \frac 1{logt}dt$ ...., then by L'Hopital's rule, ....

What does li stand for?

And what is the Cauchy principal value integral; darn it I already took real analysis and i don't know whats li

[quote="Sexynsmartjenny"]What does li stand for?[/quote]

Li = logarithm integral, which is a function defined by the integral you posted.

but what is cauchy principal value integral

It is the limit you defined, in hope of calculating the limit near some troublesome points.

I'm not quite sure if one $ \epsilon$ suffices though, I think you need two separate parameters (one from left and one from right). Having one $ \epsilon$ seems to make it sort of uniform.

Which is exactly how the Cauchy Principal Value is supposed to work: http://mathworld.wolfram.com/CauchyPrincipalValue.html

well, i am sure everything i just typed is right because not only they are straight out of a textbook but also it's a very good textbook also.  it's quite difficult for me. and thanks for the cauchy principal value theorem from jbl.  i think i understand it now.


Hi, so I am pretty sure that I made it to AIME, although the official scores are not out (probably made 108 or so on AMC12).

My goal is to score 6-8 on the AIME (last yr, I got a 5, although last-yr's AIME might be slightly easy).

What chapters of AoPS vol.2 are an absolute must?  Which chapters are less important?

Thanks.

Basically, anything thats calculus wont be too important. Tho I don't think graph theory is that useful on the AIME either.


Find all $ A$ such that $ A^2 \equal{} \begin{pmatrix} 4&0\\0&4\end{pmatrix}$.

You want [i]all[/i] of them? There are so many. Here's just one example from the multitude:
\[ A\equal{}\begin{pmatrix}\minus{}678&\minus{}8840\\52&678\end{pmatrix}\]

hello,let $ A\equal{}A\equal{}\begin{pmatrix}a&b\\c&d\end{pmatrix}$ then $ A^2\equal{}A\equal{}\begin{pmatrix}a^2\plus{}bc&ab\plus{}bd\\ac\plus{}cd&bc\plus{}d^2\end{pmatrix}$ and we have the following equation system
$ a^2\plus{}bc\equal{}4$
$ ab\plus{}bd\equal{}0$
$ ac\plus{}cd\equal{}0$
$ bc\plus{}d^2\equal{}4$
By solving this we get
$ a\equal{}2,b\equal{}0,c\equal{}0,d\equal{}2$
$ a\equal{}\minus{}2,b\equal{}0,c\equal{}0,d\equal{}\minus{}2$
$ a\equal{}\minus{}2,b\equal{}b,c\equal{}0,d\equal{}\minus{}2$
$ a\equal{}\minus{}d,b\equal{}\frac{4\minus{}d^2}{c},c\equal{}c,d\equal{}d$
$ a\equal{}\minus{}2,b\equal{}b,c\equal{}0,d\equal{}2$
$ a\equal{}2,b\equal{}b,c\equal{}0,d\equal{}\minus{}2$
Sonnhard.

Another way to characterize that: $ 2I, \minus{}2I,$ and all matrices similar to $ \begin{bmatrix}2&0\\0&\minus{}2\end{bmatrix}.$ Obviously, my ridiculous example was in the latter category.

[quote="Kent Merryfield"]You want [i]all[/i] of them? There are so many. Here's just one example from the multitude:
\[ A \equal{} \begin{pmatrix} \minus{} 678 & \minus{} 8840 \\
52 & 678\end{pmatrix}
\]
[/quote]

How did you get this example (from sketch), and is the number of $ A$ infinite (proof)?  Thanks.

(Sonnhard, have you listed all the solutions, and how did you find them?)

I just found some invertible matrix to conjugate $ \begin{bmatrix}2&0\\0&\minus{}2\end{bmatrix}$ with, and I picked an integer matrix with determinant $ 1$ so its inverse would also be an integer matrix and I wouldn't have to worry about fractions. (But that was merely a convenience - there are non-integer matrices that satisfy your condition.) I'm not sure, because I threw away the scratch work, but the other matrix might have been $ \begin{bmatrix}13&170\\1&13\end{bmatrix}.$

Yes there are infinitely many solutions. Wasn't that already evident from the sheer arbitrariness of my example?

Why ask Sonnhard how he got those? He showed you - it's right in his post. That's a completely brute force approach, requiring no knowledge other than the definition of matrix multiplication.


USAMTS
number theory
prime factorization

Do we need to prove a number is prime in USAMTS or can we just say that after we tested it or put it in a calc program it said it was prime Q.E.D.

err can't you just like.. say it's prime? just be like

"...since 3 is prime..."

i mean like some random number 1452345342953 is prime because i put it in my calculator program (which i have yet to make but i heard of them) and taht numbers a little big but i cant have one that refers to the questions

eh, i think you can just say it's prime. that'd be kinda odd to prove it

I think that's fine. You don't need to go into the details of why a [b]specific[/b] number is prime, or more generally why a [b]specific[/b] number has a given prime factorization. So for instance it's reasonable for a sentence like:

"Note that $ 2011$ is prime."

or

"The prime factorization of $ 2009$ is $ 7^2\cdot41$."

To appear in your final solution.

Relatedly, you don't need to justify basic arithmetic, so you don't have to explain why, for instance $ 123\times 456 \equal{} 59088$.

As a rule of thumb: If you had to do some work to figure something out, explain what you did in order to figure it out.

Also, you are allowed to use computer programs and calculators on these problems.  If you claim a number is prime since you factored it on a TI-89, then you could say: "Factoring (insert prime here) on a TI-89 reveals that it is prime, so..." or something to that effect.  I remember a few years ago there was a problem that I knew I could do with i think around 858 cases (I wasn't quite as good at problem solving then :oops: ).  So I made a Python program to calculate them all, and that proved the question.  In my proof I explained what my program did and what the results implied, and I got full credit for it.

This is both a question and a generalization...

So, to generalize this (not just for primes), if we say that number x has a property that can be referenced and is not specific to the problem (like, x has a prime factorization of -- or x is prime, as compared to inventing a term like "thus, x is bamboozled"), we don't need to cite sources and we can state this property...right?


the sequence {$ x_n$} is defined by $ x_1$=2; $ x_2$=12 and $ x_{n \plus{} 2}$=6$ x_{n \plus{} 1}$ - $ x_{n}$
let p be an odd prime number. let q be a prime divisor of $ x_p$. Prove that if q>3 then $ q\geq2p \minus{} 1$

Do you mean $ x_{n\plus{}2}\equal{}6x_{n\plus{}1}\minus{}x_n$ ?


Prove the following inequality for positive real numbers $ x,y,z$:

$ (xy\plus{}yz\plus{}zx)\left(\frac{1}{(x\plus{}y)^2}\plus{}\frac{1}{(y\plus{}z)^2}\plus{}\frac{1}{(z\plus{}x)^2}\right) \ge \frac{9}{4}$

[quote="Dimitris X"][quote="moldovan"]Prove the following inequality for positive real numbers $ x,y,z$:

$ (xy \plus{} yz \plus{} zx)\left(\frac {1}{(x \plus{} y)^2} \plus{} \frac {1}{(y \plus{} z)^2} \plus{} \frac {1}{(z \plus{} x)^2}\right) \ge \frac {9}{4}$[/quote]
This the well known IRAN 1996 :wink:
Muirhead kills it...[/quote]

Okie.It's old problem ;)..SOS can also kill it :P


The amount of paint needed to paint over a surface is directly proportional to the area of the surface. If 2 quarts of paint are needed to paint a square with a side of 3 ft., how many quarts must be purchased to paint a square whose side is 4 ft. 6 in. long?

[hide=" click to reveal hidden content"]

New area = $ 81/4\equal{}$ old area times $ (9/4)$, so number of quarts $ \equal{} 2$ times ($ 9/4) \equal{} 4.5$

Sorry, can't do Latex at the moment.

[/hide]


[hide="i think this is rite"]
$\frac{90t^2-15}{2t^3-3}$

$\frac{5-15}{2t-3}$

$\frac{5-5}{2t}$

$\frac{0}{2t}$

i'm actually postive that i'm wrong. oh well :?:
[/hide]

no, because you are a 5th grader and you don't know how to factor! :P

[quote="math92"]solve for t:
$\frac{5t-15}{2t^2} \cdot \frac{18t}{t-3}$[/quote]

[hide]$=\frac{3(t-3)}{2t^2}\cdot \frac{2\cdot 9 t}{t-3}$

$=\frac{45}{t}$[/hide]

[quote="math92"]simpify
$\frac{5t-15}{2t^2} \cdot \frac{18t}{t-3}$[/quote]

[hide]$=\frac{3(t-3)}{2t^2}\cdot \frac{2\cdot 9 t}{t-3}$

$=\frac{45}{t}$[/hide]

P.S.  you could hav put all those posts into one.

:huh: Huh? I'm confused! Why is the title of this topic "hi gr8sk8r" and why is the first post "hi" and why is the problem posted twice?  :?

[quote="math92"]well, gr8sk8r was on the same time i was so i wanted to catch his attention.  And the second post says solve for t which is impossible and the third post ways simplify[/quote]

 :roll: why didn't you just edit?

actually, i think he just wanted another post.

i was tired, and i was just not thinking.


Quality of IMO'2007 questions
http://www.mathlinks.ro/viewtopic.php?t=160184

I hope to read the comments. But if there is spoiler then I will give up reading the comments.

Yes there are some spoilers, not complete solutions but some keywords which can help a lot. So if you want to solve the questions yourself without any outside help I would not read this thread.


function
inequalities
integration
real analysis
real analysis unsolved

Lets say that a measurable function $f: X\to \mathbb{R}$ belongs to weak-$L^{1}$ space if there exists a real number $c$ such that for every $t>0$
\[\mu (\{x\in X: |f(x)>t|\})\leq c/t.\]
How to prove that $L^{1}\subset \text{weak}-L^{1}$ but $\text{weak}-L^{1}\not \subset L^{1}$ as $X=\mathbb{R}$ and $\mu=m$?

That $L^{1}$ is contained in weak $L^{1}$ is simply Chebyshev's inequality:
\[\mu\{|f|>t\}\le\frac{\|f\|_{1}}{t}\]
To prove this, constrict $g(x)$ such that $g(x)=t$ when $|f(x)|>t$ and is zero otherwise. Then $g(x)\le|f(x)|$ everywhere and $\int g=t\cdot\mu\{|f|>t\}\le\|f\|_{1}.$

For the failure of containment in the other direction, consider $f(x)=\frac1x$ on $(0,\infty).$ This function is in weak $L^{1}$ but not in $L^{1}.$


email
real analysis
calculus
geometry
projective geometry
search
linear algebra

[b][color=red]Σας παραθέτω το μήνυμα του συνάδελφου Αλέξανδρου Συγκελάκη και παρακαλώ να ακολουθήσουμε όλοι μας τις οδηγίες που αναγράφονται.Ευχαριστούμε και τον Αλέξανδρο και το Στέλιο Μόσχογλου για αυτή την τόσο σπουδαία προσπάθεια που ξεκίνησαν. Η συνεισφορά τους στην ολυμπιακή προετοιμασία των μαθητών μας θα είναι εξαιρετικά μεγάλη, διότι το καλό βιβλίο είναι το ισχυρότερο όπλο για αυτόν που ασχολείται με τις ολυμπιάδες.[/color][/b]

''Είμαι στην ευχάριστη θέση να σας ανακοινώσω την λειτουργία της σελίδας μας και πάλι, μετά από απουσία περίπου 2 μηνών. Μπορείτε να δείτε τη διεύθυνση της σελίδας στο πρώτο post του συγκεκριμένου μηνύματος. Να αναφέρω από τώρα ότι η σελίδα αυτή ΔΕΝ είναι η σελίδα που περιέχει τα βιβλία. Θα μπείτε όμως σε αυτή, θα ακολουθήσετε τις οδηγίες για να βρεθείτε στην εν λόγω ιστοσελίδα. Αυτό έγινε για να παραπλανήσει τους ξένους εκείνους που είδαν αυτό το post και θέλουν να αναφέρουν την λειτουργία της ιστοσελίδας μας σε εκδοτικούς οίκους με σκοπό το κλείσιμό της. Γι'αυτό τα πάντα είναι γραμμένα στα Ελληνικά!

***
Η σελίδα είναι

http://homepages.pathfinder.gr/stfu/

Παρακαλώ τον κο Στεργίου να αλλάξει το όνομα της σελίδας στο πρώτο post, να σβήσει και το κείμενο που περιέχεται ανάμεσα στους αστερίσκους στο παρόν post και τέλος να αλλάξει εαν μπορεί τον τίτλο του μηνύματος αυτού σε μικρούς χαρακτήρες χωρίς θαυμαστικά (μην καρφωνόμαστε). Επίσης γράψτε στο πρώτο μήνυμα  μερικά πράγματα από την πιο πάνω παράγραφο, για να μη νομίζουν εκείνοι που μπαίνουν ότι βρίσκονται στην ιστοσελίδα με τα βιβλία.

***

Παρακαλώ πολύ όλους για να μην κλείσει και πάλι η σελίδα (αν και αργά ή γρήγορα όταν καταλάβουν και πάλι οι εκδοτικοί οίκοι ότι βάζουμε τα βιβλία τους για κατέβασμα), να μην αναφέρουν πουθενά σε αυτό το μήνυμα την διεύθυνση της ιστοσελίδας και να μην της κάνουν διαφήμιση σε διάφορα forum κτλ κτλ. Ας την διατηρήσουμε όσο το δυνατόν πιο μυστική. Επίσης όπως θα δείτε (για τους καινούριους) όλα τα ονόματα στην σελίδα είναι στα ελληνικά για να μην καρφωθούμε!! Τις επόμενες ημέρες θα βάζω σιγά σιγά κάποια βιβλία που έχω συγκεντρώσει αυτό τον καιρό. Μπορείτε όμως να στείλετε και τα δικά σας βιβλία/σημειώσεις κτλ κτλ στο email stfu.gr@gmail.com (Να σβηστεί και αυτό εδώ και να αλλάξει στην πρώτη σελίδα αντί του diagonismoi@gmail.com). Δεν μπορεί αυτή η προσπάθεια να συνεχιστεί εαν δεν συμβάλλετε όλοι.

Για να σας ανοίξω την όρεξη βάζω στον τομέα της Γεωμετρίας, την περίφημη Ελληνική Γεωμετρία ΘΡΥΛΟ του Σ. Κανέλλου (549 σελίδες, 1η έκδοση 1970) "Ευκλείδειος Γεωμετρία Επίπεδος" που έκανα scan μία μία σελίδα. Σας συνιστώ να την εκτυπώσετε και να την διαβάσετε προσεχτικά (είναι σε πολύ καλή ποιότητα) καθώς πρόκειται για μία από τις καλύτερες γεωμετρίες που πέρασαν ποτέ από τη χώρα μας. Σίγουρα οι γονείς σας εαν ήταν σε σχολές που χρειαζόταν Μαθηματικά, θα το έχουν ακούσει! Απολαύστε το!!

Μπορούμε να ακούσουμε σχόλια και από τους μεγαλύτερους (τον κο Στεργίου ή τον κο Βήττα).

Ίσως είναι πολύ καλή ιδέα να φέρουμε σε ψηφιακή μορφή τα καλύτερα ελληνικά βιβλία που έχουμε. Το ξέρω ότι είναι κόπος αλλά σκεφτήτε ότι πριν κατεβάσετε κι εσεις αυτά τα βιβλία κάποιος άλλος κόπιασε για να τα φτιάξει. Εαν κάποιος κάνει scan κάποιο βιβλίο και δεν ξέρει πώς να το κάνει σε .pdf και να είναι μικρό σε χωρητικότητα (Φυσικά εαν κάνετε scan, να το κάνετε σε ασπρόμαυρο και προσοχή στις σελίδες να είναι ίσια για να μπορεί κάποιος που θα το κατεβάσει να το εκτυπώσει), μπορεί να μου στείλει τις εικόνες σε ένα CD με το ταχυδρομείο και θα το κάνω εγώ. (Διαφορετικά εαν έχει γρήγορο Internet μπορεί να τις συμπιέσει και να μου τις στείλει από το MSN).

Φυσικά δεν πρέπει να ξεχάσουμε και τον Στέλιο (mostel) του οποίου η βοήθεια ήταν μεγάλη στο σχεδιασμό της σελίδας και στο να επανέλθει και πάλι.''

Φιλικά,
[color=red][b]Αλέξανδρος[/b][/color]

 [color=red][b]Moderator[/b][/color]

ΔΟΚΙΜΑΣΑ τις οδηγίες και το site λειτουργεί  :lol: ! Αλέξανδρε  εύγε ! Βέβαια , είναι ευκολότερο να λύσεις το θέμα ΙΜΟ 2007(problem 3) από το να βρει ξένος τον τρόπο λειτουργίας των συνδέσμων !!! Όμως καλύτερα έτσι και είμαι με αυτό απολύτως σύμφωνος.

 Μπάμπης

ΡΕ ΠΑΙΔΙΑ ΤΙ ΕΓΙΝΕ,ΜΑΣ ΚΛΕΙΣΑΝ ΤΗ ΣΕΛΙΔΑ ΜΕ ΤΑ ΒΙΒΛΙΑ :?:

Δυστυχώς είμαστε άτυχοι!! Παρόλο που δεν την ανακάλυψαν "μάλλον" ξένοι, δεν έκλεισε μόνο η δική μας ιστοσελίδα αλλά και ΟΛΕΣ οι άλλες οι οποίες φιλοξενούνταν στον ίδιο server. Εαν περιμένετε για λίγη ώρα στη σελίδα μας, θα σας μεταφέρει αυτόματα σε μία σελίδα που γράφει ότι έκλεισε η υπηρεσία παροχής χώρου διότι απλά έκλεισε ο server.

Θα ψάξω σύντομα να βρω κάποιο άλλο χώρο να τα ξαναβάλουμε...

Φιλικά,
Αλέξανδρος

Είμαι σίγουρος  ότι το κλείσιμο οφείλεται σε Ελληνικό ... δάκτυλο. Βλέπω ότι το mediafire που έχει πλήθος βιβλίων λειτουργεί για καιρό. Εκεί γίνεται ολόκληρη διακίνηση !

 Ή μήπως έκλεισε και εκείνο ;

 Μπάμπης

ΜΗΠΩΣ ΥΠΑΡΧΕΙ ΚΑΜΙΑ ΕΞΕΛΗΞΗ?

Προτείνω:
Αν κάποιος έχει όλα τα βιβλία της ιστοσελίδας που μας έκλεισαν να μας στείλει (αν δεν βαριέται) μια λίστα με τα βιβλία μέσω private message και όλες οι συναλλαγές να γίνονται μέσω email/private message.Eτσι δεν πρόκειται να μας ανακαλύψει κανείς. :wink:

Στο mediafire  πώς βλέπεις τη σελίδα από την οποία κατεβάζεις (όχι αυτή που κάνεις upload)????

Επιτέλους βρήκα μια νέα κορυφαία σελίδα με [u]δωρεάν[/u] βιβλία για ολυμπιάδες :lol:
[url]http://trungtuan.wordpress.com/ebooks/download/[/url]
Επιπλέον και μια άλλη σελίδα με χρησιμότατα βιβλία μαθηματικων,φυσικής κτλ
[url]http://www.freebookspot.com/Default.aspx[/url]
και...καλλό διάβασμα :wink:

Δυστυχώς χθες το Springer Verlag διαμαρτυρήθηκε στο rapidshare και τα περισσότερα βιβλία τους δεν είναι πλέον διαθέσιμα για αντιγραφή! Πρόφθασα και κατέβασα καμια εκαταστή...

Τρέξτε λοιπόν!

Lancelot DuLac

Βρήκα μια καταπληκτική σειρά βιβλίων που είναι κατάληλη για όλους:μαθητές και φοιτητές ,ακόμη και για όσους ασχολουνται με μαθηματικούς διαγωνισμούς
[url]http://www.openmathtext.org/downloads.html[/url]

Δείτε και αυτό το βιβλίο (Algebra Through Problem Solving) καλό για όσους τώρα ξεκινάνε το άθλημα:
[url]http://www.thiel.edu/mathproject/atps/PDF/default.htm[/url]

Παραθέτω και εγώ κάποιεσ σελίδες με δωρεάν βιβλία και σημειώσεις μαθηματικών  :)

[url]http://de.geocities.com/darij_grinberg/[/url]

[url]http://www.math.ust.hk/excalibur/[/url]

[url]http://db.math.ust.hk/notes_download/e_notes.htm[/url]

Αξίζει να ρίξετε μια ματιά σστις παραπάνω σελίδες :wink:

An dimiourgousame ena forum tha mprousame na vaZoume oti theloume ekei eleuthera afou tha anikei se emas ti lete?

Υπάρχει νομικό θέμα. Ποιός θα αναλάβει την ευθύνη να φιλοξενεί τέτοια αρχεία. Το μόνο που μπορούμε να κάνουμε είναι να τα κάνουμε upload σε κάποιο free host όπως rapidshare

[quote="Eukleidis"]Kai na ta postaroume ta links se ena forum krifo to opoio tha einai diko mas[/quote]

Δύσκολο να είναι ή να παραμείνει κρυφό. Δεν ξέρω αν έχει αλλάξει τίποτα σε σχέση με παλιότερα, αλλά τότε μας την είχαν κλειδώσει τη σελίδα.Στο  gigapedia υπάρχουν όλα τα βιβλία που θέλουμε . Μου έδωσε ο καθηγητής κος Μπόλης 18 giga όλο με βιβλία μαθηματικών.Δεν ξέρω αν έχει νόημα πια να φτιάξουμε άλλο χώρο με ηλεκτρονικά βιβλία , αφού φιλοξενούνται όλα στο gigapedia.Αν όμως θέλετε και νοιώθετε ότι θα είναι χρήσιμος , να ανοίξω νέο sticky και να βάζουμε εκεί τον αντίστοιχο σύνδεσμο του gigapedia με ελληνικούς χαρακτήρες, ώστε να μην πέφτουν πάνω τους οι ξένοι.

 Ας μας πουν και πιο ειδικοί για τις ιστοσελίδες την άποψή του και το ξανακουβεντιάζουμε.

 Χρόνια Πολλά !

όντος θα ήταν πολύ βοηθητικό αν βάζατε ενα sticky με τα links.

Γεια σας
Θα ήθελα τη βοήθεια σας.
Ψάχνω για το βιβλίο διεθνείς μαθηματικές ολυμπιάδες 1959-2000. Πήρα τη ΕΜΕ να μου το στείλουν λενε πςω το καταργησαν.
Ξερετε μηπώς υπάρχει πουθενα εκτυπώμενο να το αγοράσω??
Gia to proigoumeno dokimaste i acrobat 9 i acrobat professional

[quote="NickNafplio"][quote="stergiu"][quote="NickNafplio"]κ Στεργιου εγω εχω κατεβασει καποια βιβλια απο τη gigapedia αλλα δεν τα ανοιγει ο acrobat και δεν μπορω να καταλαβω γιατι... μηπως εχετε καμια ιδεα για το τι θα μπορουσε να φταιει?[/quote]

Στείλε μου ένα για να δω.Μερικά είναι κλειδωμένα . Είχα και γω πρόβλημα με κάποιο, αλλά το έλυσα τελικά, χάρισ τον Αλέξανδρο.

 Μπάμπης[/quote]

Πως να σας τα στειλω? Εχετε e-mail?[/quote]

Νίκο γεια σου  ! Στείλτα στο :

 stergiu@otenet.gr

 Μπορείς όμως να μου στείλεις το σύνδεσμο του giga για να δω τι ακριβώς γίνεται.

 Μπάμπης

γράψε τα λινκς μήπως σε βοηθήσουν και άλλοι  :wink:

[quote="NickNafplio"]Το θεμα ειναι οτι ειχα φτιαξει account μονο για να κατεβασω αυτα τα βιβλια και τωρα το ξεχασα... και θελει μια μεγαλη διαδικασια για να φτιαξω καινουργιο ωστε να ξαναδω τα λινκς και δεν εχω πολυ χρονο :/

Τα βιβλια ειναι βιβλια για ολυμπιαδες φοιτητων και ειναι τα ακολουθα
1) Linear Algebra του Peter Lax
2) principles of mathematical analysis του Rudin
3) Calculus του m Spivak

Οποιος εχει χρονο και θελει ας τα κοιταξει.[/quote]

Πάντως το δεύτερο υπάρχει μεταφρασμένο και στα Ελληνικά.Το έχω και γω ,από φοιτητής ακόμα , στα αγγλικά.

 Μπάμπης

[quote="papex"]TA ΒΙΒΛΙΑ ΑΥΤΑ ΥΠΟΚΕΙΝΤΑΙ ΣΕ ΝΟΜΟΥΣ ΠΝΕΥΜΑΤΙΚΩΝ ΔΙΚΑΙΩΜΑΤΩΝ ΚΑΙ ΟΣΟΙ
ΤΑ ΔΙΑΚΙΝΟΥΝ ΧΩΡΙΣ ΤΗΝ ΑΔΕΙΑ ΤΩΝ ΚΑΤΟΧΩΝ ΤΩΝ ΔΙΚΑΙΩΜΑΤΩΝ ΤΙΜΩΡΟΥΝΤΑΙ
ΜΕ ΦΥΛΑΚΙΣΗ ΚΑΙ ΠΡΟΣΤΙΜΟ ΑΡΚΕΤΩΝ ΧΙΛΙΑΔΩΝ ΕΥΡΩ.
ΘΑ ΣΑΣ ΣΥΜΒΟΥΛΕΥΑ ΝΑ ΑΠΟΦΕΥΓΕΤΕ ΝΑ ΔΙΝΕΤΕ ΤΗΝ ΕΝΤΥΠΩΣΗ ΟΤΙ ΚΑΝΑΤΕ
ΠΑΡΑΝΟΜΗ ΔΙΑΚΙΝΗΣΗ ΕΦΟΣΟΝ ΕΙΣΑΣΤΕ ΚΑΙ ΕΠΩΝΥΜΟΙ ΕΔΩ ΜΕΣΑ.
ΦΙΛΙΚΑ[/quote]

[color=red][b]ΧΡΟΝΙΑ ΠΟΛΛΑ ![/b][/color]

Και γω αυτή την άποψη έχω. Τα περισσότερα βιβλία ολυμπιάδων, να μην πω όλα σχεδόν , υπάρχουν ήδη ηλεκτρονικά και

μεις τα έχουμε στο αρχείο μας(μας τα στέλνουν διάφοροι φίλοι από άλλες χώρες).

 Αν κάποιο μέλος μας θέλει ένα ξενόγλωσσο -εξαντλημένο κυρίως - βιβλίο, ας γράψει εδώ στο φόρουμ : ''θέλω το ΧΧΧ

βιβλίο '' και αν το έχοχυμε θα του το στείλουμε με  e-mail.Με αυτό τον τρόπο δεν κινδυνεύουμε να μας κλείσουν τη

σελίδα, για την οποία έχουμε τόσο κοπιάσει.

    Ανεξάρτητα για το αν υπάρχει ποινική ή άλλη ευθύνη με την ηλεκτρονική διακίνηση βιβλίων(φοβάμαι πως σε μεγάλες

διεθνείς σελίδες δεν μπορεί να εντοπιστούν υπεύθυνοι , διότι διαφορετικά οι μεγάλοι εκδοτικοί οίκοι κάτι θα είχαν κάνει

μέχρι τώρα) , εμείς σε αυτό το χώρο είναι ωραίο να είμαστε δίκαιοι και όσο μπορούμε νόμιμοι.Άλλωστε κάνουμε τη δουλειά

μεταξύ μας πολύ ωραία. Κανένας δεν θα χάσει από μια τέτοια θέση και συμπεριφορά .

                               Μπάμπης

ΚΥΡΙΕ ΣΤΕΡΓΙΟΥ Η ΝΟΜΟΘΕΣΙΑ ΕΙΝΑΙ ΑΡΚΕΤΑ ΑΣΑΦΗΣ ΩΣΤΕ ΚΑΝΕΙΣ ΔΕΝ
ΑΠΟΚΛΕΙΕΙ ΤΟ ΓΕΓΟΝΟΣ ΓΙΑ ΠΑΡΑΔΕΙΓΜΑΤΙΣΜΟ ΝΑ ΓΙΝΟΥΝ ΑΓΩΓΕς ΑΠΟ
ΕΤΑΙΡΕΙΕΣ ΠΟΥ ΑΠΑΣΧΟΛΟΥΝΤΑΙ ΜΕ ΤΑ ΠΝΕΥΜΑΤΙΚΑ ΔΙΚΑΙΩΜΑΤΑ Η ΚΑΙ
ΝΑ ΖΗΤΗΖΟΥΝ ΤΟ ΚΑΤΕΒΑΣΜΑ ΜΙΑ ΙΣΤΟΣΕΛΙΔΑς Η ΤΟΥ ΦΟΡΟΥΜ ΑΥΤΟΥ.
ΕΠΕΙΔΗ Η ΝΟΜΟΘΕΣΙΑ ΕΙΝΑΙ ΑΡΚΕΤΑ ΑΣΑΦΗΣ Η ΚΑΛΥΤΕΡΑ ΔΕΝ ΥΠΑΡΧΕΙ
ΕΛΛΗΝΙΚΗ ΝΟΜΟΘΕΣΙΑ ΠΟΥ ΝΑ ΚΑΛΥΠΤΕΙ ΤΑ ΑΝΤΙΚΕΙΜΕΝΑ ΟΠΩς ΑΝΑΡΤΗΣΗ
ΥΛΙΚΟΥ ΠΧ ΒΙΒΛΙΑ ΛΟΓΙΣΜΙΚΟ ΜΟΥΣΙΚΗ ΚΑΤΑΛΑΒΑΙΝΕΤΕ ΠΟΙΟΣ ΜΠΛΕΚΕΙ
ΤΩΡΑ ΜΕ ΔΙΚΗΓΟΡΟΥΣ ΚΑΙ ΑΓΩΓΕΣ. ΘΑ ΗΘΕΛΑ ΝΑ ΣΑΣ ΕΝΗΜΕΡΩΣΩ ΟΤΙ ΠΡΟΣΦΑΤΑ
ΕΚΛΕΙΣΕ ΜΙΑ ΙΣΤΟΣΕΛΙΔΑ ΕΛΛΗΝΙΚΗ ΠΟΥ ΑΝΕΒΑΖΕ ΒΙΒΛΙΑ ΟΠΩς ΘΑ ΔΙΑΒΑΣΕΤΕ
ΠΑΡΑΚΑΤΩ:

''
Αρ. πρωτ. 59 Αθήνα 20 Φεβρουαρίου 2009
Αξιότιμοι κύριοι,
Ο οργανισμός μας είναι οργανισμός συλλογικής διαχείρισης που έχει ως σκοπό του
τη συλλογική διαχείριση και προστασία των δικαιωμάτων πνευματικής ιδιοκτησίας
στα έργα του λόγου των εκπροσωπούμενων από τον οργανισμό μας συγγραφέων και
εκδοτών.
Πρόσφατα διαπιστώσαμε ότι στην ιστοσελίδα υπό τη διεύθυνση http://e-
vivl ia.blogspot .com/ παρέχεται η δυνατότητα πρόσβασης και ψηφιακής
αναπαραγωγής σε έργα των εκπροσωπούμενων από εμάς εκδοτών και συγκεκριμένα
των εκδοτών : Φιλίστωρ, Καστανιώτης, Πατάκης, Κέδρος, Αλεξάνδρεια, Μελάνι,
Αιώρα, Ψυχογιός, Λιβάνης.
Σύμφωνα με όσα γνωρίζουμε η παραπάνω διάθεση έργων του λόγου
πραγματοποιείται χωρίς την άδεια των δικαιούχων εκδοτών ή τυχόν αντιπροσώπων
τους ούτε βεβαίως με την άδεια του οργανισμού μας, ο οποίος έχει αναλάβει τη
συλλογική διαχείριση ορισμένων δικαιωμάτων και την προστασία των έργων αυτών.
Συνεπώς η εν λόγω διάθεση των έργων είναι παράνομη.
Σύμφωνα με το Ν.2121/1993 η παράνομη προσβολή των δικαιωμάτων πνευματικής
ιδιοκτησίας συμπεριλαμβανομένων των συγγενικών δικαιωμάτων επισύρει αστικές
και ποινικές κυρώσεις μεταξύ των οποίων και φυλάκιση τουλάχιστον ενός έτους.
Για τους λόγους αυτούς ΣΑΣ ΚΑΛΟΥΜΕ να αποσύρετε ή να διακόψετε την
πρόσβαση στα παραπάνω έργα όπως εμφανίζονται στην παραπάνω
ιστοσελίδα/υπηρεσία το συντομότερο δυνατό και με τον πλέον αποτελεσματικό
τρόπο.
Ευχαριστούμε για τη συνεργασία.
Για τον ΟΣΔΕΛ
Άρης Πετρόπουλος
''
ΟΠΩς ΚΑΤΑΛΑΒΑΙΝΕΤΕ ΑΡΚΕΙ ΚΑΙ Η ΑΠΕΙΛΗ ΝΑ ΜΠΛΕΞΕΙΣ ΜΕ ΜΥΝΗΣΕΙΣ
ΓΙΑ ΝΑ ΚΑΤΕΒΕΙ Η ΣΕΛΙΔΑ.
ΑΥΤΑ ΦΙΛΙΚΑ.

Σε ευχαριστούμε πολύ !

 Γι αυτό και μεις κινούμαστε στα πλαίσια του νόμιμου, ώστε να μην έχουμε μπελάδες ! Η δουλειά μας γίνεται μια χαρά και

είναι καθαρή ανοησία να μπλέξουμε χωρίς κανέναν απολύτως λόγο !

 Μπάμπης

Μήπως γνωρίζει κανείς ποια είναι τα θέματα που διαπραγματεύεται το βιβλίο "Plane Euklidean Geometry" των Gardiner και Bradley;

Δεν αγγίζει καν τα όρια του επιπέδου βαλκανιάδων , για να μην πω ...Ευκλείδη. Είναι εισαγωγικό και έχει πολλές ενότητες.

 Καλό μεν βιβλίο για αρχάριους, αλλά όχι αντάξιο του ονόματος Bradley ή gardiner , δηλαδή κατάλληλο για ολυμπιάδες !

   Μπάμπης

Μήπως μπορείτε να μου συστήσετε βιβλία με θεωρία και ασκήσεις πάνω στις πολικές, στα πλήρη τετράπλευρα,στους γεωμετρικούς μετασχηματισμούς, στις γεωμετρικές ανισότητες, στις αρμονικές δέσμες και στα ειδικά θεωρήματα Pascal, Desargues κ.τ.λ.;

Προσπαθω να ακολουθήσω τις προαναφερθείσες οδηγίες προκειμένου να πάω στο site, αλλά μετά από το 2ο βήμα αυτό είναι ανέφικτο. Συγκεκριμένα μου βγάζει αυτό:

The file you have requested does not exist, this can have various reasons:
the file has been deleted by the account owner
the file has been deleted by an administrator due to abuse
if you are using a script on your site, a redirect may be wrong
incorrect settings in scripts, like base-url etc.

πρέπει να κάνω δημιουργία λογαριασμού ή κάτι αντίστοιχο;

Τα έχουν σβήσει τώρα πια νομίζω.

Καλημέρα, μόλις τώρα προσπάθησα και τα βρήκα εντάξει(την άλγεβρα του LAX και του Rudin).
 Όποιος θέλει κάτι , ας μου γράψει pm.

Μπάμπης

Είναι κάπως πιο γενικό, αλλά ένα add της google, το scribd έχει περασμένα διάφορα ενδιαφέροντα e-books, συμπεριλαμβανομένων μαθηματικών. (μόλις κατέβασα για τα sat math 2 kai physics 2, να μάθω και καμιά αγγλική ορολογία αλλιώς δε με βλέπω καλά :P ) Κάνοντας search  έχει  αρκετά ωραία πραγματάκια, εσείς κρίνετε.

Μπορεί κάποιος που έχει την πλήρη συλλογή των βιβλίων που αναφέρονται στο αρχικό μήνυμα, να τα ανεβάσει σε έναν private tracker, όπως το κορυφαίο http://www.gamato.info ; Έτσι δεν υπάρχει περίπτωση να ξαναδιαγραφούν και μπορούν να τα πάρουν όλοι με μια απλή εγγραφή στο site. Αν δεν έχει κανείς βασικές γνώσεις upload για torrents, μπορώ να τον βοηθήσω εγώ, ας μου στείλει PM.

Φιλικά,
Ηφαιστίωνας Γκιώσης


My private messages have been disabled for almost a year. Will i ever get them back?

I private messaged Mr.Vornicu, but he hasnt replied yet...
I even tried to contact him by email, but no reply.
Is there any other way to contact him?


I don't know if this will help any of you guys, but I will say it anyways.
This procedure will help you quickly find pythagorean triples.
Pick any number, odd or even, for the shorter leg of the right triangle.
For any odd number (except 1), square the number, then find two consecutive integers that add to that squared result. i.e., $ 9^2\equal{}81$. Then, $ 81\equal{}x\plus{}(x\plus{}1)$, and we get $ x\equal{}40$. This will be the other leg of the triangle. To find the hypotenuse, simply plug it into $ x\plus{}1$, to get the hypotenuse of 41.
For any even number (except 2), square the number, then divide it by 2. Then, find two consecutive even numbers that add to half of the squared number. Once again, an example.. $ \frac{8^2\equal{}64}{2}$, then we solve $ x\plus{}(x\plus{}2)$ because we need to find consecutive even integers, then solve for x using $ 32\equal{}x\plus{}(x\plus{}2)$, to get the other leg and hypotenuse.
I hope this helped, but I want you realize that this doesn't necessarily find all of the Pythagorean triples, only ones with consecutive longer legs and hypotenuses.

$ The  Origami  Guy$

Consider two arbitrary relatively prime positive integers $ m$ and $ n$, with one odd and one even. Assume WLOG (by symmetry) that $ m > n$. Then $ x \equal{} m^2 \minus{} n^2$, $ y \equal{} 2mn$, and $ z \equal{} m^2 \plus{} n^2$ generates all primitive Pythagorean triplets.


These problems all factor. Then you just cross out the common term in the numerator and denominator and you have the  final answer. How do you know if something is factorable? Just use the determinant, $b^2-4ac$. If it turns out to be a perfect square, the trinomial is factorable. I think splitting the middle term is the easiest way to factor the ones below.

Number 1--$\frac{3x^2+5xy-2y^2}{4x^2+7xy-2y^2}$
[hide]The top--$3x^2+6xy-xy-2y^2$ into $(3x-y)(x+2y)$
the bottom--$4x^2-xy-8xy-2y^2$ into $(x+2y)(4x-y)$[/hide]

Number 2--$\frac{p^2+6p-7}{p^2-p}$
[hide] the top--$p^2+7p-p+7$ into $(p-1)(p+7)$
the bottom--just factor out p into $p(p-1)$[/hide]

Number 3--$\frac{a^2+am-an-mn}{a^2+am+an+mn}$
[hide]the top factors into $(a+m)(a-n)$
the bottom factors into $(a+m)(a+n)$ [/hide]

This looks like  homework so I didn't give the direct answers. If you need more explanation post again . :)

argg man! this aint homework! itz frm the textbook! i gotta test and i still dun understand the whole concept!!!!

Ok, well this link shows some ways to factor quadratic trinomials

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=87087

erm da problem 1 i checked with the anser key behind the book(without explanation)
is 3x-y/4x-y..help man..i am getting reali confused

herm..i kinda understand it abit alreadi..ty

Yeah for #1 I made a typo...I fixed it now. The denominator should actually be $(x+2y)(4x-y)$. So you have $\frac{(3x-y)(x+2y)}{(x+2y)(4x-y)}$ and you cross out x+2y from both the numerator and denominator (make sure you understand how this is allowed) and you have $\frac{3x-y}{4x-y}$ as your final answer.

[quote="José"]Please everybody use Latex[/quote]

We don't really enforce this so strictly in the middle school forums.  But they'll learn over time.

[quote="José"]Please everybody use Latex[/quote]
Done.


[quote="12markkram34"]$ \frac {16}{25}x^{2} \plus{} \frac {4}{9}y^{2}$

I got $ 4\left(\frac {4}{25}x^{2} \plus{} \frac {1}{9}y^{2}\right)$, but my teacher said it's Not Factorable. Which is it?[/quote]
I think when she meant factorable, it was probably take something like $ x^{2} \minus{} 4$ and factor it to $ (x\minus{}2)(x\plus{}2)$ not factoring out a number. Technically you ARE factoring

"Factoring" with one factor being an element from the field of coefficients is not to be counted as factoring, since not-null such elements are inversible (units). Otherwise there would exist no irreducible polynomials.

Since you could always factor out a $ 1$?

First of all we notice here that the coefficients are rational, hence to "factor" would mean to "factor" over the field of rational numbers.  But the set of rational numbers is not a unique factorization domain.  For example, what does factor $ \frac{1}{2}$ mean, if any?  well, I can write $ \frac{1}{2}\equal{}\frac{2}{3}*\frac{3}{4}$. Also, $ \frac{1}{2}\equal{}\frac{4}{5}*\frac{5}{8}$.  so $ \frac{1}{2}$ doesn't factor uniquely as can be seen here.  So technically, the polynomial given is factorisable over the rational numbers but the factorization is not unique.   :)

[quote="mavropnevma"]The issue of unique factorization is moot in a field, since all not-null elements are units (inversible)![/quote]But, you'll have to understand that not every country follows the same level of mathematical maturity. The topic of a field is relevantly advanced for an average high schooler in USA, as compared to Romania. Yet, you are absolutely right about units in a field. The factorization that his teacher is talking about is within $ \mathbb{Q}[x]$, the ring of polynomials over $ \mathbb{Q}$, ie polynomials with rational coefficients.

@ mavropnevma:

yes, you're right!   :lol:
Indeed all rational numbers except 0 are units


find $ 3x^2y^2$ if
$ y^2 \plus{} 3x^2y^2 \equal{} 30x^2 \plus{} 517$
x,y are integers

[hide]
$ 3x^2y^2 \plus{} y^2 \minus{} 30x^2 \minus{} 517 \equal{} 0$
$ (3x^2 \plus{} 1)(y^2 \minus{} 10) \minus{} 507 \equal{} 0$ via SFFT
$ (3x^2 \plus{} 1)(y^2 \minus{} 10) \equal{} 507 \equal{} 3*13^2$
Through intuition (I assume there isn't much you can do other than try various pairs of 507's factors) we see that
$ 3x^2 \plus{} 1 \equal{} 13$, $ y^2 \minus{} 10 \equal{} 39$
will produce integral solutions (namely $ x \equal{} \pm2$, $ y \equal{} \pm7$).
Therefore, $ 3x^2 \equal{} 12$, $ y^2 \equal{} 49$, and
$ 3x^2y^2 \equal{} 12*49 \equal{} 588$.
[/hide]

thanks for your replies,
and
@harekrishna
how can i find whether a problem is this AIME problem?


algebra
polynomial
Ring Theory
superior algebra
superior algebra unsolved

Hi,

let $ k$ be a field and let $ R: \equal{}k[{[x]}][y_1,\dotsc,y_n]$ be the polynomial ring in $ n$ variables
over the univariate power series ring $ k[{[x]}]$. I wonder whether it is true
that every maximal ideal of $ R$ must contain a polynomial or a power series only containing $ x$'s,
i.e. in every maximal ideal of $ R$ there must be an element of $ k[{[x]}]$. Does anybody know
how to prove that?

what about $ 0$ ...

by the way: are you mararo from the mp?


number theory
greatest common divisor
number theory unsolved

Assume gcd$(a,b)=10$ , what are the possible values of gcd$(a^{3},b^{4})$?

Also, if (without loss of generality) $a=2\cdot 5^{2}\cdots$, then $\text{gcd}(a^{3}, b^{4})=5000$, since you gain that extra factor of $5$, and if $a=2^{2}\cdot 5\cdots$, then $\text{gcd}(a^{3},b^{4})=2000$, since you have that extra factor of $2$ that is gained. Those are all of the possibilities, though: $1000$, $2000$, $5000$.


I don't know if I am doing this wrong.. but it just seems impossible.

Problem:
In Fig. 1, it is given <PDQ = <PAQ, AC = 7, BD = 7, PC = 6, PD = 15, DQ = 5.  Find AP, AB, and AQ.
[img]http://i75.photobucket.com/albums/i285/ngr_sqd/fig3.jpg[/img]

The problem I am encountering is that after working with the problem, we find that:
Triangle PAC is similar to Triangle PDB
    Since: <P = <P  |.:. <B = <C
              <A = <D |

Therefore:
PA / PD = AC / DB = PC / PB

But AC / DB = 7 / 7 = 1
This means that PA / PD = 1, PA = PD = 15

But in Triangle PAC, PC = 6, CA = 7.  If PA = 15, there is no way that PC and CA can connect?  Together their sum is 13.

Is this question flawed?

[b][color=red]UPDATE[/color][/b]:  I spoke with the person who had written the question, and this was actually an error they had overlooked.  A value had been modified resulting in this impossible triangle.

So I was told to proceed, solving the problem regardless.

[hide="Well, I think I can solve the problem..."]Note that $BQ=DB-QB=7-5=2$

BY AAS, we have that $APC \cong DPB$. So $AP=DP=15$

BY AA, we have that $AQB \sim DQC$.

Let $AQ=7-x$, $QC=x$:
\[\frac{AQ}{BQ}= \frac{DQ}{CQ}\]

\[\frac{7-x}{2}= \frac{5}{x}\]

\[x(7-x)=10 \]

\[x^{2}-7x+10=0 \]

\[x=2,5 \]
When $x=2$, note that triangle $\triangle QCD$ will be degenerate. Thus, $x=5$.

So $AQ=7-5 = 2$.

Now, try to use the same similar triangles and convince yourself that $AB=\frac{18}{5}$.

So the answer is:
\[AP=15 \]

\[AB=\frac{18}{5}\]

\[AQ = 2 \]
[/hide]

I'm not sure if I did something wrong, but as I said before, the triangle APC does not exist, simply because
PC = 6
AC = 7
And by proof, AP = 15

I believe because of this, multiple answers are possible.

[hide="Here is what I did."]
Since APC = DPB,
Not only AP = DP = 15
But PB = PC = 6 (by the same ratio 1/1)
Therefore AB = PA - PB
AB = 15 - 6
AB = 9.
If AB = 9, and CQD ~ BQA
Then CD / AB = 9 / 9 and CQD = BQA
If this is true, then QD = QA = 5 (by the same ratio)
[/hide]

And if your method is sound, this means that there can be more than one method of solving the question, and because of the impossible triangle, more than one answer is possible.

[hide="different?"]Triangle ACP is congruent to DBP since the ratio of their sides is 1:1. Thus, PB will equal 6, AB=9 (AP=PD=15).

We also see that CQD is similar to BQA, and since AB=CD, the two triangles are in fact congruent.

Therefore AQ=5.

However, by this method, the possibility that AQ=2 never exists. could it be because of the degenerate triangle by the flawed given condition? [/hide]

[quote="rd5493"]in the problem, u say that that

[quote]In Fig. 1, it is given <PDQ = <PAQ, AC = 7, BD = 7, PC = 6, PD = 15, DQ = 5. Find AP, AB, and AQ. [/quote]

if PC=6 and PD=15, that means that CD=9.  how can a triangles leg be larger than its hypotenuse.  i have no clue as to hw to solve it, but even though u said that it was impossible, ill say it again.  this problem cannot be solved! :huh:  :rotfl:  :|[/quote]

:dry:  Although you are the first to understand my first post, that the question has NO UNIQUE SOLUTION because the TRIANGLE DOES NOT EXIST.  It is an IMPOSSIBLE TRIANGLE.

I thank those who tried solving the problem, but as I stated, you will always get different answers depending on your method since the triangle does not exist.

Back in the first post, I said the legs sum to 13, and the hypotenuse is 15.  This is impossible.

In the update, the person who made the question has acknowledged this, and will be marking accordingly, realizing that there is space for more than one answer.


Decided to put this in here since this forum needs more activity:

AIME #10: A circle is inscribed in quadrilateral $ABCD$, tangent to $AB$ at $P$ and to $CD$ at $Q$. Given that $AP=19, PB=26, CQ=37, QD=23$, find the square of the radius of the circle.
[hide]
I wanted to know if my answer of [hide]$648$ [/hide]was correct since for some reason Kalva didn't have the answer to this problem :o [/hide]

Explain how you got your answer and we can tell you if you're right or not.

Okay, I just tried it and your answer is *very* close to mine, but not the same. Show us how you solved it and maybe we can find what's wrong.

What I did was:

Knowing that tangents from the same point are equal, I found out all four sides of the quadrilateral, by calling the other two points of tangency $X$ and $Y$, I found that $AP=19, PB=26, BX=26, XC=37, QC=37, QD=23, DY=23, YA=19$,then what I did was that by knowing that tangents to a circle form right angles, I first drew in all four radii which connect to the four points of tangency and then drew lines which become the hypotenuses to all four corners of the quadrilateral, thus forming eight right triangles, by calling the radius of the circle $x$, we see the bases of the triangles are: two $26$, two $37$, two $23$ and two $19$ and ofcourse $x$ is the height for all the triangles, so now the area of the quadrilateral is:
$\frac{2(26x+37x+23x+19x)}{2}=105x$, now using Brahmagupta's formula, we see that
$105x=\sqrt{(105-63)(105-60)(105-42)(105-45)} \Longrightarrow 11025x^2=7144200$, so
$\boxed {x^2=648}$.

I haven't tried it myself yet, but looking at your solution, why can you use Brahmagupta's?  Nowhere in the problem nor your solution does it say the quad is cyclic.

Yep, using Brahmagupta is your problem. Brahmagupta's formula is only for quadrilaterals [b]inscribed[/b] in a circle (called cyclic quadrilaterals) - not ones that are circumscribed. Amazingly your answer comes out to be [i]extremely[/i] close to my answer (probably by their design).

[hide]My solution involves trigonometry. Call the center of the inscribed circle O, and let the radius of the circle be r. Let angle OAP = $\theta_A$, OBP = $\theta_B$, OCQ = $\theta_C$, and ODQ = $\theta_D$. Notice that each of these angles is half their respective vertex, so $\theta_A + \theta_B + \theta_C + \theta_D = 180^o$.

Also notice that $\tan{\theta_A} = r/19$, $\tan{\theta_B} = r/26$, $\tan{\theta_C} = r/37$, and $\tan{\theta_D} = r/23$ (by right angle trigonometry).

So we have $\tan^{-1} \Big(\frac{r}{19}\Big) +\tan^{-1} \Big(\frac{r}{26}\Big) + \tan^{-1} \Big(\frac{r}{37}\Big) + \tan^{-1} \Big(\frac{r}{23}\Big) = 180^o$.

Taking the $\tan$ of both sides of this equation, and using the tangent sum formula $\tan {(A+B)} = \frac{\tan A + \tan B}{1 - \tan A \tan B}$ (three times), we get $\displaystyle \frac{\frac{r}{19} + \frac{r}{23}}{1 - \frac{r^2}{19 \cdot 23}} + \frac{\frac{r}{26} + \frac{r}{37}}{1 - \frac{r^2}{26 \cdot 37}} = 0$ (notice I left out some things in the tangent sum formula - namely the final denominator, because if this equals zero the denominator is irrelevent).

Solving this I get $r^2 = 647$.[/hide]

Oops  :blush:  :oops:  :surrender:, I didn't even try to prove the fact that it wasn't cyclic, by the way to  prove that a quadrilateral is cyclic, don't you use the diagonals, for example:

Say diagonals $AC$ and $BD$ intersect each other at point $P$, if $BP \cdot PD=AP \cdot PC$, then it's cyclic right? since it involves power of point?

There are a few ways to show that a quadrilateral is cyclic.

You can use the converse of power of a point.  Most common, however, is to show that opposite angles add to 180 or to show that certain angles formed by the diagonals are congruent (I don't have a picture ready so it's kinda hard to explain).  The last method I know of is the equality condition of Ptolemy's inequality.

Does anyone know a non-trig way to solve this? :)


Let a,b,c be positive number such that $ a^2\plus{}b^2\plus{}c^2\equal{}\frac{3}{4}$
Prove that
$ \frac{ab}{ab\plus{}1\minus{}a}\plus{}\frac{bc}{bc\plus{}1\minus{}b}\plus{}\frac{ca}{ca\plus{}1\minus{}c} \leq 1$

[quote="2424"][quote="8826"]Let a,b,c be positive number such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} \frac {3}{4}$
Prove that
$ \frac {ab}{ab \plus{} 1 \minus{} a} \plus{} \frac {bc}{bc \plus{} 1 \minus{} b} \plus{} \frac {ca}{ca \plus{} 1 \minus{} c} \leq 1$[/quote]
It's easy by CS! :P[/quote]
How to do it by CS? :blush:

[quote="Ji Chen"][quote="8826"]Let $ a,b,c$ be positive numbers such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} \frac {3}{4},$ Prove that

$ \frac {ab}{ab \plus{} 1 \minus{} a} \plus{} \frac {bc}{bc \plus{} 1 \minus{} b} \plus{} \frac {ca}{ca \plus{} 1 \minus{} c} \leq 1.$[/quote]
$ 1 \minus{} \sum_{cyc}\frac {ab}{ab \plus{} 1 \minus{} a} \equal{} \frac {(1 \plus{} abc)(1 \minus{} a \minus{} b \minus{} c \plus{} bc \plus{} ca \plus{} ab \minus{} 2abc)}{(ab \plus{} 1 \minus{} a)(bc \plus{} 1 \minus{} b)(ca \plus{} 1 \minus{} c)}.$

Let $ a \plus{} b \plus{} c \equal{} 3u,a^2 \plus{} b^2 \plus{} c^2 \equal{} 3v^2 \equal{} \frac {3}{4},abc \equal{} r,$ then

$ (b \minus{} c)^2(c \minus{} a)^2(a \minus{} b)^2 \equal{} \minus{} 4r^2 \plus{} 4ru(5u^2 \minus{} 3v^2) \minus{} (3u^2 \minus{} 2v^2)(3u^2 \minus{} v^2)^2\equiv f(r).$

$ 1 \minus{} a \minus{} b \minus{} c \plus{} bc \plus{} ca \plus{} ab \minus{} 2abc \equal{} v(9u^2 \minus{} 12uv \plus{} 5v^2) \minus{} 2r\geq0$

because $ v(9u^2 \minus{} 12uv \plus{} 5v^2) \minus{} u(5u^2 \minus{} 3v^2) \equal{} (v \minus{} u)(5u^2 \minus{} 4uv \plus{} 5v^2)\geq 0$ and

$ f \left[\frac {v}{2}(9u^2 \minus{} 12uv \plus{} 5v^2)\right] \equal{} \minus{} (u \minus{} v)^2(27u^4 \minus{} 36u^3v \plus{} 66u^2v^2 \minus{} 44uv^3 \plus{} 23v^4)\leq 0.$[/quote]
Yes，I think I got it!Thank you!

Using Am-Gm, it's become:
$ \frac{xy}{xy\plus{}y^2\plus{}z^2} \leq\ 1$ (obviously trues by CS and Vasc's ineq)  :wink:


What are Sterling numbers? What is their significance?

A question easily answered by a Google search.

http://mathworld.wolfram.com/StirlingNumberoftheFirstKind.html
http://mathworld.wolfram.com/StirlingNumberoftheSecondKind.html

There are some good exercises with Sterling numbers here:

[url]http://mandelbrot.org/stanfordmathcircle/stirling.pdf[/url]


Define the sequence $\{x_i\}$ as follows: $x_1=1,x_{n+1}=4^{x_n}$
Show that :$x_{n+1}-x_n\vdots (n!)^2 \forall n\geq 2$

I don't understand. What do you mean by $\vdots$? It can't be $=$, because that doesn't work for $n=2$.


i really disagree with what the government is doing right now

for every hurricane in the past, the national guard was there to help.
in hurricane andrew, the guard was 70 miles from the hardest hit city before it got hit.
this year, where is the guard?

also, bush didnt do much either. on monday, he wasnt watching the news. he was playing the guitar on his ranch with his friends and his cabinet didnt do much either.

people have been predicting that the levees wont hold for a major hurricane FOREVER. there have been NUMEROUS floods in this area. and yet, now they want help from the federal government? why didn't they pay for it themselves, with the money they had then, BEFORE this all happened? now, im not saying i think what is happened is just punishment or anything- its not the fault of the people, its the fault of their own corrupt government. i just dont think its (entirely) fair to ask people in, i dunno, North Dakota to pay for their unique problem when they had the funds AND the forewarning to do it themselves before.

But that's the levees. I think people should be giving (donating) etc to EVERYONE all that they possibly have. and i also think that, at this point, theres no other logical way to fix new orleans (which, by the way, guys, USED TO BE A SWAMP!!! IT'S SUPPOSED TO FLOOD THAT MUCH WHEN THERE'S A STORM) is by using federal funds. it still just seems rather rich of the LA gov't to be asking for money now when they had all that money AND all that warning.

but thats just me.

bleumoose you are forgetting that the Bush Administration did redraw the  funds for the repairing of the levees to (guess what) invade Iraq. Also most of the army personal and equipments are concentrated outside US, so I guess that is one of the major reasons for this late reply: [b]they weren't paying attention[/b].

I am not sure, but wasn't Lousiana a strong supporter of Bush? I guess they regret it now ...

PS It's good that most of the people got out in time though. The death toll seems to be around thousands, not hundred of thousands as it was in the Asian Tsunami last December.

when i said "they" should have paid to build up the levees beforehand, the "they" i was referring to was New Orleans' state and local governments.

its kind of a frustrating thing, all around, right now. [url=http://tinyurl.com/9zehw]Go here[/url] to see what I'm talking about. I don't think ANYONE realized on Monday, or Tuesday, that the situation in New Orleans would become this bad.

I have basically come to the conclusion that the most I can do is to help the refugees in my area and hope that everything will work out down there.

A OP-ED column from the NYTimes today:

[quote] Stuff happens.

And when you combine limited government with incompetent government, lethal stuff happens.

America is once more plunged into a snake pit of anarchy, death, looting, raping, marauding thugs, suffering innocents, a shattered infrastructure, a gutted police force, insufficient troop levels and criminally negligent government planning. But this time it's happening in America.

W. drove his budget-cutting Chevy to the levee, and it wasn't dry. Bye, bye, American lives. "I don't think anyone anticipated the breach of the levees," he told Diane Sawyer.

Shirt-sleeves rolled up, W. finally landed in Hell yesterday and chuckled about his wild boozing days in "the great city" of N'Awlins. He was clearly moved. "You know, I'm going to fly out of here in a minute," he said on the runway at the New Orleans International Airport, "but I want you to know that I'm not going to forget what I've seen." Out of the cameras' range, and avoided by W., was a convoy of thousands of sick and dying people, some sprawled on the floor or dumped on baggage carousels at a makeshift M*A*S*H unit inside the terminal.

Why does this self-styled "can do" president always lapse into such lame "who could have known?" excuses.

Who on earth could have known that Osama bin Laden wanted to attack us by flying planes into buildings? Any official who bothered to read the trellis of pre-9/11 intelligence briefs.

Who on earth could have known that an American invasion of Iraq would spawn a brutal insurgency, terrorist recruiting boom and possible civil war? Any official who bothered to read the C.I.A.'s prewar reports.

Who on earth could have known that New Orleans's sinking levees were at risk from a strong hurricane? Anybody who bothered to read the endless warnings over the years about the Big Easy's uneasy fishbowl.

In June 2004, Walter Maestri, emergency management chief for Jefferson Parish, fretted to The Times-Picayune in New Orleans: "It appears that the money has been moved in the president's budget to handle homeland security and the war in Iraq, and I suppose that's the price we pay. Nobody locally is happy that the levees can't be finished, and we are doing everything we can to make the case that this is a security issue for us."

Not only was the money depleted by the Bush folly in Iraq; 30 percent of the National Guard and about half its equipment are in Iraq.

Ron Fournier of The Associated Press reported that the Army Corps of Engineers asked for 105 million for hurricane and flood programs in New Orleans last year. The White House carved it to about 40 million. But President Bush and Congress agreed to a 286.4 billion pork-filled highway bill with 6,000 pet projects, including a 231 million bridge for a small, uninhabited Alaskan island.

Just last year, Federal Emergency Management Agency officials practiced how they would respond to a fake hurricane that caused floods and stranded New Orleans residents. Imagine the feeble FEMA's response to Katrina if they had not prepared.

Michael Brown, the blithering idiot in charge of FEMA - a job he trained for by running something called the International Arabian Horse Association - admitted he didn't know until Thursday that there were 15,000 desperate, dehydrated, hungry, angry, dying victims of Katrina in the New Orleans Convention Center.

Was he sacked instantly? No, our tone-deaf president hailed him in Mobile, Ala., yesterday: "Brownie, you're doing a heck of a job."

It would be one thing if President Bush and his inner circle - Dick Cheney was vacationing in Wyoming; Condi Rice was shoe shopping at Ferragamo's on Fifth Avenue and attended "Spamalot" before bloggers chased her back to Washington; and Andy Card was off in Maine - lacked empathy but could get the job done. But it is a chilling lack of empathy combined with a stunning lack of efficiency that could make this administration implode.

When the president and vice president rashly shook off our allies and our respect for international law to pursue a war built on lies, when they sanctioned torture, they shook the faith of the world in American ideals.

When they were deaf for so long to the horrific misery and cries for help of the victims in New Orleans - most of them poor and black, like those stuck at the back of the evacuation line yesterday while 700 guests and employees of the Hyatt Hotel were bused out first - they shook the faith of all Americans in American ideals. And made us ashamed.

Who are we if we can't take care of our own? [/quote]

Enough said.

[quote="bubala"]Enough said.[/quote]I couldn't agree more. I do wonder if by now, US has accepted any help from the International community (logistics, experience from last year's tsunami, etc.) because last time I saw someone talking about it, it was Mrs. Rice, which couldn't answer to the question "why haven't we accepted (or requested for that matter) any international help?" ... or again is Bush's personal pride coming in the way ...  :|


Suppose you adjust your garden hose nozzle for a hard stream of water. You point the nozzle vertically upward at a height of 1.5 m above the ground (See figure). When you quickly move the nozzle away from the vertical, you hear the water striking the ground next to you for another 2.0 s. What is the water speed as it leaves the nozzle?

[hide="Figure"][img]11988[/img]
You may have to click on the image to see it correctly.[/hide]

$ \minus{}1.5\equal{}v(2)\minus{}\frac{1}{2}g(2)^2 \equal{} 2(v\minus{}g) \implies v\minus{}g \equal{} \minus{}0.75 \implies \boxed{v \approx 9.05665 \frac{\text{m}}{\text{s}}}$

This problem requires two steps: the direct application of a well-known formula and the solving of a linear equation.

I think it's too straightforward for this forum  :wink:

Uhh... can you explain that formula for me?

cuz I've heard of it, but don't exactly know what it does since I forgot  :)

.The parabola marked out by an object while in projectile motion is
$ d\equal{}v_{0}t\minus{}\frac{gt^{2}}{2}$, where $ d$ is the vertical distance, $ t$ is the time, $ v_{0}$ is the intial velocity, and $ g$ is the constant of acceleration due to gravity.
This formula, though, can be applied to more then just projectile motion.  It simply models any system where an object with some initial velocity experiences some acceleration.
Deriving it can be done both with calculus and geometrically. :)

In linear motion:

Say you start with velocity $ v_o$ and accelerate with acceleration $ t$.
- Then, your velocity at time t is given by $ v(t) \equal{} v_o \plus{} at$.
- As your velocity ranges from $ v_o$ to $ v_o \plus{} at$, the average velocity is the arithmetic mean of those two, $ v_o \plus{} \frac{1}{2} at$.
- Then, [u]the net displacement[/u] over this time interval is $ \Delta d \equal{} v_{avg} t \equal{} (v_o \plus{} \frac{1}{2} at)t \equal{} v_o t \plus{} \frac{1}{2} at^2$


ye marathone moadele taebii ham rah bendazim bad nisst...
tamae tavabee $ f:\mathbb Q\rightarrow\mathbb Q$ ra biabid betorike $ f(x\plus{}y\plus{}z)\plus{}f(x\minus{}y)\plus{}f(y\minus{}z)\plus{}f(z\minus{}x)\equal{}3(f(x)\plus{}f(y)\plus{}f(z))$

khob javabe in moadele tabeyi $ f(x)\equal{}cx^{2}$ hastesh vase esbat :
1)$ x\equal{}y\equal{}z\equal{}0$ dar in soorat $ 4f(0)\equal{}9f(0)$ ke natije mide $ f(0)\equal{}0$ .
2)$ y\equal{}z\equal{}0$ ke natije mide $ f(x)\plus{}f(x)\plus{}f(\minus{}x)\equal{}3f(x)$ pas $ f(x)\equal{}f(\minus{}x)$ yani tabe zoje
3)$ z\equal{}0$ ke natije mide $ f(x\plus{}y)\plus{}f(x\minus{}y)\plus{}f(y)\plus{}f(\minus{}x)\equal{}3f(x)\plus{}3f(y)$ hala az zoj boodane tabe estefade konin be dast miad :
$ f(x\plus{}y)\plus{}f(x\minus{}y)\equal{}2f(x)\plus{}2f(y)$ .
hala inam ye moadele tabeyie maroofe Dgast ke rooye $ \mathbb{Q}$ javabesh $ f(x)\equal{}cx^{2}$ hastesh hala esbatesh :
4)farz konid $ f(1)\equal{}c$ dar in soorat gharar bedin $ y\equal{}1$ ke be dast miad $ f(x\plus{}1)\plus{}f(x\minus{}1)\equal{}2f(x)\plus{}2c$
ke inam mishe intori nevesht : $ f(x\plus{}1)\minus{}f(x)\equal{}f(x)\minus{}f(x\minus{}1)\plus{}2c\equal{}f(x\minus{}1)\minus{}f(x\minus{}2)\plus{}4c$ va age hamintori edame bedin natije mide :
$ f(x\plus{}1)\minus{}f(x)\equal{}2cx\plus{}f(1)\minus{}f(0)\equal{}c(2x\plus{}1)$ hala dobare ye rabete bazgashtie ke age hallesh konid rabeteye $ f(x)\equal{}cx^{2}$ too adade sahih be dast miad . hala vase adade gooya :
5)mikhaym sabet konim $ f(nx)\equal{}n^{2}f(x)$(n tabiyi hast) ke age ino sabet konim Dge too adade gooya ham badihie chon age bekhaim
$ f(\frac{a}{b})$ ro be dast biarim az oonjayi ke $ b$ adade sahihe (albatte mishe farz kard ke tabiyie chon tabe zoje) too rabeteye bala $ n\equal{}b$ ro gharar bedin dar in soorat baraye in adade gooya ham $ f(x)\equal{}cx^{2}$ mishe hala esbate oon rabete :
avval az hame gharar bedin $ x\equal{}y$ natije mishe $ f(2x)\equal{}4f(x)$ , bad gharar bedin $ y\equal{}2x$ be dast miad $ f(3x)\equal{}9f(x)$ va hala ba esteghra be sadegi sabet mishe $ f(nx)\equal{}n^{2}f(x)$ , Dge esbatesh ba khodetoon .

pas javabe moadele tabeyi $ f(x)\equal{}cx^{2}$ hastesh .

doroste khashi :wink:  toam soal bezar baW to ke manbae soale jabri dggg :P

yek soale asoon:

Find polynomials $ a(x), b(x), c(x)$ such that$ |a(x)|\minus{}|b(x)|\plus{}c(x) \equal{}\minus{}1$ for $ x <\minus{}1$ , $ 3x\plus{}2$ for $ \minus{}1\le x\le 0$ , $ \minus{}2x\plus{}2$ for $ x > 0$.


inequalities
algebra
polynomial
three variable inequality

[quote="siuhochung"]Using the well known substitution a replace by 1/a and so on, we have to prove
\[
\frac{{a^2 }}{{b + c}} + \frac{{b^2 }}{{c + a}} + \frac{{c^2 }}{{a + b}} \ge \frac{3}{2} + \frac{1}{4}\left( {\frac{{\left( {c - b} \right)^2 }}{{b + c}} + \frac{{\left( {b - a} \right)^2 }}{{b + a}} + \frac{{\left( {c - a} \right)^2 }}{{a + c}}} \right)
\]
Now, by rearrangement twice, we have
\[
\frac{{a^2 }}{{b + c}} + \frac{{b^2 }}{{c + a}} + \frac{{c^2 }}{{a + b}} \ge \frac{1}{2}\left( {\frac{{b^2  + c^2 }}{{b + c}} + \frac{{c^2  + a^2 }}{{c + a}} + \frac{{a^2  + b^2 }}{{a + b}}} \right)
\]
Then,
\[
\begin{array}{l}
 \frac{1}{2}\left( {\frac{{b^2  + c^2 }}{{b + c}} + \frac{{c^2  + a^2 }}{{c + a}} + \frac{{a^2  + b^2 }}{{a + b}}} \right) - \frac{1}{4}\left( {\frac{{\left( {c - b} \right)^2 }}{{b + c}} + \frac{{\left( {b - a} \right)^2 }}{{b + a}} + \frac{{\left( {c - a} \right)^2 }}{{a + c}}} \right) \\
  = \frac{1}{2}\left( {a + b + c} \right) \\
  \ge \frac{3}{2} \\
 \end{array}
\]
by am-gm.[/quote]
nice solution :)

Let $ a, b, c > 0 , abc \equal{} 1.$ The inequality

$ \frac {1}{a^3(b \plus{} c)} \plus{} \frac {1}{b^3(a \plus{} c)} \plus{} \frac {1}{c^3(b \plus{} a)} \geq \frac {3}{2} \plus{} k\left[\frac {a(b \minus{} c)^2}{b \plus{} c} \plus{} \frac {b(c \minus{} a)^2}{c \plus{} a} \plus{} \frac {c(b \minus{} a)^2}{a \plus{} b}\right]$

holds if and only if $ k\leq k_2 \equal{} 0.75822\cdots$, where $ k_2$ is  the least real root of of irreducible polynomial

$ 4194304k^8 \minus{} 31326208k^7 \plus{} 84357120k^6 \minus{} 100112384k^5 \plus{} 59959552k^4$

$ \minus{} 26256640k^3 \plus{} 10267728k^2 \minus{} 1690056k \plus{} 121039.$

With equality if $ k \equal{} k_2, a \equal{} x_2 \equal{} 3.2270\cdots$ is  the least real root of of irreducible polynomial

$ x^8 \minus{} 13x^7 \plus{} 22x^6 \plus{} 29x^5 \plus{} 6x^4 \minus{} 3x^3 \plus{} 6x^2 \plus{} 3x \plus{} 1,$

$ b \equal{} c \equal{} y_1 \equal{} 0.55666\cdots$ is  the greatest real root of of irreducible polynomial

$ y^8 \plus{} 3y^7 \plus{} 6y^6 \plus{} 7y^5 \plus{} 6y^4 \plus{} 5y^3 \plus{} 2y^2 \minus{} 3y \minus{} 1.$

Could you Ji Chen please prove those facts? I don't see where did you get that polynomial. And how did you prove that it is irreducible?

[hide="Classical Solution"]
The stronger inequality
\[\sum_{c}\frac{b^2c^2}{a(b+c)}\geq \sum_{c}\frac{ab}{2}+\sum_{c}\frac{a(b-c)^2}{4(b+c)}\]
after some calculations is equivalent to:
\[\sum_{c}a^4(b^2-c^2)^2\geq 0\]
[/hide]

I recall the dozen of inequalities I have and observed that Cauchy-Schwarz will be useful. By Cauchy-Schwarz, we have
\[\left(a(b+c)+b(a+c)+c(a+b)\right)\left(\dfrac{1}{a^3(b+c)}+\dfrac{1}{b^3(a+c)}+\dfrac{1}{c^3(b+a)}\right)\]\[ \geq\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=(ab+bc+ac)^2\]
Dividing both sides by $2(ab+bc+ac)$ is quite natural cause that will cancel the undesired term in LHS, it yields:
\[\dfrac{1}{a^3(b+c)}+\dfrac{1}{b^3(a+c)}+\dfrac{1}{c^3(b+a)} \geq \dfrac{(ab+bc+ac)}{2} \]
Note that by AM-GM \[(ab+bc+ac) \geq 3 \sqrt[3]{(a^2b^2c^2)} \; \; \;  \Rightarrow \; \; \; (ab+bc+ac) \geq 3\]
Above 2 inequalities gives us:
\[\dfrac{1}{a^3(b+c)}+\dfrac{1}{b^3(a+c)}+\dfrac{1}{c^3(a+b)} \geq \dfrac{3}{2}\]
Which is the desired result.
$\blacksquare$


calculus
derivative
calculus computations

Couldn't decipher all that writing so I'll just do it:

$ y' \equal{} 5(e^{\frac{2}{x}} \cdot 2(x^{\minus{}1})') \equal{} 5(e^{\frac{2}{x}} \cdot \minus{}2x^{\minus{}2} )$

[quote="damgyeah"]y = 5e^(2/x)
how to do derivative of this? isn't this just 5e^(2/x) * chain rule? which is -2/x^2?
but isn't there a trick when e^ something rational? (a fracton).
y = e^x the y prime is just a e^x but if y = e^(1/x) i believe there is a trick, more of a rewrite to get rid of that fraction, which i forgot.[/quote]

The trick you're looking for is the chain rule:

if $ f\left(u\left(x\right)\right)$ then $ \frac {df}{dx} \equal{} \frac {df}{du}\frac {du}{dx}$

In this case you have $ f\left(u\right) \equal{} 5e^u$ and $ u\left(x\right) \equal{} \frac {2}{x}$.

The actual calculation is trivial at this point so I'll leave it to you to finish.


geometry
geometric transformation
reflection
absolute value
combinatorics proposed
combinatorics

There are $n$ points on a circle, such that each line segment connecting two points is either red or blue.
$P_iP_j$ is red if and only if $P_{i+1} P_{j+1}$ is blue, for all distinct $i, j$ in $\left\{1, 2, ..., n\right\}$.

(a) For which values of $n$ is this possible?
(b) Show that one can get from any point on the circle to any other point, by doing a maximum of 3 steps, where one step is moving from a point to another point through a red segment connecting these points.

I hope that what follows is right:

We will work with indices modulo $n$. We also make the following association: $0 \mapsto \textrm{red}, \, 1 \mapsto \textrm{blue}$. $AB = 0$ or $AB = 1$ will mean that $[AB]$ has the corresponding color.

[b](a)[/b] Letting $j = i+k$, we see that $P_{i+2m}P_{i+2m+k}= P_{i}P_{i+k}$ and that $P_{i+2m+1}P_{i+2m+1+k}= 1-P_{i}P_{i+k}$. Thus, the color of a segment of "length" (i.e. the absolute value of the diferrence of the indices of its endpoints --- lots of "of", sorry  :blush: ) determines completely the color of all segments of length $k$. So the only contradiction that could arise is that $P_{i}P_{i+k}= 0$ implies $P_{i}P_{i+k}= 1$ (or the other way around). We have two cases (all computations are $\bmod \, n$):

[i]Case 1.[/i] $i = i+2m+1, \, i+k = i+2m+1+k \, \, \Longleftrightarrow \, \, 2m+1 = 0 \, \, \Longleftrightarrow \, \, n \equiv 1 \left( \bmod \, 2 \right)$.

[i]Case 2.[/i] $\left. i = i+2m+1+k, \, i+k = i+2m+1 \, \, \Longleftrightarrow \, \, 2m+1=-k, \, 2m+1=k \, \, \Longleftrightarrow n \equiv 2 \left( \bmod \, 4 \right) \right.$ (Don't forget that $k = j-i \neq 0$).

Thus, the only "good" $n$ are those that satisfy $n \equiv 0 \left( \bmod \, 4 \right)$.

[b](b)[/b] Make some cyclic permutations and perhaps a "mirror reflection" on the indices of the $P_{i}$'s such that $P_{0}P_{1}= 0$, where $P_{0}$ is the point from where we start our journey.

We'll prove that we can reach $2k, \, 2k+1$, $\forall k \in \overline{1, \frac{n}2-1}$ ($P_{1}$ can obviously be reached). Clearly, we have that $P_{2k}P_{2k+1}= 0$. If $P_{0}P_{2k}= 0$, then we can take the route $P_{0}\mapsto P_{2k}\mapsto P_{2k+1}$. Otherwise, $P_{0}P_{2k}= 1$, so $P_{1}P_{2k+1}= 0$. Thus, we can choose: $P_{0}\mapsto P_{1}\mapsto P_{2k+1}\mapsto P_{2k}$.

So... is it OK?

And also: what's $AMO$?

@Perfect_radio
After ten years I think may be you have made a little mistake  :oops:
I think the answer of $(a)$ is just $2|n$
You may ask $n=6$ you choose the longest diagonal will be a contradiction,but the problem didn't say you can "mod", for example,it didn't say $P_{n+1}=P_{1}$ so may be $P_{j+1}$ doesn't exist at all.
If you do this problem like this the answers will be different,and the difficulty will increase a lot,but it's a P5 so what do you think?(or you can just ignore this post and consider me as a stupid boy who couldn't figure out this problem after thinking for 10 years)
Your sincerely
$ CeuAzul $

[quote=Math1331Math]Who can resolve this proof?[/quote]

If the points aren't taken cyclicly, then part (a) of the problem makes zero sense, since any value of $n$ would work. So I think it's safe to assume that the points are taken cyclicly. I doubt part (b) would be true if the points aren't taken cyclicly since the problem would be far too weak, though I may be wrong.

(a): If $n$ is odd, then $j = i + 1$ is enough to form a contradiction by going around in a circle, since a line segment cannot be both red and blue. If $n$ is 2 mod 4, then by considering the main diagonals, the same contradiction occurs. So $n$ must be a multiple of $4$. It's easy to form a construction in this case.
(b) Suppose there are $4k$ vertices. Let $u$ be a vertex with degree $2k$ red edges and $v$ be a vertex with degree $2k-1$ red edges. Suppose for the sake of contradiction that there exists no path of length 3. Let $w$ be any point such that $v$ and $w$ are connected with a red edge. Then, there cannot exist an edge from $v$ to $v$, $u$ or any one of the possible $w$'s. This rules out $2k + 2$ vertices, so at most $2k - 2$ vertices can be connected to $v$ with a red edge. Since $v$ is of degree $2k - 1$, this is impossible. So, there must exist a path of length at most 3.


geometry
circumcircle
geometric transformation
reflection
rectangle
geometry proposed

[quote="livetolove212"]Given triangle $ ABC$, its orthocenter $ H$ and its circumcircle $ (O)$. Denote $ A_1, B_1, C_1$ the reflections of $ A,B,C$ wrt $ O$, respectively, $ P$ an arbitrary point on the plane. Let $ A_2B_2C_2$ be the pedal triangle of $ P$ wrt $ \Delta ABC, A_3,B_3,C_3$ be the reflections of $ A_1,B_1,C_1$ wrt $ A_2,B_2,C_2$, respectively. Prove that 4 points $ H, A_3,B_3,C_3$ are concyclic.[/quote]

Nice problem!

Dear Victory,
Can you post your solution?

[b]My solution:[/b]
Let $ A_4, B_4, C_4$ be the midpoints of $ BC, CA, AB$, respectively then $ A_4$ is the midpoint of $ A_3H$. Similar with $ B_4, C_4$.
We get $ HA_3//\equal{}2A_2A_4, HB_3//\equal{}2B_2B_4, HC_3//\equal{}2C_2C_4. (1)$
Construct rectangles $ OA_4A_2A_5, OB_4B_2B_5, OC_4C_2C_5$ hence $ A_5, B_5, C_5$ lie on the circle with diameter $ OP. (2)$
 But $ OA_5//\equal{}A_4A_2, OB_5//\equal{}B_4B_2, OC_5//\equal{}C_4C_2 (3)$
 From $ (1), (2)$ and $ (3)$ we claim $ H, A_3, B_3, C_3$ are concyclic. Moreover, we can show that $ \Delta A_3B_3C_3\sim \Delta ABC$.


I have another inequality. I think it's not very hard but i can't solve it
Let $ a,b,c\geq0$ and $ a\plus{}b\plus{}c\equal{}3$
$ \frac{a\sqrt{a}}{b\plus{}c}\plus{}\frac{b\sqrt{b}}{a\plus{}c}\plus{}\frac{c\sqrt{c}}{a\plus{}b}\geq\frac{3}{2}$

PS: I'm not sure it's true or not. If it's true, please solve it. Thanks

[quote="hoangnhatvn"]I have another inequality. I think it's not very hard but i can't solve it
Let $ a,b,c\geq0$ and $ a \plus{} b \plus{} c \equal{} 3$
$ \frac {a\sqrt {a}}{b \plus{} c} \plus{} \frac {b\sqrt {b}}{a \plus{} c} \plus{} \frac {c\sqrt {c}}{a \plus{} b}\geq\frac {3}{2}$

PS: I'm not sure it's true or not. If it's true, please solve it. Thanks[/quote]

 $ \sum {\frac{a^3}{b^2\plus{}c^2}}\ge\frac{\sqrt{3(a^2\plus{}b^2\plus{}c^2)}}{2}$ for any a,b,c positive.

Oh, I see. Thank you. It's a nice solution.

Another way: Jensen for $ f(x) \equal{} \frac{x\sqrt{x}}{3 \minus{} x}$

Yeah. Jensen's inequality is helpful in this problem.
I have another solution
$ \frac{a^3}{b^2\plus{}c^2}\plus{}\frac{b^3}{c^2\plus{}a^2}\plus{}\frac{c^3}{a^2\plus{}b^2}\geq\frac{(a^2\plus{}b^2\plus{}c^2)^2}{ab^2\plus{}ac^2\plus{}bc^2\plus{}ba^2\plus{}ca^2\plus{}cb^2}$
By Cauchy-Swartz, we have
$ (ab^2\plus{}ac^2\plus{}bc^2\plus{}ba^2\plus{}ca^2\plus{}cb^2)^2\leq\frac{4}{3}.(a^2\plus{}b^2\plus{}c^2)^3$
So, the inequaltity is proved.

[quote="hoangnhatvn"]
By Cauchy-Swartz, we have
$ (ab^2 \plus{} ac^2 \plus{} bc^2 \plus{} ba^2 \plus{} ca^2 \plus{} cb^2)^2\leq\frac {4}{3}.(a^2 \plus{} b^2 \plus{} c^2)^3$
[/quote]
and
[quote="Marius Mainea"]
$ \sum {\frac {a^3}{b^2 \plus{} c^2}}\ge\frac {\sqrt {3(a^2 \plus{} b^2 \plus{} c^2)}}{2}$ for any a,b,c positive.
[/quote]

Can you write narrowly how you did it? I just learn and I really want to know it :D

[quote="hoangnhatvn"]
$ 3.({ab}^2 \plus{} {ac}^2 \plus{} {bc}^2)\leq(a^2 \plus{} b^2 \plus{} c^2)^2$
[/quote]
Can you explain it?  :oops:

sorry, it's my mistake when I use Latex. Here is the right inequality:
$ 3.(a^2b^2 \plus{} a^2c^2 \plus{} b^2c^2)\leq(a^2 \plus{} b^2 \plus{} c^2)^2$


The reciprocal of 2x-3 is 2x+3. Find all possible nonnegative values of x.

[quote="sugarwergoingdown"]klebian, unfortunately your solution is wrong. :(  please try it again? ;)[/quote]
If my solution is wrong, i don't know.
But if its my answer, then simplified, you would get

[hide]$\frac{\sqrt{10}}2$[/hide]
Otherwise, I can't see my mistake

[hide]they are recipricals, so 2x+3 X 2x-3=1 Therefore, $4x^2-9=1$
Add 9 to both sides, get $4x^2=10$
Divide by 4, get $x^2=2.5x$
$sqrt{10/4}=1/2sqrt{10}$, so $x=sqrt{10}/2$[/hide]

[quote="ch1n353ch3s54a1l"][quote="pianoforte"][hide]$\frac{1}{2x-3}=2x+3$
$1=4x^2-9$
$4x^2-10=0$
$(2x+\sqrt{10})(2x-\sqrt{10})$
$x=\pm \frac{\sqrt{10}}{2}$[/hide][/quote]

[hide]it's just $x=\frac{\sqrt{10}}{2}$. The question asks for non-negative values.  :)[/hide][/quote]

[hide]darn, I forgot about that part :P [/hide]

[quote]klebian, unfortunately your solution is wrong. Sad please try it again? Wink
[/quote]

unfortunately, klebian is right, it's just not simplified enough... you never mentioned simplifing ;)
( if it was in the Sprint round, they would have mentioned to simplyfy to the simpliest radical form, but oh well...)

They wouldn't have specified it in the problem. It's a general rule that all answers must be in simplest form, so it would have been considered wrong. The only "right" answer is $\frac{\sqrt{10}}{2}$.

I believe in MC if the solution is in Radical Form, they specify to put them in the simpliest radical form

Or in this case a common fraction in simplest radical form :o  :weightlift:


AMC
AIME
geometry
trapezoid
rectangle
geometric transformation
rotation

All right, I wanted to create a geometry marathon for those who feel that they aren't as good at geometry as they'd like to be (myself included). This marathon should be for AIME level problems, and I hope that some people will make up problems to add a little variety. Maybe I'll make some too.

This marathon should follow the rules of any other marathons. If you post a solution you're sure is correct, then post a problem as well. If you just post some speculations/ideas, then dont post a new problem. If no one answers a question for two days or so, I'll post a new and perhaps easier problem.

All sorts of geometry problems are fine. I'll start us off with #8 from the AIME 2000 II.

In trapezoid $ ABCD$, leg $ \overline{BC}$ is perpendicular to bases $ \overline{AB}$ and $ \overline{CD}$, and diagonals $ \overline{AC}$ and $ \overline{BD}$ are perpendicular. Given that $ AB \equal{} \sqrt {11}$ and $ AD \equal{} \sqrt {1001}$, find $ BC^2$.

[b]EDIT: I wasn't aware of the other marathons. I'll go check those out. I've changed this marathon now to a combinatorics marathon. Please post problems.[/b]

[hide]If you draw a perpendicular AH to CD, you'll notice that AHD forms a right triangle (WOOT!). AHCB is a rectangle, so CH is equal to $ \sqrt {11}$ and AH is equal to BC. By subtraction, HD is CD-$ \sqrt {11}$. With the Pythagorean Theorem on AHD, $ BC^2 \plus{} (CD \minus{} \sqrt {11})^2 \equal{} \sqrt {1001}$ which reduces to
\[ BC^2 \plus{} CD^2 \minus{} 2\sqrt {11}CD \equal{} 990.
\]
Let us label the intersection of AC and BD as X. Now for some angle chasing! Because ABCD is a trapezoid, $ \angle BAC \equal{} \angle ACD$ and $ \angle BDC \equal{} \angle ABD$. Triangle AXB is a right triangle so $ \angle CAB \plus{} \angle DBA \equal{} 90^\circ.$ $ \angle ABC$ is a right angle, so $ \angle DBC \equal{} 90 \minus{} \angle ABD \equal{} \angle CAB.$ Similarly, $ \angle ACB \equal{} 90 \minus{} \angle CAB \equal{} \angle DBA.$ By AA, $ \triangle ABC \sim \triangle BCD.$ Using that similarity, $ \frac {CB}{DC} \equal{} \frac {AB}{BC}$ so $ BC^2 \equal{} \sqrt {11}CD.$ If we plug that back into the first equation, we get $ BC^2 \plus{} \frac {BC^4}{11} \minus{} 2\sqrt {11}BC^2 \equal{} 990$ which reduces to
\[ BC^4 \minus{} 11BC^2 \minus{} 990(11) \equal{} 0.
\]
That factors into $ (BC^2 \minus{} 110)(BC^2 \plus{} 99) \equal{} 0$ so $ BC^2 \equal{} 110$.
[/hide]

2003 AIME II Number 6: In triangle ABC, AB=13, BC=14, AC=15  and point G is the intersection of the medians. Points A', B', and C'     are the images of A, B, and C respectively, after a $ 180^\circ$ rotation about G. What is the area of the union of the two regions enclosed by the triangles ABC and A'B'C'?

i dont really think this is necissary....there is already an aime marathon...

and an interm geo marathon in the intermediate forum....

Someone should start an AIME level marathon for something other than NT and geo. (Combinatorics? Algebra?)

Edit: lol, ok, guess it's a combinatorics marathon now.

[b]Problem.[/b] A particle moves in the plane according to these rules:
1. From lattice point $ (a,b)$ it can move to only $ (a\plus{}1,b)$, $ (a\plus{}1,b\plus{}1)$, or $ (a,b\plus{}1)$.
2. It may not take any right angle turns.
Find the number of possible paths from $ (0,0)$ to $ (5,5)$.

[quote="grn_trtle"]Someone should start an AIME level marathon for something other than NT and geo. (Combinatorics? Algebra?)

Edit: lol, ok, guess it's a combinatorics marathon now.

[b]Problem.[/b] A particle moves in the plane according to these rules:
1. From lattice point $ (a,b)$ it can move to only $ (a + 1,b)$, $ (a + 1,b + 1)$, or $ (a,b + 1)$.
2. It may not take any right angle turns.
Find the number of possible paths from $ (0,0)$ to $ (5,5)$.[/quote]

I think that to solve this problem, one needs to separate the cases, and then analyze each of those.

[hide="Solution"]We need to go from (0,0) to (5,5) meaning that we need to move 5 units up and 5 units down, which can be signified by a sequence of letters:

UUUUURRRRR

The 2nd restriction essentially boils the problem down to having no U's and R's together.

Then we can look at the cases:

Case 0: 0 diagonal movements

UUUUURRRRR

This case is not possible.

Case 1: 1 diagonal movement

UUUUDRRRR

This case is possible, and there are $ 2$ ways to do it (either start off with 4 U's or 4R's)

Case 2: 2 Diagonal movements

UUUDDRRR

We can count the cases for this using a slightly-brute force method. Start off with DD.

Now, we need to place three more U's and then three R's such that they are not next to each other.

The number of ways for this (using brute force) is $ 4 + 1 + 1 + 4 + 4 + 1 + 1 = 16$

Case 3: 3 Diagonal Movements

UUDDDRR

Using some brute force again with 10 cases, I find $ 42$ more possibilities.

Case 4: 4 Diagonal Movements

UDDDDR

$ 20$ possibilties.

Case 5: 5 diagonal Movements

DDDDD

1 possibility obviously.

I find the answer then to be $ 2 + 16 + 42 + 20 + 1 = \fbox{81}$?[/hide]

EDIT: Shoot, I just checked the online solution. Im off by two!  :(  :(  :(

That's the closest I've ever gotten to solving a problem over #11, so I'm still fairly happy  :D

2nd EDIT: I should probably post a problem. Here's one, quite easy:

Given eight distinguishable rings, let $ n$ be the number of possible five-ring arrangements on the four fingers (not the thumb) of one hand. The order of rings on each finger is significant, but it is not required that each finger have a ring. Find the leftmost three nonzero digits of $ n$.

[hide="Solution"]Let the five rings be a,b,c,d,e. We're looking for things like:
\[ a|bc\parallel{}de,\]
and permutations thereof, where each $ |$ partitions our string into stuff.

There are $ 8!/3!\equal{}1680$ ways to pick the five rings, and $ \binom{8}{3}\equal{}56$ ways to pick the places for the dividers, making $ n\equal{}376320$. Therefore, the answer is $ \boxed{376}$.
[/hide]

Wow, was that an AIME problem? That was ridiculously easy. Was it a #1?

Note: When your solutions are that huge, you NEED to hide them, because all that scrolling is getting really annoying.

[b]Problem.[/b] The numbers 1, 2, ..., 8 are written randomly on a regular octahedron so that no two faces have the same number. The probability that no two consecutive numbers (where 8 and 1 are considered consecutive) share an edge is $ m/n$. Find $ m\plus{}n$.

[quote="grn_trtle"][hide="Solution"]Let the five rings be a,b,c,d,e. We're looking for things like:
\[ a|bc\parallel{}de,
\]
and permutations thereof, where each $ |$ partitions our string into stuff.

There are $ 8!/3! \equal{} 1680$ ways to pick the five rings, and $ \binom{8}{3} \equal{} 56$ ways to pick the places for the dividers, making $ n \equal{} 376320$. Therefore, the answer is $ \boxed{376}$.
[/hide]

Wow, was that an AIME problem? That was ridiculously easy. Was it a #1?

Note: When your solutions are that huge, you NEED to hide them, because all that scrolling is getting really annoying.

[b]Problem.[/b] The numbers 1, 2, ..., 8 are written randomly on a regular octahedron so that no two faces have the same number. The probability that no two consecutive numbers (where 8 and 1 are considered consecutive) share an edge is $ m/n$. Find $ m \plus{} n$.[/quote]

Sorry, my bad. My solutions tend to take up a lot of space. Yeah, that last question was a #5 surprisingly, and it was the question I answered the fastest while taking that AIME.

This new problem will require some thought...

Nah, I guess I can't solve it. Blargh. Hmm, I'm thinking of doing massive casework, but I always feel that I'm doing something wrong when I'm doing that, and its kind of boring to do so (hmm, probably why I wasn't so confident about that last super-long question)

Wow, that was a #5  :huh:

I think the actual solution for that octahedron problem uses casework, so... yeah... should have picked a better problem  :oops:

How about this?
How many ways are there to cover a 3 by 8 rectangle with 12 identical dominoes?

That is just a recursion problem! SO I would say #6.


inequalities logarithms function LaTeX calculus limit derivative

Just use homogeniztions. Will post my soln later.

[hide="@Arvind:"]Hi Arvind.
I have a request. Please do not leave short comments on problems. I won't dare say that what you say is wrong, because I've no idea what you were actually doing in detail. But you've been doing this often. For example you said "Just use Cauchy lemma and finish off" in a IRAN 96 ineq thread. In most cases, even if you were right, the method is far from obvious from your short reply. In some cases there are long discussions in a apparently hard problem and you just say about a short way which appears to me not right. Please post either full solution or don't comment.[/hide]

The inequality reversely holds.
$ LHS_{a\equal{}0.5,b\equal{}0.25,c\equal{}0.25}\approx 0.2973017787507<\frac{1}{3}$

[hide="Solution"]Let $ f(x)\equal{}\frac{(1\minus{}x)\ln x}{2}$
$ f''(x)\equal{}\minus{}\frac{1}{2x}\minus{}\frac{1}{2x^2}$
which is negative for $ x\in \mathbb R^\plus{}$
Hence $ f$ is concave.
By Jensen's inequality:
$ \ln (LHS)\equal{}\sum_{cyc} f(a)\le 3f\left(\frac{a\plus{}b\plus{}c}{3}\right)\equal{}\ln (RHS)$
Hence $ \sqrt {a^{1 \minus{} a}b^{1 \minus{} b}c^{1 \minus{} c}}\le \dfrac{1}{3}$[/hide]

Nice solution but a small doubt
Akashnil did u take the function as $ 1\minus{}x$*$ lnx/2$ becuase of the square root ???

Thx,
Befuddlers

Yes.
Remember that $ \ln t^u\equal{}u\ln t$. So:
$ \ln \sqrt{t}\equal{}\ln t^{\frac{1}{2}}\equal{}\frac{1}{2}\ln t$
Just calculate $ \ln (LHS)$ and then expand and you'll know that was the obvious function to take.

akashnil, I don not understand your solution. I don't understand the jensen's inequality as well as the $ ln$. Please explain.

[hide]I just got an idea and so said. But now I know that an idea doesn't finish it all. But in my earlier posts, wheneever I used to say Cauchy lemma, except 1, all oters I worked out and only told. I was lazy to post the full soln in latex and moreover I feel latex is difficult.[/hide]

[hide="@Arvind"][quote="Aravind Srinivas L"]akashnil, I don not understand your solution. I don't understand the jensen's inequality as well as the $ ln$. Please explain.[/quote]
It will be very heard to explain it to you. You are in 9th grade only. Do you know calculus? If you do, I'll try a bit to explain. $ \ln$ is the natural logarithm, that is $ \log_e$, where $ e$ is a mathematical constant defined to be $ \lim _{n\to \infty}\left(1\plus{}\frac{1}{n}\right)^n$ or $ \sum_{n\equal{}1}^{\infty} \frac{1}{n!}$.

[quote="Aravind Srinivas L"]I just got an idea and so said.[/quote]
It's well known that sometimes conditions like $ a\plus{}b\plus{}c\equal{}1$ provokes homogenization. No point talking about it. Also even when homogenization is used, it never completes the solution. It just helps to use known techniques because known techniques are mostly on homogenized form. If you really understand what homogenization means you would find that it is irrelevant here. The degree of the inequality is variable itself.

[quote="Aravind Srinivas L"]But in my earlier posts, wheneever I used to say Cauchy lemma, except 1, all oters I worked out and only told. I was lazy to post the full soln in latex and moreover I feel latex is difficult.[/quote]
Latex is not difficult. You should learn it if you plan to interact in this forum.[/hide]

There is a nice solution using only
$ a^ab^b\geq a^bb^a$ and am-gm

Is this the solution you were talking about? It's indeed nice. :)

[hide]
$ a^a\left(\frac{1}{3}\right)^{\frac{1}{3}}\ge a^{\frac{1}{3}}\left(\frac{1}{3}\right)^a$

$ \iff a^a\ge 3^{\frac{1}{3}\minus{}a}\sqrt[3]{a}$

similarly for $ b,c$. Hence:

$ a^ab^bc^c\ge 3^{\frac{1}{3}\minus{}a}\sqrt[3]{a}\cdot 3^{\frac{1}{3}\minus{}b}\sqrt[3]{b}\cdot 3^{\frac{1}{3}\minus{}c}\sqrt[3]{c}\equal{}\sqrt[3]{abc}\equal{}$
$ \equal{}\sqrt[3]{(a\plus{}b\plus{}c)^6abc}\ge \sqrt[3]{(3\sqrt[3]{abc})^6abc}\equal{}9abc$

$ a^ab^bc^c\ge 9abc$
$ \iff a^{1\minus{}a}b^{1\minus{}b}c^{1\minus{}c}\le \frac{1}{9}$
Taking the square root, we are done.[/hide]

[quote="pqrs"]Prove that
$ \sqrt {a^{1 \minus{} a}b^{1 \minus{} b}c^{1 \minus{} c}}\ge \dfrac{1}{3}$ for $ a,b,c\in R^ \plus{}$ such that $ a \plus{} b \plus{} c \equal{} 1$[/quote]

It should be
\[ \sqrt {a^{1 \minus{} a}b^{1 \minus{} b}c^{1 \minus{} c}}\le \dfrac{1}{3}\]
and it's quite straightforward from
\[ a^ab^bc^c\ge (abc)^\frac{a\plus{}b\plus{}c}{3}.\]
since $ abc\le \frac{1}{27}$.

Yes this was the approach I was talking about.

Btw, this can be directly solved using the stronger $ a^ab^bc^c\ge\left(\frac {a \plus{} b \plus{} c}{3}\right)^{a \plus{} b \plus{} c}$, which follows from the fact that $ f(x) \equal{} x\ln x$ is convex in $ \mathbb R^ \plus{}$.

@Arvind: If you have any bookstore/library around your home, you can probably find some books on calculus there. :)

[quote="Aravind Srinivas L"]No I do not know Calculus. Please give me some material from which I can learn functions, their types, limits, etc[/quote]

In case you are preparing for olympiads, you will not need to know calculus.

Doesn't jensen's inequality use Calculus?

no. no calculus
http://planetmath.org/?op=getobj&from=objects&id=7881

Are complex numbers used for maths olympiads? Maybe in equiangular polygons we use them . Sorry for deviation but please answer.

Yes . Thats the same doubt I had.
akashnil,  if u say not much of calculus, then what are all the topics I need to learn in Calculus? Further, remember to pm me {very old}.

[quote="Akashnil"]Not much calculus.
But don't we need to calculate 2nd derivative in order to determine if the function is convex or concave?[/quote]

It helps and is a shortcut most of the times, but is not necessary. Don't forget that the definition of convexity doesn't involve derivatives. Indeed there are convex functions that don't have a second derivative  :wink:

Regarding complex numbers and the whole equiangular polygons topic, i'm sure there are other more appropriate threads to discuss it, but in general, knowing some complex numbers (roots of unity, etc.) doesn't hurt.
They are not required though, If you can handle the ugly algebra or trigonometry you can always exclude mentioning complex numbers (in Olympiads, that is.)

here you just go to mr kk anand.
he is a genius


[quote="arqady"]For non-negative $ a,$ $ b$ and $ c$ such that $ ab \plus{} ac \plus{} bc \equal{} 3$ prove that
\[ 8(a \plus{} b \plus{} c)^2\geq9(a \plus{} b)(a \plus{} c)(b \plus{} c)
\]
[/quote]
Let $ p\equal{}a\plus{}b\plus{}c,q\equal{}ab\plus{}bc\plus{}ca\equal{}3,r\equal{}abc$
The inequality can be rewritten to
$ 8p^2 \ge 9(3p\minus{}r) \Leftrightarrow 8p^2\minus{}27p\plus{}9r \ge 0$
By Cauchy-Schwarz inequality, we have:
$ p^2\equal{}(a\plus{}b\plus{}c)^2\ge 3(ab\plus{}bc\plus{}ca)\equal{}9 \Rightarrow p\ge 3$
$ \star \ \ p\ge 5\Rightarrow 8p^2\minus{}27p\equal{}p(8p\minus{}27)>0$
$ \star \ \ 3\le p\le 5$
By Schur inequality: $ p^3\plus{}9r \ge 4pq \Rightarrow 9r \ge 12p\minus{}p^3$
So we need to prove that:
$ p^3\minus{}8p^2\plus{}15p \le 0 \Leftrightarrow p(p\minus{}3)(p\minus{}5) \le 0$
So we are done.


integration
function
calculus
logarithms
calculus computations

can anyone help me to find this,thanks before.

let $ a>1$ and $ b>1$ two reals.

calculate: $ \int_{0}^{\pi}ln\left(\frac{b\minus{}cosx}{a\minus{}cosx}\right)dx.$

[color=blue][i]clearly the integrand is well continuous on $ [0,\pi]$.
looking for a primitive seemes to be not possible.  However, we can write:
$ I=\int_{0}^{\pi}ln(\frac{b-cosx}{a-cos})dx= F(b)-F(a)$
where $ F: t\to\int_{0}^{\pi}ln(t-cosx)dx.$

ths function $ (t,x)\in\ (1,\infty)*\mathbb{R}\to\ln(t-cosx)$ is $ \mathbb{C}^{\infty},$

so we can derive under the integral sign, so that for $ t>1,$ $ F'(t)=\int_{0}^{\pi}\frac{dx}{t-cosx}dx.$

the change of variable $ u=tan(\frac{u}{2})$ give:
$ F'(t)=\int_{0}^{\infty}\frac{2du}{(1+u^{2})(t-\frac{1-u^{2}}{1+u^{2}})}$
             $ =\int_{0}^{\infty}\frac{2du}{(1+u^{2})t-(1-u^{2})}$
             $ =\frac{2}{1+t}\int_{0}^{\infty}\frac{du}{u^{2}+\frac{t-1}{t+1}}$
             ${ =\frac{2}{t+1}[\frac{1}{\sqrt{\frac{t-1}{t+1}}}arctan(\frac{u}{\sqrt{\frac{t-1}{t+1}}}})]_{0}^{\infty}$
             $ =\frac{\pi}{\sqrt{t^{2}-1}}$

and so: $ F(t)=ln(t+\sqrt{t^{2}-1})+Cste$

We deduce therefore that:
 $ I=\pi\ln(\frac{b+\sqrt{b^{2}-1}}{a+\sqrt{a^{2}-1}}).$[/i][/color]

[size=75]sorry there is some stupidity in my proof.[/size]
------------------------------------------------------------------------------------------------------------
Radouan.


Putnam
probability
induction
geometry
perimeter
expected value
college contests

An $m\times n$ checkerboard is colored randomly: each square is independently assigned red or black with probability $\frac12.$ we say that two squares, $p$ and $q$, are in the same connected monochromatic region if there is a sequence of squares, all of the same color, starting at $p$ and ending at $q,$ in which successive squares in the sequence share a common side. Show that the expected number of connected monochromatic regions is greater than $\frac{mn}8.$

I was going to post this Monday, but the forum went down. Since I saved it, I can post it now:

This is probably the hardest problem- this or B6.

Let $f(m,n)$ be the expected number of connected monochromatic regions in an $m\times n$ rectangle. We will show that $f(m+1,n)-f(m,n) > \frac{n}{8}$.
We do this by considering the new $1\times n$ strip that is added (to the top). The probability that a $1\times k$ block is, on its own, a connected monochromatic region, is $2\cdot 2^{-2k-2}=2^{-2k-1}$ if the block is in the middle and $2\cdot 2^{-2k-1}=2^{-2k}$ if the block has one end in a corner. (The block must be one color, and every adjacent square must be the opposite color). For a $1\times n$ block, the probability is simply $2\cdot2^{-2n}$
Therefore, the expected number of new $1\times k$ regions is $2^{-2k}+(n-k-1)\cdot 2^{-2k-1}+2^{-2k}=(n-k+3)\cdot2^{-2k-1}$ and the expected number of new regions is $\sum_{k=1}^{n-1} \frac{n-k+3}{2^{2k+1}}+\frac2{2^{2n}}$.

Now, we have to deal with old regions that are eliminated by merging two existing regions into one. Each such merge has the pattern: $\left[\begin{array}{c} RRR\dots RR \\ RBB\dots BR\end{array}\right]$ or its color-reverse. If the middle block in the lower row has $k$ elements, this color scheme has probability $2\cdot 2^{-2k-4}=2^{-2k-3}$, so the expected number of such "bridges" is $(n-k-1)\cdot 2^{-2k-3}$ and the total expected number of "bridges" is $\sum_{k=1}^{n-2}\frac{n-k-1} {2^{2k+3}}=\sum_{j=2}^{n-1} \frac{n-j}{2^{2j+1}}$.

We have $f(m+1,n)-f(m,n)\ge $ (number of new regions)-(number of bridges), so $f(m+1,n)-f(m,n) \ge \sum_{k=1}^{n-1} \frac{n-k+3}{2^{2k+1}} +\frac2{2^{2n}} - \sum_{j=2}^{n-1} \frac{n-j}{2^{2j+1}}$
$ = \frac{n+2}{8} +\sum_{k=2}^{n-1} \frac3{2^{2k+1}}+\frac2{2^{2n}} =\frac{n+3}{8} > \frac{n}{8}$.

If $n=1$, the details are different, but we still have $f(m+1,1)-f(m,1)=\frac12>\frac18$, so $f(m+1,n)-f(m,n)>\frac{n}{8}$ for all $m,n\ge 1$. We can therefore run induction; since $f(1,1)=1>\frac18$, $f(m,n)>\frac{mn}{8}$ for all $m,n\ge 1$.

The $\frac18$ is not quite the best possible constant; the largest error in the estimate for large $m,n$ comes from regions that are not simply connected; they are counted 0 times if they have one hole, -1 times if they have two holes, ... . The expected number of such regions is proportional to $mn$, but is very small compared to ordinary simply connected regions. All other errors in the estimate are proportional to the perimeter at worst; they become small compared to the area for large $m,n$.

Actually, my estimate can be refined; evaluating the sum exactly we have $f(m,n)\ge \frac{(m+3)(n+3)}{8}-1$, with equality when at least one of $m$ and $n$ is less than 3.

The last line should read  (m+3)(n+3)/8  - 1.
Interestingly, all solutions posted for this problem
on the internet lead to that exact expression,
which counts the number of regions if all holes are
filled in. Those solutions (and yours can be recast
that way, as well) sum over 2x2 squares, with different
weightings that can be shown to be equivalent between
the different approaches.

I don't see the $2\times2$ squares in my solution- maybe $2\times n$, but every size is needed. I tried at one point building up from individual pieces, but it would take far too many shapes to get there. This approach worked, and I wasn't going to look for anything after that.

To rephrase using 2x2 squares, instead of adding the
row all n at a time, add it square by square from left
to right.  Your solution is to count the expected number
of regions added minus the number lost (but
ignoring extra holes), but you are essentially taking
the total expectation of n sequential events considered
as drawn collectively at random, rather than generating
them one by one and adding each incremental contribution
to the total.  i.e.  Instead of counting "bridge" configurations
like

RRRRRRR
RBBBBBR

note that you only lose a region when filling in the
the last 2x2 square,

R*
BR

and the probability (when you reach *) of this being
the case is at most 1/8.

Here's a solution. Consider the underlying graph made by having a vertex centered in each square and connecting two vertices if they are adjacent and the same color. Let $E$ be the number of edges (that is adjacent same color pairs) and $E_4$ be the number of $2 \times 2$ squares (i.e. $4$-cycles). I claim that there are at least $mn - E + E_4$ connected components. The proof is simple: for each $2 \times 2$ square, delete the lowest edge. It is easy to verify that this doesn't disconnect any connected components. Now, each remaining edge (of which there are $E - E_4$) can reduce the number of connected components by at most 1 each, so there are at least $mn - E + E_4$ connected components.

Taking expectations and using linearity gives \[ \mathbb{E}[mn - E + E_4] = mn - \frac{1}{2}(m(n-1) + (m-1)n) + \frac{1}{8}(m-1)(n-1) > \frac{mn}{8} \] so we're done.

Focus for now just on the red squares. Construct a graph $G$ with vertices given by the red squares, and edges given by the adjacencies between red squares. The key idea is that for every $2\times 2$ red block, we delete the edge given by the adjacency of the bottom two red squares. This will not affect the connectedness of $G$, and furthermore, we never delete an edge twice. Thus, the expected number of edges of $G$ is
\[\bar{E}=\frac{1}{4}\cdot(m(n-1)+(m-1)n)-\frac{1}{16}(m-1)(n-1).\]
The expected number of vertices is
\[\bar{V} = \frac{mn}{2}.\]
For a graph with $V$ vertices and $E$ edges, the number of connected components is at least $V-E$, so the expected number of connected components is at least
\[\bar{V}-\bar{E} = \frac{m+n}{4}+\frac{1}{16}(m-1)(n-1).\]
This is thus a lower bound on the expected number of red components, so the total expected number of components is simply double this, which is easily seen to be more than $mn/8$, as desired.

Consider a graph where each square represents a vertex, and two vertices are adjacent by an edge if their corresponding squares share an edge and have the same color.

Recall $$\mathbb{E}[\# \textbf{connected components}] = \mathbb{E}[\# \textbf{Vertex} - \# \textbf{Edge} + \# \textbf{redundant edges}]$$

Where a redundant edge is an edge that doesn't decrease the number of components. For each pair of squares $(u,v)$ sharing an edge, let $X_{u,v}$ denote the random indicator variable that the squares $u,v$ have the same color

Note $\mathbb{E}[\# Vertex ] = mn$, $\mathbb{E}[\# Edge] = \sum_{(u,v)} \mathbb{E}[X_{u,v}] = \frac 12 (m(n-1)+n(m-1)) = mn - \frac{m+n}{2}$

One type of redundant edges occur when a 2x2 square has squares of the same color. This occurs with probability $\frac 18$, and always results in a redundant edge. Thus $\mathbb{E}[\# \textbf{redundant edges}] \ge \frac 18 (m-1)(n-1)$. Summing, we have expected number of components is at least $\frac{m+n}{2} + \frac{(m-1)(n-1)}{8} \ge \frac{mn}{8}$

Induct on $m$. For the base case of $m=1$, it is easy to see by induction on $n$ that we expect $\tfrac{n+1}{2}$ connected components. This is always greater than $\tfrac{n}{8}$, hence it suffices to show that when we go from $m$ to $m+1$ we expect to gain at least $\tfrac{n}{8}$ connected components.

Suppose we add a new row to the bottom of the grid. Create a counter and perform the following:
[list]
[*] Looking only at the new row and nothing else, add $1$ for each connected component,
[*] Now introduce the rest of the grid, and subtract $1$ for every connected component in the rest of the grid that touches a same-color connected component in the new row.
[/list]
It is evident that this counter will then measure the change in the number of connected components present in the grid, by looking at the contribution of each connected component (for example, if a red connected component is added and borders 3 old red connected components, they will be merged into a single new connected component for a net change of $-2$, which is also $1-3$). Therefore we measure the expected contribution of the two bullet points to the counter (by linearity of expectation).

For the same reason as the base case of $m=1$, we expect the first bullet point to contribute an increase of $\tfrac{n+1}{2}$. We now bound the expected contribution of the second bullet point. Split the bottom two rows of the grid into $n$ vertical $2 \times 1$ rectangles. Call a rectangle [i]good[/i] if it is monochromatic and the rectangle directly to its right is not also monochromatic of the same color (if the rightmost rectangle is monochromatic it's also good). It is clear that we can injectively map every adjacency between an "old connected component" and a "new connected component" to some good rectangle, so it suffices to count the expected number of good rectangles. The rightmost rectangle is good with probability $\tfrac{1}{2}$, and every other rectangle is good with probability $\tfrac{1}{2}(1-\tfrac{1}{4})=\tfrac{3}{8}$, so we expect a total of $\tfrac{3n+1}{8}$ good rectangles.

It follows that we expect the counter to be at least $\tfrac{n+1}{2}-\tfrac{3n+1}{8}=\tfrac{n+3}{8}$, which is enough to finish the induction. $\blacksquare$