modular arithmetic
calculus
integration
induction
floor function
algorithm
counting

[hide="counterexample"]It is known that the product of $n$ consecutive positive integers is divisible by $n!$ (I found this in The USSR Olympiad Problem Book), and after some consideration I realized (hopefully correctly..) that your proposition would imply a stronger version: that the product of $n-1$ consecutive positive integers is divisible by $n!$. This unexpected result led me to look for a counterexample, and here it is: $4\nmid\binom{4}{2}=6$.[/hide]

[hide]Well, $\binom{p}{k}=\frac{p!}{k!(p-k)!}$. If $p$ is prime, there is one factor of $p$ on top, and since $p>k$, none on the bottom, so that gives the result.[/hide]

[quote="joml88"]I believe Max meant for $p$ to be a prime.[/quote]Of course, I was just kidding :blush: :D.

[hide]$p$ is prime and $p>k$, so $(k!,p)=1$ (since the next greatest divisor of $p$ is $p$, and no factors of $k!=1\cdot2\cdot3\cdots{k}$ contain $p$). This will be used later.

\[p\mid\binom{p}{k}\Leftrightarrow{p}\mid{p}\frac{(p-1)\cdot(p-2)\cdots(p-k+1)}{k!}\]

Clearly this would be true if $k!\mid{m}$, where $m$ is the numerator of the fraction above. Since $\binom{p}{k}$ is an integer, $k!\mid{pm}$, or $pm\equiv0\pmod{k!}\Leftrightarrow{m}\equiv0\pmod{\frac{k!}{(k!,p)}}\Leftrightarrow{m}\equiv0\pmod{k!}.$ QED. I don't have much experience with writing proofs.. I hope that was alright.[/hide]

AntonioMainenti: Yes nice argument. It can be a little shorter if you know this little theorem: if $c\,\vert\, ab$ but $c\not | \,a$, then $c \,| \,b$. Since $(k!,p)=1$, $k!$ has to divide $m$. Hopefully this can add some insights to readers.

I also realised that there was an assumption in AntonioMainenti's proof, that $\binom p k$ is an integer. Prove this fact  :D

[quote="3X.lich"]I also realised that there was an assumption in AntonioMainenti's proof, that $\binom p k$ is an integer. Prove this fact  :D[/quote][quote="AntonioMainenti"]It is known that the product of $n$ consecutive positive integers is divisible by $n!$ (I found this in The USSR Olympiad Problem Book) ...[/quote]Don't tell me I have to prove that..  :D

I just read the proof of it in this book.. it's pretty nice actually.

[quote="goodyfresh741"]Proving that [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/253ede925a2ce7ba23befe261287de9b.gif[/img] is an integer is WAY easy---seriously---AS LONG AS YOU KNOW ABOUT COMPLETE SYSTEMS OF RESIDUES.  Now then, obviously, the set {p, p-1, p-2, . . ., p-k+1} forms a complete system of residues mod(k) (if you don't know what that means, look it up, I don't know how to explain it).  Therefore, one of them is congruent to 0 mod(k) so you can remove k from the numerator and denominator of p(p-1). . .(p-k+1)/(k(k-1). . .(1)).  Then, if the number on the top that k divided was Ak, you get p(p-1). . .(Ak-1)A(Ak+1). . .(p-k+1)/((k-1)(k-2). . .1), so the top is a complete residue system modulo (k-1).  So, k-1 divides some number on top, divide it out.  Continue down the list dividing out k-2, k-3, etc.  Eventually, you find that k! divides a product of k consecutive integers.  I know I didn't explain this very clearly, but if you read about complete residue systems, you will understand.[/quote]Your explanation was perfectly clear to anyone who knows what a complete system of residues mod n is. That's probably the simplest way to prove [quote="AntonioMainenti"]... that the product of $n$ consecutive positive integers is divisible by $n!$[/quote]

[quote="AntonioMainenti"][quote="blahblahblah"][quote="3X.lich"]I also realised that there was an assumption in AntonioMainenti's proof, that $\binom p k$ is an integer. Prove this fact  :D[/quote]

Because it has a combinatorial interpretation that is necessarily integral.  We can use things like this, for example, to prove Fermat's Little Theorem.[/quote]Of course it is well known that $\frac{n!}{m!(n-m)!}\in\mathbf{Z}$... can you prove it though?[/quote]

I just did.  On one hand, we have an algebraic interpretation.  On the other hand, we have a combinatorial interpretation which is necessarily integral.  Because the two things are equal, the algebraic interpretation is integral.  I could, of course, go through all the rigamarole of proving that any prime dividing the top also divides the bottom, but that's completely unnecessary by the previous remarks.

[quote="blahblahblah"][/quote][quote="AntonioMainenti"][quote="blahblahblah"][quote="3X.lich"]I also realised that there was an assumption in AntonioMainenti's proof, that $\binom p k$ is an integer. Prove this fact  :D[/quote]

Because it has a combinatorial interpretation that is necessarily integral.  We can use things like this, for example, to prove Fermat's Little Theorem.[/quote]Of course it is well known that $\frac{n!}{m!(n-m)!}\in\mathbf{Z}$... can you prove it though?[/quote][quote="blahblahblah"]

I just did.  On one hand, we have an algebraic interpretation.  On the other hand, we have a combinatorial interpretation which is necessarily integral.  Because the two things are equal, the algebraic interpretation is integral.  I could, of course, go through all the rigamarole of proving that any prime dividing the top also divides the bottom, but that's completely unnecessary by the previous remarks.[/quote]I don't see how saying we interpret something combinatorically proves anything about it..

[quote="AntonioMainenti"][quote="blahblahblah"][quote="3X.lich"]I also realised that there was an assumption in AntonioMainenti's proof, that $\binom p k$ is an integer. Prove this fact  :D[/quote]

Because it has a combinatorial interpretation that is necessarily integral.  We can use things like this, for example, to prove Fermat's Little Theorem.[/quote]Of course it is well known that $\frac{n!}{m!(n-m)!}\in\mathbf{Z}$... can you prove it though?[/quote]

Well, I just realized something.  $\binom {p}{k}$ = $\binom {p-1}{k}$ + $\binom {p-1}{k-1}$ is easily proven just by using the definition of binomial coefficients as p!/(k!(p-k)!).  Then you keep breaking down the binomial coefficients with k on the bottom until you get a number <k on the top so the coefficients are zero, and the recurrence, together with induction on the cases $\binom m 0$ = 1 and $\binom m 1$ = m for all m, gives the result that the binomial coefficient is an integer.  That proof that the binomial coeff. is an integer is about as simple as the one using the combinatorial interpretation given by blahblahblah, however, the result of my proof is weaker.

But, one can, using the same recurrence shown above, find an alternative proof of the combinatorial interpretation of the binomial coefficients, also using induction.  Of course, it's not as simple as blahblahblah's proof.

In any event, I'll give you an algebraic proof of why it's an integer.  Let $p$ be an arbitrary prime, note that exponents add under multiplication, and consider the number of times $p$ divides the numerator and denominator of $\binom{n}{m}$.  Furthermore, let $a||b=k$, where $a^k$ is the largest power of $a$ dividing $b$ (sorry for the nonstandard notation)

By a theorem of Legendre, and for sufficiently large $k$

$p\vert \vert n!=\lfloor n/p\rfloor+\lfloor n/p^2 \rfloor+\ldots +\lfloor n/p^k\rfloor$

$p\vert \vert (m-n)!=\lfloor (m-n)/p\rfloor+\lfloor (m-n)/p^2 \rfloor+\ldots +\lfloor (m-n)/p^k\rfloor$

$p\vert \vert m!=\lfloor m/p\rfloor+\lfloor m/p^2 \rfloor+\ldots +\lfloor m/p^k\rfloor$

It would be exceedingly convenient if it were true that

$\lfloor n/p\rfloor\geq \lfloor (m-n)/p\rfloor + \lfloor m/p\rfloor$

and indeed, it is true, because we can write (via the division algorithm) that:

$n=ap+q$, $m=bp+r$, $n-m=(a-b)p+(q-r)$, and we see that equality doesn't hold if $q-r<0$ (it is obvious that $0\leq q,r<p$)

Does this satisfy your requirements of 'proof'?

[quote="goodyfresh741"]Well, I just realized something.  $\binom {p}{k}$ = $\binom {p-1}{k}$ + $\binom {p-1}{k-1}$ is easily proven just by using the definition of binomial coefficients as p!/(k!(p-k)!).[/quote]

Seeing as I'm trying to bash the beauty of counting arguments into your collective heads, I'll give a counting proof for that.  Say that you have $p$ indistinguishable balls, and that only one of them is blue.  The number of ways of choosing $k$ balls from $p$ is equal to the number of ways of choosing $k-1$ balls from $p-1$ balls (given that one of the balls is blue) plus the number of ways of choosing $k$ balls from $p-1$ (given that none of the balls is blue).  These cases are exhaustive, thus:

$\binom{p}{k}=\binom{p-1}{k}+\binom{p-1}{k-1}$

EDIT: indistinguishable, not distinguishable

[quote="blahblahblah"][quote="goodyfresh741"]Well, I just realized something.  $\binom {p}{k}$ = $\binom {p-1}{k}$ + $\binom {p-1}{k-1}$ is easily proven just by using the definition of binomial coefficients as p!/(k!(p-k)!).[/quote]

Seeing as I'm trying to bash the beauty of counting arguments into your collective heads, I'll give a counting proof for that.  Say that you have $p$ distinguishable balls, and that only one of them is blue.  The number of ways of choosing $k$ balls from $p$ is equal to the number of ways of choosing $k-1$ balls from $p-1$ balls (given that one of the balls is blue) plus the number of ways of choosing $k$ balls from $p-1$ (given that none of the balls is blue).  These cases are exhaustive, thus:

$\binom{p}{k}=\binom{p-1}{k}+\binom{p-1}{k-1}$[/quote]

Don't be so harsh. . .I actually really like combinatorial/counting arguments myself.  In fact, if you look at the last line of my last submission, I said that the same recurrence can be used to prove the equivalence between the algebraic and combinatoric definitions of the binomial coefficient.  In fact, [b]the proof I was referring to is exactly the same one you just gave[/b], but I didn't actually supply the proof because I wanted to see if someone else would.  Next time, see if you can say more friendly stuff than "bash. . .counting arguments into your collective heads."  This is a forum for friendly intellectual exchange, not unfriendly exchange.  I just want to find as many possible methods to prove the same fact as we can, instead of merely sticking to one method.  Learning is better that way.

[quote="blahblahblah"][quote="goodyfresh741"]Next time, see if you can say more friendly stuff than "bash. . .counting arguments into your collective heads."[/quote]

Oh, it's not unfriendly at all - if anything, it was self-deprecating with respect to how I was being so strident as to the efficacy of counting/combinatorial arguments in this thread.

My apologies if it came across that way, as always, I'm just trying to help.

EDIT: And you should have spotted that I actually meant 'indistinguishable balls rather than distinguishable balls  :D[/quote]

Yep, I understand.  Oh, and by the way, since you seem to like both combinatorics and number theory so much, I was wondering, have you (blahblahblah) read the book [i]An Introduction to the Theory of Numbers[/i] by Hardy and Wright?  If you haven't, you should, because it has a really fun chapter on partition analysis, both combinatorial and through generating functions.  Also, it's definitely the best book ever written on number theory.  I once talked to a world-class mathematician (Stephen G. Krantz, who's published over 200 papers and thirty books) who said that he re-reads Hardy and Wright's book at least once every year.

[quote="goodyfresh741"]Yep, I understand.  Oh, and by the way, since you seem to like both combinatorics and number theory so much, I was wondering, have you (blahblahblah) read the book [i]An Introduction to the Theory of Numbers[/i] by Hardy and Wright?[/quote]

No, I haven't - I'm a first year university student, and juggling Analysis, Complex Analysis, Vector Calculus, and Discrete math courses leaves me little time to do other stuff (and I fill most of that spare time with olympiad problems)

I did buy a Dover number theory text the other day though, and will probably go over the few parts which are unfamiliar to me in the summer.  I pretty much learned number theory through the various olympiad books I have, random olympiad notes on the internet, and (to  a lesser degree) through the abstract algebra course I took this past semester, though most of that was review.  Hardy and Wright is, however, a classic, and I'm sure I'll get around to it some day.

[quote="blahblahblah"][quote="AntonioMainenti"]I don't see how saying we interpret something combinatorically proves anything about it..[/quote]

Okay, so, lets look at the number of ways of choosing $m$ objects from $n$ objects.  If you think that this quantity is not integral, you have serious problems  :D   So call this $\binom {n}{m}$, and we won't associate any algebraic properties with it yet.

Now lets use counting techniques on it.  So, the number of ways of choosing $m$ objects from $n$ without respect to order is just $n(n-1)(n-2)\ldots (n-m+1)$

Of course, we need to divide by $m!$ to avoid counting the same objects in a different order.  But after you multiply numerator and denominator by (n-m)!, you get the usual algebraic expression for $\binom{n}{m}$

One side is integral.  So the other must be also.[/quote]So you're saying that $\binom{n}{m}$ is defined combinatorically as well as algebraically. Alright then.

[quote="AntonioMainenti"]So you're saying that $\binom{n}{m}$ is defined combinatorically as well as algebraically. Alright then.[/quote]

Well, binomial coefficients have their most obvious interpretations as combinatorial quantities, don't you think?  But my proof on the second post of the second page of this thread should satisfy you, I think.


Did you mean to find the numerical value of $ sec^8(\frac{\pi}{9}) \plus{} sec^8(\frac{5\pi}{9}) \plus{} sec^8(\frac{7\pi}{9})$?

(1)

Letting the required result be $ S$ and doing the basic algebraic manipulation we get to know that:

$ sec^{8}(\frac {\pi}{9}) \plus{} sec^{8}(\frac {5\pi}{9}) \plus{} sec^{8}(\frac {7\pi}{9}) \equal{} \frac {(cos100^\circ cos140^\circ)^8 \plus{} (cos20^\circ cos140^\circ)^8 \plus{} (cos20^\circ cos100^\circ)^8}{(cos20^\circ cos100^\circ cos140^\circ)^8}$

(2)

Using these standard results:

$ cos\beta cos\alpha \equal{} cos(\beta \plus{} \alpha ) \plus{} cos(\beta \minus{} \alpha)$ (with $ \alpha < \beta$), and

$ cos\alpha cos\beta cos\gamma \equal{} \frac {1}{4} [cos(\alpha \plus{} \beta \minus{} \gamma) \plus{} cos( \minus{} \alpha \plus{} \beta \plus{} \gamma) \plus{} cos(\alpha \minus{} \beta \minus{} \gamma) \plus{} cos(\alpha \plus{} \beta \plus{} \gamma)]$

and setting $ \alpha \equal{} 20^\circ$, $ \beta \equal{} 100^\circ$ and $ \gamma \equal{} 120^\circ$ we can rewrite the result obtained at (1).

(3)

Applying the factorization of $ cos\alpha cos\beta$ wich brings $ cos\alpha cos\beta \equal{} 2[cos(\frac {\alpha \plus{} \beta}{2}) cos(\frac {\alpha \minus{} \beta}{2})]$ we find

some non friendlly results, but I will try to work this out, for now I do not have time because I am working to finish an assignment, but I will try my best to solve

this one. Best Regards!

That is all I got fo now, sorry :( .

thanks.i want the result without using calculator.
i tried but i cant continue but iam sure that there is wrong in my answer


The lengths of the legs of a right triangle are 36 cm and 48 cm. What is the number of centimeters in the perimeter of the triangle?

hello
perimeter=$ 60\plus{}36\plus{}48$=$ 144$centimeters.
thank u

The length of the hypotenuse would be 60 because this is a Pythagorean Triple, 36, 48, 60.  Then add all three together to get 144.

[hide="A bit simpler"]By observation we see that the triangle is a larger version of the $3-4-5$ right triangle.  We also note that the side lengths of this new triangle is multiplied by a factor of $12$, since $\frac{36}{3}=\frac{48}{4}=12$.  The perimeter of the triangle is $12(3+4+5)=12\times12=\boxed{144}$.  Note that we did not need to find the hypotenuse in order to solve this problem.[/hide]


I heard somewhere about even and odd functions. Can someone please explain the difference between them? Thanks in advance!

even function = $ f(x)\equal{}f(\minus{}x)$

odd function = $ f(x)\equal{}\minus{}f(\minus{}x)$

So an example of an even function would be something like $ f(x) \equal{} |x|$ and an odd function would be $ f(x) \equal{} x^3$?

yes.

making the message longer before submitting

When you want to find whether a function is even or odd, you can check to see whether

$ f(x)\equal{}f(\minus{}x)$

or $ f(x)\equal{}\minus{}f(\minus{}x)$.

But then you can also graph it. An even function is symmetric with respect to the y-axis, and an odd function is symmetric with respect to the origin.

Challenge: Give an example of
a) An even function
b) An odd function
c) One that is both
d) One that is neither

(d should be the easiest.)

When you said symmetrical with respect to the origin you'll must said it's radial symmetric :P and the other symmetri is the axial

Further extensions:

Try adding two even functions. What do you get? Try showing this in general.
Try adding two odd functions. What do you get? Try showing this in general.
Try multiplying two even functions. What do you get? Try showing this in general.
Try multiplying two odd functions. What do you get? Try showing this in general.
Try multiplying an even function and an odd function. What do you get? Try showing this in general.

Other things to note:
cos and tan are odd functions.
sin is an even function.

[quote="fishythefish"]

Challenge: Give an example of
a) An even function
b) An odd function
c) One that is both
d) One that is neither

(d should be the easiest.)[/quote]

a) $ f(x) = |x|$
b) $ f(x) = x^3$
c) impossible?
d) $ f(x) = 2x^3 - 3x^2 - 4x + 4$

Is that right?  :maybe:

Latex is messing up again...izzy, can you prove that c is impossible?

Nerd's problems:

1. $ (f + g)(x) = f(x) + g(x) = f( - x) + g( - x) = (f + g)( - x)$
$ (f + g)(x) = (f + g)( - x)$

2. $ (f + g)(x) = f(x) + g(x) = - f( - x) - g( - x) = ( - f - g)( - x) = - (f + g)( - x)$
$ (f + g)(x) = [ - (f + g)]( - x)$

3. $ (fg)(x) = f(x)\cdot g(x) = f( - x)\cdot g( - x) = (fg)( - x)$
$ (fg)(x) = (fg)( - x)$

4. $ (fg)(x) = f(x)\cdot g(x) = - f( - x)\cdot - g( - x) = (fg)( - x)$
$ (fg)(x) = (fg)( - x)$

5. $ (fg)(x) = f(x)\cdot g(x) = f( - x)\cdot - g( - x) = ( - fg)( - x) = [ - (fg)](x)$
$ (fg)(x) = [ - (fg)](x)$

Some of these can be seen better by substituting a for f+g and b for fg in the summary line.

1. $ a(x) = a( - x)$ Even+Even=Even
2. $ a(x) = - a( - x)$ Odd+Odd=Odd
3. $ b(x) = b( - x)$ Even*Even=Even
4. $ b(x) = b( - x)$ Odd*Odd=Even
5. $ b(x) = - b( - x)$ Even*Odd=Odd

[quote="Nerd_of_the_Ages"][quote="isabella2296"]
c) impossible?[/quote]

[hide="Hint"]-f(-x) = f(-x).
[hide="even bigger giveaway"]this means f(-x) is equal to it's opposite. What's the only thing that you can think of that is equal to its opposite?[/hide][/hide][/quote]

The polynomial $ f(x)\equal{}0$.


Αγαπητοί φίλοι γεια σας.

  Λοιπόν , ως χώρα πρωτοτυπίσαμε πάλι !!!

  Δύο θέματα των μεγάλων είναι από το διαγωνισμό Baltic way 2002 !. Αυτό μου υπέδειξε ο Socrates και τον ευχαριστώ. Θα ήταν πιο τίμιο να βάζανε ασκήσεις από το σχολικό βιβλίο . Τουλάχιστον έτσι ο διαγωνισμός θα ήταν πιο δίκαιος. Τι άλλο να πω. Στενοχωρήθηκα που είδα δυο θέματα να είναι από έναν τόσο γνωστό διαγωνισμό.

  Ωστόσο συγχαίρω τα παιδιά που συμμετείχαν στον Αρχιμήδη  , ανεξάρτητα από τη διάκριση ή όχι σε αυτόν.

 Καλή συνέχεια και καλά μετάλλια στις IMO και BMO 2006

 [u]Babis[/u]

Τραγική ειρωνεία ?Είχα διαβάσει τα θέματα του Baltic από το 2005,2004,2003, αλλά όχι του 2002. Τα λέμε :mad:

Λοιπόν  αν  και  πήρα  μετάλλιο  στον  αρχιμήδη  στους  μεγάλους  (χάλκινο  13 ος στην  κατάταξη) έιμαι  καθετος  σε  αυτό που  συμβαίνει  . Πρέπει  κάποιος  να  διαμαρτυρηθεί  για  αυτό  το  χάλι  . Μόλις    πληροφορήθηκα  το  γεγονός  από  τον  Σιλουανό . Λοιπόν  είδα  ότι  ούτε  τα  γράμματα  δεν  άλλαξαν  στην  γεωμετρία . Καλά  και  μετά  θέλουμε  να  ανταγωνιστούμε  την  Κίνα  και  την  Ρωσία  . Έτσι  ούτε  σε  100.000  χρόνια . Και  στην  απονομή των  βραβείων  έλεγε  ο "κ Μπόλης " οτι  τα θέματα  είναι πρωτότυπα  και έλεγε  ακόμη πώς , οτι  και  να  κάνεις  ,τα  θέματα  μοιάζουν  μεταξύ  τους . Τρίχες ακομή  και  στα  γράμματα  μοιαζουν  χάθηκε  αντι  για  Μ,Ν  να  έβαζαν  Δ , Ε και  αντί  για  χ,y να  έβαζαν  α,β . Τέλος  πάντων  χαιρετώ τον  κύριο  Στεργίου  και  εσένα  Σιλουανέ . Θα  ενημερώνω  για τα  νεότερα  από  τους  προκριματικούς  . Λοιπόν  τι  λέτε από   ποίο  διαγωνισμό  θα  βάλουν  φέτος θέματα στους προκριματικούς  μήπως απο  την ΙΜΟ  :mad:   :mad:

Θα βάλλουν από τη Shortlist 2005 BMO που κυκλοφορεί στο forum ;)
kai άντε να μπει και κανένα από την ΙΜΟ shortlist ,

ΛΟΙΠΟΝ ΚΥΡΙΕΣ ΚΑΙ ΚΥΡΙΟΙ ΟΙ ΠΟΛΥ "ΕΞΥΠΝΟΙ ΚΑΙ ΠΡΩΤΟΤΥΠΟΙ" ΚΥΡΙΟΙ ΤΗΣ ΕΜΕ ΒΑΛΑΝ ΤΑ ΘΕΜΑΤΑ ΑΠΟ ΤΟΝ BALTIC WAY  ΕΔΩ ΠΟΥ ΦΤΑΣΑΜΕ ΑΠΟ ΤΟΥ ΧΡΟΝΟΥ ΘΑ ΤΟΥΣ ΓΡΑΦΟΥΜΕ ΣΤΙΣ  ΛΥΣΕΙΣ
ΘΕΜΑ 1
(BALTIC WAY 2002 ΠΡΟΒΛΗΜΑ 1 )ΚΑΙ ΤΕΛΕΙΩΣΑΜΕ ΤΑ ΣΟΥΡΓΕΛΑ  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:
ΧΑΙΡΕΤΙΣΜΑΤΑ ΣΤΟΝ ΦΙΛΟ ΜΟΥ ΤΟΝ ΣΙΛΟΥΑΝΟ
ΓΕΙΑ ΣΑΣ
P.S. ΞΈΡΕΤΕ ΠΟΥ ΜΠΟΡΩ ΝΑ ΒΡΩ ΟΛΑ ΤΑ ΘΕΜΑΤΑ ΟΛΩΝ ΤΩΝ ΔΙΑΓΩΝΙΣΜΩΝ ΣΕ ΟΛΕΣ ΤΙΣ ΧΡΟΝΙΕΣ ΑΝΑ ΤΟΝ ΚΟΣΜΟ?

Αμβρόσιε γεια !

  Η ερώτηση σου είναι καλή αλλά πολύ αισιόδοξη ! Χαίρομαι πάντως που την κανεις  , γιατί δείχνει ότι έχεις πολύ όρεξη. Λοιπόν  , ας το δούμε λιγάκι. Θέλεις να μάθεις που μπορείς να βρεις όλα τα θέματα , όλων των διαγωνισμών , όλου του κόσμου και όλων των εποχών !
    α) Τι να τα κάνεις ; Και εγώ που δεν είμαι μαθητής , παρόλο που θα ήθελα να τα έχω , μου φαίνεται ότι θα μου ήταν άχρηστα. Όχι διότι δε θα ήταν καλά , αλλά διότι δε θα προλάβαινα ποτέ να δω ούτε τις εκφωνήσεις .
    β)  Τα νέα θέματα των τελευταίων 10 ετών ( για ολυμπιάδες εννοώ) δεν είναι παραπάνω από 200. Πολλά βάζω , αλλά ας είναι. Αν μπορούσαμε να έχουμε αυτά τα θέματα με τις λύσεις τους και τα σχετικά σχόλια , να είσαι σίγουρος ότι μελετώντας τα θα μπορούσαμε να λύσουμε και όλα τα άλλα.
   γ) Συγκεντρωμένα  δε μπορούμε να τα βρούμε πουθενά. Πολλά παλια θέματα με τις λύσεις τους υπάρχουν στο CRUX . Έχω όλα τα τεύχη από το 1975, αλλά δεν τα έχω ξεφυλλίσειάκόμα όλα. Πολλά θέματα υπάρχουν στο διαδύκτιο, αλλά θέλει πολύ χρόνο να τα κατεβασεις. Σε λίγο θα κυκλοφορήει και το βιβλίο του φίλου Dushan Dukic με άλλους , στο οποίο θα υπάρχουν όλα τα ΙΜΟ θέματα με τα shortlist(λυμένα). Όποιος το θέλει , ας μου τηλεφωνήει γιοα νατο βρούμε .... πιο φτηνό!
   δ) Πάντως , ότι βιβλία έχω  που κάποιος μαθητής νομίζει ότι του χρειάζεται ,  μπορώ να του τα δώσω(φωτοτυπία)

   ε) Αυτό που συνιστώ, όμως είναι τα εξής¨
  ι) Μέχρι το τέλος της Γ γυμνασίου πρε΄πει να έχει κανείς τελειώσει με τη βασική άλγεβρα, όλη την επιπεδομετρία , τη στοιχείδη αριθμοθεωρία και τα βασικά combinatorics
 ii)  Στην Α΄ λυκείου πρέπει να ξεμπλέξει με τα ειδικά θεωρήματα και θέματα(αλλά με πολύ καλή κατανόηση και γερές εφαρμογές)
  ιιι) Στη Β ΄λυκείου να μελετήσει και να λύσει όλα τα θέματα των σοβαρών διαγωνισμών : ΙΜΟ , ΒΜΟ , Baltic Way, Διαγωνισμός πόλεων και θέματα εθνικών ολυμπιάδων από χώρες όπως: Ρωσία( και όλες τις πρώην ανατολικές χώρες -Βουλγαρία , Ρουμανία κλπ) , Κίνα, Αμερική, Καναδάς , Ιραν κλπ.και να χτενίσει το βιβλίο του Engels:Problem Solving Strategies.

 Μόνο  έτσι μένουν πιθανότητες για χρυσό.Για αυτό το ...τιμημένο παλεύουμε !

 Αυτά για τώρα.
  Καλή δύναμη και καλή μελέτη σε όλους.

 Μπαμπης

 ( Και μη ξεχνατε : Τα χρόνια της εφηβείας περιμένουν να τα ζήσετε. Βρείτε και για αυτά λίγο χρόνο. Ρωτήστε το Σιλουανό να σας πει ποιος του το είπε αυτό. Ο γιος του έχει πάρει μετάλιο κια η γνώμη του αξίζει  !)

ΑΓΑΠΗΤΕ ΚΥΡΙΕ ΣΤΕΡΓΙΟΥ
ΣΩΣΤΑ ΟΣΑ ΛΕΤΕ ΑΛΛΑ ΔΕΝ ΜΠΟΡΕΙ ΚΑΘΕ ΑΞΙΟΤΙΜΟΣ ΚΥΡΙΟΣ ΠΟΥ ΘΕΛΕΙ ΝΑ ΘΕΩΡΕΙΤΑΙ ΠΡΩΤΟΤΥΠΟΣ ΝΑ ΚΑΤΑΔΙΚΑΖΕΙ ΤΟΣΑ ΠΑΙΔΙΑ ΠΟΥ ΕΡΧΟΥΝ ΟΡΕΞΗ ΚΑΙ ΑΓΑΠΗ ΓΙΑ ΤΟ ΑΝΤΙΚΕΙΜΕΝΟΕΓΩ ΠΡΑΓΜΑΤΙΚΑ ΘΕΩΡΩ ΕΝΤΕΛΩΣ ΑΧΡΗΣΤΗ ΤΗ ΣΥΛΛΟΓΗ ΟΛΩΝ ΤΩΝ ΠΡΟΒΛΗΜΑΤΩΝ ΑΛΛΑ ΕΔΩ ΠΟΥ ΦΤΑΣΑΜΕ  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  ΚΑΙ ΠΡΕΠΕΙ ΕΠΙΣΗΣ ΝΑ ΠΡΟΣΕΧΟΥΝΕ ΤΙ ΛΕΝΕ ΓΙΑ ΝΑ ΜΗΝ ΓΙΝΟΝΤΑΙ ΑΚΡΙΒΩΣ ΣΟΥΡΓΕΛΑ  :rotfl:  :rotfl:  :rotfl:
ΣΑΣ ΕΥΧΑΡΙΣΤΩ ΓΙΑ ΤΗΝ ΑΠΑΝΤΗΣΗ ΣΑΣ ΚΙ ΕΛΠΙΖΩ ΚΑΠΟΙΟΣ ΝΑ ΠΑΡΕΙ ΧΡΥΣΟ ΣΤΗΝ ΙΜΟ ΕΠΙΤΕΛΟΥΣ  :)  :)


Sorry, this is kinda off topic, but it would be even more off-topic in any other forum.

Is there anyway to change pdf-viewers? When I press the adobe button it returns a ghostgum file. (ofcourse it returns a ghostgum file when I press the ghostgum button). What if I want to change it to something else? (ex. Bluebeam)

I don't use WinEdt so no idea if it helps but try this thread: [url=http://tech.groups.yahoo.com/group/WinEdt/messages/8663?threaded=1&m=e&var=1&tidx=1]alternative PDF viewer in WinEdt 5.3[/url]


Find all functions $f: \mathbb N \to \mathbb N$ such that
\[f(x+f(x^2+y))=2yf(x+y)\]
holds true for all positive integers $x$ and $y$.

[quote="irantst"]find all functions f:N--N
f(x+f(x^2+y))=2yf(x+y)
 :lol:[/quote]
There are no such functions. It seemed obvious to me but I needed a rather heavy method to show it. I hope (not so much :) ) that somebody will find an elementary one.

I consider, as usual in Mathlinks, that $ 0\notin\mathbb N$

Let $ P(x,y)$ be the assertion $ f(x\plus{}f(x^2\plus{}y))\equal{}2yf(x\plus{}y)$

1) $ f(a)\equal{}f(b)$ and $ a,b>1$ implies $ a\equal{}b$
$ a>1$ and $ P(1,a\minus{}1)$ $ \implies$ $ f(1\plus{}f(a))\equal{}2(a\minus{}1)f(a)>0$
$ b>1$ and $ P(1,b\minus{}1)$ $ \implies$ $ f(1\plus{}f(b))\equal{}2(b\minus{}1)f(b)>0$
So $ a>1$ and $ b>1$ and $ f(a)\equal{}f(b)$ $ \implies$ $ a\minus{}1\equal{}b\minus{}1$ and so $ a\equal{}b$
Q.E.D.

2) $ f(x^2\plus{}2y)>f((x\plus{}2y\minus{}1)^2\plus{}1)$ $ \forall x,y\in\mathbb N$
$ P(x,2y)$ $ \implies$ $ f(x\plus{}f(x^2\plus{}2y))\equal{}4yf(x\plus{}2y)$
Let $ u(x)\equal{}x\plus{}f(x^2\plus{}1)\ge 2$ (notice that $ u(x)>x$)
$ x\plus{}2y\minus{}1>0$ and so $ P(x\plus{}2y\minus{}1,1)$ $ \implies$ $ f(u(x\plus{}2y\minus{}1))\equal{}2f(x\plus{}2y)$
$ x\plus{}2y\minus{}1>0$ and $ u(x\plus{}2y\minus{}1)\minus{}y>x\plus{}2y\minus{}1\minus{}y>0$ and  so :
$ P(u(x\plus{}2y\minus{}1)\minus{}y,y)$ $ \implies$ $ f(u(x\plus{}2y\minus{}1)\minus{}y\plus{}f((u(x\plus{}2y\minus{}1)\minus{}y)^2\plus{}y))\equal{}2yf(u(x\plus{}2y\minus{}1))$ $ \equal{}4yf(x\plus{}2y)$

So $ f(x\plus{}f(x^2\plus{}2y))\equal{}f(u(x\plus{}2y\minus{}1)\minus{}y\plus{}f((u(x\plus{}2y\minus{}1)\minus{}y)^2\plus{}y))$
We have $ x\plus{}f(x^2\plus{}2y)>1$ and $ u(x\plus{}2y\minus{}1)\minus{}y\plus{}f((u(x\plus{}2y\minus{}1)\minus{}y)^2\plus{}y)>u(x\plus{}2y\minus{}1)\minus{}y>x\plus{}2y\minus{}1\minus{}y\ge 1$
So we can apply the pseudo-injectivity (see 1. above) and :

$ x\plus{}f(x^2\plus{}2y)\equal{}u(x\plus{}2y\minus{}1)\minus{}y\plus{}f((u(x\plus{}2y\minus{}1)\minus{}y)^2\plus{}y)$
$ \iff$ $ x\plus{}f(x^2\plus{}2y)\equal{}(x\plus{}2y\minus{}1)\plus{}f((x\plus{}2y\minus{}1)^2\plus{}1)\minus{}y\plus{}f((u(x\plus{}2y\minus{}1)\minus{}y)^2\plus{}y)$
$ \iff$ $ f(x^2\plus{}2y)\equal{}y\minus{}1\plus{}f((x\plus{}2y\minus{}1)^2\plus{}1)\plus{}f((u(x\plus{}2y\minus{}1)\minus{}y)^2\plus{}y)$ $ >f((x\plus{}2y\minus{}1)^2\plus{}1)$
Q.E.D.

3) There are no solution
Let $ m\equal{}\inf(f(\mathbb N\backslash\{1,2,3,4,5\}))$ and $ a\ge 6$ any positive integer such that $ f(a)\equal{}m$
3.1) If $ a$ is odd
Let $ b\equal{}\frac{a\minus{}1}2\ge 3$
Using property established in 2. above with $ x\equal{}1$ and $ y\equal{}b$, we get $ f(a)>f(4b^2\plus{}1)$ and so $ f(4b^2\plus{}1)<m$
And since $ 4b^2\plus{}1\ge 37$, this is impossible (see definition of $ m$)

3.2) If $ a$ is even
Let $ b\equal{}\frac a2\minus{}2\ge 1$
Using property established in 2. above with $ x\equal{}2$ and $ y\equal{}b$, we get $ f(a)>f((2b\plus{}1)^2\plus{}1)$ and so $ f((2b\plus{}1)^2\plus{}1)<m$
And since $ (2b\plus{}1)^2\plus{}1\ge 10$, this is impossible (see definition of $ m$)

Q.E.D.

Hence no solution for this equation.

tanks
It was very nice soloution
very good
 :)

yes it's very nice
pco is god of algebra


Prove that the equation $ 12t\minus{}x^2\equal{}y^z\minus{}1$ has no solutions for $ x,y,z$ and $ t$ prime numbers.

[hide]
My first N.T problem  :)
[/hide]

we have

$ y^z \plus{} x^2 \equal{} 12t \plus{} 1 \Rightarrow x \equal{} 2$  or $ y \equal{} 2$ because $ y$ and $ x$ haven't the same parity

If $ y \equal{} 2$ it means that $ x^2 \equal{} 12t \plus{} 1 \minus{} 2^z$.But    $ 12t \plus{} 1 \minus{} 2^z\equiv \minus{} 3; \minus{} 7 (\mod12)$ .Hence $ x^2 \equiv 9,5 (\mod12) \Rightarrow x^2 \equiv 5 (\mod12)$ because $ x$ is prime and $ x\neq3$.when $ x \equal{} 3$ it's clearly impossible because we get $ 2^z \equal{} 12t \minus{} 8 \equal{} 4(3t \minus{} 2)$  . we have $ x\equiv 1;5;7;11 (\mod12)$ it means that $ x^2\neq 5 (\mod12)$. Hence $ x \equal{} 2$
and so the equation becomes as $ y^z \equal{} 12t \minus{} 3 \equal{} 3(4t \minus{} 1)\Rightarrow 3^{z \minus{} 1} \equal{} 4t \minus{} 1$ and so $ 3^{z \minus{} 1}\equiv \minus{} 1 (\mod4)$ which is not true if $ z > 2$ and if $ z \equal{} 2$ we get $ t \equal{} 1$  Which is a contradiction

Correct me if i'm wrong !! Because Nowdays my mind is not OK  :D


An equilateral triangle is cut from a piece of material in square shape with dimension 1mx1m. Show that the area of the wasted material ( that is, the area of the figure remained after taking out the triangle) is at least $(4-2\sqrt{3})$$m^2$.

[hide="hint"]
Finding the least=minimizing the waste=maximizing the area of the triangle.
[/hide]

Let me re-state the question:
Prove that the area of the largest equilateral triangle inscribed in a square with dimension 1mx1m is $(2\sqrt{3}-3)m^2$

[hide="hint again"]
Put the triangle in such a position that one of the vertices is also one of the vertices of the square.
[/hide]

well argument for reddevilz's hint:  [hide] If one makes the triangle have vertices where none of the three is also a vertex of the square, then we can make the area of the triangle bigger by placing one vertex of the triangle at one vertex of the square, and then extending the side opposite that vertext to make the triangle equilateral [/hide]

Proof:  good luck to me on LaTeX: [hide] First, we note that there is only one such equilateral traingle that shares one vertex with the square.  It s easy to see that if we pick a shared vertex WLOG, then the triangle must be similar over the diagonal of the square from that vertex to the opposite vertex, and there is only one equilateral triangle like that.

Now, we let the distance from an adjacent vertex of the square to the vertex of the triangle nearest it be x.

We now see that the length of a side of the triangle can be expressed as $x=\sqrt{x^2+m^2}=\sqrt{2}(m-x)$.

Expanding both sides and doing a little algebra, we get $x^2-4xm+m^2=0$.

By the quadractic formula and some more algebra, we see that $x=2m\pm m\sqrt{3}$.  We reject the plus, for then $x\ge{m}$, which is impossible.

Now, we know that the area of the triangle is $\frac{(m^2+x^2)\sqrt{3}}4$, and doing some algebra, we find that yields the desired result.
[/hide]

[quote]It s easy to see that if we pick a shared vertex WLOG, then the triangle must be similar over the diagonal of the square from that vertex to the opposite vertex[/quote]

Hmmm.
The meaning of the above paragraph is a bit difficult to understand.  :roll:
Besides, the m stands for metre.  :)

well, since I didn't label any points, I had to write a sort-of weird explanation.  It does make sense, though.  I mean that the draw a diagonal from the shared vertex to the opposite vertex of the square, the triangle must be symmetric over that diagonal.

M stand for meter?  Oh--I thought it was a constant, so I treated it as such.

[img]http://www2.hkedcity.net/sch_files/a/lpl/lpl-lsk/public_html/max_triangle.gif[/img]
A Geometry proof
[hide]
WLOG, we may assume the required triangle $\triangle EFG$ touches three sides of the square ABCD as shown in the figure, for otherwise $\triangle EFG$  is not the largest possible.
Drop a perpendicular From G to EF and let the foot be H.
Note that HFCG is then a cyclic quadrilateral. ( opp. angles supp.)
Therefore, $\angle GCH = \angle GFH = 60^0$.
With a similar argument, $\angle GDH = \angle GEH = 60^0$.
So, $\triangle DHC$ is an equilateral triangle.
We have shown that [b]EF must pass through the fixed point H[/b], no matter how $\triangle EFG$ varies under the given  condition.
Clearly, the size of $\triangle DEF$ is largest when E collides with A. In such a configuration, $\angle DEG =\angle FEB = 15^0$.

What remains to show is some calculations with trigonometry.
[/hide]

An algebraic proof:
[hide]
Put the square on the complex plane and assume the vertices to be represented by
$\pm\frac12\pm\frac12i$. Assume the vertices of the equilateral triangle touches the
sides of the square at $\lambda+\frac12i, -\frac12+ui,\frac12+vi$.(positively oriented)
Then, area of the triangle
$=\frac{i}4{\left|\begin{array}{ccc}
1&1&1\\
{\lambda+\frac12i}& -\frac12+ui&{\frac12+vi}\\
{\lambda-\frac12i}& {-\frac12-ui}&{\frac12-vi}\end{array}\right|}$
$=\frac12[\frac12-\frac12(u+v)+\lambda(u-v)]$---(1)
Since the triangle is equilateral,
$(\lambda+\frac12)+\omega(-\frac12+ui)+\omega^2(\frac12+vi)=0$
where $\omega=-\frac12+\frac{\sqrt{3}}2i$ is the cube root of unit.
Expanding and re-arranging terms, we have
$(\lambda-\frac{\sqrt{3}}2u+\frac{\sqrt{3}}2v)+
i({\frac12-\frac{\sqrt{3}}2-\frac{u}2-\frac{v}2)=0}$.
Equating real and imaginary parts,
$(\lambda-\frac{\sqrt{3}}2u+\frac{\sqrt{3}}2v)=0$ and
$({\frac12-\frac{\sqrt{3}}2-\frac{u}2-\frac{v}2)=0}$.
Solving equations,
$u+v=1-\sqrt{3}$---(2) and
$u-v=\frac2{\sqrt{3}}\lambda$---(3).
Sub. (2) and (3) into (1),
Area of the triangle is given by
$f(\lambda)=\frac2{\sqrt{3}}(\lambda^2+\frac34)$.
Note that $\lambda$ is maximum when $v=-\frac12$,
in this case $\lambda=\sqrt{3}-2$ and
$f(\lambda) =2\sqrt{3}-3$ which is the maximum area.

[/hide]


ratio
geometry
analytic geometry
similar triangles

In $ \triangle ABC$, medians $ AE$ and $ CD$ are drawn intersecting at $ F$. Segment $ BG$ is drawn intersecting $ AE$ at $ H$ and $ G$ is the midpoint of $ CD$. Find the ratio of the area of $ \triangle FGH$ to $ ABC$.

[hide="Solution"]
Draw the line from $ D$ parallel to $ \overline{AE}$ and call the points it intersects $ \overline{BG}$ and $ \overline{BE}$ points $ I$ and $ D'$, respectively. Then using the given facts, the properties of medians and centroids, and similar triangles give the diagram. Also by similar triangles, $ \frac{a\plus{}2b}{3a\plus{}b}\equal{}\frac{\frac y2}{\frac y2\plus{}\frac y4}\equal{}\frac23\implies3a\plus{}6b\equal{}6a\plus{}2b\implies3a\equal{}4b\implies\frac a{2b}\equal{}\frac23$
$ \implies\frac a{a\plus{}2b}\equal{}\frac25\implies\frac{[FGH]}{[FCE]}\equal{}\frac{[FGH]}{\frac16[ABC]}\equal{}\frac a{a\plus{}2b}\cdot\frac{\frac x6}{\frac x6\plus{}\frac x2}\equal{}\frac25\cdot\frac14$
$ \implies\frac{[FGH]}{[ABC]}\equal{}\boxed{\frac1{60}}$.
[/hide]

[hide="Alternate Solution"]
Assume $ \bigtriangleup ABC$ is an equilateral triangle with side length 1. (It shouldn't matter, since we're looking for a ratio in the end)

One way to approach this problem would be to set it on a coordinate plane. Since we are dealing with medians, it should be easy to find the equation of the lines binding $ \bigtriangleup FGH$ and find their intersections.

Set $ C$ at $ (0,0)$, $ B$ at $ (1,0)$, and $ A$ at $ (\frac{1}{2},\frac{\sqrt3}{2})$.

Points $ D$ and $ E$ are then at $ (\frac{3}{4},\frac{\sqrt3}{4})$ and $ (\frac{1}{2},0)$, respectively.

The equation of $ CD$ is then found to be $ y\equal{}\frac{x\sqrt{3}}{3}$
The equation of $ AE$ is then found to be $ x\equal{}\frac{1}{2}$.

$ G$ is at the midpoint of segment $ CD$, so $ G$ is at the point $ (\frac{3}{8},\frac{\sqrt3}{8})$.

The equation of $ BG$ is then found to be $ y\equal{}\minus{}\frac{x\sqrt3}{5}\plus{}\frac{\sqrt3}{5}$

The points $ F$ and $ H$ can be determined by finding the intersections of $ CD$ with $ AE$ and $ BG$ with $ AE$, respectively.

We find $ F$ to be $ (\frac{1}{2},\frac{\sqrt{3}}{6})$ and $ H$ to be $ (\frac{1}{2},\frac{\sqrt{3}}{10})$.

Shoelacing, we get the area of $ \bigtriangleup FGH$ to be $ \frac{\sqrt3}{240}$.
The area of $ \bigtriangleup ABC$ is $ \frac{\sqrt{3}}{4}$.

$ \frac{\frac{\sqrt3}{4}}{\frac{\sqrt{3}}{240}}\equal{}\boxed{\frac{1}{60}}$.
[/hide]


function
algebra
domain
cyclic function
AMC

The function $ f$ has the property that for each real number $ x$ in its domain, $ 1/x$ is also in its domain and
\[ f(x) \plus{} f\left(\frac {1}{x}\right) \equal{} x.
\]What is the largest set of real numbers that can be in the domain of $ f$?
$ \textbf{(A) } \{ x | x\ne 0\} \qquad \textbf{(B) } \{ x | x < 0\} \qquad \textbf{(C) }\{ x | x > 0\}\\
\textbf{(D) } \{ x | x\ne \minus{} 1 \text{ and } x\ne 0 \text{ and } x\ne 1\} \qquad \textbf{(E) } \{ \minus{} 1,1\}$

[hide="Solution"]
First we notice that $\frac{1}{x}$ is a cyclic function; i.e., $\frac{1}{\frac{1}{x}}=x$. Therefore, if we substitute $\frac{1}{x}$, we get \[ f\left(\frac{1}{x}\right)+f(x)=\frac{1}{x}. \] Hence $x=\frac{1}{x} \implies x^2=1$, giving us the domain $\{-1,1\}$, which corresponds to answer choice $\boxed {(E).}$
[/hide]

very nice question, i know a bunch of people who put A as answer just because of 1/0 but the answer is E:
[hide="solution"]
replace $x$ by $\frac{1}{x}$ we get:

$f(\frac{1}{x})+f(x)=\frac{1}{x}$
subtract this from the equation given, and you get:

$x-\frac{1}{x}=0$ which is only ture if $x=\pm 1\Rightarrow \boxed{E}$[/hide]

[hide]
f(x) + f(1/x) = x
f(1/x) + f(x) = 1/x
x = 1/x
x = +/- 1

This question was trivial once you saw that, but it probably scared people away because properties of functions are a lost art in school curriculums today.
[/hide]

JB

Yes! I got it!

I made an educated guess.  ;)  :rotfl:

[hide="Answer"]Say $x=\frac{1}{y}$, then $f\left(\frac{1}{y}\right)+f(y)=\frac{1}{y}$, and $f(y)+f\left(\frac{1}{y}\right)=y$. Therefore, we have $y=\frac{1}{y}$, so $y=\pm 1$. $\boxed{E}$[/hide]

Here's a quick counter example. Plug in $ x \equal{} 2$ and $ x \equal{} \frac {1}{2}$ into the equation to obtain the following two equations:

$ f(2) \plus{} f\left(\frac {1}{2}\right) \equal{} 2 \\
f\left(\frac {1}{2}\right) \plus{} f\left(2\right) \equal{} \frac {1}{2}.$

This implies that $ 2 \equal{} \frac {1}{2}$ which is not true. Therefore, $ x \equal{} 2$ and $ x \equal{} \frac {1}{2}$ can't be in the domain of $ f$ because if they were, they would yield a contradiction.

If we set $x\to \frac{1}{x}$, we get $f(x)+f\left(\frac{1}{x}\right)=\frac{1}{x}$, so $x=\frac{1}{x}$.

Thus, $x\in \boxed{\textbf{(E)}\ \{-1,1\}}$.

Clearly the function $f(1)=\frac{1}{2}$ and $f(-1)=-\frac{1}{2}$ works.


number theory
prime numbers
number theory proposed

Consider the prime numbers $n_{1}<n_{2}<...<n_{31}$
Prove that if 30 divides $(n_{1})^4+(n_{2})^4+...+(n_{31})^4$, then there are three
consecutive primes in the colection
 :D  :D  :D


induction
combinatorics unsolved
combinatorics

There are 42 guests in a club. Given any31 of them there exist a couple (men and women) who had slept together. Prove that one can find 12 man a1,a2,a3...,a12 and 12 woman b1,b2,b3....,b12 such that ai slept with bi for every i=1,2,3....,12.

Actually I modified a little bit :cool: Thats why I like this problem!

Hi Pascual!
Excellent  problem!

We will prove following statement:
Suppose biparate graph G contains m nodes, and each n-tuples of nodes contains at least one edge, where n <= m. Then it is possible to choose m-n+1 pairwise disjoint edges.

(0) Base of induction: m=n is obvious
(1) Let A be set of men and B be set of women. If |A|>=n (|B|>=n) then we take n-tuple of men (women) and there are no edges. Thus |A|<n and |B|<n. It follows that |A|>=m-n+1 and |B|>=m-n+1.
(2) Suppose each subset A' of men is connected to at least |A'| women. Then using Holl's theorem we obtain that there are complete matching between all men and subset of women, so we are done, since |A|>=m-n+1.
(3) Suppose there are subset A' of men which is connected to at most |A'|-1 women. We find such minimal subset A'. Let x be arbitrary node from A', then each subset X of A'\{x} is connected to at least |X| women (due to minimal choice of A'). And again using Holl's theorem we find |A'|-1 pairwise disjoint edges from A'\{x} to subset B' of women, where |B'|=|A'|-1.
(3.1) If |B'|>=m-n+1 then we are done.
(3.2) Suppose |B'|<=m-n. Consider subgraph G'=A+B-B'. We have |G'|=|A+B-B'|>=m-(m-n)=n. Each n-tuple of G' contains edge (inherited property). Hence we can apply induction proposition. Thus we will find |G'|-n+1 pairwise disjoint edges in G'. It is clear that men in these edges are not in A', since all edges from A' lead to B'. It follows that obtained edges pairwise disjoint with edges which were obtained in step (3).
(3.3) Let's sum up all obtained pairwise disjoint edges: |B'|(from step (3)) + (|G'|-n+1)(from step (3.2)) = |B'|+((m-|B'|)-n+1)=m-n+1.

[quote] Myth doesn't live here anymore[/quote]

Hmmm.. Myth seems to live here from time to time. And that's an excellent news :)

Pierre.

An easier solution than Myth's one:

If in any group of 31 persons there is a couple it means that the males are not more of 30 so in the club there are at least 12 females and every female have slept with at least one male because if not we have a set of 30 men and a woman who didn't sleep with anybody against the hypothesi of problem.

Is it correct?

Because maybe two girls receive the same Boy, the 12 pairs must be disjoint!


Prove that $ (0,1)\cap \mathbb{Q}$ and $ [0,1]\cap \mathbb{Q}$ are homeomorphic. (under subspace topology)

Hm, I can't think of a bijection. Obviously there should be one since they're both countable...

Here's one, let $ \{r_n: n\in\mathbb{N}\}$ the set of rational numbers in $ (0,1)$. We define;
$ f(r_n)=\begin{cases} 0 & n=1 \\[1ex] 1 & n=2 \\[2ex] r_{n-2} & n\neq 1,2 \end{cases}$

yeah i thought of that one too. not sure if its continuous though

[hide="hint"]try to show that they are both homeomorphic to $ \mathbb{Q}$[/hide]


function
induction
geometric series
number theory unsolved
number theory

Find, as a function of $\, n, \,$ the sum of the digits of
\[ 9 \times 99 \times 9999 \times \cdots \times \left( 10^{2^n} - 1 \right),  \]
where each factor has twice as many digits as the previous one.

The answer is $9*2^n$ Now if anyone can prove it ...  :P
by the way is there some kind of formula for the sum of the digits of a number .. ? or ... :noo:

Sorry,  the proof is invalid. I made a stupid mistake.

I'll try to post a solution

[hide]Proof by induction that the function is 9*2^n

It works for n=0

Let G(n)=9*99*9999...*(10^2^n-1)

Let F(n) be the sum of the digits of that number

Now for the inductive step
G(N+1)=G(N)(10^2^(N+1)-1)

10^2^(2n+1) has 2^(n+1)+1 digits (and more digits than G(n) which would have less than 2^(n+1)-1 by the sum of a geometric series formula)

Now we can see G(N+1)=10^2^(N)*G(N)-G(N)

The sum of the digits of 10^2^(N)*G(N)= F(N)

Since the last digit of G(a) is 1 or 9, it is greater than 0.  When we subtract G(N) from 10^2^(N)*G(N) the last number of G(N) decreases by 1 and the number after it increases by 9, so that really is an increase of 8.

Now when you carry over several digits like 1000000-234233, it becomes 999990+10-234233

With the 8 added on we have 9*2^(N+1)  when we subtract F(N)=9*2^(N) from that we have 9*2^(N)

add that to 9*2^(N)we have have 9*2^(N+1) so for any number N, it works for N+1[/hide]

Consider the polynomial$$(X-1)(X^2-1)(X^4-1) \cdots (X^{2^n} - 1).$$The main idea is to connect the coefficients of this polynomial to the sum of digits of the resulting number.

Notice that for every $0 \leq k \leq 2^{n+1} - 1$, the coefficient of $X^k$ is either 1 or $-1$; in particular, it is 1 if the number of zeroes in the binary representation of $k$ is even and $-1$ otherwise.

Furthermore, check that every $-1$ coefficient contributes $9-1=8$ to the sum of the digits and every coefficient of 1 contributes 1 to the sum of the digits. However, it is well-known that exactly half of the integers from 0 to $2^{n+1} - 1$ have an even number of zeroes, so the total sum is simply $\boxed{9 \cdot 2^n}$.

[hide=very weird sol]
$9*99*9999*...*(10^{2^n}-1) = (10-1)(10^2-1)(10^4-1)...(10^{2^n}-1)$
Let $x=10$
$(x-1)(x^2-1)(x^4-1)...(x^{2^n}-1)$ when expanded the exponents that have an even number of zeroes in binary (witih leading zeroes) would have a coeffieicent of $1$ while those wih an odd number of zeroes would have a coefficient of $-1$ because the zeroes represent multiplying by $-1$ instead of a power of $x$ and the $-1$ is multiplied an odd number of times

To get $8$ the pos/neg of itself and the two cofficients after must be + - -. For this to happen the number's binary must end with $11$ or $1001$ or $100001$ or .....
The ways to end with $11$ are $2^{n-1}$ and to have an even number of zeroes there are $2^{n-2}$. Similarly, the ways to end with $1001$ is $2^{n-4}$ and this repeats until there it goes to $2$ and then $1$ if $n$ is odd. So use geometric sequence to get $2\frac{(4^{(n-1)/2}-1)}{3}+1=2\frac{(2^{n-1}-1)}{3}+1=\frac{2^n+1}{3}$
If $n$ is even its just a normal geometric sequence so its $2^{n-2}+2^{n-4}+...+1=\frac{4^{n/2}-1}{3}=\frac{2^n-1}{3}$
Because it has to divide $9$, the number of $8$s is the same as the number of $1$s and the total sum is the number above mutliplied by $9$

To get $9$ pos/neg of itself and the next coefficients must be -+. This happens if the binary ends with $1$ or $100$ or $10000$ or ....
The ways to end with $11$ is $2^n$ and to have an even number of zeroes, there are $2^{n-1}$ ways. Similar logic applies for other possible endings so the total ways is $2^{n-1}+2^{n-3}+...+1$
If $n$ is odd, use geometric sequence to get $\frac{4^{(n+1)/2}-1}{3}=\frac{2^{n+1}-1}{3}$
If $n$ is even, then its $2^{n-1}+2^{n-3}+...+2+1$, so it becomes $\frac{2^{n+1}+1}{3}$
The total sum is also the above number multiplied by $9$

So the total becomes $\frac{2^{n+1}+2^n}{3} = \frac{2*2^n+2^n}{3}=\frac{3*2^n}{3}=2^n$, but multiply it by $9$ to get $\boxed{9*2^n}$


Express $ 0.4\overline5$ as a common fraction.


[quote="gijsbijs"]What kind of math/algebra do I need to learn for this?[/quote]
It depends on what list of formulas you're talking about.

For a statistics study like AP Statistics, Intermediate Algebra is enough to understand them (see the list of formulas on the AP Test [url=http://www.collegeboard.com/prod_downloads/student/testing/ap/statistics/ap07_stat_frq.pdf]here[/url], pg. 3-5. Just to warn you, this is a pdf.).

Thanks for your reply!

..and for the PDF.

I appreciate it!.

It's difficult for me to define the list of formula's, as I don't know them. :lol:

But I guess it's rather basic to start with. Like formula's such as the ones for Standard deviation, poisson etc etc..

And I want to learn them fundamentally.  I mean, you don't have to explain them, if I see such formula's I need to know what to "do" with them immediately.

So the type of algebra that is used for these kind of formula's is what I need to learn.

I will answer by paragraph.

[quote="gijsbijs"]Hi,

I'm very interested in Probability and Statistics, it really caught my attention.
But I quit school very early and thus, lack a solid mathematical background.

I bought a few elementary math books to study and it went well.
However much of the type of formula's you see in statistics weren't covered at all.

I do not want to go straight into a statistics book, first want to lay down a solid foundation.

There seem to be so many "type" of mathematics, that I really don't know what kind of book I need to buy.

So, my question is: What kind of mathematics is best to learn when after that you want to get into Probability and Statistics?

Let me give you some info on what I'm looking for.
On the wikipedia page about Standard Deviation:
http://en.wikipedia.org/wiki/Standard_deviation

If you scroll down a bit you see the following formula:
[img]http://upload.wikimedia.org/math/5/c/3/5c3c0e2a4a7cdb91fa08f54dc2e43722.png[/img]

Now I understand the second part of it but not the first part. (the explaination about this formula is actually on the page, but I want develop understanding about these formula's so that I don't need an explaination, except about the variable of course.)
I want to understand these kind of formula's. What kind of math/algebra do I need to learn for this?[/quote]

Greetings. Welcome to AoPS.

To know probability, you must know counting. The Introductory and Intermediate Counting and Probability books are very useful for that. Links: http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=202       and       http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=302

It's good that you are reading books. :)

To gain a solid mathematical background, see links above.

By "type", I assume you mean "field". Such as "Algebra", "Combinatorics", "Number Theory", etc. If you want to buy a book, and you don't know what's in it, try to find an excerpt. The Intro and Intermediate C&P have excerpts, for example.

The answer to your question is above: Counting and Probability(combinatorics).

I see the formula for the average of N numbers before me.

To understand formulae, you can't just memorize it and move on. You need to apply it in unexpected ways. This is covered in great detail in the C&P books.

For the algebra, I think that these formulae only need  knowledge in basic algebra... :maybe:

Great!
Thanks so much for your recommendations!

I'll check them out.


[quote="biffanddoc"]Which sprint did just one person get right? THe one I thought was hardest was the one about sequences. Which ones were challenged and more than one answer accepteD?[/quote]

Mark Zhang was the only person in the whole contest to get #30 right. So for all you people who think Mark isn't as good as people think......EAT THAT!!!!

[quote="Magixmaniak"]Number 30 was the sprint question that only Mark Zhang got right. Everyone unfolded the net the wrong direction and got 4root29 instead of 2root113.  Btw, does anyone know a good way to solve number ten on the team? I heard that only one team got that question right and it was the question that my team (Texas) missed.[/quote]
Congratulations to Mark on number 30!  Actually, I think most people just didn't get to this problem in the time allowed.  [I know one member of my State's team had sqrt(452) when time was called.  This person, and probably many others, may have solved this problem if the previous 29 problems hadn't taken so long.]

[hide="Hint on Team Number 10"]I noticed a lot of "partition" problems on this year's competitions -- I.e., a problem in which "partitions" led to a relatively simple solution.  National Team #10 is such a problem.[/hide]

Number 10 was the one that only NJ got right. Ian solved it for us. (Ian's a teammate, not a proctor -_- )

Our team would have gotten #10 right but I didn't get all partitions of 10 soon enough.

10 was where there was a four by 4 grid. Pints are taken from each row, how many split the area into 2.  Look at it this way: call the distances between two points to be 1.
For the first row, call the distance from the left side, x, the second row is y and so on.
Now there are just three trapezoids:
(x+y)/2+(y+z)/2+(z+w)/2=9/2
x+2y+2z+w=9. 0 <= x,y,z,w <= 3  There are 28 ordered quadruplits that satisfy this

We got 14... not 28. Maybe my teammate forgot to flip it or something.

The problem had a 4 x 4 array of dots.  You had to pick one dot from each row such that connecting them (from top to bottom) in such a way that when you shade everything to the left of the line connecting the dots is half of the total area within the array.

When I first started this problem I started by doing the partitions of ten, but I noticed that 4,1,1,4 did not yield half shaded, although 1,4,1,4 did.  I thought that might make doing it by partitions may still take too long as you would have to get the partitions and then check them.

I decided that I would find the ways where you started on the top left point and find out how many paths starting there would cut the figure in half(area wise).   However many there were there would be the same number for starting at the top right dot and just be the mirror image.  the same is true for starting at the 2nd dot from the top left versus the snd from top right.  In the end it really didn't take that long to find the ways from the first two dots in the top row and then double it.

Did that make any sense?  I won't give my answer for awhile(assuming I can find the paper I wrote it on during the flight home :) )

Congratulations to Ian and New Jersey on number 10!

[hide="A complete solution based on partitions"]
nat mc did some algebra (above) showing the area of three trapezoids.  I'll borrow his algebra, as it clearly explains the conditions on the partitions:
$x+2y+2z+w=9,\qquad 0\le x,y,z,w \le 3$

I will allow a, b, c, and d to be the elements of the partition, such that:

$a=x,\qquad a\in \{0,1,2,3\}\\
b=2y, \qquad b\in \{0, 2, 4, 6\}\\
c=2z, \qquad c\in \{0, 2, 4, 6\}\\
d=w, \qquad d\in \{0, 1, 2, 3\}$

Now I list all four-element partitions of 9 consisting of the elements 0, 1, 2, 3, 4, or 6:

$6+3+0+0\\
6+2+1+0\\
6+1+1+1\\
4+4+1+0\\
4+3+2+0\\
4+3+1+1\\
4+2+2+1\\
3+3+3+0\\
3+3+2+1\\
3+2+2+2$

We know that $b$ and $c$ must be even, so we discard partitions that don't contain at least two even elements.  We are left with:

$6+3+0+0\\
6+2+1+0\\
4+4+1+0\\
4+3+2+0\\
4+2+2+1\\
3+2+2+2$

Note that each partition contains exactly one odd element.  The odd element must be assigned to $a$ or $d$ -- This always gives us two choices.

Elements larger than 3 must be assigned to $b$ or $c$ -- In every case, this gives us one or two choices.

The remaining elements are assigned to the remaining variables -- In every case, this gives us one or two choices.

Now, we tally the choices:

$
6+3+0+0:\qquad 2\times 2\times 1=\hbox{4 choices} \\
6+2+1+0:\qquad 2\times 2\times 2=\hbox{8 choices} \\
4+4+1+0:\qquad 2\times 1\times 1=\hbox{2 choices} \\
4+3+2+0:\qquad 2\times 2\times 2=\hbox{8 choices} \\
4+2+2+1:\qquad 2\times 2\times 1=\hbox{4 choices} \\
3+2+2+2:\qquad 2\times 1\times 1=\hbox{2 choices}$

$4+8+2+8+4+2=28$ total choices.[/hide]

YEs, that is exactly my soultion.  It actually doesn't have to do with partitions of 10.

[quote="Magixmaniak"]Number 30 was the sprint question that only Mark Zhang got right. Everyone unfolded the net the wrong direction and got 4root29 instead of 2root113.  Btw, does anyone know a good way to solve number ten on the team? I heard that only one team got that question right and  it was the question that my team (Texas) missed. I think this year's test wasn't too difficult except for the team. The sprint was fairly difficult and the target had no killers but did have several difficult ones.

 I'm disappointed at myself for not doing so well. I bombed the sprint (stupid careless mistakes). :(  :(  :(  :(  :(  :(  :(  :(  :(  :(  :([/quote]

I heard that Mark only got five target, otherwise he could easily have gotten 2nd.

I know that David Benjamin only got a 7 target and only a 21 sprint, Nathan only got a 6 target and Pardha only got 6 target.  They should alll have gotten perfect target an at least 25 sprint.

if i was on a team i wouldve gotten 78 points maybe


given real numbers a,b,c,x,y and z with

(a+b+c)(x+y+z)=3

(a^2+b^2+c^2)(x^2+y^2+z^2)=4

prove that ax+by+cz>=0[/i]

[quote="computer"]given real numbers $ a,b,c,x,y$ and $ z$ with

$ (a\plus{}b\plus{}c)(x\plus{}y\plus{}z)\equal{}3$

$ (a^2\plus{}b^2\plus{}c^2)(x^2\plus{}y^2\plus{}z^2)\equal{}4$

prove that $ ax\plus{}by\plus{}cz\geq0$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=70625


A three-digit integer contains one of each of the digits $ 1$, $ 3$, and $ 5$. What is the probability that the integer is divisible by $ 5$?

$ \textbf{(A)}\  \frac{1}{6}  \qquad
\textbf{(B)}\   \frac{1}{3}  \qquad
\textbf{(C)}\   \frac{1}{2}  \qquad
\textbf{(D)}\  \frac{2}{3}   \qquad
\textbf{(E)}\   \frac{5}{6}$

[quote="PowerOfPi"]A three-digit integer contains one of each of the digits $ 1$, $ 3$, and $ 5$. What is the probability that the integer is divisible by $ 5$?

$ \textbf{(A)}\ \frac {1}{6} \qquad \textbf{(B)}\ \frac {1}{3} \qquad \textbf{(C)}\ \frac {1}{2} \qquad \textbf{(D)}\ \frac {2}{3} \qquad \textbf{(E)}\ \frac {5}{6}$[/quote]

The only two integers divisble by 5 with these 3 digits are 135 and 315. Thus, the answer is $ \frac{2}{6}\equal{}\frac{1}{3}$, or $ \boxed{B}$.

[hide="Alternatively..."]
The integer is divisible by $ 5$ if and only if its last digit $ 5$.  This has probability $ \dfrac{1}{3}$, or $ \boxed{\textbf{(B)}}$.
[/hide]

There are 3! ways to order the digits around, and 2 ways for the last digit to be 5. So there is a 2/6=1/3 chance so the answer is B.


geometry
3D geometry
tetrahedron
geometric transformation
reflection
symmetry
parallelogram

Given a regular tetrahedron with side length 1. Find a closed curve of minimal length inside it intersecting all its faces.

As far as I understand, the problem asks us to find the minimal possible length of a polygonal lin e$XYZT$ where $X \in Int(ABC),Y \in Int(BCD), Z \in INT(CDA), T \in Int(DAB)$.

The first idea that comes to mind is the reflexion. suppose $XYZT$ is such a minimal path (it's true by Weierstrass Theorem). Now if we reflect $Z$ wrt $BCD$ to get $Z_1$ and let $XZ_1$ intersect the interior of $\delta BCD$ in $Y_1$ (it always does) then $Y=Y_1$ or otherwise just replace it. Hence we start reflecting: reflect $A$ wrt $BCD$ to get $A_1$ then $XY$ intersects $CDA_1$ in $Z$'s reflection $Z_1$. Reflect again $B$ wrt $A_1CD$ to get $B_1$ then $T_1$ the intersection of $XY$ with $DA_1B_1$ corresponds to $T$ in $DAB$. And finally reflect wrt $A_1b_1D$ to get $C_1$ and $X_1$ in $A_1B_1C_1$ correspoding to $X$ in $ABC$. Thus the total length of our line is $XX_1$ and we just have to minimize it (any choice of $X$ produces a valid arrangement, this is because as the map $X\rightarrow X_1$ is affine, the locus of $Y$ is the triangle with vertices when $X$ is $A,B,C$ which is clearly inside $BCD$ hence $Y$ is inside $BCD$ and we conclude analogously for $Z,T$). Now it's convenient to note $u=\overline{AD}, v=\overline{BD}, w=\overline{CD}$ then if $\overline{DX}=xu+yv+zw, x+y+z=1$ then $\overline{XX_1}=x\overline{AA_1}+y\overline{BB_1}+z\overline{CC_1}$. Now squaring this, we note that the $\overline{AA_1},\overline{BB_1},\overline{CC_1}$ are equal in magnitutde and have equal products in maginitude by symmetry wrt $D$ therefore we need to maximize $xy+yz+zx$ which gives $x=y=z=\frac 13$ so $X$ is the center of $\delta ABC$ and analogouly for $Y,Z,T$ thus giving us the answer $\frac 43$

We can prove that $X,Y,Z,T$ are centers in a different way. Indeed, suppose $X,Y,Z,T$ produce such a minimal arrangement and suppose wlog that $X$ is not the center.Produce a transformation $A\rightarrow B, B \rightarrow C, C \rightarrow D, D \rightarrow A$ (this is like an affine tranformation). Then $X$  goes to $X'$ inside $BCD$ and so do $Y$, $Y', Z'$. We prove $X'=Y, Y'=Z,Z'=T,T'=X$. Suppose that $X'\neq Y$. Then let $M,N,P,Q$ be midpoints of $X'Y, Y'Z, Z'T, T'X$.
Then $MN+NP+PQ+QM\leq \frac{XY+YZ+ZX++ZT+X'Y'+Y'Z'+Z'T'+T'X'}2=XY+YZ+ZT+TX$. By the minimality, we conclude that equalities hold everywhere which means that $X'YY'Z, T'XX'Y$ are parallelograms thus $X'Y||Y'Z||X'Y$ since it's impossible because the planes $ABC,BCD,DAB$ do not contain three parallel lines. Hence $X'=Y$ and so on, thus $XY=YZ=ZT=TX$. Again asa above, if $\overline {DX}=x\overline{DA}+y\overline{DB}+z\overline{DC}$ then $\overline {DY}=x\overline{DB}+y\overline{DC}+z\overline{DD}$ which means
$\overline {YX}=x\overline{BA}+y\overline{CB}+z\overline{AC}$ and like above this has minimal magnitude when $X$ is the center of the face.


Find all positive integer solutions of the equation $ x^2\plus{}3\equal{}28y^3$.
Is it solvable in $ Z[\omega]$?

What you mean by $ \omega$ ? just be more specefic PLZ

The $ \omega$ is pretty standard notation for the cubic roots of unity.
[url=http://en.wikipedia.org/wiki/Roots_of_unity]Roots of unity[/url]


Let $ x,y,z$ be non-negative real numbers with sum $ 3$. Let $ a,b$ be two constants. Find the maximal and minimal value of the following expression
\[ \sqrt{x^{2}\plus{}ax\plus{}b}\plus{}\sqrt{y^{2}\plus{}ay\plus{}b}\plus{}\sqrt{z^{2}\plus{}az\plus{}b},\]
in each of two following cases
(i). $ a,b\ge 0.$
(ii). $ b\ge 0, 4b\ge a^{2}.$

[hide="Never Mind, Wrong"][hide="Not sure"](i) Using Jensen's, this simplifies to:
$ \sqrt{(x\plus{}y\plus{}z)^{2}\plus{}a(x\plus{}y\plus{}z)\plus{}b}$
We can substitute 3 for x+y+z:
$ \sqrt{(9\plus{}3a\plus{}b}$
The maximum is infinite, while the minimum is $ 3$.[/hide][/hide]

Jensen's inequality doesn't allow you to simplify anything. It is an inequality, as its name suggests. In particular, that means it can at best only help you prove either the max or min, but definitely not both.

Oops, I misinterpreted  it as an equality.... I accidentally replaced the $ \Rightarrow $ with an $ \equal{}$...


Real number $ x, y, z$ satisfy $ x^{2}\equal{}y\plus{}2$,  $ y^{2}\equal{}z\plus{}2$, $ z^{2}\equal{}x\plus{}2$.
 If $ xyz$ if negative, find all possible values of $ x^{2}\plus{}y^{2}\plus{}z^{2}$.
The problem is from Indian TST 1998.

If $ |x|>2$, then $ y>|x|,z>|y|\equal{}y,x>|z|$ contradition. Therefore exist t, suth that $ x\equal{}2cost$. It give $ y\equal{}4cos^{t}\minus{}2\equal{}2cos2t$ and
$ z\equal{}2cos4t,x\equal{}2cos8t$. Therefore $ 8t\equal{}t\plus{}2k\pi, t\equal{}\frac{2k\pi}{7}$ or $ 8t\equal{}\minus{}t\plus{}2k\pi, t\equal{}\frac{2k\pi}{9}.$
$ x^{2}\plus{}y^{2}\plus{}z^{2}\equal{}x\plus{}y\plus{}z\plus{}6$ and in first case $ xyz>0$ for any k. In second case or x=y=z=-1 and $ x^{2}\plus{}y^{2}\plus{}z^{2}\equal{}3$ or
$ x^{2}\plus{}y^{2}\plus{}z^{2}\equal{}6\plus{}2(cos\frac{2\pi}{9}\plus{}cos\frac{4\pi}{9}\plus{}cos\frac{8\pi}{9}$. In last case $ x,y,z$ are roots of $ x^{3}\minus{}3x\equal{}\minus{}1$, therefore by Viet theorem $ x\plus{}y\plus{}z\equal{}0$ and $ x^{2}\plus{}y^{2}\plus{}z^{2}\equal{}6.$

Thank you very much, Rust. Your solution is very nice.  :)


inequalities
function
inequalities unsolved

Let $ x,y,z,\alpha\geq 0$ and $ x\leq y\leq z$. Prove that

$ (\alpha \minus{}1)[(x\minus{}z)y^{\alpha}\plus{}(y\minus{}x)z^{\alpha}\plus{}(z\minus{}y)x^{\alpha}]\geq 0$

[b]1.[/b] The inequality holds clearly for $ \alpha\in\{ 0,1\}$ (we obtain the equality).

[b]2.[/b] Suppose that $ \alpha\in (1,\infty)$. Then the above inequality can be equivalently written in the form
$ (x - z)y^{\alpha} + (y - x)z^{\alpha} + (z - y)x^{\alpha}\ge 0.$
From the fact that $ 0 < x\le y\le z$ we deduce the existence of the real numbers $ K,\, L$ such that $ 1\le K\le L$, $ y = Kx$, $ z = Lx$. This implies

$ \begin{array}{rcl} (x - z)y^{\alpha} + (y - x)z^{\alpha} + (z - y)x^{\alpha} & \ge & 0 \\
(x - Lx)(Kx)^{\alpha} + (Kx - x)(Lx)^{\alpha} + (Lx - Kx)x^{\alpha} & \ge & 0 \\
K^{\alpha}(1 - L) + K(L^{\alpha} - 1) - L^{\alpha} + L & \ge & 0. \end{array}$

But $ 1\le K\le L\quad\Rightarrow\quad\exists M\ge 1: \, L = MK\quad\Rightarrow$

$ \begin{array}{rcl} K^{\alpha}(1 - L) + K(L^{\alpha} - 1) - L^{\alpha} + L & \ge & 0 \\
K^{\alpha}(1 - MK) + K(M^{\alpha}K^{\alpha} - 1) - M^{\alpha}K^{\alpha} + MK & \ge & 0 \\
K^{\alpha}(M^{\alpha} - M) + K^{\alpha - 1}(1 - M^{\alpha}) + M - 1 & \ge & 0. \end{array}$

Define the function $ f(K)$ as $ f(K): = K^{\alpha}(M^{\alpha} - M) + K^{\alpha - 1}(1 - M^{\alpha}) + M - 1$ for $ K\ge 1$. Then
$ \frac {\mathrm{d}}{\mathrm{d}K}f(K) = K^{\alpha - 2}\left (\alpha K(M^{\alpha} - M) + (\alpha - 1)(1 - M^{\alpha})\right ).$
Now we prove the fact that $ \frac {\mathrm{d}}{\mathrm{d}K}f(K)\ge 0$ which is equivalently with

$ \begin{array}{rcl} \frac {\mathrm{d}}{\mathrm{d}K}f(K) & \ge & 0 \\
\alpha K(M^{\alpha} - M) & \ge & (\alpha - 1)(M^{\alpha} - 1) \\
K & \ge & \frac {(\alpha - 1)(M^{\alpha} - 1)}{\alpha (M^{\alpha} - M)},\qquad M > 1. \end{array}$
(The case M=1 is evident.)

The last inequality holds by the fact that the $ \mathrm{RHS}\le 1$, which is equivalent to

$ \begin{array}{rcl} \mathrm{RHS} & \le & 1 \\
M^{\alpha} - 1 - \alpha (M - 1) & \ge & 0. \end{array}$

Set $ g(M): = M^{\alpha} - 1 - \alpha (M - 1)$, $ M \ge 1$. Using the inequlity $ \frac {\mathrm{d}}{\mathrm{d}M}g(M) > 0$ we deduce

$ \inf_{M > 1}g(M) = g(1) = 0.$

Finally, this implies that for $ M > 1$

$ K\ge\frac {(\alpha - 1)(M^{\alpha} - 1)}{\alpha (M^{\alpha} - M)}\quad\Leftrightarrow\quad\frac {\mathrm{d}}{\mathrm{d}K}f(K) > 0\quad\Rightarrow\quad\inf_{K\ge 1}f(K) = f(1) = 0.$

Hence

$ f(K)\ge 0\quad\Leftrightarrow\quad (\alpha - 1)\left [(x - z)y^{\alpha} + (y - x)z^{\alpha} + (z - y)x^{\alpha}\right ]\ge 0$.

[b]3.[/b] For $ \alpha\in (0,1)$ we can use the same technique.

$ (x \minus{} z)y^{\alpha} \plus{} (y \minus{} x)z^{\alpha} \plus{} (z \minus{} y)x^{\alpha} \ge 0 \Longleftrightarrow (y \minus{} x)z^{\alpha} \plus{} (z \minus{} y)x^{\alpha} \ge (z \minus{} x)y^{\alpha} \Longleftrightarrow mz^{\alpha} \plus{} nx^{\alpha} \ge y^{\alpha}$

Here $ m \equal{} \frac {y \minus{} x}{z \minus{} x} \ge 0$ and $ n \equal{} \frac {z \minus{} y}{z \minus{} x} \ge 0$. Besides that,

$ m \plus{} n \equal{} 1$
$ mz \plus{} nx \equal{} y$

Being $ f(x) \equal{} x^{\alpha}$, we know that the inequality is equivalent with

$ mf(z) \plus{} nf(x) \ge f(mz \plus{} nx)$.

And here the problem ends.


Hi,

I've seen some problem that deals with medians intersecting each other (not all three of them so cevians doesn't work, or at least to me because I'm not sure on how to use that one neither) and finding the ratio.

How would you do problem like this one?

I'm not sure on assigning values or variables so here's just this:

Draw a triangle ABC such that medians from angle A intersects BC at F and medians from angle B intersects AC at E. Let the midpoint of EF be Z. The ray AZ intersects FC at Q. Find the ratio of AZ:ZQ.

How would you do ones like those?

[hide]
Using the area ratios......
$AQ:ZQ=[AFC]:[ZFC]$
and $[AFC]=\frac12 [ABC]$, $[ZFC]=\frac12 [CFE]= \frac14 [CBE]=\frac18 [ABC]$.
So, $AQ:ZQ=4:1$
Thus, $AZ:ZQ=3:1$.[/hide]

Using Menelaos
 $\triangle FCE\cap AQ $ obtain $\frac{QF}{QC}\frac{CA}{EA}\frac{EZ}{ZF}=1$ and thus 2QF=QC
$\triangle AQC\cap FE$ obtain $\frac{AZ}{ZQ}\frac{QF}{FC}\frac{CE}{EA}=1$ and thus $\frac{AZ}{ZQ}=\frac{FC}{QF}=3$


Find k min st $\forall{a,b,c>0}$,we have :
$\frac{a+b+c}{3}-\sqrt[3]{abc}\leq k\sqrt{a^{2}+b^{2}+c^{2}-ab-bc-ca}$ :)

With $k\geq\sqrt{\frac{2}{3}}$,I have a good solution :)

Let c$\Rightarrow{0}$ and a=b .We have :k$\geq{\frac{2}{3}}$
And when $k\geq{\frac{2}{3}}$...???
Wait for all of you :)


calculus
vector
topology
geometry
Gauss
linear algebra
Functional Analysis

What books would you recommend (for a beginner, obviously) on the topic?

[url=http://www.amazon.com/Vector-Calculus-Linear-Algebra-Differential/dp/0971576637/ref=sr_1_1?ie=UTF8&s=books&qid=1222399264&sr=8-1]Hubbard's Vector Calculus, Differential Forms, and Linear Algebra: An Integrated Approach[/url] has a superb introduction to Manifolds (as well as good introductions into other subjects including Differential Geometry, Functional Analysis and Probabiity theory) in the third chapter.

 If you eventually decide to get the book, you should buy the most recent edition although it's more expensive; unlike some other books, Hubbard's texts do change quite a bit edition to edition and the third edition is the best content-wise and error-wise.

 [url=http://www.amazon.com/Calculus-Manifolds-Approach-Classical-Theorems/dp/0805390219/ref=pd_bbs_sr_1?ie=UTF8&s=books&qid=1222399335&sr=1-1]Spivak's Calculus on Manifolds[/url] is also good and is much shorter than the Hubbards text (it cover less tangetial material, aka the good stuff). The definitions and theorems are also clearer in Spivak's text than in Hubbard's.

Lastly, An interesting thing to note is that Hubbard's text covers all the topics that Spivak does in addition to some other subjects.

I do very much love the Hubbard text, but it doesn't go very far. I think the greatest exposition on this subject are the two John Lee texts. This would either be [i]An Introduction to Topological Manifolds[/i] (which I would probably skip if you've had lots of topology) and [i]An Introduction to Smooth Manifolds[/i].

Note that I didn't actually look these titles up, but I'm 99% certain that is what they are called. These start from basic topology and build very nicely through some pretty difficult topics in manifold theory. There are many, many exercises and unlike books such as Boothby, there is a lot of writing. Everything is explained thoroughly with good examples. I wish I had found these when struggling through Boothby! They are both GTM books.

As a side note, I don't think any presentation of differential forms is as clear and reader friendly as in Hubbard. Even if you decide on a different book, I would read this before trying one of the more abstract versions. My only gripe about this book is that the notation is so non-standard that is takes a little bit to convert your thinking when you first encounter the standard way it is done.

@emjay285: Do you know when Volume 2 is coming out? There has been rumor for awhile (at least a year) that there is going to be one.

On the Lee texts- they were designed around the year-long graduate course here at the University of Washington, and I really liked that course. You have the names right, and the second edition should be out by now.

It's worth pointing out that there are many different types of introductions to manifolds:  There's calculus on manifolds (which emphasizes differential forms and stokes theorem), Riemannian geometry (which emphasizes the metric properties of manifolds, includes stuff like Gauss-Bonnet), and differential topology (which emphasizes things like degrees of vector fields, hairy ball theorem, etc).  And then there's the phrase "differential geometry" which may or may not include topics from all of the above, and could also includes things like the Frobenius Theorem or vector bundles.  Any of these should give the basic definition of a manifold, but from there they may go different ways.

Regardless of emphasis, just about every book comes in two flavors:  (1) submanifolds of $ \mathbb{R}^n$, and (2) abstract manifolds.  If you're looking for a book, decide first which approach you want.  It might be better for some to start with (1) because it's easier to understand, but others might want to cut to the chase and start with (2), since you don't technically need to understand (1) first.  It's a matter of taste and your level of mathematical maturity.  For example, a grad student should definitely start with (2), while even a very advanced high school student has absolutely no reason to do so.


modular arithmetic
number theory
algebra unsolved
algebra

Let $ k$ be a positive integer and $ p$ be a prime. I dont know how to prove that the group $ \mathbb{Z}_{p^k}^*$ is cyclic. Do you know some related results? Thank you very much for your help!

This is not quite true: The group is cyclic only if $ p$ is an odd prime, or if $ p\equal{}2$ and $ k \equal{} 1, 2$. The group $ \mathbb{Z}_8^\times$ is not cyclic, as you can see if you notice that $ 1^2 \equiv 3^2 \equiv 5^2 \equiv 7^2 \equiv 1 \pmod{8}$.

This is a standard result in Number Theory, usually called something like "existence of primitive roots mod n". It was first proved by Gauss. Here's the [url=http://en.wikipedia.org/wiki/Primitive_root_modulo_n]Wikipedia[/url]article. The proof is available in most books on number theory and in some places on-line. If you can't find it, I'll give you an outline.


In how many zeros does the number 100! end?

I think [hide]24, because there are 24 instances of 5 in the product - the 20 multiples of five, plus the 4 multiples of 25.  There are at least as many 2s.  Every zero at teh end will result from multiplying a 5 by a 2.[/hide]

I got [hide]24, 4 numbers :le: 100 that are divisible by 25 (2 fives) for a total of eight and 16 numbers :le: 100 that are divisible by exactally 1 five for a total of 16+8=24 fives since there are more five's than two's there are 24 zeros.[/hide]


[quote="kunny"]we can set $x=\cos A,\ y=\cos B,\ z=\cos C,\ \ 0<A,B,C<\frac{\pi}{2}.$[/quote]
What would you do with this condition: $x^2+y^2+z^2+2xyz=1(*)$?
[hide="hide"] I have a solution, but it's not really good, I think. I wonder what did they do to prove that: $(*) \implies A,B,C$ are 3 angles of an acute-angle triangle?
[/hide]

Did you read my question carefully, kunny?
[quote]... wonder what did they do to prove that:...[/quote]

Sorry for that I didn't read [hide="hide"].[/hide]

[quote="Lovasz"]If $x,y,z>0$ satisfy $x^2+y^2+z^2+2xyz=1$. Prove that:
\[ 1+xyz=x\sqrt{(1-y^2)(1-z^2)}+y\sqrt{(1-z^2)(1-x^2)}+z\sqrt{(1-x^2)(1-y^2)}  \][/quote]

 http://www.mathlinks.ro/Forum/viewtopic.php?t=23888. Check it! :D

Oh, an algebraic solution is the one I need, very nice! :) .

Indeed this is very simple. I was moved by your solution, fuzzylogic.


If only the positive integers from $ 1$ through $ 49$, inclusive, are written on a piece of paper, what is the sum of all the digits that are written on the paper?


function
algebra
domain
topology
absolute value
real analysis
real analysis theorems

Consider g: R^2 -> R, which is continuous.
Could $ g^{\minus{}1}(\left[ 0,1 \right])$ equal any of the subsets of R^2 below?

a) {(x, y) : x^2 + y^2 <1}
b) {(0, 0), (1, 1)}
c) x-axis

I honestly have no idea how I should approach the problem... I know that taking the function of a compact set gives you a compact set, but I'm not sure if that is relevant here.

Any help/hint would be great... Thank you.

[quote="jmerry"]From the topological definition of continuity, the inverse image of a closed set is closed. Therefore, (a) is impossible. On the other hand, (b) and (c) are closed sets. Can you construct examples for those two?[/quote]

Thank you. I think I got (c) with g(x) = { 0 if x=0, and 1/x if x $ \neq$ 0}. Then $ g^{ \minus{} 1}(\left[0,1\right]) \equal{} x$.

For some reason I'm still having trouble coming up with a g for (b), though.
I guess it's the fact that the domain would be restricted to {(0, 0), (1, 1)} which is giving me a hard time...

Bad example. That's not a continuous function, it has the wrong inverse image, and it's not even written as a function of two variables.

As an intermediate step (easier than (b), harder than (c)), try this (d): find a continuous function $ f$ from $ \mathbb{R}^2$ to $ \mathbb{R}$ such that $ f^{\minus{}1}([0,1])\equal{}\{(0,0)\}$.

By the way, there's no need to quote the entire post immediately preceding yours. You're assumed to be responding to it anyway.

Sorry, I feel like I'm missing something.

$ f^{ \minus{} 1}(\left[0, 1\right]) \equal{} \left\{ (0, 0) \right\}$, so $ f(0, 0) \equal{} \left[0, 1\right]$.

I know there exists such function according to the theorem you gave, but I just don't understand how I could create one.

Off-topic, but just to note that the baseball player referenced in your user name did attend the same high school as jmerry.

Carry on.

No, we shouldn't have $ f(0,0)\equal{}[0,1]$. That doesn't make sense. We should have $ f(0,0)$ be a number [i]in[/i] $ [0,1]$.
What number? Well, note that the inverse image of the open interval $ (0,1)$ is an open set $ U$ in $ \mathbb{R}^2$. In all the examples we're trying to build, the inverse image of the closed interval $ [0,1]$ is a set with empty interior- so any open set contained in it will be empty. Thus $ f(0,0)$ can't be in the open interval, and must be either $ 0$ or $ 1$.

But isn't that like saying $ f^{\minus{}1}(1/2)\equal{}\left\{ (0, 0) \right\}$ and $ f(0, 0) \neq 1/2$?

Okay, after a night's sleep I think I understand better what you said regarding $ f(0, 0)$ being either $ 0$ or $ 1$

For (d), would $ f(x,y)\equal{}\minus{} | x |$ work? Then $ f(0,0) \equal{} 0$ and $ f^{\minus{}1}( [ 0, 1]) \equal{} (0, 0)$, right?

No, the zero set of $ f(x,y)\equal{}\minus{}|x|$ is $ \{(x,y): x\equal{}0\}$, the $ y$-axis. If you flip it to $ \minus{}|y|$, it's an example for $ c$.
To get only the origin, you'll need something like an absolute value involving both variables.

No, that gets you a parabola. To get the two points, get one and multiply two versions of that; I would go with $ \minus{}(x^2\plus{}y^2)((x\minus{}1)^2\plus{}(y\minus{}1)^2)$.

oh, I see... I thought I really got it this time  :roll:

Thanks for being so patient with me. I have more similar problems to work on so I'll try those on my own.


geometry
geometric transformation
real analysis
real analysis unsolved

The following problem was given in a Russian Olympiad ($n$ years ago  :) ) and none of the participants solved it. The interesting thing is that the author had no solution himself and, as far as I am aware, this problem still remains unsolved. Here it is:

$\{a_n\}_{n=1}^{\infty}$ and $\{b_n\}_{n=1}^{\infty}$ are decreasing sequences of positive numbers. Define the sets

$M = \{A \subset \mathbb{N}\mid \sum_{n\in A}a_n < \infty\}$
and
$L = \{B \subset \mathbb{N}\mid \sum_{n \in B}b_n < \infty\}$.

Is it true that $M = L$, if we know that there is a bijection $\varphi:\mathbb{N}\to\mathbb{N}$, such that $M = \{\varphi(B)\mid B\subset L\}$?

Have you seen any solutions of this problem? Or may be you can solve it yourselves ... I tried, but unsuccessfully...

Regards,
jivko

[quote="orl"][quote="jivko777"]The following problem was given in a Russian Olympiad ($n$ years ago  :) ) and none of the participants solved it. [/quote]
More details on $n$ please. In other words EXACT source please.   :)[/quote]
It is a paranoia.
This problem is from problem book, not exact source known. Jivko gave as much information as it is possible.

[quote="Myth"]This problem is from problem book, not exact source known. Jivko gave as much information as it is possible.[/quote]

I know a person with paranoia with respect to thread naming.  :D I didn't know that it is math book.  :lol: Better take your time and post the missing ALL-Russian MO problems...  ;)

The answer is "Yes". Hint: if $\{a_k\}$ and $\{b_k\}$ are two sequences of positive numbers tending to $0$ (but not necessarily decreasing), then the inclusion $M\subset L$ is equivalent to existence of a constant $C>0$ and a convergent positive series $\sum_k c_k<+\infty$ such that $b_k\leq Ca_k+c_k$ for all $k$. ;)

To: orl

Didn't get your point about the "missing ALL-Russian MO Problems", orl, but hope you got what you needed. Sadovnichi and Konyagin are the authors of the book (please see the source field of my posting) and its name is Russian Olympiads for University Students (sorry if my translation from Russian makes no sense to you). There is no solution to this problem in the book, otherwise I would have known it and (thus) post the problem in a different section. I would never take the mickey out of you, mate, when it comes to maths problems!

To: fedja
Many thanks for your hint, fedja! Did you solve the problem yourself, or already knew the solution? If latter, could you pls post the source you used - I'm just curious to see it. Many thanks once again!

Regards,
jivko

[quote="jivko777"]
To: fedja
Did you solve the problem yourself, or already knew the solution?
jivko[/quote]
I solved this one myself  :)

[quote="jivko777"]

Didn't get your point about the "missing ALL-Russian MO Problems", orl, but hope you got what you needed. Sadovnichi and Konyagin are the authors of the book (please see the source field of my posting) and its name is Russian Olympiads for University Students (sorry if my translation from Russian makes no sense to you). There is no solution to this problem in the book, otherwise I would have known it and (thus) post the problem in a different section. I would never take the mickey out of you, mate, when it comes to maths problems!
[/quote]

Don't worry. I thought you are referring to the ALL-Russian MO for school students. Thus I was wondering why you neither gave the exact year when the problem was used on the contest nor stated that there is no solution available.  :)


How do you prove that f(x)=sin(1/x) is continuous on [0,1]?

It isn't. It's continuous on $(0,1],$ but not at $0.$

if it cant even be continuous how can you hope it to be differentiable?


find all functions that satisfies the given conditions
$f: Z \longrightarrow Z$ and $f(n)-f(n+f(m))=m$

Are you sure about this? I got that there are no such functions, but I could be wrong, I guess..

Put $m=-f(a),n=a-f(m)$ for some $a$. We get $f(n)=0$. Now make $m=n$ s.t. $f(m)=f(n)=0$ in the opriginal relation to get $n=0$, so $f(0)=0$.

In the original relation, make now $n=0$. We get $f(f(m))=-m\ (*)$. Put $f(m)$ instead of $m$. We find $f(n-m)=f(n)-f(m)$. $(*)$ shows that $f(-m)=-f(m)$, so $f$ is additive: $f(m+n)=f(m)+f(n)$. This means that we have $f(n)=kn$ for some constant integer $k$, so $-n=f(f(n))=k^2n$, which is absurd...

If I am right then solution is following.
Put $n=f(m)$. We get $f(f(m))=m$. Put $n=2f(m)$ $\Rightarrow$ $f(2f(m))=2m$ and so you. You can easily check that $f(kf(m))=km$ for all $k\in\mathbb{Z}$. Put $m=f(1)$, then we obtain $f(k)=kf(1)$. Substituting it into initial equation we see that $mf^2(1)=m$, so $f(1)=\pm 1$ and there are two solutions: $f(n)=n$ and $f(n)=-n$.

sinankaral is right about the question and the solution is there is no solution.. :P

yes sdogan is right... there is no such function...

[quote="grobber"] Put $f(m)$ instead of $m$. We find $f(n-m)=f(n)-f(m)$[/quote]

I get $f(n+m)=f(n)-f(m)$  :? How did you manage to get what you get ?

We have $f(n)-f(n-m)=f(n)-f(n+f(f(m)))=f(m)$. What exactly is wrong here? The first equality follows from $f(f(m))=-m$, and the second one follows from replacing $m$ with $f(m)$ in the initial relation $f(n)-f(n+f(m))=m$.

Thanks  :) Anyway I know what caused the difference - I think about equation in really silly way  :blush:


By linguistic problems i mean problems that appear on NACLO and IOL. Those questions often ask for some sort od explanation. However the true explanation is " hmm i think this is the rule for the problem ... it is true in every case they've given .... so it must be true'.
So I'm interested in how do you write solutions?  Does anybody know how are this solutions graded?

This might seem a strange question but i want to know is there a guideline of some sort.

Can you find an old exam and a set of model answers for it?  Since I'm not affiliated with those exams, I don't know what the graders' expectations are.  But, I could probably help show you what's significant about the way the model answers were written up.

Hey thanks for replaying so fast.  :lol:

Here is the exam (with model answers) [url]http://iol6.linguistics-bg.com/zadachi/IOL6en.pdf[/url].

All right.  Which questions in particular would you like to look at?

Sure.  How would you have answered the question, and how would you have written it up?  (I'm not trying to be obnoxious; it would just help me help you out if I had a better idea of how you think about these problems and what you already know. :) )


@lomos_lupin: That's a very beautiful solution. Good job! :)

@killmesoft: for inequalities, you need to prove some lemmas sometimes and then finish the problem with well-known ineq(am-gm, muirhead, jensen, schur, etc). For problems in general, you usually need to find  a way to simplify the given question to a problem you know how to do, and then finish it... this comes with experience(usualy)

[quote="killmesoft"]what should i learn to solve these kinds of proving questions? i don't get any of the questions posted around here.[/quote]

Unless your crazy like eylot, you don't start on these things, you start somewhere basic. I really do reccomend all of the books that they sell here:
http://www.artofproblemsolving.com/Books/AoPS_B_Texts.php

I really only started to learn how to problem solve, after working through the counting book they have here. Its only after working through all of aops 1, and a bit of aops 2, I feel like I can 'attempt' inequalities like the one above.

Still working on the question above. Very fun so far, hoping I can finish it soon ( without hints !)

[quote="killmesoft"]anyone has scanned or [b]pdf[/b]ed copy of these books?[/quote]

Man, I highly suggest you [b]buy[/b] these books. If you have no intention on buying them, and plan on getting them through some [i]underhanded[/i] ways, then I highly suggest you don't talk about it here!

I hope you noticed, but the site that this form is on, wrote, produces and sells the books. One of the Admins who works down in california with them ( Nstao, right? ), trolls this form often.

Heh, just trying to keep you out of trouble, and learning math instead! :D

hosted on aops servers, different frame I believe.

[quote="killmesoft"]huh? i thought this was mathlinks forum.[/quote]

Yes... mathlinks and AOPS are twinned... when you post on one, it appears as a post on the other.  Essentially, the databases are merged into one database with 2 interfaces.  Of course, most of us use mathlinks, since it has a better skin and no ads.

[quote="killmesoft"]oh btw, is it possible for non-olympiad participant like me to participate in putnam competition?[/quote]

Way to totally hijack the thread with an off-topic comment!   :o

I just posted some info about putnam in one of the other threads.  Check it out.

Damn! nice solution.

I kind of had a long one with two cases, where if $x^{2}_{1}+x^{2}_{2}+\cdots+x^{2}_{n}$ was greater than or less than one. Used CS. Not even sure if it was right. I'll post it later, too see how it stands up.

How does one think up a solution like lupins? Where did the insight come from?

[quote="rem"][quote="rzsolt"]How does one think up a solution like lupins? Where did the insight come from?[/quote]
Experience[/quote]

ah think i figured out the insight in his solution, by pretending to work backwords. Here is what I meant I was asking for:

Taking it all to the LHS , and noticing that mMn = Mm + Mm + ... n times

we have

$x_{1}^{2}+Mm+x_{2}^{2}+Mm+\cdots x_{n}^{2}+Mm\leq  0$

All the symmetry makes us think that $(x_{i}^{2}+Mm) \leq  0$ for all $i$. Finaly we would notice that it looks almost like a perfect square, so we are basicly done then.

Then working backwords, we have to sol.

I find that idea of how we got the solution, just as helpfull as the actual answer!  :)

Yes, lomos_lupin's solution is nice.  Here's an alternative solution:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=17066[/url].

[quote="rem"][quote="rzsolt"]How does one think up a solution like lupins? Where did the insight come from?[/quote]
Experience[/quote]

don't worry, no one can think of these solutions without having to do a sufficient amount of questions.


Bác nào ủng hộ thì vào đây !  :)

Các bác có thể post theo mẫu sau :Bôi đen rồi copy vào bài post (Cho nó thống nhất)

[code][b][color=darkblue]BĐT.số thứ tự[/color] (Tác giả)[/b]
[hide]Đề bài[/hide]
[color=darkblue][b]Lời giải[/b][/color]
[b]Lời giải 1 (Tác giả)[/b]
[hide]............[/hide][/code]

[size=150][b]MỤC LỤC[/b][/size]
[hide][url=http://www.mathlinks.ro/Forum/viewtopic.php?p=260986#p260986]Bài 1[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=260994#p260994]Bài 2[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=261723#p261723]Bài 3[/url]
[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=261738#p261738]Bài 4[/url][/hide]

[b][color=darkblue]BĐT.1[/color] (vasc)[/b]
 [hide]Cho $\displaystyle  a,b,c$ là số thực dương.Chứng minh rằng :

 $\displaystyle  a^2+b^2+c^2 \geq \sqrt {3(a^3b+b^3c+c^3a)} $[/hide]
[b][color=darkblue]Lời giải:[/color][/b]
[b]Lời giải 1 (Anh Cuong & Sasha)[/b]
[hide] Giả sử rằng $\displaystyle a=\min{a,b,c}$

 Đặt $\displaystyle  x=b-a,y=c-a$

 BĐT đã cho tương đương với

 $\displaystyle g(a,x,y)=(5x^2+5y^2-xy)a^2+(x^3+x^2y+4xy^2+y^3)a+x^4+y^4+2x^2y^2-3x^3y \ge 0$

 Rõ ràng nếu $\displaystyle  x,y$ đồng thời = 0 ta có ĐPCM.
 Ngược lại giả sử $\displaystyle y>0$ khi đó ta có thể giả sử rằng $\displaystyle  y=1$ (Do BĐT đồng bậc)
 Từ đó

 $ \displaystyle \Delta=(x^3-5x^2+4x+1)^2-4(x^2-x+1)(x^4-3x^3+2x^2+1)$
   $                             =-3(x^3-x^2-2x+1)^2 \le 0$

 Chú ý rằng Hệ số của $a^2$ : $5x^2+5y^2-xy \ge 0$

 Suy ra $g(a,x,y) \ge 0$[/hide]
[b] Lời giải 2 (Stefan V.)[/b]
 [hide]Ta có
  $\displaystyle \left(a^2+b^2+c^2\right)^2-3\left(a^3b+b^3c+c^3a\right)$
 $\displaystyle  =\frac{1}{2}\left( \left( a^{2}-2ab+bc-c^{2}+ca\right) ^2+\left(b^{2}-  2bc+ca-a^{2}+ab\right)^2+\left( c^{2}-2ca+ab-b^{2}+bc\right)^2 \right) \geq 0 $

 Dấu bằng xảy ra khi và chỉ khi

$\displaystyle  a^{2}-2ab+bc-c^{2}+ca = 0$

$\displaystyle  b^{2}-2bc+ca-a^{2}+ab = 0$

$\displaystyle  c^{2}-2ca+ab-b^{2}+bc  =0$

 Có 2 trường hợp

1.$\displaystyle  a=b=c$

2.$\displaystyle a:b:c=\sin^2\frac{4\pi}{7}:\sin^2\frac{2\pi}{7}:\sin^2\frac{\pi}{7}$ (?)

----------------------------------
Mong các bác hoàn thiện [?][/hide]

[b][color=darkblue]BĐT.2[/color] (Viet Math)[/b]
 [hide]Cho $\displaystyle a,b,c,n$ là các số thực không âm.
          Chứng minh rằng :

 $\displaystyle \displaystyle \frac{na+b}{na+c}+\frac{nb+c}{nb+a}+\frac{nc+a}{nc+b} \leq   \frac{a}{b}+\frac{b}{c}+\frac{c}{a}$[/hide]
[b][color=darkblue]Lời giải:[/color][/b]
[b]Lời giải 1 (Viet Math)[/b]
[hide] Đặt
       $X=\displaystyle\frac{\frac{a}{b}+n}{n+1}$

       $Y=\displaystyle\frac{\frac{b}{c}+n}{n+1}$

       $Z=\displaystyle\frac{\frac{c}{a}+n}{n+1}$

 Dễ thấy $\displaystyle XYZ \ge 1$

 Ta có

 $\displaystyle\frac{a}{b}-\frac{nc+a}{nc+b}=\frac{nc(a-b)}{b(nc+b)}
                               =\frac{nc}{nc+b}\cdot(\frac{a}{b}-1)$

 $\displaystyle =\frac{n}{(n+1)Y}\cdot(n+1)\cdot\frac{\frac{a}{b}+n-(n+1)}{n+1}=\frac{n(X-1)}{Y}$

 Do đó BĐT trên tương dương với :

 $\displaystyle\frac{X}{Y}+\frac{Y}{Z}+\frac{Z}{X} \ge \frac{1}{X}+\frac{1}{Y}+\frac{1}{Z} $  (*)

 Áp dụng BĐT [b]Cauchy[/b] ta có :

$\displaystyle\frac{X}{Y}+\frac{X}{Y}+\frac{Z}{X} \ge 3\sqrt[3]{\frac{XZ}{Y^2}}
=3\sqrt[3]{\frac{XYZ}{Y^3}} \ge \frac{3}{Y}$

(Vì $\displaystyle XYZ \ge 1$)

 Tương tự

$\displaystyle 2\frac{Y}{Z}+\frac{X}{Y} \ge \frac{3}{Z}$

$\displaystyle 2\frac{Z}{X}+\frac{Y}{Z} \ge \frac{3}{X}$

 Từ đó ta có (*)

 Suy ra điều phải chứng minh
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Xin lỗi  :P  nhưng vì cái này em rành lời giải (Làm mẫu)[/hide]

[b][color=darkblue]BĐT.3[/color] (Mahbub)[/b]
 [hide]Cho $\displaystyle x_1,x_2,\cdots,x_n$ là các số thực dương thỏa mãn $\displaystyle x_1x_2\cdots x_n =1$
 Chứng minh rằng :
$\displaystyle \frac{1}{n-1+x_1}+\frac{1}{n-1+x_2}+...+\frac{1}{n-1+x_n}\leq 1$[/hide]
[color=darkblue][b]Lời giải[/b][/color]
[b]Lời giải 1 (prowler)[/b]
[hide]
  Giả sử rằng $a_1 \le a_2 \le \cdots a_n$
  Suy ra $x_1 \le 1,x_1x_2 \le 1 \cdots x_1..x_{n-1}\le 1$

[/hide]

[b][color=darkblue]BĐT.4[/color] (hungkhtn)[/b]
[hide]Cho $\displaystyle a,b,c$ là số thực dương.Chứng minh rằng:

$\displaystyle \displaystyle \frac{a^3}{2a^2+b^2}+\frac{b^3}{2b^2+c^2}+\frac{c^3}{2c^2+a^2} \ge \frac{a+b+c}{3}$[/hide]
[color=darkblue][b]Lời giải[/b][/color]
[hide]Đang tổng hợp[/hide]

[b][color=darkblue]BĐT.5[/color] (hungkhtn)[/b]
[hide]Cho $a,b,c$ là các số thực dương.Chứng minh rằng:

$\displaystyle \frac{a}{b^2+c^2}+\frac{b}{a^2+c^2}+\frac{c}{a^2+b^2} \ge \frac{4}{5}(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a})$[/hide]
[color=darkblue][b]Lời giải[/b][/color]
[hide]Đang tổng hợp[/hide]

[b][color=darkblue]BĐT.6[/color] ()[/b]
[color=darkblue][b]Lời giải[/b][/color]

[b][color=darkblue]BĐT.7[/color] ()[/b]
[color=darkblue][b]Lời giải[/b][/color]

[b][color=darkblue]BĐT.8[/color] ()[/b]
[color=darkblue][b]Lời giải[/b][/color]

[b][color=darkblue]BĐT.9[/color] ()[/b]
[color=darkblue][b]Lời giải[/b][/color]

[b][color=darkblue]BĐT.10[/color] ()[/b]
[color=darkblue][b]Lời giải[/b][/color]

[b][color=darkblue]BĐT.11[/color] ()[/b]
[color=darkblue][b]Lời giải[/b][/color]

[b][color=darkblue]BĐT.12[/color] ()[/b]
[color=darkblue][b]Lời giải[/b][/color]

[b][color=darkblue]BĐT.13[/color] ()[/b]
[color=darkblue][b]Lời giải[/b][/color]

[b][color=darkblue]BĐT.14[/color] ()[/b]
[color=darkblue][b]Lời giải[/b][/color]

[b][color=darkblue]BĐT.15[/color] ()[/b]
[color=darkblue][b]Lời giải[/b][/color]

Anh VietMath lam` tot' qua' .The' nay` thi` the' he bon em moi' co' tai` lieu ma` xem chu'.Nhung em hoi? may bai` sau dau ca? roi` a

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CONG NHAN ANH VIET ANH DINH THAT .ANH LAM TRANG NAY CO TOON COONG LAM K0


Is there a difference between mod and modulo (if that's how you spell it)?

T-H-A-T isn't how you spell I-T. ;)

No, there's no difference between mod and modulo. Unless you mean mod as in moderator, or mod as in 1960s fashions. Except for the spelling of course.

I don't know about any other meanings, but those are used for the same idea. "mod" is just an abbreviation for "modulus." As a sidenite, "modulo" is the ablative declension form of "modulus," so "modulo $n$" is equivalent to "in modulus $n$."

we write $4\equiv4\pmod{7}$
We say 4 is congruent to 4 modulo 7.
We don't want to write, $4\equiv 4 (\text{(modulo} 7)$.
That is bulky. mod is an abreviation of modulo.


induction
function
algebra proposed
algebra

Let $f: R \rightarrow R$ such that :

$f(1)=1$
$f(a+b)=f(a)+f(b)$ for all $a,b \in R$ and $f(x)f(\frac{1}{x})=1$ for all $x \in R_0$

Show that $f(x)=x$ for all $x \in R$

It's easy to show that $f(x)=x$ for $x\in\mathbb{Q}$ in a fairly traditional manner:

By an easy induction, $f(nx)=nf(x)$ for $x\in\mathbb{R}$ and $n\in\mathbb{Z}$, and hence (i) $f(x)=x$ for $x\in\mathbb{Z}$, (ii) $f\left(\frac{p}{q}\right)=p.f\left(\frac{1}{q}\right)=\frac{p}{f(q)}=\frac{p}{q}f(1)=\frac{p}{q}$ for $p,q\in\mathbb{Z}$, $q\ne0$.

I've seen lots of different ways to extend this to the reals by assuming things like monotonicity or continuity, but I'm not sure how you'd do it from the information you have here.  Are you sure it's sufficient?

IMO Shortlist/Longlist 1989, proposed by Ireland.

Yeah, it's enough :).

We find that $f\left(x+\frac 1x\right)=f(x)+f\left(\frac 1x\right)$ for every non-zero $x$, and, since the numbers of the form $a+\frac 1a$ are precisely those in $U=(-\infty,-2]\cup[2,\infty)$, the hypothesis tells us that $f$ maps $U$ into $U$.

Now try to prove (it's a pretty nice exercise) that an additive function is either continuous (and thus of the form $f(x)=kx,\ \forall x$), or has a graph which is dense in $\mathbb R^2$. Since the observation in the paragraph above implies that the graph of $f$ cannot be dense, the only possibility is that it's linear, and $f(x)=x,\ \forall x$ follows immediately.

Proposed by Ireland? Shame on me then, for not solving it...

Nice proof grobber - I've seen the result you used before, but I hadn't thought of applying it like that.


For some reason, I am missing my AMC Problem Series message board archive, is there a limit for the archives?

Thank you!

Can you give me more information?  Which class -- AMC 10 or AMC 12?

I enrolled in AMC 10 (around January of this year).


inequalities
calculus
derivative
function
calculus computations

let $ f(x)$ is derivative on $ \left[ {0; \plus{} \infty } \right]$ and $ f(0) \equal{} f(1) \equal{} 1$.; $ f'(x) \ne f(x), \forall x \in \left( {0; \plus{} \infty } \right)$
show that $ f(x) < e^x$ , ($ x \ge 0$)

Just cosider the function $ g(x)\equal{}e^{\minus{}x}f(x)$, then $ g(0)>g(1)$, hence  $ g'(x)<0$ :wink:

[quote="leshik"]Just cosider the function $ g(x) \equal{} e^{ \minus{} x}f(x)$, then $ g(0) > g(1)$, hence  $ g'(x) < 0$ :wink:[/quote]

is it right?  :wink:

Certainly. Consider $ g'(x)\equal{}e^{\minus{}x}(f'(x)\minus{}f(x))$. By the first condition of the problem $ g'(x)$ does not vanish ($ f'(x)\ne f(x)$),hence (recalling Darboux property of the derivative) it has steady sign. So our function $ g$ is either increasing or decreasing. But $ g(0)\equal{}1<g(1)$, thus our function is strictly decreasing and therefore $ f(x)e^{\minus{}x}\equal{}g(x)<g(0)\equal{}1$


find all function $ f: R \minus{} \minus{} \minus{} > R$  such that
$ f(xy \plus{} f(x)) \equal{} xf(y) \plus{} f(x)$

[hide]$ x \equal{} \minus{}1, y \equal{} f(\minus{}1)$ gives $ f(\minus{}f(\minus{}1)\plus{}f(\minus{}1)) \equal{} f(0) \equal{} f(\minus{}1) \minus{} f(f(\minus{}1))$
$ x \equal{} \minus{}1, y \equal{} 0$ gives $ f(f(\minus{}1)) \equal{} f(\minus{}1) \minus{} f(0)$
From these two equations we conclude that $ f(0) \equal{} f(\minus{}1) \plus{} f(0) \minus{} f(\minus{}1)$, so $ f(0) \equal{} 0$.

Suppose $ f(a) \equal{} f(b)$. Then $ af(b) \plus{} f(a) \equal{} f(ab \plus{} f(a)) \equal{} f(ab \plus{} f(b)) \equal{} bf(a) \plus{} f(b)$. So $ (a\plus{}1)f(a) \equal{} (b\plus{}1)f(a)$, and either $ a \equal{} b$ or $ f(a) \equal{} f(b) \equal{} 0$.

$ y \equal{} 0$ gives $ f(f(x)) \equal{} f(x)$. If $ f(x)$ is nonzero, then from the above we conclude that $ f(x) \equal{} x$. So for every point $ f$ is either $ 0$ or the identity. $ f$ being one of these on all points works.

Suppose $ c$ is a nonzero real such that $ f(c) \equal{} 0$. From $ x \equal{} c$ we know that $ f(cy) \equal{} cf(y)$. From $ y \equal{} 1$, $ f(1) \equal{} 0$. Put $ y \equal{} 1$ into the original equation to get $ f(x\plus{}f(x)) \equal{} f(x)$. Suppose $ f(x) \neq 0$. Then we get $ x \plus{} f(x) \equal{} x \Rightarrow f(x) \equal{} 0$, contradiction. So $ f$ must be zero at all points if it's zero at more than just $ 0$.

The solutions are $ f(x) \equal{} 0$ and $ f(x) \equal{} x$.[/hide]

[hide="Question"]Excuse me, but why can we conclude that f(0) = 0 from f(0) = f(-1) + f(0) - f(-1)?[/hide]

To be honest,I didn't try to solve this problem.I've just read [b]Carcul[/b]'s question and was trying to find another way to prove that $ f(0)\equal{}0$.
Suppose to the contrary that $ f(0)\neq 0$ and substitute $ y\equal{}0$ in the main equation:
\[ f(f(x))\equal{}x\cdot f(0)\plus{}f(x)
\]
From where it follows that $ f$ is injective function.Indeed,if $ f(x)\equal{}f(y)$,then
$ x\cdot f(0)\equal{}y\cdot f(0)$,but as we suppose $ f(0)\neq 0$,hence $ x\equal{}y$.
Now substitute $ x\equal{}y\equal{}0$,therefore,
$ f(f(0))\equal{}f(0)$ combining it with injectivity we conclude that $ f(0)\equal{}0$,contradiction.