How many of the first one hundred positive integers are divisible by all of the numbers $2,3,4,5$?

$\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 3 \qquad \text{(E)}\ 4$

[quote="4everwise"]How many of the first one hundred positive integers are divisible by all of the numbers $2,3,4,5$?

\[ \text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 3 \qquad \text{(E)}\ 4  \][/quote]

[hide]
4 is a multiple of 2 so we can just find numbers that are divisible by {4,3,5}

$4\cdot 3\cdot 5 = 60$
There is 1 multiple of 60 in 100

so B
[/hide]

[hide="solution"]For a number to be divisible by a set of numbers, the number must a multiple of the lcm. The lcm of $5,4,3,2$ is $60$. The only multiple of $60$ less than is $100$ is $60$. So the answer is $\boxed{\text{B: }1}$[/hide]

ccy: your solution is right, answer choice is wrong.

[quote="4everwise"]How many of the first one hundred positive integers are divisible by all of the numbers $2,3,4,5$?

\[ \text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 3 \qquad \text{(E)}\ 4  \][/quote]

[hide]
The number must contain 2^2, because of the 4.
We also need a 3 in this number.
Finally, we need a 5 in this number, so we have:
4*3*5=60.
Any multiples of this number are bigger than 100 so our answer is:
B, 1.
[/hide]

[hide]
the only number that works is 3*4*5=60, the 2 cancels out because 4 is divisible by it.
So, the answer is $B$[/hide]

[hide]The least common multiple is 60, so 60 is the minimum number that works. The next number be a multiple of 60, and the smallest multiple of 60 other than itself is 120, which is greater than 100. THus the answer is just one number. [/hide]

[quote="4everwise"]How many of the first one hundred positive integers are divisible by all of the numbers $2,3,4,5$?

\[ \text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 3 \qquad \text{(E)}\ 4  \][/quote]

[hide="Solution"]
LCM = 60

100/60 rounded down = 1

[b]B[/b]
[/hide]


geometry
3D geometry
pyramid
geometry proposed

Let the pyramid $O.ABC$ while $OA, OB, OC$ are perpendicular to each other. Line the perpendicular $OH$ to the plane $(ABC)$. Let $\alpha=\angle HOA, \beta=\angle HOB, \gamma=\angle HOC$. Prove that
      $cotg\alpha.cotg\beta.cotg\gamma \le \frac{\sqrt{2}}{4}$

It looks everybody is not interested in solid geom :huh:


geometry
circumcircle
Euler
conics
hyperbola
incenter
geometry solved

Let $O$ and $H$ denote the circumcenter and the orthocenter of a triangle $ABC$, respectively. Prove that the nine-point circles (Euler circles) of the triangles $AOH$, $BOH$, $COH$ have two common points.

This is obvious : the three circles go through the midpoint of OH and through the center of the rectangular hyperbola going through A,B,C,O,H (Jerabek hyperbola)

Why is it obvious that all three circles pass through the centre of the Jerabek hyperbola?  I don't see it.

[quote="grobber"]The nine-point circle is the locus of the centers of the rectangular hyperbolas through $A,B,C$ (or, in other words, the locus of the centers of the hyperbolas through $A,B,C,H$). I think this is what Pierre used.[/quote]

Ok, thanks, this result does look familiar.  Is there a nice proof of it?

The fact that the NP-circle of a triangle inscribed in a rectangular hyperbola goes through the center of the hyoerbola is a  very classical old theorem due to Jean-Victor Poncelet (1817)
The original synthetic proof is available on line in french (Page 511) at
http://visualiseur.bnf.fr/Visualiseur?Destination=Gallica&O=NUMM-90214
Note that, in your statement, you can take any point instead of O
Some related classical results :
1) The center of the rectangular hyperbola going through A,B,C,M lies on the circumcircle of the pedal triangle of M wrt ABC
2) The center of the rectangular hyperbola going through A,B,C,M lies on the circumcircle of the cevian triangle of M wrt ABC
It follows from 1) that the common points of the NP-circle and the pedal circle of M are the centers of the rectangular circumhyperbolae going through M and through M* (isogonal conjugate of M), because M and M* have the same pedal circle.
Hence the pedal circle of M touches the NP-circle <=> the conic through A,B,C,M,M* is a rectangular hyperbola <=> O,M,M* lie on a same line (as O = H*, the isogonal conjugate of a rectangular circumhyperbola is a line through O)
Now, if M is the incenter or an excenter, we have M = M*; hence the NP-circle touches the incircle (at the center of the rectangular circumhyperbola going through I) and each excircle  (at the center at the center of the rectangular circumhyperbola going through the corresponding excenter). So, we can see, that the Feuerbach theorem is an immediate consequence of 1)


[quote="Moubinool"]$u(n)=\sqrt{1+\sqrt{2+\sqrt{3+\sqrt{...+\sqrt{n}}}}}$

1/ Prove that $u(n)$ is convergent

2/ Prove that the limit verify $L \sim 1.757932...$[/quote]

1).It is readily that there exists a constant $C$ s.t. $n \leq C^{2^n}$ which implies the boundenss assertion of the sequence $u(n)$, to be accurate the following does:
$u(n)=\sqrt{1+\sqrt{2+\sqrt{3+\sqrt{...+\sqrt{n}}}}} \leq \sqrt{C^{2^1}+\sqrt{{C^{2^2}}+\sqrt{C^{2^3}+\sqrt{...+\sqrt{C^{2^n}} }}}} =C \sqrt{1+\sqrt{1+\sqrt{1+\sqrt{...+\sqrt{1}}}}}$

$($ we used there the boundness of the sequence $ a(n)=\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{...+\sqrt{1}}}}})$
It is left to remind that our sequence $u(n)$ is increasing.

${u_n=\sqrt {1+\sqrt {2+\sqrt {3+\sqrt {\ldots +\dot {\dot {\dot {\dot {\sqrt n}}}}}}}}},\ n\in N^*$

$1^{\circ}.\ x>0\Longrightarrow \sqrt x\le \frac{1+x}{2}.$

$2^{\circ}.\ u_n\le 2\cdot \sum \limits_{k=1}^{n+1} \frac {k}{2^k}+\frac {n}{2^n},\ n\in N^*.$

$3^{\circ}.\ \sum \limits_{k=1}^n \frac {k}{2^k}=2-\frac {n+2}{2^n},\ n\in N^*.$

From the relations $1^{\circ}-3^{\circ}$ results that the sequence $(u_n)_{n\in N^*}$ is convergent.


[quote="xtar"]Let $a,b,c$ real numbers. Prove that

$27(a+b)^2(b+c)^2(c+a)^2 \ge 64(ab+bc+ca)^3$[/quote]

We make use of the equality $9(a+b)(b+c)(c+a)\geq8(ab+bc+ca)(a+b+c)$
(a very [i]in[/i]  one  :lol: )
 so :

$81(a+b)^2(b+c)^2(c+a)^2\geq64(ab+bc+ca)^2(a+b+c)^2\geq 3\cdot 64 (ab+bc+ca)^3$

since $(a+b+c)^2\geq3(ab+bc+ca)$....

Nice ineq, xtar!!   :lol:
 :)

[quote="socrates"]
We make use of the equality $9(a+b)(b+c)(c+a)\geq8(ab+bc+ca)(a+b+c)$
(a very [i]in[/i]  one  :lol: )
 so :

$81(a+b)^2(b+c)^2(c+a)^2\geq64(ab+bc+ca)^2(a+b+c)^2\geq 3\cdot 64 (ab+bc+ca)^3$

since $(a+b+c)^2\geq3(ab+bc+ca)$....

Nice ineq, xtar!!   :lol:
 :)[/quote]

This solution is valid for $a,b,c>0$, but $a,b,c$ be real numbers.
.

Hey, this is my inequality from JBMO roumanian test. Actually, my problem was given in this form:

  Let $a, b, c>0$ such that $(a+b)(b+c)(c+a)=1.$ Prove that the following inequality holds:

                                                   \[ ab+bc+ca\leq\frac{3}{4}. \]

  I have posted it once, and Darij gave a beautiful solution different from my solutions. ;)


Μετρίστε τις δυνάμεις σας , απολαύστε τα θέματα  :)

1)Δείξτε ότι για όλους τους πραγματικούς $ x_1,x_2,...,x_n$ έχουμε
\[ \sum_{i \equal{} 1}^n \sum_{j \equal{} 1}^n ij \cos {(x_i \minus{} x_j)} \geq 0.
\]
2)Θεωρούμε τρίγωνο $ ABC$ και τον περιγεγραμμένο του κύκλο $ K$ . Θεωρούμε τους κύκλους $ k_1,k_2,k_3$ που εφάπτονται εσωτερικά στον
$ K$ στα σημεία $ A_1,\ B_1 ,\ C_1$ και στις πλευρές $ BC,CA,AB$ στα $ A_2,\ B_2 ,\ C_2$ αντίστοιχα έτσι ώστε τα $ A,A_1$ ναι είναι σε διαφορετικό ημιεπίπεδο της $ BC$ ,τα $ B,B_1$ ναι είναι σε διαφορετικό ημιεπίπεδο της $ AC$ ,τα $ C,C_1$ ναι είναι σε διαφορετικό ημιεπίπεδο της $ AB$ . Αν οι $ A_1A_2,B_1B_2,C_1C_2$ τέμνουν τον $ K$  στα $ A',B',C'$ , να προσδιορίσετε τις γωνίες του $ A'B'C'$ συναρτήσει των γωνιών του $ ABC$ .

3)Σε μια χώρα που υπάρχουν 2 αεροπορικές εταιρίες , 6 πόλεις συνδέονται ανά δύο με αεροπορικά ταξίδια μετ' επιστροφής που πραγματοποιούνται από ακριβώς μία από τις δύο αεροπορικές εταιρίες . Αποδείξτε ότι υπάρχουν 4 πόλεις Α,Β,Γ,Δ έτσι ώστε καθεμιά από τις διαδρομές Α <-> Β ,Β <-> Γ ,Γ <-> Δ, Δ <-> Α
να εκτελούνται από την ίδια αεροπορική εταιρεία .

4)Έστω $ n$ ένας ακέραιος , $ n\geq 2$. Βρείτε όλα τα σύνολα $ A$ με $ n$ ακέραια στοιχεία ώστε το άθροισμα των στοιχείων καθενός μηκενού υποσυνόλου του  $ A$ να μη διαιρείται με  $ n \plus{} 1$

[hide]Στην (1), το άθροισμα ισούται με $ \left(\sum\limits_{i\equal{}1}^n i \cos(x_i)\right)^2 \plus{} \left(\sum\limits_{i\equal{}1}^n i \sin(x_i)\right)^2$[/hide]

Μια αποδειξη του 3του.

[hide]Συνολικα υπαρχουν 6*5/2=15 πτησεις. Απο περιστεροφωλια, θα υπαρχουν τουλαχιστον 8 με την αεροπορικη εταιρια Α.

Τωρα αρκει να δειξουμε οτι:

Αν εχουμε ενα γραφημα με 6 κορυφες και 8 ευθειες που να ενωνουν τις κορυφες υπαρχει σιγουρα ενα τετραπλευρο.
Τωρα αν υποθεσουμε οτι τη i κορυφη τη ακομπανε α_i ευθειες θα ισχυει οτι:

(α_1+α_2+...+α_6)/2=8, αρα παλι απο περιστεροφωλια θα υπαρχει μια κορυφη την οποια την αγγιζουν τουλαχιστον 3 ευθειες. Εε, ζωγραφιστε ενα γραφηματακι τωρα και δειτε οτι σιγουρα θα υπαρχει ενα τετραπλευρο.

Και η αποδειξη ολοκληρωθηκε. :lol: [/hide]

[quote="userresu"]Παιδια να ρωτήσω κάτι, αυτά είναι επιπέδου Αρχιμήδη;[/quote]

Όλες οι παραπάνω ασκήσεις απαιτούν εντελώς στοιχειώδεις γνώσεις και η καθεμιά δεν έχει λύση πάνω από 5-6 γραμμές ... Επομένως θεωρώ ότι είναι κοντά στο επίπεδο του αρχιμήδη . Από εκεί και πέρα κάθε άποψη για τη δυσκολία των θεμάτων είναι υποκειμενική !

[quote="userresu"]ΟΚ thanks. Βασικά απλώς ρώτησα, επειδή δεν ξέρω ποιο είναι το επίπεδο στον Αρχιμήδη.  :)[/quote]
Δες τα προηγουμενα θεματα και θα καταλαβεις...
Τα θεματα αυτα ειναι πολυ ωραια αλλα εχω την εντυπωση λιγο πιο κατω απο Αρχιμηδη.....

4) [hide]Τα ζητουμενα συνολα ειναι αυτα των οποιων τα στοιχεια αφηνουν ολα το ιδιο μη μηδενικο υπολοιπο με το $ n \plus{} 1$. Πραγματι, εστω $ a_i, i \equal{} 1,2,...,n$ τα στοιχεια του συνολου. Εστω οτι δεν ισχυει ο παραπανω ισχυρισμος, τοτε χωρις βλαβη της γενικοτητας μπορουμε να εχουμε $ a_1 \neq b_2 (mod(n \plus{} 1))$. Θεωρουμε τους αριθμους $ a_1, a_2, a_1 \plus{} a_2, a_1 \plus{} a_2 \plus{} a_3, ....., a_1 \plus{} a_2 \plus{} ... \plus{} a_n$, αν δυο απο αυτους ειναι ισοτιμοι $ (mod(n \plus{} 1))$ τοτε προφανος απο την υποθεση αυτοι δεν θα ειναι οι $ a_1, a_2$ αρα η απολυτη τιμη τις διαφορας τους θα ειναι ενα αθροισμα καποιων διαφορετικων μεταξυ τους αριθμων $ a_i$ και θα ειναι $ 0(mod(n \plus{} 1))$, το οποιο ειναι ατοπο. Αρα οι παραπανω $ n \plus{} 1$ αριθμοι ειναι ανα 2 διαφορετικοι μεταξυ τους $ (mod(n \plus{} 1))$ και επομενος ενας απο αυτους ειναι $ 0(mod(n \plus{} 1))$, το οποιο ειναι επισεις ατοπο.
Q.uite E.asily D.one[/hide]

[quote="filosofimenos"]Παιδια, η αποδειξη μου στο 3το ειναι καλα θεμελιωμενη. Θελω να πω, το τελευταιο βημα που αναφερω ως προφανες απο το γραφημα ειναι οντως προφανες ή θελει περετερω θεμελιωση :?:[/quote]

Αν έγραφες μόνο ότι έγραψες πιο πάνω, προφανώς δεν είναι προφανές. Αν σχεδίαζες το γράφημα ίσως βοηθούσε. Εγώ πάντως χρειάστηκα να σχεδιάσω αρκετά γραφήματα. (Αυτές οι πλευρές είναι μπλε, άρα χωρίς βλάβη γενικότητας αυτές είναι κόκκινες, άρα αυτές είναι μπλε άρα ... άρα άτοπο.)

[quote="NickNafplio"][quote="socrates"][quote="Demetres"]Για την (4) ίσως να έχετε προσέξει ότι όλα τα σύνολα της μορφής $ A \equal{} \{a_1,\ldots,a_n\}$ με $ a_1 \equal{} a_2 \equal{} \dots \equal{} a_n \bmod (n \plus{} 1)$ και $ a_1 \neq 0 \bmod (n \plus{} 1)$ δουλεύουν. Μπορείτε να δείξετε ότι δεν υπάρχουν άλλα;[/quote]

n=3, A={2,6,10}...  :huh:[/quote]

Τι ακριβως ενοεις? Παντος ειναι $ 2 \plus{} 6 \equal{} 8 \equiv 0 (mod4)$[/quote]

Δικό μου το λάθος. Είχα πει πως [b]όλα[/b] το σύνολα αυτής της μορφής δουλέυουν, ενώ όπως παρατήρησε ο socrates δεν συμβαίνει.

[quote="Demetres"][/quote][quote="NickNafplio"][quote="socrates"][quote="Demetres"]Για την (4) ίσως να έχετε προσέξει ότι όλα τα σύνολα της μορφής $ A = \{a_1,\ldots,a_n\}$ με $ a_1 = a_2 = \dots = a_n \bmod (n + 1)$ και $ a_1 \neq 0 \bmod (n + 1)$ δουλεύουν. Μπορείτε να δείξετε ότι δεν υπάρχουν άλλα;[/quote]

n=3, A={2,6,10}...  :huh:[/quote]

Τι ακριβως ενοεις? Παντος ειναι $ 2 + 6 = 8 \equiv 0 (mod4)$[/quote][quote="Demetres"]

Δικό μου το λάθος. Είχα πει πως [b]όλα[/b] το σύνολα αυτής της μορφής δουλέυουν, ενώ όπως παρατήρησε ο socrates δεν συμβαίνει.[/quote]

Πρέπει επιπλέον $ (n+1, a_1\ mod(n))=1$... Συμφωνείτε?:)

[quote="socrates"]
Πρέπει επιπλέον $ (n \plus{} 1, a_1\ mod(n)) \equal{} 1$... Συμφωνείτε?:)[/quote]

Σύμφωνοι.

[quote="Demetres"][quote="filosofimenos"]Δημητρη θα ηθελα να δω τη λυση σου ή αν δεν εχεις χρονο ενα outline για το πως αποδεικνυω το τελευταιο βημα στη αποδειξη μου...[/quote]

Μπορώ να υποθέσω ότι οι 12,13 και 14 είναι κόκκινες. Από τις 25,35,45 τουλάχιστον δύο πρέπει να είναι μπλε. Ομοίως και για τις 26,36,46. Χωρίς βλάβη γενικότητας οι 35 και 36 είναι μπλε. Από τις 25,26 τουλάχιστον μια είναι κόκκινη. Χωρίς βλάβη γενικότητας η 25. Άρα η 45 μπλε. Άρα η 46 κόκκινη. Άρα η 26 μπλε. Από τις 15 και 16 τουλάχιστον μια είναι κόκκινη. Από συμμετρία μέχρι στιγμής μπορώ να υποθέσω η 15. Άρα 23 μπλε. Αλλά τώρα η 24 δεν μπορεί να είναι ούτε κόκκινη ούτε μπλε.[/quote]

Δημητρη πολυ ωραια λυση :coolspeak:


(i) By considering the equation $ x^{2}\plus{}x\minus{}a\equal{}0$, show that the equation $ x\equal{}(a\minus{}x)^{\frac{1}{2}}$ has one real solution when $ a\ge 0$ and no real solutions when $ a<0$.

Find the number of distinct real solutions of the equation

$ x\equal{}((1\plus{}a)x\minus{}a)^{\frac{1}{3}}$

in the cases that arise according to the value of $ a$.

(ii) Find the number of distinct real solutions of the equation

$ x\equal{}(b\plus{}x)^{\frac{1}{2}}$

in the cases that arise according to the value of $ b$.

[b](i)[/b] $ x\equal{}\sqrt{a\minus{}x} \Leftrightarrow \begin{cases} x \ge 0 \\ x^2\plus{}x\minus{}a\equal{}0 \end{cases}$

For $ a<0$, $ x^2\plus{}x\minus{}a\equal{}0$ has no non-negative roots.
For $ a>0$, $ x^2\plus{}x\minus{}a\equal{}0$ has exactly one non-negative root.

$ x\equal{}\sqrt [3] {(1\plus{}a)x\minus{}a} \Leftrightarrow (x\minus{}1)(x^2\plus{}x\minus{}a)\equal{}0$
Put $ x\equal{}1$ into $ x^2\plus{}x\minus{}a\equal{}0$ we get $ a\equal{}2$.

So for $ a\equal{}2$ there are $ 2$ distinct real roots.
For $ a<\minus{}\frac{1}{4}$ there is only one real root.

For $ a\equal{}\minus{}\frac{1}{4}$ there are two distinct real roots.

For $ a>\minus{}\frac{1}{4}$ with $ a \neq 2$ there are three distinct real roots.

[b](ii)[/b] $ x\equal{}\sqrt{b\plus{}x} \Leftrightarrow \begin{cases} x \ge 0 \\ x^2\minus{}x\minus{}b\equal{}0 \end{cases}$

For $ b > 0$ or $ b\equal{}\minus{}\frac{1}{4}$ , there is one distinct non-negative root for the equation $ x^2\minus{}x\minus{}b\equal{}0$ , i.e. one distinct real solution for the original equation.

For $ \minus{}\frac{1}{4} < b \le 0$, there are two distinct non-negative roots for the equation $ x^2\minus{}x\minus{}b\equal{}0$, i.e. two distinct real solution for the original equation.

For $ b<\minus{}\frac{1}{4}$, there are no real roots for the equation $ x^2\minus{}x\minus{}b\equal{}0$, i.e. no real solution for the original equation.


limit
trigonometry
real analysis
real analysis unsolved

Let  $e$ be Napier's constant. [color=green]  [b]
Does there exists a positive number $\alpha$ such that   $\displaystyle \lim\limits_{n\to \infty}\displaystyle n^{\alpha}\sin\left(2\pi e^{\pi} n! \right)$  exists and is finite ? [/b] [/color]

Another problem which looks like this one was proposed last year at "Unirea contest" 11th grade.
Find $a\in \mathbb{R}$ such that $n\{an!\}$ is convergent as $n \to \infty$. Is easy to prove that theis problem is eqiuvalent to:
Find $a\in \mathbb{R}$ such that $n \sin (2a\pi n!)$ is convergent as $n \to \infty$.
Try this one too.

[quote="xirti"]Another problem which looks like this one was proposed last year at "Unirea contest" ...
 [/quote]
Thanks for interest !  Let us remark, that many years ago (1955-1970 ? ),  Donald J. Newman has published an interesting problem  in Amer.Math.Monthly, namely to find  limit (of the sequence)   $\left( n \sin \left(2\pi e n !\right) \right )_{n\ge 1}$  .  Why this is an interesting  question :  because the limit is  $2\pi$  and [b] therefore Napier'constant  $e$  is an irrational number.  [/b]
[color=green]  However ,  here is another (more) difficult question ! [/color]


In the absence of ATP, glucose-6-phosphate is unstable at pH7, in the presence of enzyme glucose-6-phosphatase, it undergoes hydrolysis:

glucose-6-phosphate + H2O -> glucose + phosphate

Calculate delta G0' at 25C for the hydrolysis of glucose-6-phosphate.

I was going to use formula: delta G = delta G0' + RT ln K

But I don't have other information. Is there another approach or am I missing some information given here?

I'm not sure what you mean by "But I don't have other information."  As stated, there is no way to solve this problem without 1) going out and doing an experiment or 2) getting some very, very good computer software (I don't even know if Gibb's free energies can be predicted by computer.  That's an interesting side question).

It looks to me like you need standard Gibb's free energies of formation, or even better, the equilibrium constant for the reaction.

In the former case, you can just apply a Hess's law-like approach:  Take the difference in the total Gibb's energy of formation of products minus total of reactants.  Alternatively, try the formula $\Delta G^{o}=-RT \text{ln }K.$  (Note that your formula has a typo:  the K you have should be a Q.  This is because Q gives the current reaction position, while K is fixed as the equilibrium position.)

Thank you so much.
 Where can I find the delta G value for glucose-6-phosphate? It is not in the table at the back of the text book. In the internet, I could only find delta G0' which is what I am suppose to look for.

How can I find Q when there is no concentration given in this homework problem?

Ok I think you are confusing your terms, delta G0' versus delta G0' of formation.  You will not find any values for delta G online because this number depends intimately on reaction conditions, such as concentration and temperature.  However, you should be able to find delta G0' of formation for the species in the reaction.  This number is the free energy released or absorbed (depending on the sign) when forming 1.0 mol of a species from its elements, in their most stable form, at $25^{o}$ C.

On the other hand, delta G0' of the reaction (which you supposed to find) is the energy released by 1.0 mol of [i]this specific reaction[/i] at $25^{o}$ C.  There is quite a fine distinction between the two concepts.


Putnam
limit
integration
trigonometry
calculus
college contests

Evaluate $ \lim_{n\to\infty} \int_0^1 \int_0^1 \cdots \int_0^1 \cos ^ 2 \left\{\frac{\pi}{2n}(x_1\plus{}x_2\plus{}\cdots \plus{}x_n)\right\} dx_1dx_2\cdots dx_n.$

[hide]
This integral evaluates to $ \frac{1}{2}$ for all $ n$, and thus the limit is $ \frac{1}{2}$. First note that $ \cos^2{\frac{\pi}{2n}(x_1 \plus{}...\plus{} x_n)} \equal{} \frac{1}{2} \plus{} \frac{1}{2}\cos{\frac{\pi}{n}(x_1 \plus{}...\plus{} x_n)}$. We will show that the inner $ k$ integrals evaluate to $ \frac{1}{2} \plus{} \frac{1}{2} \left(\frac{2n}{\pi} \sin{\frac{\pi}{2n}}\right)^k \cos{\left(\frac{k \pi}{2n} \plus{} \frac{\pi}{n}(x_{k\plus{}1} \plus{}...\plus{} x_n)\right)}$. This is readily verified for $ k\equal{}1$. Now assume this is true after $ k$ integrals have been evaluated. We need to show that $ \int_0^1{\cos{\left(\frac{k \pi}{2n} \plus{} \frac{\pi}{n}(x_{k\plus{}1} \plus{}...\plus{} x_n)\right)} dx_{k\plus{}1}} \equal{} \left(\frac{2n}{\pi} \sin{\frac{\pi}{2n}}\right) \cos{\left(\frac{(k\plus{}1) \pi}{2n} \plus{} \frac{\pi}{n}(x_{k\plus{}2} \plus{}...\plus{} x_n)\right)}$, which is also readily verified and shows that our formula holds for $ k\plus{}1$, completing induction.

So after all $ n$ integrals have been evaluated, we are left with $ \frac{1}{2} \plus{} \frac{1}{2} \left(\frac{2n}{\pi} \sin{\frac{\pi}{2n}}\right)^n \cos{\frac{\pi}{2}} \equal{} \frac{1}{2}$.
[/hide]

That's correct answer. We can evalute the integral by Hint: $ \sin ^ 2 \theta \plus{}\cos ^ 2 \theta \equal{}1$.

By the substitution $x_i\to1-x_i$, we can denote
$\int_{0}^{1}\int_{0}^{1}\cdots\int_{0}^{1}\cos^2\left(\frac{\pi}{2n}(x_1+x_2+\cdots+x_n)\right)dx_1\,dx_2\,\cdots dx_n= \int_{0}^{1}\int_{0}^{1}\cdots\int_{0}^{1}\sin^2\left(\frac{\pi}{2n}(x_1+x_2+\cdots+x_n)\right)dx_1\,dx_2\,\cdots dx_n $

By the fact that $\sin^2x+\cos^2x=1$, we have
$\lim_{n\to\infty}\int_{0}^{1}\int_{0}^{1}\cdots\int_{0}^{1}\cos^2\left(\frac{\pi}{2n}(x_1+x_2+\cdots+x_n)\right)dx_1\,dx_2\,\cdots dx_n=\frac{1}{2}$


Ερώτηση που μου έκανε ένας φίλος φυσικός και την βρίσκω ενδιαφέρουσα
1) Οι συμβολισμοί $ y'(t)$ και $ (y(t))'$ διαφέρουν σε κάτι ? Αν όχι γιατί? Αν ναί σε τι?
2) Σε μια διαφορική εξίσωση ,πχ της ΑΑΤ, τι πρέπει να γράφουμε a)$ y''(t)\plus{}\omega^2y(t)\equal{}0$ ή b)$ (y(t))''\plus{}\omega^2y(t)\equal{}0$ ή είναι το ίδιο πράγμα?
Περιμένω τις απόψεις σας

r_boris, τα παρακάτω είναι γραμμένα υπό το πρίσμα της προσωπικής αισθητικής. Προφανώς, συμβολικά μπορούμε να γράφουμε ό,τι θέλουμε και σίγουρα δε θα δημιουργηθούν παρανοήσεις.

(αυτά τα λέω για να μη δοθεί η εντύπωση ότι ό,τι γράφω έρχεται από κάποια ιερά βίβλο ;) )

[quote="r_boris"]Ερώτηση που μου έκανε ένας φίλος φυσικός και την βρίσκω ενδιαφέρουσα
1) Οι συμβολισμοί $ y'(t)$ και $ (y(t))'$ διαφέρουν σε κάτι ? Αν όχι γιατί? Αν ναί σε τι?
[/quote]

Κατ'εμέ, η σωστή ερμηνεία της παραγώγου είναι ως τελεστής που δρά πάνω σε συναρτήσεις. Όταν δρα πάνω στη συνάρτηση $ y$, τότε παίρνεις τη συνάρτηση $ y'$; άρα, ο σωστός συμβολισμός είναι $ y'(t)$, δηλώνοντας την τιμή της $ y'$ στο σημείο $ t$. Ο άλλος συμβολισμός έχει αρκετά προβλήματα που πηγάζουν κυρίως από το γεγονός ότι συμπεριφερόμαστε στο σύμβολο $ y(t)$ σαν να είναι συνάρτηση και λέμε "έστω συνάρτηση $ y(t)$" αντί να λέμε "έστω συνάρτηση $ y: A\to B$ με $ y(t) \equal{} \ldots$".

Κατ'αρχάς, η $ y(t)$ δεν είναι, αυστηρά μιλώντας, συνάρτηση; είναι απλώς η τιμή της $ y$ στο $ t$. Άρα, τίθεται θέμα σύγχυσης όταν γράφουμε $ (y(t))'$: εννοούμε την $ y'(t)$ ή εννοούμε $ 0$ (παράγωγος της σταθεράς συνάρτησης $ c \equal{} y(t)$ για την τιμή του $ t$ που δηλώνουμε)? Επιπλέον, από τη στιγμή που χρησιμοποιούμε το $ t$ ως μεταβλητή, θα πρέπει να μπορούμε να θέσουμε όποια τιμή του $ t$ θέλουμε στο συμβολισμό και να διατηρεί το νόημά του. Δηλαδή, θα καταλήγαμε να μιλάμε για πράγματα όπως $ (y(3))'$ εννοώντας $ y'(3)$ κλπ. Δηλαδή ο συμβολισμός αυτός δεν μπορεί να παραμείνει συνεπής για συγκεκριμένες τιμές του $ t$.

Αν θέλεις, το πρόβλημα που προσπαθώ να περιγράψω είναι ότι ο συμβολισμός $ (y(t))'$ δεν έχει την τελεστική ακολουθία που θα θέλαμε. Αυστηρά μιλώντας, σου λέει: πρώτα υπολόγισε το $ y(t)$ και μετά παραγώγισε τον αριθμό που προκύπτει. Εσύ θέλεις το ανάποδο: πρώτα υπολόγισε την παράγωγο συνάρτηση και μετά υπολόγισε την στο σημείο $ t$.

Τώρα, αυτά είναι φυσικά ψιλά γράμματα και δύσκολο να τεθεί θέμα παρανόησης :) Πάντως, σίγουρα θα έλεγα ότι είναι πολύ πιο σωστός ο συμβολισμός $ y'(t)$.

[quote="r_boris"]
2) Σε μια διαφορική εξίσωση ,πχ της ΑΑΤ, τι πρέπει να γράφουμε a)$ y''(t) \plus{} \omega^2y(t) \equal{} 0$ ή b)$ (y(t))'' \plus{} \omega^2y(t) \equal{} 0$ ή είναι το ίδιο πράγμα?
Περιμένω τις απόψεις σας[/quote]

See above.

Cheerio,

Durandal 1707

Συμφωνώ στην ερμηνεία της παραγώγου ως τελεστή
Ενα άλλο σκεπτικό προέρχεται από την δράση ή την μη δράση του τελεστή σε συνθετες συναρτήσεις π.χ $ (y(2t))''\equal{}4y''(2t)$εδώ θεωρήθηκε οτι έδρασε ο τελεστής της παραγώγου 2 φορές ενω γράφοντας $ y''(2t)$ δεν πρέπει να βγεί ενα τεσσάρι έξω άρα ο τελεστής δεν εδρασε. Επίσης  η εξίσωση $ (y(t))''\plus{}\omega^2y(t)\equal{}0$ είναι διαφορετική από φυσικής σκοπιάς από την $ y''(t)\plus{}\omega^2y(t)\equal{}0$ διότι στην πρώτη εκκρεμεί η δράση του τελεστή αρα το $ (y(t))''$ δεν παριστάνει την επιτάχυνση αλλά το $ y''(t)$ την παριστάνει. Αν σκεπτικό είναι να βρούμε την μετατόπιση y(t) ίσως καλύτερη έκφραση της εξίσωσης είναι αυτή που περιέχει το $ (y(t))''$
Απάντησα με το attachment
Εσείς τι νομίζετε?

Προσωπικά, θεωρώ ότι η παράγωγος είναι ένας τρόπος έκφρασης του τύπου μιας συνάρτησης σε σχέση με έναν άλλο τύπο. Γι' αυτό θεωρώ ότι ίσως ο σωστότερος συμβολισμός είναι ο $ \frac{d}{dx}f(x)$. Αλλά ο συμβολισμός $ f'(2t)$ δεν είναι κατ' ανάγκη λάθος, αφού είναι σύμβαση για τη γραφή (o η σύνθεση) $ (fog)'(t), g(t)\equal{}2t$.

Εγώ πάλι θα συμφωνήσω με τον Durandal. Γενικά πιστεύω ότι συμβολισμοί της μορφής $ (y(t))'$ προέρχονται κυρίως από τον τρόπο με τον οποίο θέλουμε να γράφουμε τον chain rule: $ (f(g(t)))' \equal{} f'(g(t)) g'(t)$. Προσωπικά δε μου αρέσει, αλλά και το να καθόμαστε να τα γράφουμε όλα εντελώς formally γίνεται επίσης κουραστικό. Φαντάσου να έπρεπε κάθε φορά που θες να γράψεις $ (f(2t))'$ να έπρεπε να θέσεις $ h(t) \equal{} 2t$ και να πεις $ (f \circ h)'(t) \equal{} f'(h(t)) h'(t)$...

maniopas, δεν ξέρω αν έχεις ασχοληθεί ποτέ με multivariable calculus αλλά εκεί θα δεις ότι το Leibniz notation είναι αρκετά άσχημο και επίσης προκαλεί αρκετή σύγχυση - πάλι, κατά τη γνώμη μου.

Εν πάσει περιπτώσει, quot hominess, tot sententiae. :) Αρκεί να καταλαβαινόμαστε μεταξύ μας.

EDIT: Ώπα μου ξέφυγε κάτι... Maniopas όταν γράφουμε $ f'(2t)$ [b]δεν[/b] εννοούμε (ούτε κατά σύμβαση) αυτό που έγραψες.

Νομίζω ότι Το μπέρδεμα γίνεται στο εξής σημείο
έστω ότι έχουμε την $ y''(t)\plus{}4y(t)\equal{}0$ αν βάλουμε όπου $ t$ το $ 2t$ τι θα γίνει τελικά ?

A)$ 4y''(2t)\plus{}4y(2t)\equal{}0$ οπότε δικαιολογείται καλύτερα Το $ (y(t))''$ ή
Β)$ y''(2t)\plus{}4y(2t)\equal{}0$ οπότε δικαιολογείται καλύτερα Το $ y''(t)$ ?

[quote="r_boris"]
Νομίζω ότι Το μπέρδεμα γίνεται στο εξής σημείο
έστω ότι έχουμε την $ y''(t) \plus{} 4y(t) \equal{} 0$ αν βάλουμε όπου $ t$ το $ 2t$ τι θα γίνει τελικά ?

A)$ 4y''(2t) \plus{} 4y(2t) \equal{} 0$ οπότε δικαιολογείται καλύτερα Το $ (y(t))''$ ή
Β)$ y''(2t) \plus{} 4y(2t) \equal{} 0$ οπότε δικαιολογείται καλύτερα Το $ y''(t)$?
[/quote]

Hands down το Β.

Για μένα δεν υπάρχει μπέρδεμα: η $ y'$ είναι μία καλά ορισμένη συνάρτηση την οποία την υπολογίζεις όπου θέλεις. Το $ y'(2t)$ δεν έχει να κάνει με το $ \frac {d}{d(2t)} y(2t)$ (γραμμένο με μία μικρή κατάχρηση συμβολισμού για να γίνει κατανοητό το τι εννοώ πιο γρήγορα και να μη γράφω σύνθεση με την $ h(t) \equal{} 2t$).

Ας ορίσουμε για κάθε $ h > 0$ τη συνάρτηση μετατόπισης κατά $ h$: $ T_h: x\mapsto x \plus{} h$ και τον τελεστή $ I_h$ που δρα πάνω σε συναρτήσεις $ y$: $ I_h(y) \equal{} \frac {1}{h}(y\circ T_h \minus{} y)$. Τότε, (υπό την κατάλληλη ύπαρξη ορίου στην κατάλληλη τοπολογία και με την κατάλληλη έννοια και blah blah blah) η συνάρτηση $ y'$ είναι η $ \lim_{h\to 0} I_h (y)$.

Άρα, η $ y'$ δεν έχει να κάνει με τη μεταβλητή της συνάρτησης, το $ '$ δεν αναφέρεται σε παραγώγιση ως προς χρόνο ή ως προς όποια άλλη μεταβλητή. Αναφέρεται στην [i]παράγωγο[/i] συνάρτηση.

Υπό αυτό το πρίσμα, δε μου φαίνεται ιδιαίτερα κομψός ο συμβολισμός $ (y(t))'$ γιατί ο τόνος $ '$ δε μεταφέρει τη μεταβλητή ως προς την οποία γίνεται η παραγώγιση. Αν είναι να γλυτώσουμε το $ \omega$, θα έβρισκα πολύ καλύτερο, πιο περιγραφικό και χωρίς αμφισημίες το συμβολισμό Leibniz, εφαρμοσμένο στη rescaled έκδοση $ \tau \equal{} t/\omega$ του χρόνου στη συνάρτηση: $ \tilde{y}(\tau) \equal{} y(t/\omega)$.

My tuppence :)

Cheerio,

Durandal 1707


calculus
integration
derivative
calculus computations

Can anyone direct me to a post where the formula for integration by parts has been derived.Or if somebody knows then please derive.

Thanks in advance.

Let $ F(x)\equal{}\int f(x)\ dx\plus{}C$, by the differentiation of Product,

$ \{F(x)g(x)\}'\equal{}F'(x)g(x)\plus{}F(x)g'(x)\equal{}f(x)g(x)\plus{}F(x)g'(x)$

$ \Longleftrightarrow \boxed{\int f(x)g(x)\ dx\equal{}F(x)g(x)\minus{}\int F(x)g'(x)\ dx}$

Thanks kunny.You make it look so simple.

It is simple. Just take your normal product rule for differentiation and rearrange the terms.


Show that the collection of all finite subsets of N is countable. What goes wrong if we try to use the diagonal argument to show that it is uncountable?

first,

[b]A countable union of countable sets is countable.[/b]

use this as follows. let $ I$ be the set of all ordered pairs of natural numbers. $ I$ is countable. for each $ i\in I$, let $ S_i \equal{} \{ U|\forall x\in U, x\in \mathbb{N} \wedge a\leq x \leq b \}$, where $ i\equal{}(a,b)$.

then the collection of all finite subsets of $ \mathbb{N}$ is $ \bigcup_{i\in I} S_i$, which is a countable union of countable sets, and hence countable.

[quote="Pythagorean12"]Show that the collection of all finite subsets of N is countable.[/quote]We know that every natural number has a dual expansion $ \sum_{i \equal{} 0}^\infty 2^{a_i}$ where $ a_i \equal{} 1$ for only finitely many $ i\in \mathbb{N}$. Hence, there is a bijection $ f$ from $ \mathbb{N}$ into such sequences $ (a_i)_{i\in \mathbb{N}}$. Define $ F(n) \equal{} \{i \in \mathbb{N} | (f(n))_i \equal{} 1\}$, then $ F$ is a bijection from the natural numbers into the finite subsets of $ \mathbb{N}$.

[quote="Pythagorean12"]What goes wrong if we try to use the diagonal argument to show that it is uncountable?[/quote]Depends on what argument you exactly mean. There is the argument that if you have a bijection $ f: \mathbb{N} \rightarrow P(\mathbb{N})$ then you can take $ A: \equal{} \{n | n \not\in f(n)\}$, then there must be $ \mathbb{N} \ni a \equal{} f^{ \minus{} 1} (A)$, and we have $ a \in A \Rightarrow a \not\in A \Rightarrow a \in A \Rightarrow \bot$. However, if you replace $ P(\mathbb{N})$ by $ P_{\mbox{finite}}(\mathbb{N})$, then this argument is not sufficient, since the $ A$ you get will be infinite.


algorithm
number theory
Euclidean algorithm

I posted this over [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=250886]here[/url] to poke at a crank. It's neat enough that I'd like to pose it for the forum in general.

$ 835905841\equal{}28380^2\plus{}5521^2$
$ 835905841\equal{}26340^2\plus{}11921^2$

Factor $ 835905841$.

No brute force searching allowed; the point is to use the above information to factor it faster than we could for a generic integer of its size.

[hide="Brute Force"]
$ 28201 * 29641$
[/hide]

How is one supposed to do this? I can't think of any easy ways to factor numbers of this form given the information.

I read through the thread you posted, and it seems as if this problem was posed in order to root out the stupidity of some crank who wanted to show he had a new method of "factoring" things like this.

I realize that basically everyone in that thread owned him, but I don't think I ever saw an actual method to doing problems like this...

[hide="Minor hint"]$ (a^2\plus{}b^2)(c^2\plus{}d^2) \equal{} (ac\minus{}bd)^2 \plus{} (ad\plus{}bc)^2$[/hide]

using mellowmelon's hint:
$ 835905841\equal{}26380^2\plus{}5521^2\equal{}26340^2\plus{}11921^2$

Multiply to get:
$ 835905841^2\equal{}(26380^2\plus{}5521^2)(26340^2\plus{}11921^2)$
$ 835905841^2\equal{}(a^2\plus{}b^2)(c^2\plus{}d^2)\equal{}(ad\minus{}bc)^2\plus{}(ac\plus{}bd)^2$

These are pythagorean triples hence we have that:
$ 3|(ad\minus{}bc)^2, 4|(ac\plus{}bd)^2, 5|835905841$

I dunno where I made a mistake..... :blush:
Help please.

Really, every Pythagorean triple is a multiple of (3, 4, 5)?  7 and 13 will be very unhappy to hear that they aren't prime any more (think (5, 12, 13), (7, 24, 25)).  In fact, the way you're using the hint can't be the right direction: it would require us to be able to factor a sum of squares to be useful!

[hide="A different way to use the hint"]Write $ 835905841 \equal{} 28380^2 \plus{} 5521^2 \equal{} (ad \plus{} bc)^2 \plus{} (ac \minus{} bd)^2$ and $ 835905841 \equal{} 26340^2 \plus{} 11921^2 \equal{} (ac \plus{} bd)^2 \plus{} (ad \minus{} bc)^2$.  Now try to solve.[/hide]

[hide="An additional remark"] Ideally, to solve this problem you will need to rediscover the Euclidean algorithm in the Gaussian integers.  The comments [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=219017]here[/url] might be useful. [/hide]

It's been two weeks, so here's a bump.

The factorization is as a product of two large primes- I had a list available for unrelated reasons. MellowMelon's hint was the key to constructing both pairs of squares, and t0rajir0u's remark is the solution I thought of.


$A$ is a set of divisors of $210^{2006}$ such that if $x;y$ $\in$ $A$ then $x\not|y$ and $y\not|x$.

Find, with proof, the maximum possible number of elements of $A$.

The answer is ${4015\choose 3} - 4\sum_{i = 2007}^{4012} {4014 - i \choose 2}$ (which of course can be simplified), unless I've made a mistake.  This is the number of solutions to the equation $4012 = a + b + c + d$ where $a, b, c, d$ are nonnegative integers no larger than 2006.  They correspond to divisors of $210^{2006}$ in which the sum of the exponents is 4012.  This is clearly a lower bound (since no one of these divides another), and I believe that an argument similar to the proof of Sperner's Theorem relying on symmetric chains will show that it is maximal.  (In general, it is not too hard to show that $k = 2n$ maximizes $|\{(a, b, c, d) | a + b + c + d = k \textrm{ and } 0 \leq a, b, c, d \leq n\}|$.)


I am interested in forming such a team, but I can't find people at my school that are interested. Is anyone on AOPS interested in forming a team? (prefferably with me on it :))?

I will join the team if
1)I decided to participate in ASHMIC
2) if you do not afraid that I mess up for the team

I will probably join the team on the same conditions a Beta. But I think it will be more fun if it is with a school team.


function
algebra
functional equation
algebra open

[hide]Let $ y \equal{} 0$, then $ f(f(x)) \equal{} xf(0) \plus{} f(x)$. Knowing this,

$ f(f(xy \plus{} f(x))) \equal{} f(xf(y) \plus{} f(x))$

$ (xy \plus{} f(x))f(0) \plus{} f(xy \plus{} f(x)) \equal{} f(xf(y) \plus{} f(x))$

$ xyf(0) \plus{} f(x)f(0) \plus{} xf(y) \plus{} f(x) \equal{} f(xf(y) \plus{} f(x))$

$ x(yf(0) \plus{} f(y)) \plus{} f(x) ( f(0) \plus{} 1) \equal{} f(xf(y) \plus{} f(x))$

$ xf(y) \plus{} f(x) \equal{} f(xf(y) \plus{} f(x))$

$ f(x) \equal{} x$

Now we test it to see it it satisfies the functional equation,

$ f(xy \plus{} f(x)) \equal{} xf(y) \plus{} f(x)$

$ xy \plus{} x \equal{} xy \plus{} x$

It works!

$ \therefore f(x) \equal{} x$[/hide]

[quote="modularmarc101"]Let $ y \equal{} 0$, then $ f(f(x)) \equal{} xf(0) \plus{} f(x)$. Knowing this,

$ f(f(xy \plus{} f(x))) \equal{} f(xf(y) \plus{} f(x))$

$ (xy \plus{} f(x))f(0) \plus{} f(xy \plus{} f(x)) \equal{} f(xf(y) \plus{} f(x))$

$ xyf(0) \plus{} f(x)f(0) \plus{} xf(y) \plus{} f(x) \equal{} f(xf(y) \plus{} f(x))$

$ x(yf(0) \plus{} f(y)) \plus{} f(x) ( f(0) \plus{} 1) \equal{} f(xf(y) \plus{} f(x))$

$ xf(y) \plus{} f(x) \equal{} f(xf(y) \plus{} f(x))$

$ f(x) \equal{} x$

Now we test it to see it it satisfies the functional equation,

$ f(xy \plus{} f(x)) \equal{} xf(y) \plus{} f(x)$

$ xy \plus{} x \equal{} xy \plus{} x$

It works!

$ \therefore f(x) \equal{} x$[/quote]
$ f(x) \equal{} 0$?

hello!
I propose my solution  :)
Let $ H(x,y): f(xy+f(x))=xf(y)+f(x)$
if f is a constante function so we get easily that: $ f(x)=0$  $ \forall$ $ x\in{\mathbb{R}}$.
Let $ f\neq0$ and $ f(0)=a$
$ H(0,0): f(a)=a$
$ H(a,y): f(a(y+1))=a(f(y)+1)$
$ H(a,0): f(a)=a(a+1)$
so we get $ a^{2}=0$==>$ a=f(0)=0$
$ H(x,0): f(f(x))=f(x)$
consider now a sequence $ (x_{n})$ such that: $ x_{0}=x$ and $ f(x_{n})=x_{n+1}$.
so the f.e become: $ x_{n+2}=x_{n+1}$
if we solve this equation $ x^{2}=x$ we get $ x=0$ or $ x=1$.
so : $ x_{n}=\alpha (1)^{n}$ ==> $ (\alpha=x)$
then: $ f(x)=x$ $ \forall$ $ x\in{\mathbb{R}}$

Hi Hamza  :lol:

i think you are not in need of to find $ f(0)$

$ x\equal{}0 \implies f(f(x))\equal{}f(x)$ the afterpart is as yours :)

Let $P(x,y)$ be the assertion that $f(xy+f(x))=xf(y)+f(x)$
$P(0,0)\implies f(f(0))=f(0)$ so $f(f(f(0)))=f(0)$
$P(f(0),0)\implies f(f(f(0)))=f(0)f(0)+f(0)$
So $f(0)^2=0\implies f(0)=0$
$P(x,0)\implies f(f(x))=f(x)$
$P(f(1),1)\implies f(f(1)+f(1))=f(1)^2+f(1)$
$P(1,f(1))\implies f(f(1)+f(1))=f(1)+f(1)$
So $f(1)[f(1)-1]=0\implies f(1)=1\text{ or }0$

CASE 1: $f(1)=1$
$P\left(\frac{1}{x},x\right)\implies f\left(1+f\left(\frac{1}{x}\right)\right)=\frac{f(x)}{x}+f\left(\frac{1}{x}\right)$
$P\left(1,f\left(\frac{1}{x}\right)\right)\implies f\left(1+f\left(\frac{1}{x}\right)\right)=1+f\left(\frac{1}{x}\right)$
So $f(x)=x, \forall x\in\mathbb{R}$
It is easy to check that this indeed is a solution.

CASE 2: $f(1)=0$
$P(f(x),x)\implies f(xf(x)+f(x))=f(x)^2+f(x)$
$P(x,f(x))\implies f(xf(x)+f(x))=xf(x)+f(x)$
So $f(x)[f(x)-x]=0\implies f(x)=0\text{ or }x\forall x\in\mathbb{R}$
Suppose $\exists a\in\mathbb{R}\text{ and }a\neq 0 \text{ st. }f(a)=a$
$P\left(a,\frac{1-a}{a}\right)\implies 0=af\left(\frac{1-a}{a}\right)+a\implies f\left(\frac{1-a}{a}\right)=-1$
So $f\left(\frac{1-a}{a}\right)\left[f\left(\frac{1-a}{a}\right)-\frac{1-a}{a}\right]=0\implies 1=0$ CONTRADICTION
So $\forall x\neq 0$ we have $f(x)\neq x$
So $\forall x\in\mathbb{R}$ we have $f(x)=0$
It is easy to check that this indeed is a solution

So $f(x)=x, \forall x\in\mathbb{R}$ or $f(x)=0, \forall x\in\mathbb{R}$

[quote="Thjch Ph4 Trjnh"]Find all functions $ f: R\rightarrow R$ such that:
$ f(xy\plus{}f(x))\equal{}xf(y)\plus{}f(x)$, for any $ x,y\in R$.[/quote]
My solution:
We have: $f(xy+f(x))=xf(y)+f(x)$.
$f(xy+f(y))=yf(x)+f(y)$.
If $f(x)=f(y)$ then $xf(y)=yf(x)$.
$\Rightarrow f(x)=f(y)=0$ or $x=y$. (*)
Let $x=y=0$ we have: $f(f(0))=f(0)$.
Using (*), we obtain $f(0)=0$.
Let $y=0$: $f(f(x))=f(x)$.
$\Rightarrow f(x)=x$ or $f(x)=0$.
Assume $f(x_0)=0$ for $x_0\ne 0$.
Let $y=x_0$: $f(xx_0+f(x))=f(x)$.
Using (*): $f(x)=0$ or $x_0+f(x)=x$.
$\Rightarrow f(x)=0$.
Thus if $f(x_0)=0$ for $x_0\ne 0$ then $f(x)\equiv 0$, else $f(x)\equiv x$.


calculus
integration
function
trigonometry
logarithms
algebra
partial fractions

I think this sum can not be evaluated in simple way...  :wink:
We even can`t calculate the sum:
$ \sum_{k\equal{}1}^{n}{\frac{1}{k^2\plus{}k\plus{}1}}$

although there we can bound sum of reciprocal of squares using integration (this will involve the inverse tangent function), there is a simple way to sum              1/(1+x^2+x^4). observe 1+x^2+x^4=1+2x^2+x^4-x^2=(x^2+1)^2-x^2=(x^2+x+1)(x^2-x+1). then partial fractions and telescope. that's the sequence of ideas.
edit:also observe the identity: (x-1)^2+(x-1)+1=x^2-x+1

We can represent sum as:
$ \sum_{k \equal{} 1}^{n}{\frac {1}{k^4 \plus{} k^2 \plus{} 1}} \equal{} \frac {1}{2}\cdot \frac {n \plus{} 1}{n^2 \plus{} n \plus{} 1} \plus{} \frac {1}{2}\cdot \sum_{k \equal{} 1}^{n \minus{} 1}{\frac {1}{k^2 \plus{} k \plus{} 1}}$
But I think we can`t calculate the last sum in genereal case not usung high math.

[quote="Akashnil"]Observe that: $ n^4 \plus{} n \plus{} 1 \equal{} (n^2 \plus{} n \plus{} 1)(n^2 \minus{} n \plus{} 1)$
This means: $ \frac {1}{n^4 \plus{} n^2 \plus{} 1} \equal{} \frac {1}{2n}(\frac {1}{n^2 \minus{} n \plus{} 1} \minus{} \frac {1}{n^2 \plus{} n \plus{} 1})$

is incorrect.  You should only end up with

$ \frac {1}{n^4 \plus{} n^2 \plus{} 1} \equal{} \frac {1}{2}(\frac { \minus{} n \plus{} 1}{n^2 \minus{} n \plus{} 1} \plus{} \frac {n \plus{} 1}{n^2 \plus{} n \plus{} 1})$

After a bit more work I ended up with

$ \minus{}\frac {1}{2}\sum_{n \equal{} 1}^{N}\frac {1}{n^2 \minus{} n \plus{} 1} \plus{} \frac {N}{N^2 \plus{} N \plus{} 1} \plus{} \frac {1}{2}(1 \minus{} \frac {1}{N^2 \plus{} N \plus{} 1})$

however, I cannot see how to add the finite sum.

Let`s find the sum:
$ \sum_{k\equal{}1}^{\infty}{\frac{1}{k^2\plus{}k\plus{}1}}\equal{}\minus{}1\plus{}\sum_{k\equal{}0}^{\infty}{\frac{1}{k^2\plus{}k\plus{}1}}\equal{}\minus{}1\plus{}\sum_{k\equal{}0}^{\infty}{\frac{1}{(k\plus{}\frac{1}{2})^2\plus{}(\frac{\sqrt{3}}{2})^2}}$
We know that:
$ \cos(\phi)\equal{}\prod_{k\equal{}0}^{\infty}{\left( 1\minus{}\frac{4\phi ^2}{(2k\plus{}1)^2 \pi^2} \right)}$
Now:
$ \ln (\cos(\phi))\equal{}\sum_{k\equal{}0}^{\infty}{\ln\left( 1\minus{}\frac{4\phi ^2}{(2k\plus{}1)^2 \pi^2} \right)}$
Now differentiate both sides by $ \phi$:
$ \tan(\phi)\equal{}\sum_{k\equal{}0}^{\infty}{\frac{8\phi}{(2k\plus{}1)^2 \pi^2\minus{}4\phi^2}}$
Now put $ \phi\equal{}\pi\alpha\cdot i$, we will get:
$ \tan(\pi\alpha\cdot i)\equal{}i\cdot \tanh(\pi\alpha)\equal{}i\cdot \sum_{k\equal{}0}^{\infty}{\frac{8\pi\alpha}{(2k\plus{}1)^2 \pi^2\plus{}4\pi^2\alpha^2}}\equal{}\frac{2\alpha i}{\pi}\cdot \sum_{k\equal{}0}^{\infty}{\frac{1}{(k\plus{}\frac{1}{2})^2 \plus{} \alpha^2}}$
So, we have:
$ \sum_{k\equal{}0}^{\infty}{\frac{1}{(k\plus{}\frac{1}{2})^2 \plus{} \alpha^2}}\equal{}\frac{\pi}{2\alpha}\cdot \tanh(\pi\alpha)$
In our case $ \alpha\equal{}\frac{\sqrt{3}}{2}$. We get:
$ \sum_{k\equal{}0}^{\infty}{\frac{1}{(k\plus{}\frac{1}{2})^2\plus{}(\frac{\sqrt{3}}{2})^2}}\equal{}\frac{\sqrt{3}\pi}{3}\cdot \tanh\left(\frac{\sqrt{3}\pi}{2}\right)$
Now we have:
$ \sum_{k\equal{}1}^{\infty}{\frac{1}{k^2\plus{}k\plus{}1}}\equal{}\frac{\sqrt{3}\pi}{3}\cdot \tanh\left(\frac{\sqrt{3}\pi}{2}\right)\minus{}1$
And our basic problem:
$ \sum_{k\equal{}1}^{\infty}{\frac{1}{k^4\plus{}k^2\plus{}1}}\equal{}\frac{1}{2}\cdot \sum_{k\equal{}1}^{\infty}{\frac{1}{k^2\plus{}k\plus{}1}}\equal{}\frac{\sqrt{3}\pi}{6}\cdot \tanh\left(\frac{\sqrt{3}\pi}{2}\right)\minus{}\frac{1}{2}\approx 0.3990736408$

[quote="Winged Potato"]
[quote="Akashnil"]Observe that: $ n^4 \plus{} n \plus{} 1 \equal{} (n^2 \plus{} n \plus{} 1)(n^2 \minus{} n \plus{} 1)$
This means: $ \frac {1}{n^4 \plus{} n^2 \plus{} 1} \equal{} \frac {1}{2n} (\frac {1}{n^2 \minus{} n \plus{} 1} \minus{} \frac {1}{n^2 \plus{} n \plus{} 1})$
[/quote]
is incorrect. You should only end up with ...
[/quote]
I have made a mistake in the first line: it should be $ n^4 \plus{} n^2 \plus{} 1$ instead of $ n^4 \plus{} n \plus{} 1$
But I dont see a problem elsewhere.
$ n^4 \plus{} n^2 \plus{} 1 \equal{} (n^4 \plus{} 2n^2 \plus{} 1) \minus{} n^2 \equal{} (n^2 \plus{} 1)^2 \minus{} n^2 \equal{}$
$ (n^2 \plus{} 1 \plus{} n)(n^2 \plus{} 1 \minus{} n) \equal{} (n^2 \plus{} n \plus{} 1)(n^2 \minus{} n \plus{} 1)$
$ \frac {1}{2n} (\frac {1}{n^2 \minus{} n \plus{} 1} \minus{} \frac {1}{n^2 \plus{} n \plus{} 1}) \equal{} \frac {1}{2n}\frac {(n^2 \plus{} n \plus{} 1) \minus{} (n^2 \minus{} n \plus{} 1)}{(n^2 \minus{} n \plus{} 1)(n^2 \plus{} n \plus{} 1)}$
$ \equal{} \frac {1}{2n}\frac {2n}{(n^2 \minus{} n \plus{} 1)(n^2 \plus{} n \plus{} 1)} \equal{} \frac {1}{(n^2 \minus{} n \plus{} 1)(n^2 \plus{} n \plus{} 1)}$
$ \equal{} \frac {1}{n^4 \plus{} n^2 \plus{} 1}$


group theory
abstract algebra
superior algebra
superior algebra solved

If H is a proper subgroup of a finite group G, then G is not the union of all conjugates of H

Here you go:
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=Union+conjugates&t=22729


KT: 2 1 7 7 7 2    26
CS: 2 7 0 7 0 2   18
DC: 2 1 1 7 7 7   25
YS: 0 7 7 7 2 3   26
ZY: 0 1 6 7 7 7   28
JW: 2 4 7 7 2 0   22
Total: 145

As you can see, 1 was our HARDEST problem. Very strange IMO this is. 4 was easiest, follwed by 3, due to our not so secret inequality-whack training sessions. And of course, our three observers (Me too!) helped in coordination, so our poor team leader didn't have to stay up till 4am to go throught the solutions. 5 is third easiest, and 2 and 6 are the hardest except for 1.

What do we owe this weird score distribution to? Who cares! We got 145 points!

Congratulations to the Singapore contestants for their excellent performance. Some of you really owe us a treat. Others... we were regrettably unable to help.

But it's over now, so to those who may be reading this, it is already a great achievement to represent Singapore. Even if you did not really get what you wanted, the experiences will remain with you for a lifetime, whereas no one would care about the colour of your medal several years down the road. You were there, in Mexico, in 2005, and you were hit by a hurricane. (Ok, not yet, but soon I believe)

[quote="Yossarian"]Congratulations to the Singapore contestants for their excellent performance. Some of you really owe us a treat. Others... we were regrettably unable to help.

But it's over now, so to those who may be reading this, it is already a great achievement to represent Singapore. Even if you did not really get what you wanted, the experiences will remain with you for a lifetime, whereas no one would care about the colour of your medal several years down the road. You were there, in Mexico, in 2005, and you were hit by a hurricane. (Ok, not yet, but soon I believe)[/quote]Congrats! You know what they say now: "I have survived both the IMO and the Hurricane in Mexico"  :rotfl:

:clap: Congratulations to the Singapore team! :clap2:

Yep, those are excellent results. A pity about Qn 1, but hey, let's not sweat the small stuff.

Y'know, if our score continues to be strictly increasing from year to year....  :D  :first:

[quote="Valiowk"]A big thank you to our team leader, Dr Tay, our deputy team leader, Mr Koh, and Calvin, Julius, Kiat and Gary! :)[/quote]

Go to sleep already! Got to get up early for Chichen Itza.

Just got back from the jury meeting. We have 4 Silvers and 2 Bronzes. Silver is at 23, the rest is irrelevant. Ha ha.

Edit: Hey, I'm named behind Calvin. No fair! Ok, ok, just kidding.

And we are ranked 14th!

And may I once again draw attention to our well done Q3? Ok, so the solutions were all ugly-argly whack-majorisation, but points are points.

And I must also add, we are NOT a former communist state.

So, I am sure the next few observers for Slovenia 2006 will do an even better job, and get us into the top 10 (2004:18th-2005:14th-2006:10th)? Ha ha. Just kidding. I doubt it will ever happen, but please prove me wrong!


[quote="ragnarok23"]An $n\times n$ magic square is a square grid in which the numbers $1$ through $n^{2}$ are arranged in a way such that all the columns, rows, and diagonals add up to the same number. Find this number for a $5\times 5$ magic square.[/quote]
[hide="answer"]The sum of all the numbers is 325. Each of the columns have no overlap of numbers. Since there are five of them they each have to add up to 65 since the sum of all the columns is the sum of all the numbers.[/hide]
[hide="generalization"]Using the same logic, for an n by n square the magic number is $\frac{n^{3}+n}{2}$[/hide]


$ 8bc\plus{}2b^{2}\minus{}3a^{2}\plus{}4ac\equal{}887$

$ 4b\minus{}5a\plus{}8c\plus{}5ab\equal{}\minus{}396$

Given the above 2 equations
Were suppose to find out what does (a+b+c)

It's surely under-determined.  Take, for example, $ b \equal{} 0$; in that case there are already two solutions that give different values of $ a \plus{} b \plus{} c$.

Suppose a,b,c are positive integers with a odd

The given Options are

a) 14
b) 21
c) 25
d) 32
e) 43


It took 2.3 million stones with a total weight of 6 million tons to construct the Great Pyramid. To the nearest tenth, what is the number of tons in the mean weight of the stones?


Show that the number of divisors of $ 111...1$ (1992 digits) is even.

$ x=\underbrace{11...111}_{1992\, 1's}\equiv 11\equiv 3\mod 4$. But no perfect square modulo $ 4$ is equivalent to $ 3$. Hence $ x$ is not a perfect square. Consequently, $ x$ has an even number of divisors.

[hide="Solution"]
Let $ N=111...1$, with $ 1992$ digits. Then, $ \frac{N}{3}= 037037...037$ with $ \frac{1992}{3}=664$ repetitions of the block $ 037$. But, the sum of this number's digits is not divisible by three, thus $ 3$ divides $ N$ only once, and thus the number of divisors of $ N$ is even.[/hide]

You could just say that it is $ 1992\equiv 3\mod 9$ instead of doing the whole dividing thing.


function
trigonometry
calculus
integration
abstract algebra
real analysis
real analysis unsolved

If $ f(x)$ is a $ 2\pi$ periodic function, and $ f(x) \in C(\mathbb{R})$

assume it's fourier series is : $ f(x) \sim \frac {a_0}{2} \plus{} \sum_{n \equal{} 1}^{\infty}(a_n\cos nx \plus{} b_n\sin nx)$

if the above seires convergence everywhere(pointwise). Show that: It's sum function is : $ f(x)$

[color=blue]I want to estimate the below integral:

$ |f(x_0) \minus{} S_n(x_0)| \equal{} \left|\frac {1}{\pi}\int_{ \minus{} \pi}^{\pi}(f(x_0) \minus{} f(x_0 \plus{} t))\frac {\sin(n \plus{} \frac {1}{2})t}{2\sin\frac {t}{2}}dt\right|$

If it tends to $ 0$ as $ n \rightarrow \infty$. we can get the conclusion. but i can't do that.  :maybe:

actually, i have no idea to use the condition: $ \frac {a_0}{2} \plus{} \sum_{n \equal{} 1}^{\infty}(a_n\cos nx \plus{} b_n\sin nx)$ pointwise convergence everywhere :maybe: [/color]

What this calls for is Abel's theorem - or perhaps the corresponding Abelian theorem for, say, Cesaro summability.

Theorem (Cesaro summability of Fourier series in $ \mathbb{C}:$ Suppose $ f\in C(\mathbb{T}).$ Define $ S_N(f)(x) \equal{} \frac {a_0}2 \plus{} \sum_{n \equal{} 1}^N(a_n\cos nx \plus{} b_n\sin nx).$ Define $ \sigma_N(f)(x) \equal{} \frac1{N \plus{} 1}\sum_{n \equal{} 0}^NS_n(f)(x).$ Then $ \sigma_n(f)(x)\to f(x)$ pointwise everywhere; in fact, $ \sigma_n(f)\to f$ uniformly on $ \mathbb{T}.$

Theorem (Abelian theorem for Cesaro summability). For a series $ \sum_{n \equal{} 0}^{\infty}a_n,$ define $ S_N \equal{} \sum_{n \equal{} 0}^Na_n$ and $ \sigma_N \equal{} \frac1{N \plus{} 1}\sum_{n \equal{} 0}^NS_n.$ If $ S_N\to L$ (the series converges), then $ \sigma_N\to L$ (the series is Cesaro summable, and to the same sum).

Put these two theorems together and you have what you want.

One could also use Abel summability rather than Cesaro summability. For a given $ f,$ and for $ 0\le r < 1,$ define $ M_r(f)(x) \equal{} \frac {a_0}2 \plus{} \sum_{n \equal{} 1}^{\infty}r^n(a_n\cos nx \plus{} b_n\sin nx).$ Then $ M_r(f)(x)\to f(x)$ as $ r\to1^{ \minus{} }$ pointwise everywhere for continuous $ f,$ and we can use the original Abel's theorem.

(I typed this much up before I saw alekk's post; yes, I'm talking about the same thing he is. Using Abel rather that Cesaro summability would invoke the Poisson kernel rather than the Fejer kernel.)


Inscribe a sphere in a regular tetrahedron of side length 1. What is its radius?

[hide]Draw a line from the vertex to the center of a triangle. Since the length is 2/3 the height its $\frac{\sqrt{3}}{3}$. Then draw a line from the center of the sphere to the vertex, the length of this is $\frac{\sqrt{6}}{3}-x$(x=radius). Now drawt the radius of the sphere  (from the center of the sphere to the center of the face) lenght=x. So we now have a right triangle:

$(\frac{\sqrt{6}}{3}-x)^{2})^{2}=x^{2}+(\frac{\sqrt{3}}{3})^{2}$

$x=\frac{\sqrt{6}}{12}$[/hide]


[quote="athunder"]Ok so like this is the question:

$ (\frac {sin^2\theta}{2} \plus{} \frac {2}{cos^2\theta})^{\frac {1}{4}}$+$ (\frac {cos^2\theta}{2} \plus{} \frac {2}{sin^2\theta})^{\frac {1}{4}}$=$ 68^{\frac {1}{4}}$

So, i am pretty sure this looks like an AM-GM inequality but how do you get it into a position like that?

Any help is best.[/quote]

This should be greater than or equal to, btw.

You can use AM-GM directly on the two terms on the left side and get

$ (\frac{sin^{2}\theta}{2}\plus{}\frac{2}{cos^{2}\theta})^{\frac{1}{4}} \plus{}(\frac{cos^{2}\theta}{2}\plus{}\frac{2}{sin^{2}\theta})^{\frac{1}{4}}\ge 2((\frac{sin^{2}\theta}{2}\plus{}\frac{2}{cos^{2}\theta})(\frac{cos^{2}\theta}{2}\plus{}\frac{2}{sin^{2}\theta}))^{\frac{1}{8}}\equal{}2(2\plus{}\frac{sin^{2}\theta cos^{2}\theta}{4}\plus{}\frac{4}{sin^{2}\theta cos^{2}\theta})^{\frac{1}{8}}\equal{}2(2\plus{}\frac{sin^2(2\theta)}{16}\plus{}\frac{16}{sin^2(2\theta)})^{\frac{1}{8}}$.
If we let $ x\equal{}sin^2(2\theta)$, and then consider the expression $ \frac{x}{16}\plus{}\frac{16}{x}$. We know that $ 0\le x\le 1$, because that is the range of sine.
It seems like the minimum of this expression is when $ x\equal{}1$, and indeed, we can prove this by showing that $ \frac{y}{16}\plus{}\frac{16}{y}>\frac{1}{16}\plus{}16$ if $ 0<y<1$.

Therefore, $ 2(2\plus{}\frac{sin^2(2\theta)}{16}\plus{}\frac{16}{sin^2(2\theta)})^{\frac{1}{8}}\ge 2(2\plus{}\frac{1}{16}\plus{}\frac{16}{1})^{\frac{1}{8}}$. After some algebra, this turns out to be equal to $ (68)^{\frac{1}{4}}$, so we are done.


Prove that the number of pairs $ \left(\alpha,S\right)$ of a permutation $ \alpha$ of $ \{1,2,\dots,n\}$ and a subset $ S$ of $ \{1,2,\dots,n\}$ such that
\[ \forall x\in S: \alpha(x)\not\in S\]
is equal to $ n!F_{n \plus{} 1}$ in which $ F_n$ is the Fibonacci sequence such that $ F_1 \equal{} F_2 \equal{} 1$

we find the number of such permutation and subset in order of subset:
for each subset with $ k$ elements we should pick these $ k$ elements and put them in $ n \minus{} k$ place and fill other blanks with other elements so the nimber is:
$ \sum_{k \equal{} 0}^n\binom{n}{k} \binom{n \minus{} k}{k}k!(n \minus{} k)!$=$ n!\sum_{k \equal{} 0}^n\frac {(n \minus{} k)!}{(n \minus{} 2k)!k!}$=$ n!\sum_{k}\binom{n \minus{} k}{k}$=$ n!f_{n \plus{} 1}$

as we know that $ f_{n \plus{} 1} \equal{} \sum_{k}\binom{n \minus{} k}{k}$


this just in....
the US team won 5 gold medals (Oleg. Tiankai, Aaron, Tony, Alison) and a silver (Matt) in Athens, coming in 2nd overall. China was 1st and Russia was 3rd.

Wow, that's awesome! How many students received perfect scores and who?

[quote="texas137"][quote="confuted"]I still think I should have won a gold medal at IMO. It would have been good.[/quote]

Confuted - Are you Matt Ince? If so, congratulations! Any medal at all at the IMO is fabulous.  And if you're not, well, congrats are still well deserved by anyone on this list who was involved in the IMO.[/quote]
No, but I still think I should have won a gold medal at IMO.

[quote="keta"]How many students received perfect scores and who?[/quote]

no perfect scores among the Americans. The top US score was 40. I don't know about anyone else. Hopefully a full list of results will be available soon.

[quote="Kyptonite"]I read on Mathlinks that Jacob Tsimermann (Canada) got a perfect.

Two russians, Dubashinskij Mikhail and Badzjan Andrej also got perfects.[/quote] LOL, well in fact you read on this site (which is the same!!) ...
Oh and btw, there is a special forum for the results of this imo, so I am moving this overthere ...

[quote="Valentin Vornicu"]Oh and btw, there is a special forum for the results of this imo, so I am moving this overthere ...[/quote]

Valentin, I think moving this topic was a bad idea.  Obviously, the people who posted to this thread do not read the "Advanced --> IMO" threads.  However, they do read the "US Contests" threads.

The posting was of special interest and importance to original AoPS members, who were almost entirely contained in North America.

At my last count, there were 78 low-level forums on the combined board versus about a dozen on the previous AoPS board.  Many AoPS members (who are U.S. middle school and high school students and their parents) will never venture into the Mathlinks "Advanced" forums.

Please show restraint when moving a topic like this.  Thank you.

[quote="Kyptonite"]I read on Mathlinks that Jacob Tsimermann (Canada) got a perfect.

Two russians, Dubashinskij Mikhail and Badzjan Andrej also got perfects.[/quote]

Congrats to everyone.

BTW, is Andrej Badzjan 'andrew' on mathlinks? Also, did the vietnam perfect scorer from last year also get full score this year? (presumably he should have as last years was much harder).

Hi ,
      To Vinoth : andrew's real name is Sergej , he is the instructor of Russian IMO team.

      Btw, I came to know from BMOC ( British Mathematical olympiad ) site that the 4th perfect scorer is from Hungary . Does any one know who ?

 Can any one post scores of Vietnam , China and Bulgaria individually. I would be very obliged .

If andrew=Andrey Badzyan, then my congratulations for him for the great result on this IMO! The Yunhao Fu solution of all problems is already posted here. I personally would be very glad to see Andrey Badzyan's solutions! Is it possible? Also congratulations to all Russian team! But can anyone tell me, why Nadezhda did not get the perfect score???

Hi all,
         I am really sorry for posting the wrong name of  andrew , actually Sergej is real name of fagot a new Russian member of mathlinks . He is the instructor of Russian IMO team.

Sorry I was really confused  :(     . So andrew is your full name Andrey Badzyan , I am dying to know  :D

There were actually 4 perfect scorers:

Bela Andras Racz (Hungary);
Jacob Tsimerman (Canada);
Andrey Badzyan and Mykhail Dubashinsky (Russia).

 1. China                     220
 2. USA                       212
 3. Russia                    205
 4. Vietnam                   196
 5. Bulgaria                  194
 6. Taiwan                    190
 7. Hungary                   187
 8. Japan                     182
 9. Iran                      178
10. Romania                   176
11. Ukraine                   174
12. Korea                     166
13. Belarus                   154
14. India                     151
15. Israel                    147
16. Poland                    142
17. Moldova                   140
18. Singapore                 139
19. Mongolia                  135
20. UK                        134
21= Brazil                    132
21= Canada                    132
21= Kazakhstan                132
21= Serbia and Montenegro     132
25. Germany                   130
26. Greece                    126
27. Australia                 125
28. Georgia                   123
29. Colombia                  122
30. Hong Kong                 120
31= Slovakia                  119
31= Turkey                    119
33. South Africa              110
34. Czech Rep                 109
35. Thailand                   99
36. Armenia                    98
37. Mexico                     96
38. FRANCE                     94
39. Argentina                  92
40. Croatia                    89
41. Morocco                    88
42= Belgium                    86
42= Macau                      86
44. Estonia                    85
45. Uzbekistan                 79
46. Sweden                     75
47. Azerbaijan                 72
48. FYROM                      71
49= Italy                      69
49= Slovenia                   69
51. Lithuania                  65
52= Kyrgystan                  63
52= Latvia                     63
54. Indonesia                  61
55= Albania                    57
55= Spain                      57
55= Switzerland                57
58. New Zealand                56
59= Austria                    55
59= Norway                     55
61. Netherlands                53
62. Turkmenistan               52
63= Cyprus                     49
63= Finland                    49
63= Peru                       49
66. Ireland                    48
67. Uruguay                    47
68. Denmark                    46
69. Puerto Rico                43
70. Bosnia and Hercegovina     40
71. Luxembourg                 36
72. Iceland                    35
73. Malaysia                   34
74. Sri Lanka                  33
75. Tunisia                    31
76. Trinidad and Tobago        29
77. Portugal                   26
78. Cuba                       17
79. Phillipines                16
80. Venezuela                  15
81. Ecuador                    14
82= Mozambique                 13
82= Paraguay                   13
84. Kuwait                      5
85. Saudi Arabia                4

There is a mistake for Turkey.
Its score is 118 (not 119) and its rank is 32th.


There are three suspects for a murder: Adams, Brown, and Clark.
Adams says "I didn't do it. The victim was an old acquaintance of
Brown's. But Clark hated him." Brown states "I didn't do it. I didn't
even know the guy. Besides I was out of town all that week." Clark
says "I didn't do it. I saw both Adams and Brown downtown with
the victim that day; one of them must have done it." Assume that
the two innocent men are telling the truth, but that the guilty man
might not be. Who did it?

There are 3 possibilities right? Ok... this is a series of guess and check.. The final answer is Adam..

waiittt.
i thought azn inzasion and jonathanchou was the same person...

maybe i stand corrected..

whos azn?

Duh. It's my bro. aznkid711 is me. but i dont really use it anymore.

going back to the original topic, i'm pretty sure Brown did it.

It is pretty much impossible to make sure, but I think it was Brown.

Well Brown's and Clark's statements contradict, so it must be one of them, therefore Adam was telling the truth. Because Brown's statement contradicts Adams, who is telling the truth, then Brown must've done it.


Construct two diameter AB and CD such that $ AB\perp CD$. Construct incenter I of triangle ABC. A line through I and perpendicular to CD intersects the bisector of angle AOC at $ I_1$. Construct $ (I_1, I_1I)$. Similarly for other circles and we are done.


algebra
polynomial
quadratics
irrational number
algebra unsolved

Prove that for every irrational number $ a$, there are irrational  real numbers $ b$ and $ b'$ such that $ a\plus{}b$ and $ ab'$ are both rational
while $ ab$ and $ a\plus{}b'$ are both irrational.

If the minimal polynomial of $ a$ is cubic or of higher degree (or doesn't exist), this is true of any choice of $ b, b'$ such that $ a \plus{} b, ab'$ are rational.  If $ a$ has quadratic minimal polynomial $ x^2 \minus{} cx \plus{} d \equal{} 0$, pick $ b, b'$ such that $ a \plus{} b \neq c$ and $ ab' \neq d$.


[quote="thanhnam2902"]Find $ x,y \in R$ of this system:
$ \{\begin{array}{l} \frac {3}{{2x - y}} - \frac {6}{{x + y}} = - 1 \\
\frac {1}{{2x - y}} - \frac {1}{{x - y}} = 0 \\
\end{array}$[/quote]

$ \frac {1}{{2x - y}} = \frac {1}{{x - y}} \Leftrightarrow 2x-y=x-y \Leftrightarrow x=0$

$ \frac {3}{{- y}} - \frac {6}{{ y}} = - 1 \Leftrightarrow y=9$

So $ (x,y)=(0,9)$ is the solution.


Define a sequence of reals by $ x_{1}= 1$ and $ x_{n+1}= x_{n}+\sqrt[3]{x_{n}}$ for $ n \ge 1$. Prove that there exist $ a,b\in R$ so that $ \lim_{n\to\infty}\frac{x_{n}}{ab^{n}}=1$ :D

Ofc $ a,b\neq 0$
Notice that sequence $ \{x_{n}\}$ is increasing and has a infinite limit $ +\infty$, so it's must be  $ b>1$ if we want to have a finite $ \lim_{n}\frac{x_{n}}{ab^{n}}$

Now let $ \sigma_{n}=\frac{x_{n}}{b^{n}}$ for $ b>1$. We have $ b\sigma_{n+1}=\sigma_{n}+\sqrt[3]{\frac{\sigma_{n}}{b^{2n}}}$

To resolve our problem, we can proof that the exists $ b$, such that sequence $ \{\sigma_{n}\}$ has a finite limit $ a\neq 0$. But if $ \{\sigma_{n}\}$ has a finite limit, limit will be always $ 0$... becouse:
Let $ \lim_{n}\sigma_{n}=\sigma$, we'll have:
$ b\sigma=\sigma+\lim_{n}\sqrt[3]{\frac{\sigma}{b^{2n}}}$
$ b\sigma=\sigma+0$
$ \sigma=0$
I hope that I havent done any mistake...

In other words:  since $ x_{n}\to \infty$, the given statement could be correct if and only if $ \lim_{n \to \infty}\frac{x_{n+1}}{x_{n}}= b > 1$.  However, $ \lim_{n \to \infty}\frac{x_{n+1}}{x_{n}}= \lim_{n \to \infty}\frac{x_{n}+\sqrt[3]{x_{n}}}{x_{n}}= \lim_{n \to \infty}1+x_{n}^{-\frac{2}{3}}= 1$, a contradiction.


abstract algebra
group theory
superior algebra
superior algebra unsolved

Let $ X\equal{}\mathbb{R}\setminus\{\minus{}1\}$, and define the operation in $ X$ as $ x*y\equal{}x\plus{}y\plus{}xy$.

Let $ A\equal{}\{x\in X : x>\minus{}1\}$.

Define $ X/A$?

Clearly, $ X, *$ is an abelian group. If $ x, y > \minus{}1$, then $ x*y \equal{} (x\plus{}1)(y\plus{}1)\minus{}1 > \minus{}1$, so $ A$ is a subgroup of $ X$. Since $ X, *$ is abelian, it is also a normal subgroup.

Now, we can take the quotient group $ X \slash A$. Take a fixed $ a < \minus{}1$. We will show that $ aA \equal{} \left\{x \in X : x < \minus{}1\right\}$, implying that there are only two cosets of $ A$ in $ X$, so $ X \slash A \cong \mathbb{C}_2$. Indeed, $ a*x \equal{} (a\plus{}1)(x\plus{}1)\minus{}1$, so for a given $ b < \minus{}1$, we can always find a unique $ x > \minus{}1$ such that $ (a\plus{}1)(x\plus{}1)\minus{}1 \equal{} b$, namely $ x \equal{} \frac{b\plus{}1}{a\plus{}1} \minus{}1 > \minus{}1$.

Or we could just notice that $ X$ is isomorphic to the nonzero reals under multiplication, and $ A$ corresponds to the subgroup of positive numbers.


calculus
derivative
parameterization
function
Support
integration
real analysis

How to define the weak partial derivatives i.e. distribution partial derivatives for two parameter function $f: I^{2}\to \mathbb{C}$ if $I^{2}=[0,1]\times [0,1]$ and $f\in L^{2}(I^{2})$? And how to define the Sobolev space in $I^{2}$.

$f \in L^{2}(I^{2})$ is locally integrable, so defines a distribution $[f]$ in the following way: if $\phi: I^{2}\to \mathbb R$ is a smooth function with compact support, one defines
\[([f], \phi) = \int_{I^{2}}f(x,y) \phi(x,y) dx dy\]
Suppose for a moment that $f$ is also a smooth function. Then, integration by parts shows that
\[([\frac{\partial f}{\partial x}], \phi) = \int_{I^{2}}\frac{\partial f}{\partial x}(x,y) \phi(x,y) dx dy=-\int_{I^{2}}f(x,y) \frac{\partial \phi}{\partial x}(x,y) dx dy=-([f], \frac{\partial \phi}{\partial x})\]
Hence, if $D$ is a distribution, one defines its partial derivative as:
\[(\frac{\partial D}{\partial x}, \phi) =-(D, \frac{\partial \phi}{\partial x})\]

Now, suppose that there exists functions $f,g \in L^{2}$, such that
\[-([f], \frac{\partial \phi}{\partial x})=([g], \phi)\]
for all compactly supported smooth functions $\phi$. Then one says that $g$ is the weak partial derivative of $f$. The Sobolev space $H^{1}(I^{2})=W^{1,2}(I^{2})$ is the set of functions $f \in L^{2}(I^{2})$ that have weak derivatives $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ also in $L^{2}(I^{2})$.


Do you guys remember the thing on the AIME form where we put AMC 10/12 and score? What's that for? I have a strange feeling that I put it in wrong by a point or something. Will it affect me at all?

Also, I'm just curious...which one did you put if you took both 10 and 12?

I remember the director saying that the score bubbled on the AIME only provides a weak identity check and that they will calculate people's USAMO index by information in their database.

Hmmmm? I don't even remember seeing a place to put my AMC score.

You must've missed it.

http://www.unl.edu/amc/e-exams/e7-aime/e7-1-aimearchive/2008-aa/2008%20AIME%20AF,1+2.pdf

Hmmm I'm positive I did not see that...whatever not like it matters.

Yeah, its always there. There's a box with two bubbles: AMC 10 or AMC 12

Then there's another box that asks you to enter your score on it. I think it was on the bottom on the other side of the answering part. The side of the paper with your name and such.

[quote="firecricket91"][quote="mchoi815"]i did 12, even though im a 9th grader, i figured 6 pts didnt matter.

132/9
vs
138/9[/quote]
you might even qualify through the 12 with that 222 index so it doesnt even matter what  your 10 score was, and that is good for <=10th graders because that opens up 1 more spot.

what if you accidentally misbubble your score as the AMC 12 when you intended for AMC 10 (due to careless  bubbling, not intentional)?[/quote]

I think that they'll recognize the mistake, since they can put 2 and 2 together when they realize your name doesn't coincide with your highest score.  I wouldn't worry too much about it, as previously said, it's mainly for identification purposes.

RisingStar worte:
[quote]the director saying that the score bubbled on the AIME only provides a weak identity check and that they will calculate people's USAMO index by information in their database.[/quote]

This is correct.  This information gives our office a quick, but weak, identification check which is used for a fast first sort.  It is never taken to be definitive, and we understand that incorrectly bubbled scores are frequent.

It is far more important to correctly and consistently bubble your name and school identification information.  By "consistently" I mean: Do not use a nickname on one test and a given name on another test, use the same name identification on both tests.

Steve Dunbar
MAA Director,
American Mathematics Competitions


Probability with Combinations Part 4

I've been invited to participate in the National MATHCOUNTS Countdown round.  See how I do!

What happened with Probability with Combinations Part 3 ?

I don't really get how he does 6-choose-2 over 11-choose-2. Please explain :(


$ tan(90\minus{}x) \equal{} \frac{1}{tanx}$

$ \tan(90\minus{}x)\neq\frac{1}{\tan(x)}$
You need to specify degrees. All angles are assumed to be radians unless told otherwise.
[hide="Proof"]
$ \tan(\pi/2\minus{}x)\equal{}\dfrac{\sin(\pi/2\minus{}x)}{\cos(\pi/2\minus{}x)}\equal{}\dfrac{\cos(x)}{\sin(x)}$ By definition of tangent and co-functions.

$ \tan(\pi/2\minus{}x)\equal{}\dfrac{\cos(x)}{\sin(x)}\equal{}\cfrac{1}{\cfrac{\sin(x)}{\cos(x)}}$ By algebra

$ \tan(\pi/2\minus{}x)\equal{}\dfrac{\cos(x)}{\sin(x)}\equal{}\cfrac{1}{\tan(x)}$ By the definition of tangent. Qwedbert.

[/hide]

[hide="Geometric interpretation"]
Draw a right triangle with sides $ a$ and $ b$ opposite from angle $ \theta$, such that $ \cot\theta\equal{}\frac ba$. Rotate the triangle around to form a rectangle. The complementary angle to $ \theta$ is $ 90^\circ\minus{}\theta$, and the sides remain $ b$ and $ a$, so we have $ \tan(90^\circ\minus{}\theta)\equal{}\frac ba$.
[/hide]


geometry
circumcircle
trigonometry
geometry proposed

in triangle $ ABC$ let $ H$ be foot of perpendicular from $ A$ to $ BC$  and $ T$ and $ T'$ be foot of prpendiculars from $ H$ to $ AB$ and $ AC$ in order , if $ 2OT \equal{} AC$ prove that $ 2OT' \equal{} AB$.

Using the stewart theorem we obtain $ R^2(\overline{AT} \plus{} \overline{TB}) \equal{} \overline{AB} (\overline{TO}^2 \plus{} \overline{AT} \cdot \overline{TB}) \ \Longleftrightarrow \ \overline{TO}^2 \equal{} R^2 \minus{} \overline{AT} \cdot \overline{TB} \equal{} R^2 \minus{} \overline{TH}^2$

so we have $ 4R^2 \minus{} 4\overline{TH}^2 \equal{} \overline{AC}^2 \ \Longleftrightarrow \ \frac{R^2}{\sin^2{\beta}} \minus{} \overline{AH}^2 \cot^2 \beta \equal{} R^2 \ \Longleftrightarrow \ R \equal{} \overline{AH}$

equally we can say that $ 2\overline{OT}' \equal{} \overline{AB} \ \Longleftrightarrow \ R \equal{} \overline{AH}$

Then $ 2\overline{OT} \equal{} \overline{AC} \ \Longleftrightarrow \ R \equal{} \overline{AH}\ \Longleftrightarrow \ 2\overline{OT}' \equal{} \overline{AB}$

[size=134][color=darkblue][b]Nice problem ![/b][/color][/size]

[quote="Amir.S"] [color=darkred]In $ \triangle ABC$ let denote $ \|\begin{array}{ccc}
D\in BC & , & AD\perp BC\\\\
X\in AB & , & DX\perp AB\\\\
Y\in AC & , & DY\perp AC\end{array}$ . Prove that  $ OX = \frac b2\ \Longleftrightarrow\ OY=\frac c2$ .[/color][/quote]

[color=darkblue][b][u]Proof.[/u][/b] Suppose w.l.o.g. that $ 90^{\circ}\not\in\{B,C\}$ . Denote the power $ p_w(X)$ of the point $ X$ w.r.t. the circumcircle $ w=C(O,R)$ .

Observe that $ \|\begin{array}{c}
DX=h_a\cdot|\cos B|\\\\
DY=h_a\cdot|\cos C|\end{array}$  and $ \|\begin{array}{c}
p_w(X)=\overline {XA}\cdot\overline {XB}=OX^2-R^2\ \Longleftrightarrow\ OX^2=R^2-h_a^2\cos^2B\\\\
p_w(Y)=\overline {YA}\cdot\overline {YC}=OY^2-R^2\ \Longleftrightarrow\ OY^2=R^2-h_a^2\cos^2C\end{array}$ . Thus,

$ OX=\frac b2$ $ \Longleftrightarrow$ $ OX^2=R^2\sin^2B$ $ \Longleftrightarrow$ $ R^2-h_a^2\cdot \cos^2 B=R^2\sin^2B$ $ \Longleftrightarrow$  $ R=h_a$ .

$ OY=\frac c2$ $ \Longleftrightarrow$ $ OY^2=R^2\sin^2C$ $ \Longleftrightarrow$ $ R^2-h_a^2\cdot \cos^2 C=R^2\sin^2C$ $ \Longleftrightarrow$  $ R=h_a$ .

In conclusion, $ \boxed {\ OX=\frac b2\ \Longleftrightarrow\ OY=\frac c2\ \Longleftrightarrow\ h_a=R\ \Longleftrightarrow\ bc=2R^2\ \Longleftrightarrow\ \cos (B-C)=2\sin^2\frac A2\ }$ .[/color]


geometry
MATHCOUNTS
trigonometry
circumcircle
inradius
area of a triangle
Heron\u0027s formula

I dont have a specific problem, but in general I am not comfortable with the general A=1/2BH method because when im given a mathcounts problem with all the sides and am asked to find the area, i resort to Herons formula even though I know that there is an easier way.

Thanks in advance :maybe:

It depends on the numbers.  Like in the case of 13,14,15, it just so happens that one of the altitudes is 5, which you kinda have to use a little wishful thinking or algebra.  Go with what's easier for you.  If it takes you longer to find what you need for an easier formula than with the direct, more complicated one, then in a contest, use the complicated one.  In practice, try finding more than one solution.

I think Heron's formula works just fine for me, seeing as it's pretty fast already compared to finding the height of a triangle from the Pythagorean Theorem.

maybe look for angles.
LIke if there are angles that are OR are multiples of 30,60,or 90
You might wanna try cutting the triangle up into smaller triangles.

Oh yeah, why should you be uncomfortable with $ A\equal{}\frac{bh}{2}$?  Surely not all area of triangle problems give you all the sides and make you resort to herons.

And cutting the triangle up into smaller triangles is generally good if you try to find the height, or a needed side length.

Well Heron's formula is nice, but if you have the luxury of a calculator, then we have $ A\equal{}\frac{1}{2}ab\sin{C}$.  I remember reading somewhere that there are thousands of ways to find the area of a triangle.

[quote="JRav"]Well Heron's formula is nice, but if you have the luxury of a calculator, then we have $ A \equal{} \frac {1}{2}ab\sin{C}$.  I remember reading somewhere that there are thousands of ways to find the area of a triangle.[/quote]
i bet there is. also, if you know the circumradius (or you have a triangle inscribed in a circle), use (abc)/4R where R is the circumradius and a,b, and c are the sides.
or if you know the inradius and you have the side lengths, just do rs, where r=inradius and s=semiperimeter.
but usually the classic 1/2bh is the most powerful, especially in mathcounts.

[quote="JRav"]Well Heron's formula is nice, but if you have the luxury of a calculator, then we have $ A \equal{} \frac {1}{2}ab\sin{C}$.  I remember reading somewhere that there are thousands of ways to find the area of a triangle.[/quote]

Well, if you have the luxury of a calculator, you don't even need to find sines of angles when given the three sides.  Just make a program for the formula on your TI beforehand and use that on the day of the test on like the target or team round when this problem crops up.  I usually find this problem fairly simple even with Heron's formula.  Hope that helps! :D


search calculus integration LaTeX \/closed

I'd like to post this integral:

\[ \int_0^{\infty}\frac{x^a\minus{}1}{x^3\minus{}1}dx;\quad 1<a<2\]

in the Calculus forum but would like to try first searching if it's been worked before.  If I enter the latex string \int_0^{\infty}\frac{x^a-1}{x^3-1} in the search criteria, it comes back with too many look-alikes (12 pages).

Is there any way to better search for specific math codes to find integrals that closely match this integral?

Thanks,
Shaw

It's an impossible task for a computer (at least today). You will have to restrict yourself to manual search with keywords (for example in your case they would be "int, infty, dx". Other than that you have to manually look through the results.

Ok. Thanks for replying.  Just don't want to annoy anyone with a redundant thread if I can help it.

[quote="shawtoos"]Just don't want to annoy anyone with a redundant thread if I can help it.[/quote]Are you sure about that statement? http://www.mathlinks.ro/viewtopic.php?p=1094018

. . . dang it . . . sorry about that.   I was wrong.


hi,
prove the identity
\[ \sum_{k\equal{}1}^{2^{n}}{\lfloor log_{2}(k) \rfloor} \equal{} (n\minus{}2)2^{n} \plus{} n \plus{} 2,\]
$ \forall n \in \mathbb{N}$.

\[ \sum_{k \equal{} 1}^{2^{n}}{\lfloor log_{2}(k) \rfloor} \equal{}n\plus{}\sum_{k\equal{}1}^{n\minus{}1}k2^k\equal{} n\plus{}(\frac{1\minus{}x^n}{1\minus{}x})'_{x\equal{}2}\equal{}(n \minus{} 2)2^{n} \plus{} n \plus{} 2,
\]
$ \forall n \in \mathbb{N}$.[/quote]


geometry
trigonometry
MATHCOUNTS
quadratics
ratio
trig identities
Law of Cosines

[hide="hint"]
Break it up into 10 triangles.
[/hide]

[quote="ln(dx/dy)"]A regular pentagon is inscribed in a circle that has an area of
144 pi.  What is the area of the pentagon?[/quote]

[hide="my explanation which probably used more work than I should have done"][hide="how I started"]radius=12
split the pentagon  into 5 triangles.
the angle in between the legs of 12 is 72[/hide]
[hide="law of cosines"]$a^2=12^2+12^2-2*12*12*cos(72)$
$a=14$[/hide]
[hide="area of triangle using herons formula"]area=sqrt(19*7*7*5)
=68.3[/hide]
[hide="area of pentagon"]68.3*5=[hide="the final answer"][b]3431.6[/b][/hide][/hide][/hide]

[quote]Break it up into 10 triangles.
[/quote]

That doesn't work.

[quote]my explanation which probably used more work than I should have done[/quote]

You did do more work than you should have done. (and I think you also got the wrong answer)

jli - [hide]one way to check if there was any typo/error is to make sure that the answer  is less than  $144 \pi $ )or  less than about 460 (pi  is just a little more than 3 but less than 10/3 ) The area  of pentagon should be less than  $144 \pi $ (This is one way to double check for   error ) so 3431 is in error .. answer is about 342.4 [/hide]

That's what I got too.

Can anyone generalizes a way without using trig?

[hide="Here_is_answer_folks"]Since the center of the pentagon is also the center of the circle, we have this:

Five congruent isosceles triangles with two congruent legs that have length of 12.

Since it's circle, 360/5 = 72 is the angle that is made by two radius. Apply the sin area formula:

\[\sin 72 * \frac {1}{2} * 12 * 12 = 68.4760917\]

Since there are five of them, multiply by five, and you'll get:

\[342.3803459\]
[/hide]

Assume this came from real MC problems, I must note that there must be a non-trig way although I'm not going to find this out.

It is up to you, MC participant, to solve without using trig since the correct answer is given.

[quote="ln(dx/dy)"]A regular pentagon is inscribed in a circle that has an area of
144 pi.  What is the area of the pentagon?[/quote]
ln:  If this is a MATHCOUNTS-level problem, please provide a reference to a WorkOut or competition on which this, or a similar problem appears.

There is a non-trig solution.  [hide="hint"]The method involves nothing more than similar triangles and quadratic equations -- The Golden Ratio appears pervasively.[/hide]
But I would rate this method above that of a National Masters Round.  Barring a reference, I suggest this thread be bumped.

Just to add to RCV,   the question can be bumped up ... the answer, if anyone is interested is
[hide]$90 \sqrt {10+2 \sqrt {5} }    $
 which is about  $342.38..$

(Easy if you know by heart things like
 $\cos  {(2\pi /5)}  =  \frac {\sqrt {5} - 1} {4} $
(which is exactly half of the golden ratio)  and  you use above to calculate
 $\sin {(2\pi /5)}  $

Of course, if you don't know these things by heart you have to use calculator  etc..:)[/hide]

[quote="AznXtreme418"]y can't u break it up into 5 triangles and use proportions to find the area for each one of them?[/quote]

You can't do that because they wouldn't be equilateral.  It's a pentagon, not a hexagon. It would work for a hexagon.


Hi all,

As you may have noticed we have upgraded our Forums, and as of now many new features stay at your disposal.

:arrow: First of all, we have increased security measures drastically, so, as of now, there is going to be no guest posting in any of the forums, and also guest users cannot see the memberlist, or your profile.

:arrow: Users have now more choices at hand. Use the [b]options[/b] menu, located in the header (above forum) section to select your viewing options. For example, if you want to remove this annoucement from the index page, just go to [b]board layout[/b] and select the [b]No[/b] option of [b]Display Board announces box[/b].

You can now choose, in the [b]Reading Topics[/b] menu, the option to [b]Keep messages unread[/b] when you leave the Forum, or not to keep them.

New emoticons have been introduced, but they mainly have the same codes. [hide="Click here to see some of them"] :)  :(  :?  :D  :lol:  ;)  :blush:  :P  :o  :roll:  :oops:  :cool:  :huh:  :no:  :love:  :wacko:  :what?: [/hide]
The link you see above that expands some text is the new "spoiler", or "[b]hide[/b]" text feature. Just enclose any text between the tags $\textrm{[hide}\textrm{]}$ and $\textrm{[}\slash \textrm{hide]}$ and it automatically will be transformed into a simple link. All the hidden text will be revealed by clicking on the link.

:arrow: You have now a [b]Favorites[/b] place to store your favorite topics. In each topic, just in the right upper corner, next to the Huge title link of the topic, there are 3 links: Add (remove) to watched topics (this will send you an email each time someone posts in that topic), Add (remove) from Favorites, and if you select the Keep Unread Topics options, you'll also have an option to mark that topic unread (for future reading).

:arrow: Users on dial up can now select from their profile [b]not to view other people's signatures or avatars[/b], thus speeding up the page loading process.

:arrow: Adding attachements can be done in the same way as before, just that this time you first have to push the attachement button to get the menu part visible.

In the post part you will find the quote button below the actual post (it will be on the right hand side eventually ;) ).

Also each topic will have a default icon (which for now is set to none), which is open to debate as of what it should mean (we can make it mean that the problem is unsolved yet, etc.).

$\LaTeX$ Now long formulae can be clicked on, and a pop-up window with the LateX code of the formula will be shown. Same happens if you just leave your mouse above a formula for 1-2 seconds.

:arrow: Some other features are introduction of Forums Watch (emails you each time something gets posted in that forum) the link being located in the down-left corner of each forum view, and in the same area there is a Delete your cookies link if you have problems that might need this.

:oops: Finally, we have lost part of our moderators list. Discussion about moderating, and requests for moderator status are to be done in [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=20383]this Topic[/url].

:ninja: To post bugs, please use the [url=http://www.mathlinks.ro/Forum/index.php?f=145]Support Forum[/url]

:alien: Further more all tests of the new features should be done this the [url=http://www.mathlinks.ro/Forum/index.php?f=224]test forum[/url]

:?: Any other questions, address them in this topic. Have fun $f^{2}(x)=\frac{1}{2}$ :D

Wow, this sure is different. It's cool though.

Looks like it took a ton of work. Congrats and thanks admin!  :)  8)

[quote="Scrambled"]just curious: did they like send a virus or something or did they hack at the server[/quote]Sorry, security details are not publically available :D

[quote="Valiowk"][quote="Valentin Vornicu"]For example, if you want to remove this annoucement from the index page, just go to [b]board layout[/b] and select the [b]No[/b] option of [b]Display Board announces box[/b].[/quote]

That doesn't seem to work :? .[/quote]Hehe, it won't work for a couple of days just to get people to actually read this :) :P

[quote="Valentin Vornicu"]Hehe, it won't work for a couple of days just to get people to actually read this :) :P[/quote]

:D :D :D

Is it possible to change topic page layout?
In my opinion, new layout is quite uncomfortable, because I see many unnecessary information in the page header and this information take too large area.
And the same I can say about "Post a reply" page. It is nonsense that I everytime need to scrool page to see text-field for reply.

I am sure interface must be more compact.

[quote="darij grinberg"][quote="Myth"]Is it possible to change topic page layout?
In my opinion, new layout is quite uncomfortable, because I see many unnecessary information in the page header and this information take too large area.[/quote]

Myth, can you say exactly what objects you find unnecessary? The only thing I would call superfluous is

"You last visited on Today, at 4:05 pm
The time now is Sat Dec 11, 2004 4:10 pm
All times are GMT + 1 Hour"

But this doesn't take too much space. All the rest is rather to the point, I'd say, isn't it?

  darij[/quote]

I think the box at the bottom that shows you who was on in the last 24 hours might be extraneous. But it might not matter once the admins figure out the + and - thing.

Yay, it's back :P .

*figures out on my own*

If all else fails, read the instructions :P .

[quote="Scrambled"]can i test new features on here?[/quote]

there's a board for that

I noticed that moderators don't have "Moderator" in their title...

[quote="Valentin Vornicu"][quote="yif man12"]That only happens when you're in a forum you moderate.

Edit: Whoops, my bad. It doesn't appear in specific forums either.[/quote]Being a moderator is not really a show-off thing ;) ... we'll consider making it appear in forum in which one has moderating rights.[/quote]

True, I was just wondering if it was going to be a change or not.  :D
I have no preference either way on the topic, even though I lean towards not having the titles.

I haven't been on here in a while and I was really surprised by the new forum.  Wow.  Nice work!!!!!!!! :D

[quote="Valentin Vornicu"]I've just added a small feature: when you click on the big title of a topic (the one in the center, the really BIG one) a prompter comes up with the link for that topic. I hope you find it usefull.

Also each post has the same prompter showing up when clickin on the #1, or #n, where n is the ordinal of the post in that topic.[/quote]

[b]THANK YOU !![/b]

  Darij

[quote="Peter VDD"][quote="Valentin Vornicu"]I've just added a small feature: when you click on the big title of a topic (the one in the center, the really BIG one) a prompter comes up with the link for that topic. I hope you find it usefull. [/quote]

Nice... only it doesn't seem to happen here? Perhaps because I'm using firefox? Or am I just too tired to see it at 2 am? :([/quote]I have tested this feature with Firefox 1.0, Opera 7.56 and IE 6.0 (in fact all the settings of this site are tested with the above browsers - the recommended resolution are 1024x768 and 1280x1024, but the site also renders without horizontal scrolling bar on 800x600 too on the MathLinks style).

Maybe you have not enabled java script?

Meanwhile another thing has changed: you can now use the options menu (go to options -> board layout) to remove the global announcements from the index forum). So people actually reading this can get it off ;)

[quote="byrdgyrl"]I love the new forum.  Thank you so much for redoing it.  Quite a job! :lol:[/quote]We're glad you do like it.

Another feature: on the artofproblemsolving url the advanced and college categories are collapsed by default, and on the mathlinks url the problem solving and contest categories are collapsed by default.
If you wish to have all of the expanded by default for now you have to join the usergroup "See all cats expanded" which is an open usergroup. You will automatically then see them all expanded.
In the future the collapsible script will be back, but for now this solution will have to do.

Finally after days of work, I've managed to bring back the collapsible categories, better and improved (board wise).

As before you can use the big config button to save a prefered state, or to expand/collapse all categories at once.

All bugs should be directed in the Support forums, NOT in this thread. Thanks.

Another small mention: it seems that in Firefox 1.0 by double clicking on a categories' name, you obtain the same effect as before the collapsing occured. Same effect can be obtained in IE by right clicking on the link and choosing the first option in the menu, "open" - or, in mathlinks style, using on e of the 4 links located in the upper-left side of the page.

almost every message board that I enjoy has upgraded these weeks  :D

For the foreign users (and not only) we currenly have installed some scripts to detect your language (based on your geographical location and your browsers' settings). It might be possible that the settings are not correct, in this case just use the bolded English Version button located in the left of the forum index page or the language dropdown fast change menu in the lower left corner of the main page.

Also, starting now, clicking on the name of a category will take you directly to that category, instead of making that category roll in / out ;)
Furthermore, now clicking on the URL and Img buttons when posting a message will reveal a more friendly way to input URLs and Images links.

:D Man this really cool :ninja: this  my fist day on this website :rotfl:  :ninja:
[/hide]this in case you dont know i like fiber[hide][/hide]

[quote="The Math Mage"]:D Man this really cool :ninja: this  my fist day on this website :rotfl:  :ninja:[/quote]Then welcome and have a nice stay ;)

Wow, that was quite a rush.  I joined one day after [hide="you first posted in this thing."] I hope that checks out when you are all reading it to the left.[/hide]  You're right, the forum does rock; I learned some new features, but I guess I'll have to go find that [url=http://www.artofproblemsolving.com/Forum/index.php?f=145]Support page[/url]* to mention that my 1024 x 768 monitor here (Firefox-- ME and AoPS to mention it) doesn't display the names below the (bottom) quotes clearly--enough to see unless you already know the name of the person cited... tisk, tisk.

* That "when you click on the big title of a topic" thing that worked well in this form didn't seem to in the Support forum at least.  (It just reloaded the page.)

[quote="The Math Mage"]:alien: thank you , and to you
[hide]I'm still kind of estatic about this hide thing[/hide][/quote]

estatic isn't a word. you probably meant ecstatic which means ebullient or very happy  ;)

I would mind my own business and stop spamming on dead threads (hey that rhymes  :P )

I am new to AoPS Forums, and have a couple of questions/doubts (probably more will turn up):

1.  I had sent a private question to a 'Moderator' in another section of AoPS Forums, as I did not want to clutter up the forum with stuff of little interest to the members.  Now I find the question has appeared on my [b]'Outbox' [/b]list when I check [b]"Private Messages" [/b]- but it does not appear in the [b]'SentBox' [/b]list.

Does this mean the question has NOT reached the Moderator to whom it was addressed?   If so, how to process that question so it reaches the Moderator (and gets into my 'Outbox' list?)

2.  I seem to have forgotten the second question I have, shall put it up later, if I may.

3.    I now feel the question I had asked the Moderator in that other section may well be appropriate to this section as a whole, so I reproduce it below.  It is somewhat long, but I crave your indulgence!

++++

[color=red]Admin edit: <snipped boring advertizing attempt>[/color]

Yours truly
G.S. Chandy

Just have patience. If a message is in the Outbox it means it is waiting for him to read it. Once the message is read, it will appear in the Sentbox.

OK, on re-reading my posting just sent, I remembered question No. 2 that I had forgotten then.  Here it is:

>> Members and moderators at these forums are identified by 'Rank' ("Navier Stokes Equation"; "Riemann Hypothesis" etc) and by stars.  What do these mean?  Where are these things explained?

Thanks in advance for clearing these doubts.

GSC

This has been discussed before,

here [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=26876&postorder=asc&highlight=5000&start=20[/url]


A circle is inscribed inside a square. The square is inturn inscribed inside an equilateral triangle. If the side of an equilateral triangle is s what is the area of the circle and square in terms of s?

This problem seems to be way too easy for the advanced section. Sounds more like getting started material =/.

Too lazy to do the actual math, but...

[hide]

Circle of radius r inside a square, so the square has side 2r. Inscribed in the equilateral triangle, there will be a mini-equilateral triangle on top and two 30-60-90 triangles on the sides.

[/hide]

[hide]

Let the "top" equilateral triangle have side length x, then the smaller right triangle has side x opposite a 60 degree angle, and hypotenuse s-x.

Therefore,  :sqrt: 3 / 2 = x / (s-x)

so x = 3s / (3 + 2 :sqrt: 3)

[/hide]


Can someone pleeeeeeeeease post some 4-grade problems :(

OK HERE'S AN EASY ONE :mad:

Problem $ 1$

You have $ 8899*22*66$ apples.
If you eat $ 20$ apples a day,
in how many days will you eat all of them

binomial_4eva, in that case you would have to round up if there is a decimal because you are missing some of the apples by rounding down.
[hide="So it is "]
$ 646067.4 \Rightarrow 646068 \ \text{days}$
[/hide]

Can someone please send some 6th grade math Olympiads questions pleaseeeeeeeeeeee

ummm...i dont know what 6th grade level is cuz i skiped 6th and 7th grade math and my 5th grade math teacher stunk.

heres a couple i guess should be around that dificulty

1. find 1+2+3.....100

2. how many primes are less than 100

3. how many factors does 36 have

4. 11x/4=99. solve for x.

Uh,

What is the surface area of a cone with height 4, and radius 3?

I can't think of anymore :P

that just reminded me of something. during mathcounts practice one day, we were talking about surface area, volume, area, and all those other formulas. my teacher said. find the surface area of a- and my friend said pinecone. so now we make a big deal about finding the surface area of a pinecone. i know its irelavant.

do they have exponents in moems?

1. find the value of $ \sqrt{x^{3}+2^{y}}$ if x=5 and y=2

2. what is the area of the largest circle that can be inscribed in a 4*4 square

Do you think we should post the solutions to the problems we are posting?

[hide="Solution for my problem"]
[img]http://library.thinkquest.org/20991/media/geo_cone1.gif[/img]

There are two circles in a cone. One is the base, and the other has the height as its radius.

[b]First circle[/b]
Since the circle has radius $ 3$  you get $ 3^{2}\pi=9\pi\approx28.27$

[b]Second circle[/b]
The circle radius (height) is $ 4$,you get$ 4^{2}\pi=16\pi\approx50.27$

Adding the areas together you get $ \boxed{78.54}$
[/hide]

yeah i guess so.
[hide="hints"][hide="set 1"]
1. use the sum formula n(n+1)/2
2. count them i guess
3. express it in the form $ a^{m}\plus{}b^{n}$.
4. solve it normally.
[/hide]
[hide="set 2"]
1. plug them numbers in
2. draw it out. find the ratio of the raius of the circle to the side of the square[/hide][/hide]

[hide="answers"][hide="set 1"]
1. 5050
2. 26
3. 9
4. 36
[/hide]
[hide="set 2"]
1. $ \sqrt{129}$
2. $ 4\pi$ or 12.56
[/hide]
[/hide]

More problems:

What is product of the divisors of 16?

[hide="Solution"]
First, write out the factors of 16.
\begin{eqnarray*}1,16\\ 2,8\\ 4,4\end{eqnarray*}
So you have $ (1\cdot16)(2\cdot8)4=16^{2}\cdot4=\boxed{1024}$
[/hide]

Or you could just count the powers of 2 in the divisors-
$ 2^{0\plus{}1\plus{}2\plus{}3\plus{}4}\equal{} 1024$

True,but the solution is bit easier to follow.

i think school competitions from mathcounts will help.

easier ones
http://www.mathcounts.org/webarticles/articlefiles/510-07_Sc_Sprint.pdf
http://www.mathcounts.org/webarticles/articlefiles/510-07_Sc_Countdown.pdf

a bit tougher
http://www.mathcounts.org/webarticles/articlefiles/510-07_Sc_Target.pdf
http://www.mathcounts.org/webarticles/articlefiles/510-07_Sc_Team.pdf

answers. sorry there are no solutions
http://www.mathcounts.org/webarticles/articlefiles/510-07_Sc_AnswerKey.pdf

Not really MATHCOUNTS usually involves guess and check,while MOEMS is bit of algebra, geometry etc.

no, not realy. at chapter, theres a lot of fast arithmatic involved. for school competitions you just need to know basic formulas.

Can someone send some math Olympiads questions  please :lol:


Solve the equation
$ log_{3}(2^{log_{3}(2^x\plus{}1)}\plus{}1)\equal{}log_{2}(3^x\minus{}1)$.

[quote="SUPERMAN2"]Solve the equation
$ log_{3}(2^{log_{3}(2^x \plus{} 1)} \plus{} 1) \equal{} log_{2}(3^x \minus{} 1)$.[/quote]

Let $ 2^x\plus{}1\equal{}3^k$, $ 2^k\plus{}1\equal{}3^y$, then the equation
$ \Leftrightarrow log_3(2^k\plus{}1)\equal{}log_2(3^x\minus{}1)$
$ \Leftrightarrow y\equal{}log_2(3^x\minus{}1)$
$ \Leftrightarrow 2^y\plus{}1\equal{}3^x$

Now $ \begin{cases} 2^x\plus{}1\equal{}3^k ...........(1) \\ 2^k\plus{}1\equal{}3^y ...........(2) \\ 2^y\plus{}1\equal{}3^x .............(3) \end{cases}$

If $ x<k$, then from $ (1)$ and $ (2)$ we get $ k<y$, then from $ (2)$ and $ (3)$ we get $ y<x$, so $ x<k<y<x$, contradiction.
If $ x>k$, then similarly we get $ x>k>y>x$, contradiction.
So it's easy to see that $ x\equal{}y\equal{}k$.

Now it becomes to solve $ 2^x\plus{}1\equal{}3^x \Leftrightarrow (\frac{2}{3})^x\plus{}(\frac{1}{3})^x\equal{}1$
Clearly $ x\equal{}1$ is a solution. As $ L.H.S.$ is strictly decreasing, $ x\equal{}1$ is the only solution.


I took these examples out of the chapter i am working on in my book. To be quite honest, im lost on it.

Thx for the help in advance guys....

1.  Simplify and evaluate:
(42x3y-2z-3/x-5y8z-9)2

2.  Convert to scientific notation (here you may use x as multiplication):
0.00083

3.  For f(x) = 6x – 7, find f(2).

4.  Perform the indicated operations:
(-x2 – 6x + 3) – (-9x2 + x – 1) – (x2 + 5x – 2)

5.  Use FOIL to multiply:
(8x – 1)(2x + 5)

6.  Multiply:
(x4 – 4x2 – 5)(x2 – x + 2)

please do not post the same thing twice.
we have answered these questions on the other forum.

Im sorry...I didnt realize that they are both the same forums....I thought it was 2 different ones all together..


How many numbers are palindromes that are also primes between 100 and 200?

[hide="solution"]
The number has to be in the form $\text{1A1}$ or $101,111,121,...191$. $101$ is prime. $111$ isn't because it is a multiple of $3$. $121=11^2$. $131$ is prime. $141$ isn't because it is a multiple of $3$. $161$ is a multiple of $7$. $151$ is prime. $171$ is a multiple of $3$. $181$ is prime. $191$ is prime. That's a total of $\boxed{5}$
The fastest way to see if those numbers are prime is knowing that only odd numbers but not $5$ can be factors. Also know that the last digit of multiples repeats every $10n$ where $n$ in the factor. An example is to find if $161$ is prime quickly, add up the digits and see it is not a multiple of $3$, and know that $91$ is a multiple of $7$, and the last digit repeats every $70$ numbers so $91+70=161$ so $161$ is a multiple of $7$. Also know that you only have to check the odd numbers up to $\sqrt{n}$ for $n$.[/hide]
EDIT: 91+70=161


:lol:   :play_ball:

What is the sum of the sequence 1^5 + 2^5 + 3^5 + 4^5............+ n^5?

See http://www.mathlinks.ro/Forum/viewtopic.php?t=36189 and probably some other topics.

  Darij

See also http://www.mathlinks.ro/Forum/viewtopic.php?t=59905 .