modular arithmetic
Summation
number theory
Number theoretic functions
IMO Shortlist

[hide="Solution"]
[b]Lemma[/b]: Consider a necklace with $ n$ beads, of two colors, say black and white.  $ a_n$ is the number of "good necklaces," ($ a_1a_2a_3\cdots a_n$), so that $ a_ia_{i \plus{} 1}a_{i \plus{} 2}\cdots a_{i \minus{} 1}$ are all distinct for $ 1\le i\le n$.

[i]Proof[/i]: We proceed inductively.  Indeed, it is true for prime $ n$ since there are $ 2^n$ ways to order the necklace, but we must take out two possibilities since these are the ways to have all of the beads being the same color.  Now, we prove it for $ n$ in general.  Clearly, if a necklace was not good, then it would there would have to be a repeating block in it, of length $ k$ (minimal length $ k$), so that $ k|n$.  There are $ a_k$ such blocks.  Then, there are $ \sum_{k|n, k\ne n} {a_k}$ bad necklaces.  Yet, there are $ 2^n$ necklaces total, so $ 2^n \minus{} \sum_{k|n, k\ne n} {a_k} \equal{} a_n$ good necklaces, so the lemma is proven.

Now, notice that there are $ n$ possible values of $ i$, so we can shift our good necklace $ n$ times.  This implies that for each good string, we can find $ n$ "shifts" of this good string.  Hence, there are $ n$ "shifts" per good string, which implies that $ n$ is a factor of the number of good strings.  This means that $ n|a_n$. [/hide]

A different proof by Mobius inversion formula:
Let $p$ a prime such that $p^k\| n$ then I prove $p^k\mid a_n$, the problem follows easily.
$\displaystyle a_n=\sum_{d\mid n}\mu(d)2^{\frac{n}{d}}=\sum_{i=0}^k\left(\sum_{d\mid n\wedge \upsilon_p(d)=i}\mu(d)2^{\frac nd}\right)=\sum_{i=0}^k\left(\sum_{d\mid\frac{n}{p^k}}\mu(d\cdot p^i)2^{\frac n{p^id}}\right)=$
$\sum_{d\mid \frac{n}{p^k}}\mu(d)2^{\frac nd}+\sum_{d\mid \frac{n}{p^k}}\mu(pd)2^{\frac n{pd}}=\sum_{d\mid \frac{n}{p^k}}\mu(d)\left(2^{p^k}\right)^{\frac n{p^kd}}-\sum_{d\mid \frac{n}{p^k}}\mu(d)\left(2^{p^{k-1}}\right)^{\frac n{p^kd}}\equiv $
$\displaystyle \equiv \sum_{d\mid \frac{n}{p^k}}\mu(d)\left(2^{p^{k-1}}\right)^{\frac n{p^kd}}-\sum_{d\mid \frac{n}{p^k}}\mu(d)\left(2^{p^{k-1}}\right)^{\frac n{p^kd}}\equiv 0\pmod{p^k}$

Notice that $a_{p^k}=2^{p^k}-2^{p^{k-1}}=2^{p^{k-1}}(2^{p^{k-1}(p-1)}-1)$ which is divisible by $p^{k}$  for $k \ge 0$ by Euler. Then show $a_{mk}=a_ma_k$ for $m,k$ rel. prime by expanding the sum of $a_d$ over all divisors  $d$ of $mk$. Thus $n|a_n$ for all $n$.

I don't think $a_{mk}=a_ma_k$ is necessarily true?

Anyways here is a sol by me and [b]vincenthuang75025[/b].

By Mobius inversion, we have that $a_n=\sum_{d\mid n} \mu (d)\cdot 2^{n/d}$. Let $n=\prod p_i^{e_i}$ have $k$ prime divisors, $n'=\dfrac{n}{\prod_{p\mid n} p}$. Finally, define $X=2^{n'}$. Then $\displaystyle a_n=\sum_{S} (-1)^{k-|S|}X^{\prod_{p\in S} p}$ where $S$ is a subset of $\{p_1,...,p_k\}$. Then note that $X^{ap_i}\equiv X^a\pmod{p_i^{e_i}}$ (note this follows since $X=2^b$ where $p_i^{e_i-1}\mid b$). On the other hand, $a_n=\sum_{S} (-1)^{k-|S|}\left(X^{\prod_{p\in S} p}-X^{p_i\cdot \prod_{p\in S} p}\right)$ where $S$ is a subset of $\{p_1,...,p_k\} \slash \{ p_i \}$. Then we see that $p_i^{e_i}$ divides $a_n$. Considering this for all primes that divide $n$, we are done.

lol

Note that $a_n$ is the number of elements of $\mathbb{F}_{2^n}$ that are not elements of $\mathbb{F}_{2^m}$ for any $m<n$. Now, group those elements by minimal polynomial and note that all of those polynomials have degree $n$, so we are immediately done.

First we have $a_{p^k} = 2^{p^k} - 2^{p^{k-1}} \equiv 0 \pmod{p^k}.$ Now assume that $p^k|a_{sp^k}$ for all $1 \le s \le m-1$. Now note that
$$a_{mp^k} = 2^{mp^k} - \sum_{d|mp^k, d \neq mp^k} a_d = 2^{mp^k} - \sum_{d|mp^{k-1}} a_d - \sum_{d|m, d \neq m} a_{p^k \cdot d}. $$
The hypothesis gives $$p^k|\sum_{d|m, d \neq m} a_{p^k \cdot d}$$ and because of the condition $$\sum_{d|mp^{k-1}} a_d  = 2^{mp^{k-1}}.$$
Therefore we have $a_{mp^k} \equiv 2^{mp^k} - 2^{mp^{k-1}} \equiv 0 \pmod{p^k}$.

Thus by induction we proved that $p^k|n$ implies $p^k|a_n$ which finishes the problem.

By Möbius inversion, $a_n=\sum_{d\mid n}^{} \mu(\frac{n}{d})2^d$. It is not hard to show that $a$ is multiplicative. From here, it suffices to consider $n=p^e$ for some prime $p$ and non negative $e$. Observe that $$\sum_{d\mid n}^{} \mu(\frac{p^e}{d})2^{d} = 2^{p^e}-2^{p^{e-1}}\equiv2^{p^{e-1}}-2^{p^{e-1}}\equiv 0 \mod p^e$$ and the conclusion follows. $\square$

A detailed proof with a detailed motivation can be found on https://journeyinmath.wordpress.com/2017/07/25/i-learned-how-to-do-addition/.

[quote=bora_olmez][quote=orl]Define sequence $ (a_n)$ by $ \sum_{d|n} a_d \equal{} 2^n.$ Show that $ n|a_n.$[/quote]

How does one prove multiplicity here? Could someone please spell it out. Thanks![/quote]

Sorry for double posting but I have another question.
Could someone possibly write the first couple (maybe 10) terms of the sequence because I don't actually think that the sequence I have is multiplicative.
e.g, $a_2 = 2$, $a_3 = 6$ but $a_6 = 54$ therefore $a_6 = a_2 \cdot a_3$ does not hold, yet $gcd(2,3)=1$ obviously.

@#12,#13: You are completely right, the sequence is not at all multiplicative. Basically (by Möbius inversion) this would be equivalent to $2^n$ being multiplicative which is of course wrong).
So the solutions in #4, #10, #11 are not correct!


Solve in $\mathbb{N}$:

$6xy+6xz+6yz=5xyz$

[hide]$6xy + 6xz + 6yz = 5xyz$ is rewritten as $\frac 6x + \frac 6y + \frac 6z = 5$ then several applications of Simon's Factoring trick will yield the different solutions. Some I've found are $(2, 12, 4), (2, 6, 6), (3, 4, 4), (3, 6, 3)$ and all their permutations.[/hide]


geometry
incenter
circumcircle
cyclic quadrilateral
geometry unsolved

Lemma: Let ABC a triangle and I its incenter and AI meet the circumcircle at D. So DI = DA = DB. Proof of the Lemma: only angle chasing.
Proof: $ E: AC \cap BD$, $ J: AL \cap BD$, $ K: DM \cap AC$, AJ meet the circumcircle at P and LM meet BD in I and AC in N.
Clearly $ \angle AJE \equal{} \angle AKD$ and for the Lemma PL = PM and $ \angle GML \equal{} \angle GLM$ and so considerng the triangle $ \triangle ILJ$ and $ \triangle KMN$ we have that $ \angle JIL \equal{} \angle MNK$ but $ \angle KML \equal{} 90 \minus{} \angle JIL \equal{} 90 \minus{}  \angle MNK \equal{} \angle KLM$.

[quote="nguyentrang"]In cyclic quadrilateral $ ABCD$, denote by $ L,M$ the incenters of triangles $ BCA,BCD$ respectively. The perpendiculars from $ L,M$ to the lines $ AC,BD$ respectively intersect at $ R$. Show that the triangle $ LMR$ is isosceles.[/quote]
[hide="Solution"]
Let $ LR$ meet $ AC$ at $ P$ and $ MR$ meet $ BD$ at $ Q$.  Notice that \[ \angle BLC \equal{} 90 \plus{} \frac {\angle BAC}{2} \equal{} 90 \plus{} \frac {\angle BDC}{2} \equal{} \angle BMC\] since $ ABCD$ is cyclic and $ L$ and $ M$ are incenters, so $ LMCB$ is cyclic.  Now, \[ \angle BLP \equal{} \angle CLP \plus{} \angle BLC \equal{} 90 \plus{} \frac {\angle BAC}{2} \plus{} 90 \minus{} \angle LCB \equal{} 180 \plus{} \frac {\angle BAC}{2} \minus{} \frac {\angle ACB}{2}\]  Similarly, we can find that $ \angle QMC \equal{} 180 \plus{} \frac {\angle BDC}{2} \minus{} \frac {\angle DBC}{2}$.  Since $ BLMC$ is cyclic, $ \angle BLM \equal{} 180 \minus{} \angle MCB \equal{} 180 \minus{} \frac {\angle BCD}{2}$, so \[ \angle RLM \equal{} \angle PLB \minus{} \angle BLM \equal{} \frac {\angle BCD \plus{} \angle BAC \minus{} \angle ACB}{2}\] and we can find that $ \angle RML \equal{} \frac {\angle BDC \plus{} \angle ABC \minus{} \angle DBC}{2}$.  Now, \[ RM \equal{} RL\iff \angle RLM \equal{} \angle RML\iff \frac {\angle BDC \plus{} \angle ABC \minus{} \angle DBC}{2} \equal{} \frac {\angle BCD \plus{} \angle BAC \minus{} \angle ACB}{2}\]  Since $ \angle BDC \equal{} \angle BAC$, we have that we must show that \[ \angle ABC \minus{} \angle DBC \equal{} \angle BCD \minus{} \angle ACB\iff \angle ABC \plus{} \angle ACB \equal{} \angle BCD \plus{} \angle DBC\iff 180 \minus{} \angle BAC \equal{} 180 \minus{} \angle BDC\iff \angle BAC \equal{} \angle BDC\] which is true since $ ABCD$ is cyclic.  Hence, $ \triangle RLM$ is isoceles.  [/hide]


Hi,

How would I draw a diagram like [img]http://upload.wikimedia.org/wikipedia/commons/6/62/SnakeLemma01.png[/img] on these forums?

Thanks!

Use the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=diagxy]diagxy package[/url] - see [url=http://sixthform.info/steve/wordpress/?p=28]here[/url] for examples.

Hi,

btw. if you want to draw commutative diagrams with LaTeX in general you could also use the xy-pic package or amscd, both packages are providing macros for commutative diagrams.
At the moment I'm preferring TikZ, even though it's a huge graphicx package providing a lot of different features, but that can be also an advantage. With TikZ you could define nodes, edges with arrows, matrices of math nodes, chains of nodes and more. Have a look at this blog posting containing a short demonstration: [url=http://texblog.net/latex-archive/maths/tikz-commutative-diagram-edges-over-under/]TikZ: Commutative diagrams with crossing edges[/url] or here: [url=http://texblog.net/latex-archive/maths/pgf-tikz-commutative-diagram/]Chains with labeled edges[/url].

Stefan

So how exactly would one use these packages on these forums? Also, are there any guides for how to code in these packages?

You can't use any packages on the forum that haven't already been loaded. They are for use in a document. You could
1. use the built-in LaTeX picture environment, for example
$ \setlength{\unitlength}{0.5cm}
\begin{picture}(5,2) \put(5,0){\vector( \minus{} 1,0){5}}
\put(0,0){\vector(1,1){2}}
\put(2,2){\vector(3, \minus{} 2){3}}
\end{picture}$
(for more on this see the picture environment in [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=lshort-english]A (Not So) Short Introduction to LaTeX2e[/url]
or
2. draw the picture outside the forum and copy it here (as you did in your post)
or
3. use Asymptote which is supported by the forum

here's a nice overview of several diagram drawing packages for latex http://www.jmilne.org/not/CDGuide.pdf including (tikz, diagxy, diagrams, xymatrix)

my personal preference is the diagrams package for simple diagrams, and xymatrix package for more complicated ones.

[quote]Hey shallow, I can't really follow the link you gave. I scrolled down to diagrams and clicked on the link it gave to download the "macros" for the package diagrams. What do I do with the .tex it gave me? :\[/quote]
I believe the link you are referring to is http://www.paultaylor.eu/diagrams/

From there, you should download the most recent macro file, at the present time it is version 3.93. Copy this file to your macro directory (this will depend on your implementation of TeX).

I realize this probably doesn't answer your question, but it really depends on your brand of TeX, try looking for a 'macros' or 'misc' directory under your TeX directories. If all else fails, I believe you can just rename the file to 'diagrams.sty' and put it in the same folder as the .tex file you are writing.


What is the LPP of the set {-3, -2, -1, 0, 1, 2, 3}?

Correct, although I have no clue why you knew the problem asked for the product of three numbers and not two but oh well.

What will the answer be if you can choose as many numbers as you want?

good point i have no clue either. which year is that from?
the second question is -36.

:lol:, correct.

This is from the Number Sense Stretch in the 2003-2004 MATHCOUNTS School Handbook.

Think about it..."Great to be here..."

Hint: MAML Olympiad

Answer: [hide]Husetts[/hide]

Hehe.  I live right near him, in Massac.  Although I'm at college, so I don't know if that counts.  Really I live in NY.  But you weren't asking about me.  Never mind.

Nope, I'm at Harvard.  I don't actually have to declare a major until the end of this year or next, but barring unexpected happenings, I'm planning to major in math.


The ethyl acetate concentration in an alcoholic solution was determined by dilutiing a 10.00-mL sample to exactly 100 mL. A 20.00-mL portion of the diluted solution was refluxed with 40.00 mL of 0.04672 M KOH. The reaction proceeds as follows:
[b]                   CH3COOC2H5 + OH^-1 ---> CH3COO^-1 + C2H2OH.[/b]
After cooling, the excess OH^-1 was back-titrated with 3.41 mL of 0.05042 M sulfuric acid. Calculate the grams of ethyl acetate per 100 mL of the original sample.

c",)

mole H+ =mole OH-
we can find excess KOH from hydrolysis reaction
and then find gram of ester convert it into ratio in original sample END


algebra
polynomial
linear algebra
matrix
induction
linear algebra unsolved

If $p$ is a prime number and $A,B\in\mathcal l M_n(\Re)$ such that $(AB-BA)^p=I_n$ then $p|n$.

It is equivalent to proving that if $\epsilon_0,\epsilon_1...\epsilon_{n-1}$ are roots of the equations of $x^p=1$ such that $\epsilon_0 +\epsilon_1+...+\epsilon_{n-1}=0$ then $p|n$ or at least i think so.

I have found the problem`s original text: Let $p$ be a prime number. Prove that $\exists A,B\in \mathcal l M_n(\Re)$ such that $(AB-BA)^p=I_n$ if and only if $p|n$.

I would very much appreciate a sollution or some hints into proving this problem :blush: .

A hint from the proposer ;) : for one implication, use the irreducibility of the polynomial $1+X+X^2+...+X^{p-1}$ in the ring of polynomials with coefficients in $Q$. For the other implication, use the fact that for any matrix $A$ with trace 0 there are matrices $X,Y$ such that $A=XY-YX$. In order to prove the last assertion, prove by induction that a matrix with zero trace is similar to a matrix with nul diagonal.

For the one-side implication:
as Harazi said, if we denote by $\alpha=cos(\frac{2\pi}{p})+isin(\frac{2\pi}{p})$-the primitive $p$'th root of unity, then we can write that the eigenvalues of $C=AB-BA$, $\epsilon_i=\alpha^{k_i}$, for $i=0,...n-1$ with some integer $0 \leq k_i$, the as $trC=0$ we can conclude that the polynomial $f(x)=\sum_{i=0}^{n-1}x^{k_i}$ vanishes at $\alpha$. From the other hand since $\alpha$ is a root of $g(x)=X^{p-1}+...+1$ which is irredicuble in $\mathbb{Z}[X]$ $g(x)$is a minimal polynomial of $\alpha$ over $\mathbb{Z}[X]$, thus $g(x) | f(x)$ and $n=f(1)=kp=kg(1)$.
Am i right?

sorry if this is a silly qn, but i'm thinkin how the soln to this problem will ever exist. Trace of (AB-BA)^p is always 0, how can it equal a matrix whose trace is n? or may b i got it wrong, from which field are u takin the entries of the matrix?

$tr((AB-BA)^p)$ is not the same with $(tr(AB-BA))^p$ example $\left ( \begin{array} {cc} -1&0\\0&1 \end{array} \right )$


Αν η συνάρτηση $ f$ είναι συνεχής στο $ [a,b]$ παραγωγίσιμη στο $ (a,b)$ με $ f(a)\equal{}f(b)\equal{}0$ να δείξετε ότι υπάρχει $ j$ που ανήκει στο $ (a,b)$ τέτοιο ώστε $ f'(j)\equal{}kf^2(j)$ για κάθε πραγματικό αριθμό $ k$

Den eimai sigouros oti katalava to provlhma swsta...Efoson les gia ka8e pragmatiko k, mporw na 8esw k=0, opote pairnw f'(j)=0.
Estw f(x)=x^2-x. Einai sunexhs sto [0,1] ws poluwnimikh, einai paragwgisimh sto (0,1) me paragwgo f'(x)=2x-1 kai isxuei f(0)=f(1)=0. Prepei na exw f'(j)=0, ara 2j-1=0 ara j=1/2, pou anhkei sto (0,1) alla gia j=1/2 kai k diaforo tou mhden h sxesh p egrapses den isxuei!
Ekana kapou la8os?Mhpws den katalava swsta to provlhma?

Rolle στην f(x).e^[(-k).(ololk. απο α εως χ f(t)dt)] sto [α,b]

[quote="xp"]Rolle στην $ h(x)\equal{}f(x)e^{(\minus{}k)\int_{a}^x f(t)dt}$ sto [α,b][/quote]

Nice  :lol:

Μπορεί κάποιος να βοηθήσει στις ασκήσεις;
1)Έστω η συνάρτηση $ f$ που είναι συνεχής στο$ [0,1]$  . Να βρείτε το $ \lim_{t \to \infty } \int^{1}_{0}f(x)x^tdx$
2)Έστω οι συναρτήσεις $ f$ και $ g$  που είναι συνεχείς στο$ A$ με $ [k,l]<\equal{}A$  $ [a,b]<\equal{}A$
Αν $ \int^{b}_{a}f(x)dx>\int^{b}_{a}g(x)dx$ και $ \int^{l}_{k}f(x)dx<\int^{l}_{k}g(x)dx$ να δείξετε ότι οι $ f$ και $ g$ έχουν ένα τουλάχιστον κοινό σημείο

Για την πρώτη σκέψου το θεώρημα μέγιστης και ελάχιστης τιμής (αφού η $ f$ είναι συνεχής στο $ [0,1]$) και το όριο βγαίνει 0.

Για την δεύτερη δεν καταλαβαίνω τί σημαίνει $ [k,l]<\equal{}A$ $ [a,b]<\equal{}A$. Μήπως ήθελες να γράψεις κάτι άλλο? Μήπως το $ \leq$ που βάζεις σημαίνει υποσύνολο, δηλαδή $ [k,l]\subseteq A$ και  $ [a,b] \subseteq A$ ;

Παρεπιπτώντως το υποσύνολο στo $ LaTeX$ γράφεται με την εντολή \subseteq

Αλέξανδρος

Aπο ΘΜΤΟΛ υπάρχουν χ1 και χ2 ώστε (β-α)(f(χ1)-g(χ1))>0 kai (k-l)(f(x2)-g(x2))<0 kai κατοπιν Θ.Βοizano....

[quote="mostel"][quote="xp"]Aπο ΘΜΤΟΛ ...[/quote]

Πρέπει να 'ναι εκτός ύλης[/quote]
H απόδειξη του ΘΜΤΟλ είναι πολύ απλή. Κάνε ΘΜΤ σε μια παράγουσα Αν παρόλα αυτά δεν θέλεις πήγαινε με "άτοπο" θα βγει πολύ εύκολα!

Cretaman δεν καταλαβαίνω την λύση σου στην πρώτη άσκηση. Μπορείς να μου την εξηγήσεις;
Όσο για την απόδειξη του ΘΜΤΟΛ δεν νομίζω ότι είοναι δύσκολη στην απόδειξή της.

Αφού η $ f$ είναι συνεχής στο $ [0,1]$ άρα έχει μέγιστη και ελάχιστη τιμή $ M, m$ αντίστοιχα οπότε $ m\leq f(x) \leq M$ για κάθε $ x\in[0,1]$.

Οπότε $ mx^t \leq f(x) x^t \leq Mx^t$ ($ x^t\geq 0$ στο $ [0,1]$) και ολοκληρώνοντας στο $ [0,1]$ παίρνουμε

$ m\frac {1}{t \plus{} 1} \leq \int_0^1 f(x) x^t dx \leq M \frac {1}{t \plus{} 1}$ και $ \displaystyle\lim_{t\to \plus{} \infty} \frac {1}{t \plus{} 1} \equal{} 0$,

άρα από το κριτήριο παρεμβολής έχουμε  $ \displaystyle\lim_{t\to \plus{} \infty} \int_0^1 f(x) x^t dx \equal{}0$.

Αλέξανδρος

[quote="mostel"]Δε μιλάω περί δυσκολίας στην απόδειξη. Απλώς δε πιστεύω να μπει κάτι πανελλαδικές που να λύνεται μόνο με ΘΜΤΟΛ

Στέλιος[/quote]

 [i]Εκτός και αν το βάλουν ως πρώτο υποερώτημα και ακολουθήσουν τα άλλα ως εφαρμογές. Θα πρόκειται βέβαια για διαστροφή , αν σκεφτεί κανείς πόσες χιλιάδες ωραία πράγματα υπάρχουν για τις εξετάσεις.
 Αλλά σε αυτή τη χώρα , με όσα βλέπετε  , μην αποκλείετε τίποτα ! Γι αυτό  , αν κυνηγάτε το άριστα , καλό είναι να ρίξετε μια ματιά στην απόδειξη του ΘΜΤΟΛ και σε δυο εφαρμογούλες , ώστε αν χρειαστεί , να μην χαθούν 5-6 μοριάκια.(Στο σχολείο βέβαια ποτέ δεν το λέω!)

Μπάμπης[/i]

Αν $ f(x)\equal{}\int^{b}_{a}sin(2tx)dt$ τότε να δείξετε ότι $ f'(x)\equal{} 2\int^{b}_{a}cos(2tx)tdt$

[quote="gregg"]Αν $ f(x) \equal{} \int^{b}_{a}sin(2tx)dt$ τότε να δείξετε ότι $ f'(x) \equal{} 2\int^{b}_{a}cos(2tx)tdt$[/quote]
Προκύπτει άμεσα από την "παραγώγιση κάτω από το σύμβολο του ολοκληρώματος" αλλά αυτό βρίσκεται στην ύλη του Α έτους. Αλλιώς με τον ορισμό, τριγωνομετρική ταυτότητα, αλλαγή μεταβλητής και ΘΜΤΟΛ

Να αποδείξετε ότι:
$ \int 1/(1\plus{}t^20^n dt\equal{}t/2(n\minus{}1)(1\plus{}t^2)^(n\minus{}1)\plus{}(2n\minus{}3)/2(n\minus{}1)\int 1/(1\plus{}t^2)^(n\minus{}1)dt$


[i]Let a,b,c ,d be positive real numbers such that a+b+c+d=4
  Prove that
           $ a^2bc\plus{} b^2cd\plus{}c^2da\plus{}d^2ab \leq 4$[/i]
   Y An Forever

[quote="nkht-tk14"][i]Let a,b,c ,d be positive real numbers such that a+b+c+d=4
  Prove that
           $ a^2bc \plus{} b^2cd \plus{} c^2da \plus{} d^2ab \leq 4$[/i]
   Y An Forever[/quote]

$ d \leq c \leq b \leq a \minus{} > bcd \leq acd \leq abd \leq abc$

$ a^2bc \plus{} b^2cd \plus{} c^2da \plus{} d^2ab \leq (ab \plus{} bc)(ac \plus{} bd) \leq \frac{(ac \plus{} ab \plus{} bc \plus{} bd)^2}4 \leq 4$

[quote="nkht-tk14"][i]Let a,b,c ,d be positive real numbers such that a+b+c+d=4
  Prove that
           $ a^2bc \plus{} b^2cd \plus{} c^2da \plus{} d^2ab \leq 4$[/i]
   Y An Forever[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=100960

[quote="Algadin"][quote="Greenstyle."]

$ d \leq c \leq b \leq a$[/quote]
[color=red]we can't have it. Okie[/color]
[size=150][color=blue]you are chicken!!![/color][/size] :rotfl:  :rotfl:[/quote]

sorry

Suppose that $ (x,y,z,u)$  is a permutation of $ (a,b,c,d)$ and $ u \leq z \leq x \leq y$
$ x^2yz \plus{} y^2zu \plus{} z^2ux \plus{} u^2xy \leq (xy \plus{} yz)(xz \plus{} yu) \leq \frac{(xy \plus{} yz \plus{} xz \plus{} yu)^2}4 \leq 4$


number theory
greatest common divisor
function
relatively prime
number theory unsolved

Let $ a;n$ be integer positive numbers;$ a>1;n>1$ satisfying $ n|a^n\minus{}1$
Prove that
$ gcd(n;a\minus{}1)>1$

[quote="chien than"]Let $ a;n$ be integer positive numbers;$ a > 1;n > 1$ satisfying $ n|a^n \minus{} 1$
Prove that
$ gcd(n;a \minus{} 1) > 1$[/quote]
Let $ p|n$ $ p\in \mathbb P$ ,and $ p$ min,$ h\equal{}ord_p (a)$thus $ h|gcd(n;p\minus{}1)$
Then $ p\equal{}2$ and $ 2|a\minus{}1$ it is obivious to find $ gcd(n;a\minus{}1)\ge 2$Done :wink:

sorry,i have some mistake in my last post,and now i will fix it
Since $ h|gcd(n;p\minus{}1)$ we have $ h\equal{}1$thus $ p|a\minus{}1$ because $ n\ge 2$ so $ gcd(n;a\minus{}1)>1$done
Now i think my proof is true :blush:
[quote="Fermat's Little Turtle"]What are the most useful properties of the order of numbers ?[/quote]
I cant understand your question(order of number is very useful)

[quote="Fermat's Little Turtle"]You don't understand it? :S
If it is so useful, could you explain the reasoning behind it. How it is used, etc.[/quote]
Oh guy,
if $ p|a^n,p|a^k$ thus $ ord_p(a)|gcd (n,k)$.Do I understand right your queston

Order is a concept that comes naturally out of this question.  The order of an integer $ a$ relative to another integer $ p$ relatively prime to it is the smallest integer $ h$ such that $ a^h \equiv 1 \bmod p \Leftrightarrow p | a^h \minus{} 1$, and in fact if $ p | a^k \minus{} 1$ then it follows immediately that $ h | k$ from some straightforward reasoning with groups.

In general, computing $ \text{ord}_p(a)$ is difficult.  Its most straightforward property is that it divides the [url=http://en.wikipedia.org/wiki/Carmichael_function]Carmichael function[/url] $ \lambda(p)$.

[quote="Allnames"]
Since $ h|gcd(n;p \minus{} 1)$ we have $ h \equal{} 1$
[/quote]
 why? for example $ n\equal{}10,p\equal{}5$ than $ 2|gcd(10,4)$

[quote="Jure the frEEEk"][quote="Allnames"]
Since $ h|gcd(n;p \minus{} 1)$ we have $ h \equal{} 1$
[/quote]
 why? for example $ n \equal{} 10,p \equal{} 5$ than $ 2|gcd(10,4)$[/quote]
No,$ p$ is smallest divisor of $ n$,you can see above


$ x$ and $ y$ are real numbers.
$A)$ If it is known that $ x \plus{} y$ and $ x \plus{} y^2$ are rational numbers, can we conclude that $ x$ and $ y$ are also rational numbers.
$B)$ If it is known that $ x \plus{} y$ , $ x \plus{} y^2$ and $ x \plus{} y^3$ are rational numbers, can we conclude that $ x$ and $ y$ are also rational numbers.

a) No, for example $ x\equal{}\minus{}\frac{\sqrt5}2$ and $ y\equal{}\frac{1\plus{}\sqrt5}2$.

b) Assume there exist irrational numbers $ x,y$ such that $ x\plus{}y,x\plus{}y^2,x\plus{}y^3$ are rational numbers. $ (x\plus{}y^2)\minus{}(x\plus{}y)\equal{}y^2\minus{}y$ is rational, let $ y^2\minus{}y\equal{}a$, $ a\in\mathbb{Q}$. We have $ y\equal{}\frac{1\pm\sqrt{1\plus{}a^2}}2$. So $ \sqrt{1\plus{}a^2}$ is irrational. But
\[ (x\plus{}y^3)\minus{}(x\minus{}y)\equal{}y^3\minus{}y\equal{}\frac{(4\plus{}3a^2)(1\pm\sqrt{1\plus{}a^2})}8\minus{}\frac{1\pm\sqrt{1\plus{}a^2}}2\equal{}\frac{3a^2(1\pm\sqrt{1\plus{}a^2})}8.\]This number must be rational, let $ \frac{3a^2(1\pm\sqrt{1\plus{}a^2})}8\equal{}q$, $ q\in\mathbb{Q}$. So $ \pm\sqrt{1\plus{}a^2}\equal{}\frac{8q}{3a^2}\minus{}1$. Left hand side is irrational, right hand side is rational, we reach a contradiction.

Thus, if $ x\plus{}y,x\plus{}y^2,x\plus{}y^3$ are rational numbers, we can conclude that $ x$ and $ y$ are also rational numbers.

[quote="theSA"]$ x$ and $ y$ are real numbers.
A-)if it is known that $ x \plus{} y$ and $ x \plus{} y^2$ are rational numbers, can we conclude that $ x$ and $ y$ are also rational numbers.[/quote]

No : $ x\equal{}\minus{}\frac{\sqrt{2}}{2}$ and $ y\equal{}\frac{1\plus{}\sqrt{2}}{2}$ $ \implies$ $ x\plus{}y\equal{}\frac{1}{2}$ and $ x\plus{}y^2\equal{}\frac{3}{4}$

[quote="theSA"]$ x$ and $ y$ are real numbers.
B-)if it is known that $ x \plus{} y$ , $ x \plus{} y^2$ and $ x \plus{} y^3$ are rational numbers, can we conclude that $ x$ and $ y$ are also rational numbers.[/quote]

Yes :
$ A\equal{}(x\plus{}y^2)\minus{}(x\plus{}y)\equal{}y(y\minus{}1)$ is rational.
$ B\equal{}(x\plus{}y^3)\minus{}(x\plus{}y^2)\equal{}y^2(y\minus{}1)$ is rational.

If $ A\equal{}0$, $ y\equal{}0$ or $ y\equal{}1$ and $ y$ is rational, so is $ x\equal{}(x\plus{}y)\minus{}y$
If $ A\neq 0$, $ \frac{B}{A}\equal{}y$ is rational, so is $ x\equal{}(x\plus{}y)\minus{}y$

.... But I dont think these problems need to be in an olympiad section ...

Yeah, this one is quite easy. I'm sure I have seen this problem (or a very similar one) before. :D

a)No, $x=2-\sqrt{2},y=\sqrt{s}+\frac{1}{2}$
b)$y^3-y$ is rational and $y^2-y$ is rational.
$y^3-y=y(y^2-y+1)\in\mathbb{Q},y^2-y+1\in\mathbb{Q}$, so $y\in\mathbb{Q}$ and from $(x+y)\in\mathbb{Q}$ we get $x\in\mathbb{Q}$.


inequalities
geometry
trigonometry
trig identities
Law of Cosines
geometry proposed

For a convex hexagon $ ABCDEF$ with an area $ S$, prove that:
\[ AC\cdot(BD\plus{}BF\minus{}DF)\plus{}CE\cdot(BD\plus{}DF\minus{}BF)\plus{}AE\cdot(BF\plus{}DF\minus{}BD)\geq 2\sqrt{3}S
\]

very nice and difficult probem. in the official solution there is proved a lemma which I didn't know and I didn't solve the problem but it's nice and learnable ;)

silver  :( . I could have 25 point without any problem

[quote="thaithuan_GC"]I am very happy if someone gives a solution or hint. Thanks.[/quote]
no, let's better wait till someone else will come up with own solution. I did solve the problem, but it is approximately the same as the official, actually, the lemma used there was in one of our TST's.

Lemma.
Let the incircle of BDF touch DF, FB, BD at X, Y, Z respectively. Then there exists a point O such that $ OB \leq \frac {2}{\sqrt {3}} BY$, $ OD \leq \frac {2}{\sqrt {3}} DZ$, $ OF \leq \frac {2}{\sqrt {3}} FX$.
Proof.
[hide]
Let u=DZ, v=FX, t=BY. It's not hard to see that there exists r such that the circles centered at B, D, F with radii rt, ru, rv have a point in common. Call this point O. We claim $ r \leq \frac {2}{\sqrt {3}}$. Since $ \angle BOD + \angle DOF + \angle FOB = 360$ one of these angles is at least 120, say without loss of generality $ \angle FOD$. Then using the Law of Cosines on $ \Delta FOD$,
\begin{eqnarray*} \frac { - 1}{2} = \cos 120 \geq \cos \angle FOD & = & \frac {OF^2 + OD^2 - FD^2}{2OF\cdot OD} \\
& = & \frac {r^2 v^2 + r^2 u^2 - (u + v)^2}{2uvr^2} \end{eqnarray*}
Now if $ r > \frac {2}{\sqrt {3}}$ this rearranges to $ 2uv > u^2 + v^2$ contradicting the Arithmetic Mean- Geometric Mean inequality. So $ r \leq \frac {2}{\sqrt {3}}$ and O satisfies the conditions.[/hide]
Now,
[hide]
$ \begin{eqnarray*} AC \cdot (BD + BF - DF) & = & 2 AC \cdot BZ \\
& \geq & 2 AC \cdot \frac {\sqrt {3}}{2} BO \\
& \geq & 2 \sqrt {3}\left[ABCO\right]$.
Adding up similar inequalities gives the result.
[/hide]
Some notes on motivation:
[hide]
1. The BD+BF-DF, etc. in the problem reminds me of the length of a tangent to the incircle, so I rewrite it with BZ, etc.
2. I would very much like the left-hand side to be a sum of areas that add to the area of the hexagon. This gives motivation to somehow come up with a point O such that $ AC(BD + BF - DF) \geq 2\sqrt {3} \left[ABCO\right]$; then we could add cyclically.
3. The area of ABCO is at most $ BO \cdot AC$. So this motivates us to come up with the relation in the lemma.
[/hide]

[quote="silversheep"]
3. The area of ABCO is at most $ BO \cdot AC$. So this motivates us to come up with the relation in the lemma.
[/quote]is this also true for non-convex quadrilateral? how to prove it?

[quote]3. The area of ABCO is at most $ BO \cdot AC$. So this motivates us to come up with the relation in the lemma.
is this also true for non-convex quadrilateral? how to prove it?[/quote]
Suppose ABCO is oriented counterclockwise. We use directed angles and areas (counterclockwise is positive). Let BO and AC intersect at P. We have, using A=1/2bh,
$ [ABC]\equal{}\frac{1}{2} AC \cdot BP \sin(\angle APB)$
and
$ [COA]\equal{}\frac{1}{2} AC \cdot OP \sin(\angle CPO)$
Add these together and use that sin is less than 1.

thanks for your reply silversheep, i didn't realize that obvious thing. :blush:

[quote="silversheep"]It's not hard to see that there exists r such that the circles centered at B, D, F with radii rt, ru, rv have a point in common.[/quote]i cannot 'see' it, can you prove it? :oops:

[quote]It's not hard to see that there exists r such that the circles centered at B, D, F with radii rt, ru, rv have a point in common.
i cannot 'see' it, can you prove it? [/quote]
Consider circles centered at B, D, F with radii rt, ru, and rv. As r increases from 0, the circles expand so eventually there will be a point in XYZ on the boundary or in the interior of all three circles. Take the least r such that this is true. Then if all three circles do not have a point in common, there is a point P strictly inside all three circles. Decreasing r by a sufficiently small amount P will still be inside all three circles, contradicting the minimality of r.

[quote=silversheep]
3. The area of ABCO is at most $ BO \cdot AC$. So this motivates us to come up with the relation in the lemma.
[/quote]
Actually, its area is at most $\frac{BO \cdot AC}{2}$ since $max(sin \alpha)=1$, isn't it ?

My solution is pretty much the same as the one in #7, so I'll just expand on the magical point $X$ (and how I came to construct it). Rather than intersecting circles centered at $B$, $D$, and $F$ with radii $rt$, $ru$, $rv$, consider the Apollonius circles of $DF$, $FB$, $BD$ going through $X$, $Y$, and $Z$, respectively. To prove these concur at one point, take the internal intersection of two of these circles, and using the equal ratios, it's clear that this point lies on the third Apollonius circle as well.


Find all pozitive integers x,y,z which satisfy : x divides y+z, y divides x+z, z divides x+y.

Suppose  $x\leq y\leq z$
We know that $z | x+y$ so $x+y=z$ or $x=y=z$
We know that $y | x+z$ so from here we get:
$y | 2x$ so $x=y$ or $2x=y$
All the options sattisfy the condition $x | y+z$
So the solutions are permutations of:
$(x,x,x),(x,x,2x),(x,2x,2x),(x,2x,3x)$ for positive integer $x$

Pavel25: I'm afraid x,2x,2x is not a solution.

Let x=ad, y=bd, z=cd, where a,b,c are pairwise coprime. Then we have
$abc|a+b+c$
For the max M and min m of a,b,c
$Mm^2\leq 3M$ yields m=1, so wlog a=1
$bc|b+c+1$
We have
$bc\leq b+c+1$, so
$(b-1)(c-1)\leq 2$. If the product is 1, the sols are 2 and 2 (doesn't satisfy the divisibility condition) , if the product is 2, the sols are 3 and 2.
so $(x,y,z)\in\{(d,d,d),(d,d,2d),(d,2d,3d) \}$.[/quote]

If x divides y + z, then x divides x + y + z.  Similarly, both y and z divide x + y + z.

So, let a = (x + y + z)/x, b = (x + y + z)/y, and c = (x + y + z)/z, so a, b, and c are positive integers.  Furthermore,
\[ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{x + y + z}{x + y + z} = 1. \]
From here, it's not hard to solve.


modular arithmetic
algebra
binomial theorem
number theory unsolved
number theory

Let $p$ be a prime and $a,b$ be positive integers such that $a \equiv b \pmod p.$ Prove that if $p^x \| a-b$ and $p^y \| n,$ then $ p^{x+y} \| a^n - b^n.$

Hello stephencheng, we've been asked to prove that $ x \plus{} y$ is the highest exponent of p in $ a^{n} \minus{} b^{n}$

There are 2 things that need to be shown:
1. $ p \parallel{} \frac {a^{np} \minus{} b^{np}}{a^n \minus{} b^n}$ for any $ n$

2. $ p \nmid \frac {a^n \minus{} b^n}{a \minus{} b}$ when $ (n,p) \equal{} 1$

If we'll put $ n \equal{} p^ym, (p,m) \equal{} 1$, then the result will follow from the factorization:
$ a^n \minus{} b^n \equal{} (a \minus{} b)\frac {a^m \minus{} b^m}{a \minus{} b}\prod_{k \equal{} 0}^{y \minus{} 1} \frac {a^{mp^{k \plus{} 1}} \minus{} b^{mp^{k \plus{} 1}}}{a^{mp^{k}} \minus{} b^{mp^{k}}}$

We'll start with the 2nd part: $ \frac {a^n \minus{} b^n}{a \minus{} b} \equal{} \sum_{i \equal{} 0}^{n \minus{} 1} a^ib^{n \minus{} 1 \minus{} i} \equiv$
because $ a \equiv b \pmod p$:
$ \equiv \sum_{i \equal{} 0}^{n \minus{} 1} b^ib^{n \minus{} 1 \minus{} i} \equiv \sum_{i \equal{} 0}^{n \minus{} 1}b^{n \minus{} 1} \equal{} nb^{n \minus{} 1} \neq 0 \pmod p$

Now the second proposition: in the same manner like the 2nd part, it follows that $ p | \frac {a^{np} \minus{} b^{np}}{a^n \minus{} b^n}$.
I notice that you forgot to mention that $ (a,p) \equal{} (b,p) \equal{} 1$ and $ x \ge 1$. Anyway, put $ a^n \equal{} s, b^n \equal{} t, s \minus{} t \equal{} h$ and use the binomial theorem to get:

$ \frac {1}{p}\frac {a^{np} \minus{} b^{np}}{a^n \minus{} b^n} \equal{} \frac {s^p \minus{} t^p}{p(s \minus{} t)} \equal{} \frac {(t \plus{} h)^p \minus{} t^p}{ph} \equal{}$
$ \frac {1}{p}\sum_{i \equal{} 0}^{p \minus{} 1} t^ih^{p \minus{} i \minus{} 1} \binom{p}{i} \equiv \sum_{i \equal{} 0}^{p \minus{} 1} t^ih^{p \minus{} i \minus{} 1}i^{ \minus{} 1} \binom{p \minus{} 1}{i \minus{} 1}$
But $ h \equiv 0 \pmod p$, so the only non-zero term in the last sum occurs when $ i \equal{} p \minus{} 1$ and equals $ t^{p \minus{} 1} \equiv 1 \neq 0 \pmod p$ by FLT, because $ (t,p) \equal{} 1$ (because $ (b,p) \equal{} 1$). Thus $ p^1$ is the highest exponent of $ p$ in $ \frac {a^{np} \minus{} b^{np}}{a^n \minus{} b^n}$.

I used the identity $ \binom{p}{i} \equal{} \frac {p}{i}\binom{p \minus{} 1}{i \minus{} 1}$

aadil: Sorry, but would you please stop posting that kind of comment¿ Nothing you posted hasn't already been said before in a much more explicite way. You can post such a thing as a hint if it hasn't been said before, or to give an alternative solution; but not that way, please.


Here is a very famous (and very delightful, too) problem:

Does there exist a continuous function $f: \mathbb{R} \to \mathbb{R}, f \not\equiv 0$, such that
\[
f(x) + f(2x) + f(3x) = 0
\]
for all $x \in \mathbb{R}$?

Have fun!  ;)

--Vesselin

It is a cool problem, and was also posed in a thread called [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=19406]Continuous function[/url].  I think we basically solved the question there, although maybe we missed a more elegant way to show that a certain complex number existed.


LaTeX
inequalities unsolved
inequalities

[quote="EhsanNamdari"]Given a,b,c,d >0 and a+b+c+d = 4;prove that:
∑1/(a^2+1)≥2. :([/quote]

$ 0 < x < 4$

we have

$ \frac {1}{x^2 \plus{} 1}\geq 1 \minus{} \frac {x}{2} \Longleftrightarrow \frac {x(x \minus{} 1)^2}{2(x^2 \plus{} 1)}\geq 0$

so

$ \sum{\frac {1}{a^2 \plus{} 1}}\geq 4 \minus{} \frac {a \plus{} b \plus{} c \plus{} d}{2}\equal{} 4 \minus{} 2 \equal{} 2$.

Hi :D  !
Try to use LaTeX .
[hide]$ \Leftrightarrow \sum \frac{a^2}{a^2\plus{}1}\leq 2$

But we have $ \sum \frac{a^2}{a^2\plus{}1} \leq  \sum \frac{a^2}{2a}\equal{}\sum \frac{a}{2}\equal{}2$[/hide]


Diophantine equation
number theory unsolved
number theory

Prove that the equation $ x^3+y^3=z^4-t^2 $ has an infinite number of solution in the  set of positive integer numbers so that $ gcd(x,y,z,t)=1 $

I believe you can construct such a sequence using the identity

\[(1 + 2 + ... + n)^2 = 1^3 + 2^3 + ... + n^3 = \left(\frac{n(n + 1)}{2}\right)^2\]

Indeed, take, take $t = \frac{(n - 2)(n - 1)}{2}$, $x = n - 1$ and $y = n$.

If we take $n$ such that $\frac{n(n + 1)}{2}$ is a square, we are done, since then we can take $z = \sqrt{\frac{n(n + 1)}{2}}$.

Now, it is easy to see that there exist infinitely many $n$ such that $\frac{n(n + 1)}{2}$  is a square since the Pell equation $(2n + 1)^2 - 2(2k)^2 = 1$ has infinitely many solutions in integers. Hence we are done.

i posted a similar problem :

  http://www.artofproblemsolving.com/Forum/viewtopic.php?p=307296#p307296   :!:


calculus
integration
trigonometry
calculus computations

Evaluate

\[\int_0^{\frac{\pi}{4}} \frac{x\cos 5x}{\cos x}dx\]

$-\frac{\pi}{4}+\frac{\pi^2}{32}+\frac{1}{4}$

However (as usual) I dont have a very elegant solution I used the fact that
[hide]$\cos(5x)=16 \cos(x)^5-20 \cos(x)^3+5 \cos(x)$[/hide]


function
logarithms
real analysis
calculus
integration
Functional Analysis
real analysis unsolved

I need to find a function that is in L^1 but not in L^p for p not equal to 1.

Let's say we're working on the line.

Try $f(x)=\frac1{x\ln^{2}x}$ for $0<x<\frac12$ and zero otherwise.

This is in $L^{1}(\mathbb{R})$ but not in $L^{p}$ for $p>1.$

You ought to be able to construct a similar example in any non-atomic measure space (every point has measure zero).

With that example, I was only concerned with the Banach space $L^{p}$ spaces - that is, with $p\ge 1.$ If you're also considering $p<1,$ then try this example on $(0,\infty):$

$g(x)=\frac1{x(1+\ln^{2}x)}.$

Can someone provide an example that is more constructive and one that helps explain the idea behind what is going on here.

Hmmmm - I thought I was being constructive. Those are specific examples. (And no one is going to give you anything that's significantly simpler - these are as simple as it's going to get.) Learning such subjects as Lebesgue measure and integration does not mean that we ever abandon our knowledge of calculus.

Suppose $p>1.$ Note that for $|y|<1,\ |y|^{p}<|y|$ That means that in places where $|f|$ is small, it is easier for $f$ to be in $L^{p}$ than in $L^{1}.$ But if $|y|>1,\ |y|^{p}>|y|.$ Hence where the values of the function are large, it is harder for $f$ to be in $L^{p}$ than in $L^{1}.$

So, for $p>1,$ to find a function that is in $L^{1}$ but not $L^{p}$, we need only concern ourselves with a local example (whose important action takes place in a bounded region) that goes to infinity. It has to go to infinity slowly enough to be integrable - but just barely so.

We understand that $\frac1x$ is the borderline case for integrability of powers of $x.$ The function $\frac1x$ is itself not integrable in a neighborhood of zero, and $\frac1{x\ln x}$ is still not integrable, but $\frac1{x\ln^{2}x}$ is - just barely - integrable. Raise that to any power $p>1$ and you have $\frac1{x^{p}\ln^{2p}x}$ which is adversely comparable to $\frac1{x^{s}}$ for $1<s<p$ and thus not integrable.

The second example employed a related idea for dealing with $p<1$ and small values of the function on unbounded intervals.

Basic knowledge of sizes of functions, common estimates, and straightforward calculus: it's fundamental to the study of analysis.


The length of the perimeter of a right triangle is 60 inches and the length of the altitude perpendicular to the hypotenuse is 12 inches. Find the length of the sides.

[hide="hint"]set up equations and solve:
$a^{2}+b^{2}=c^{2}$
$a+b+c=60$
$ab/2=6c$[/hide]


Find $ a,b$ such that $ Max|\sqrt {x} \plus{} ax \plus{} b|$ is smallest possible, where $ a,b$ are 2 real numbers and $ x \in [0;16]$

[quote="lalalala"]Find $ a,b$ such that $ Max|\sqrt {x} \plus{} ax \plus{} b|$ is smallest possible, where $ a,b$ are 2 real numbers and $ x \in [0,16]$[/quote]

1) your multi-account postings aren't funny

2) let $ u \equal{} \sqrt {x}$

so this problem reduces to finding the smallest value of $ Max|au^2 \plus{} u \plus{} b|$ with $ u \in [0,4]$

it's a quadratic; it's not hard, but I'm leaving for HMMT in a couple minutes, so I'll leave the rest up to somebody else.

If I'm not mistaken, all of these problems posted by lalala and clone accounts are from an ongoing Vietnamese competition and should be locked.


[quote="zabelman"]What is the IMC?[/quote]

Here's the website: http://www.imc-math.org/

Probably because it's in Europe, and it's relatively expensive to get to Europe from America but not so expensive to move around Europe.  (Rail is cheap.)

Hm, but still it is the "international mathematics competiton for undergraduates" you'd think it'd look nice or something on a CV and that people would be willing to sponsor US participation.

Oh well, I can understand your reasoning.

[quote="Idiot_without_a_village"]Anyone have any information regarding US participation in the IMC and why it's virtually nonexistant?[/quote] there is already a topic, can't give you a link right now, but it's in the college playground. you can find out more there.

and american universities DO send teams to participate. Princeton will send a Romanian team (as usual :P) and I can't say more about others ... but there is at least one doing it :P

This is the link to the place in the forum where IMC is discussed
http://www.mathlinks.ro/Forum/index.php?f=79


The ratio of $ x$, $ y$, and $ z$ is $ x: y: z \equal{}2: 3: 5$. Also $ x\plus{}y\plus{}z\equal{}100$. If $ y\equal{}ax\minus{}10$, find the value of $ a$.

[hide]
$ x: y: z\equal{}2: 3: 5$.
$ x\plus{}y\plus{}z\equal{}100$.
Hmmm... $ \{20,30,50\}$ kinda jumps out.  In that case, $ 20a\minus{}10\equal{}30$ so $ a$ must be $ 2$.[/hide]

to solve it algebraicaly, put everything in temrs of one variable. i chose x.
$ x\plus{}\frac{3} {2} x\plus{}\frac{5} {2}x\equal{}5x\equal{}100$ so $ x\equal{}20$ $ y\equal{}$ yadayadayada ect. then do what archimedes1 did


geometry
cyclic quadrilateral
geometry unsolved

In a cyclic quadrilateral $ ABCD$, the midpoint of diagonals $ AC$ and $ BD$ are $ E$ and $ F$, respectively. Prove that $ \angle AEB\equal{} \angle AED$ implies $ \angle BFA\equal{} \angle BFC$.

Singapore TST 2007, Turkish MO 1999.
http://www.mathlinks.ro/Forum/viewtopic.php?t=6557
http://www.mathlinks.ro/Forum/viewtopic.php?t=50002


Using Geometry for Probability Part 2

We use area to solve a probability problem.

He is soooooooo cheap... Yet sooo funny... I figured it out halfway through the video.....

The content wasn't bad but the game was so cheesy.  I knew what he picked after he explained the rules.  The chances of winning the game were actually 0/1

Even though Mr. Rusczyk always find a way to win (he is just that awesome), the videos are enough for me :) I'd like him to play one of my games MUWAHAHHAHAHA

Only if i picked $2-.\overline{9}$ which equals $1.00...1$ but in fraction form it equals $2-\frac{9}{9}=1$


There are $ n$ candidates on a table. Petrik and Mikola alternately take candies from the table according to the following rule. Petrik starts by taking one candy; then Mikola takes $ i$ candies, where $ i$ divides $ 2$, then Petrik takes $ j$ candies, where $ j$ divides $ 3$, and so on. The player who takes the last candy wins the game. Which player has a winning strategy?

[hide="Is this correct?  :maybe:"]
[b]Case: [i]n is even[/i][/b]
Mikola can always ensure (by picking up one candy/an odd number of candies every time it is her turn) that when its Petrik's turn, he has to pick up an even number of candies to win, which is impossible and thus, she has the winning strategy.

[b]Case: [i]n is odd[/i][/b]
Mikola can always ensure (by picking up an even number of candies once and odd number of candies every other time it is her turn) that when its Petrik's turn, he has to pick up an even number of candies to win, which is impossible and thus, she has the winning strategy.
[/hide]

we have to cases :
1- n is odd : in this case player 2 wins because player 1 can only choose odd numbers and player 2 has to do this strategy " make the remaining candidates even" because he can choose 1 and 2.
2- n is even : in this case player 2 wins with same strategy.


Let a,b,c are real numbers such that $ a,b,c\in [\minus{}1;1]$ and $ a\plus{}b\plus{}c\plus{}abc\equal{}0$.
Prove that
$ \frac {1}{1\minus{}x}\plus{}\frac {1}{1\minus{}y}\plus{}\frac {1}{1\minus{}z}\plus{}4xyz\ge 3$

[quote="thegod277"]Let a,b,c are real numbers such that $ a,b,c\in [ \minus{} 1;1]$ and $ a \plus{} b \plus{} c \plus{} abc \equal{} 0$.
Prove that
$ \frac {1}{1 \minus{} x} \plus{} \frac {1}{1 \minus{} y} \plus{} \frac {1}{1 \minus{} z} \plus{} 4xyz\ge 3$[/quote]
[b] what is x,y,z?[/b]

no one can prove this?It is very nice. :)

[quote="thegod277"]Let a,b,c are real numbers such that $ a,b,c\in [ \minus{} 1;1]$ and $ a \plus{} b \plus{} c \plus{} abc \equal{} 0$.
Prove that
$ \frac {1}{1 \minus{} a} \plus{} \frac {1}{1 \minus{} b} \plus{} \frac {1}{1 \minus{} c} \plus{} 4abc\ge 3$[/quote]
Setting $ a \equal{} \frac {1 \minus{} x}{1 \plus{} x},b \equal{} \frac {1 \minus{} y}{1 \plus{} y},c \equal{} \frac {1 \minus{} z}{1 \plus{} z}$, where $ x,y,z \ge 0$ then we must have
\[ \sum (1 \minus{} x)(1 \plus{} y)(1 \plus{} z) \plus{} (1 \minus{} x)(1 \minus{} y)(1 \minus{} z) \equal{} 0
\]
or
\[ xyz \equal{} 1
\]
The inequality becomes
\[ \sum \frac {1}{1 \minus{} \frac {1 \minus{} x}{1 \plus{} x}} \plus{} \frac {4(1 \minus{} x)(1 \minus{} y)(1 \minus{} z)}{(1 \plus{} x)(1 \plus{} y)(1 \plus{} z)} \ge 3
\]

\[ \sum \frac {1 \plus{} x}{2x} \plus{} \frac {4(1 \minus{} x)(1 \minus{} y)(1 \minus{} z)}{(1 \plus{} x)(1 \plus{} y)(1 \plus{} z)} \ge 3
\]

\[ xy \plus{} yz \plus{} zx \plus{} \frac {8(xy \plus{} yz \plus{} zx \minus{} x \minus{} y \minus{} z)}{x \plus{} y \plus{} z \plus{} xy \plus{} yz \plus{} zx \plus{} 2} \ge 3
\]
If $ xy \plus{} yz \plus{} zx \ge x \plus{} y \plus{} z$, it is trivial.
If $ x \plus{} y \plus{} z \ge x \plus{} y \plus{} z$, it suffices to prove
\[ xy \plus{} yz \plus{} zx \plus{} \frac {8(xy \plus{} yz \plus{} zx \minus{} x \minus{} y \minus{} z)}{2(xy \plus{} yz \plus{} zx) \plus{} 2} \ge 3
\]

\[ (xy \plus{} yz \plus{} zx)(xy \plus{} yz \plus{} zx \plus{} 1) \plus{} 4(xy \plus{} yz \plus{} zx \minus{} x \minus{} y \minus{} z) \ge 3(xy \plus{} yz \plus{} zx \plus{} 1)
\]

\[ (xy \plus{} yz \plus{} zx)^2 \plus{} 2(xy \plus{} yz \plus{} zx) \minus{} 4(x \plus{} y \plus{} z) \ge 3
\]

\[ x^2y^2 \plus{} y^2z^2 \plus{} z^2x^2 \plus{} 2(xy \plus{} yz \plus{} zx) \minus{} 2(x \plus{} y \plus{} z) \ge 3
\]

\[ \sum \left( \frac {1}{x^2} \plus{} \frac {2}{x} \minus{} 2x \minus{} 1\right) \ge 0
\]
It can be proved by Mixing variables method :)


S(n) is the sum of digits of n
Find
$ S(S(S(S(9^{9^9 } )))$

[quote="mathwizarddude"][hide]$ S(n)\equiv n\mod9$; $ 9^{9^9} < 10^{9^9}\implies s(9^{9^9}) < 9^{10}$

$ 9^{10} < 10^{10}\implies s_2 \equal{} s(s(9^{9^9})) < 90$

$ s_3 < 17$

$ s_4\le9$ with equality when $ s_3 \equal{} 9$ Clearly $ s_4\neq0,s_4\equiv0\mod9\implies s_4 \equal{} 9$[/hide][/quote]
yes, that's right. :D


If you placed three British 10-penny coins together in a triangular pattern, what would the distance be from the center of one coin to the center of another? A 1969 British 10-penny coin has a diameter of about 2.8 cm.

[hide]
Approximately 2 times the radius = the diameter, 2.8 cm.[hide][/hide][/hide]

[quote="gr8sk8r"][hide]they have to be at least twice the radius apart, so that would equal the diameter which is about 2.8.[/hide][/quote]
not only do they have to be at least, it has to be equal


can anyone help me with this problem( please show work!)

find the mystery number _ _[size=150],[/size] _ _ _ [size=200].[/size]_ _ _ _ _

clues:
1 the digits in the ten thousands, thousands ,and tenths places are the same
2 the hundreds digit is 1 more than the thousands digit
3 the digit in the ten thousandths place is the sum of the digits in the thousands and ten thousands places
4 no odd digit occurs more than once
5 the digit in the hundred thousandths place is the sum of the digits in the tenths and hundredths laces
6 the part of the number to the right of the decimal has no repeated digits.
7 the only odd digits appear in the thousandths, hundreds, and ones places
8 the digit in the ones place is the sum of the digits in the tens and hundreds places
9 the digits in the tens and hundredths places are the same and are half of the digit in the tenths place
10 the digit in the tenths and thousandths places have exactly 3 factors

I don't want to do this, but here are things that I noticed.

10. the digit in the tenths and thousandths places have exactly 3 factors
[hide]Those have to be perfect squares less than 10.(not including 1.)
So they are either 4 or 9.[/hide]

1. the digits in the ten thousands, thousands ,and tenths places are the same
[hide]They have to be even number because of #4.[/hide]

My suggestion is that you just go through all of the clues and match them up.

I don't think this is really hard to solve, so just take your time to solve it...
Sorry I couldn't help..but reading my hint should give you tenths and thousandths digit for sure... and it should make it easier..

can you help me more? i still don't get it

[quote="DonkeyKong"]I don't want to do this, but here are things that I noticed.

10. the digit in the [b]tenths[/b] and thousandths places have exactly 3 factors
[hide]Those have to be perfect squares less than 10.(not including 1.)
So they are either 4 or 9.[/hide]

1. the digits in the ten thousands, thousands ,and [b]tenths[/b] places are the same
[hide]They have to be even number because of #4.[/hide]
[/quote]

thus, the tenths, ten thousands, thousands, and thousandths places are all 4

so far we have 44, _ _ _.4_4_ _

#2 states that hundreds=thousands +1, so hundreds=5
#3 states ten thousandths=thousands + ten thousands, so its 4+4=8

CHECKPOINT

44,5_ _.4_48_

BUT...now #s 6&10 contradict! so the thoundths must be 9

44,5_ _.4_98_

#9 says tens=hundredths=.5tenths, or 2
#8 says ones=tens+hundreds=7

CHECKPOINT

44,527.4298_

the last number is either 1 or 3 or 6

BUT...#7 states that the last # cant be odd, so its 6!

MYSTERY #: 44,527.42986

To crack these kind of problems, I think, is that you have to catch the hints that gives you definite range...


Let $f(x)=x^2\sin x+x\cos x+x^2+\frac12$.

Complete the square and obtain, for $x\ne -\frac{\pi}2+2\pi n$ and $n\in\mathbb{Z}$,

\[ f(x)=(1+\sin x)\left(\left(x+\frac12\frac{\cos x}{1+\sin x}\right)^2+\frac14\right)>0  \]

This holds because $1+\sin x>0$ for any real $x\ne -\frac{\pi}2+2\pi n$, otherwise it is zero, and

$\left(x+\frac12\frac{\cos x}{1+\sin x}\right)^2+\frac14>\frac14$

If $x=-\frac{\pi}2+2\pi n$, $f(-\frac{\pi}2+2\pi n)=\frac12>0$.

Thus, $f>0$ for all reals $x$.

Or try proving the ineq:
$x^2-\frac{1}{x\sqrt{x^2+1}}+\frac{1}{2}>0$, I think.

I don't know why you guys have complexified, but this is really simple, so posting the solution after 1 year since I had solved it.....

Why can't we put this expression in am-gm ?

x^2sinx +xcosx+x^2+(1÷2)
=x^2(1+sinx)+xcosx+(1/2)

D=b^2-4ac
 =(cosx)^2-2(1+sinx)
 =1- (sinx)^2-2(1+sinx)
 =-(1+sinx)^2<0
Hence, the sign of the equation is the same as the sign of the coefficient of x^2, which is+ve
Hence, x^2sinx + xcosx+ x^2 +(1/2)>0  (Proved)

[quote=Rushil]Show that for any real number $x$:
\[ x^2 \sin{x} + x \cos{x} + x^2 + \frac{1}{2} > 0 . \][/quote]

$let,$ $f(x)= x^2(\sin x + 1) + x \cos x + \frac{1}{2} $
$ D^2 = - (\sin x + 1)^2 \le 0$ $for$ $all$ $x$
We conclude that the $graph$ the equation does not touch the $x$ $axis$
Putting $x=0, f(x)=\frac{1}{2} > 0$
The conclusion is immediate.

Nice solution! Congratulations!  :thumbup:

[quote=booth][quote=Singular]The discriminant is $-(\sin x + 1)^2$.[/quote]

Your discriminant depends on $x$ :noo: .[/quote]

This is NOT a problem! The essence it is, that the discriminant can not be negative!
If you want, we can write this: ${{x}^{2}}\sin x+x\cos x+{{x}^{2}}+\frac{1}{2}>0\Leftrightarrow (\sin x+1)\left( {{\left( x+\frac{\cos x}{2(\sin x+1)} \right)}^{2}}+1 \right)>0$

[quote=Kuroshio][quote=Rushil]Show that for any real number $x$:
\[ x^2 \sin{x} + x \cos{x} + x^2 + \frac{1}{2} > 0 . \][/quote]

$x^2(\sin x + 1) + x \cos x + \frac{1}{2} > 0 $
Coefficient of $x^2 = \sin x + 1 \ge 0$
$ D^2 = - (\sin x + 1)^2 \le 0$
We conclude that the $graph$ the equation is an $upward$ $parabola$ not touching the $x$ $axis$
The conclusion is immediate.[/quote]

It is not a parabola because the equation is not a quadratic equation.
You can't directly conclude that it is always positive just by saying that the discriminant is negative.
However you could have written it like this :
$$\left(x-\frac{-\cos(x)+\sqrt{-(\sin(x)+1)^2}}{2\sin(x)+1}\right)\left(x-\frac{-\cos(x)-\sqrt{-(\sin(x)+1)^2}}{2\sin(x)+1}\right)$$
And then you can equate it to $0$ to get $(2x\sin(x)+x+\cos(x))^2=-(\sin(x)+1)^2$
Which implies that both of them are 0 and then show that they can't be 0 simultaneously
Which implies that the graph never crosses the x-axis and it lies on any one side of the x-axis. Then put a value of x in the original equation and show that it is positive.


Show that for all constants a and b, and all real values of x,
$ | a\ sin(x)+b\ cos(x) | \leq  \sqrt{ a^{2}+b^{2}}$

Is it possible to have equality? If so, for what value of  x is the bound achieved?

we have $ |a\sin x+b\cos x|\le \sqrt{a^{2}+b^{2}}$
$ \iff (a\sin x+b\cos x)^{2}\le a^{2}+b^{2}$
$ \iff a^{2}\sin^{2}x+b^{2}\cos^{2}x+2ab\sin x\cos x\le a^{2}+b^{2}$
$ \iff a^{2}-a^{2}\cos^{2}x+b^{2}-b^{2}\sin^{2}x+2ab\sin x\cos x\le a^{2}+b^{2}$
$ \iff a^{2}\cos^{2}x+b^{2}\sin^{2}x-2ab\sin x\cos x\ge 0$
$ \iff (a\cos x-b\sin x)^{2}\ge 0$

with equality if and only if $ a\cos x=b\sin x\iff \tan x=\frac{a}{b}$
that is $ x=\arctan\frac{a}{b}$.

[quote="ashrafmod"]Show that for all constants a and b, and all real values of x,
$ | a\ sin(x)+b\ cos(x) | \leq \sqrt{ a^{2}+b^{2}}$[/quote]
If $ a,b=0$ the equality is trivial.

Otherwise,
$ a\sin x+b\cos x = \sqrt{a^{2}+b^{2}}\left( \frac{a}{\sqrt{a^{2}+b^{2}}}\sin x+\frac{b}{\sqrt{a^{2}+b^{2}}}\cos x\right) = \sqrt{a^{2}+b^{2}}\left( \sin x \cos \alpha+\cos x \sin \alpha \right) = \sqrt{a^{2}+b^{2}}\sin (x+\alpha)$.

The inequality follows immediately since $ |\sin (x+\alpha)| \leq 1$

Note: This also answers the question when equality occurs. but please feel in the details yourself.

$ (a\sin x+b\cos x)^{2}+(a\cos x-b\sin x)^{2}=a^{2}+b^{2}\Longrightarrow (a\sin x+b\cos x)^{2}\leq a^{2}+b^{2}$
$ \Longleftrightarrow |a\sin x+b\cos x|\leq \sqrt{a^{2}+b^{2}}$. The equality holds when $ a\cos x=b\sin x \Longleftrightarrow (a,\ b)\parallel (\cos x ,\sin x)$.


Little Johnny painted on a sheet of paper a polygon with N(3<=N<=10000) vertices. Then, for every edge of the polygon, he drew the middle point of the edge. After that, he went to school. When he came back, he found out that his brother had erased the polygon (both the edges and the vertices). The only thing left were the middle points of the edges of the polygon. Help Johnny redraw his polygon. The polygon may contain self-intersections. Although in many geometric problems, self-intersections only make things more difficult, in this case, they make things a lot easier.

The problem seemed slightly ambiguous, so I googled for it. The main point you missed from the actual question: you are given the midpoints in the actual order they were on the polygon.
Thus, if the midpoints are (a1,b1), (a2,b2), .., (an,bn), and the first vertex of the polygon is at (x,y), the vertices proceed as follows:
$(x,y)\\(2a_1-x, 2b_1-y)\\(2a_2-2a_1+x,2b_2-2b_1+y)\\(2a_3-2a_2+2a_1-x,2b_3-2b_2+2b_1-y)$ etc.
If n is even, we end up with $2a_n - 2a_{n-1} + 2a_{n-2} + .. - 2a_1 + x = x$, ie that particular sum must be 0, and the similar sum for the y coordinate must be 0. If they both are, then there are infinitely many such polygons, choose whatever x, y you like. If not, there are none.
If n is even, we get $2a_n - 2a_{n-1} + 2a_{n-2} + .. + 2a_1 - x = x$, ie x = that sum/2. Similarlr for y; this time you get a unique polygon.


The inner point $ X$ of a quadrilateral is [i]observable[/i] from the side $ YZ$ if the perpendicular to the line $ YZ$ meet it in the colosed interval $ [YZ].$ The inner point of a quadrilateral is a $ k\minus{}$point if it is observable from the exactly $ k$ sides of the quadrilateral. Prove that if a convex quadrilateral has a 1-point then it has a $ k\minus{}$point for each $ k\equal{}2,3,4.$

First, we will prove that this convex quadrilateral has exactly 1 angle <90*
it isn't hard , we will just see some cases
1)exactly 3 angles are bigger than 90
2)exactly 2
We use these facts
1.if the two angles which has common side are $<=90$ then every inner point is observable from that side
2.if one is greater but the second is less,  when we outline  height  from the vertex of greater angle to common side
then all inner points of small quadrilateral are observable from that side
other facts are same to these ones .So if you draw picture you will understand everything ))


calculus
derivative
inequalities
inequalities proposed

For a finite number of positive variables $ a,b,c,...$ with product $ 1$ and naturals $ n > m$, prove

$ \prod\frac {a^n \plus{} a^{n \minus{} 1} \plus{} \cdots \plus{} 1}{n \plus{} 1}\ge\prod\frac {a^m \plus{} a^{m \minus{} 1} \plus{} \cdots \plus{} 1}{m \plus{} 1}$

It is easy to prove that by using derivatives or Karamata.
BUT [b]wanted:[/b][i] the most elegant proof that uses [b]only AM-GM [/b]:D[/i]

[hide="hint"]$ \frac{a^n\plus{}\cdots\plus{}1}{n\plus{}1}\geq \frac{a^{\frac{n\plus{}m}{2}}\plus{}\cdots\plus{}a^{\frac{n\minus{}m}{2}}}{m\plus{}1}$ for every positive a. :wink: [/hide]

[quote="Albanian Eagle"][hide="hint"]$ \frac {a^n \plus{} \cdots \plus{} 1}{n \plus{} 1}\geq \frac {a^{\frac {n \plus{} m}{2}} \plus{} \cdots \plus{} a^{\frac {n \minus{} m}{2}}}{m \plus{} 1}$ for every positive a. :wink: [/hide][/quote]
Doesn't it equivalent to the problem?
I mean to ask how to use AM-GM in the (hint)  :D

If u use the hint you can easily get the inequality posted.

on the other side the hint can be proven by cross multiplication and a few simplifications. Let me remind you that $ a^{r\plus{}s}\plus{}1\geq a^r\plus{}a^s$  :wink:

[quote="Albanian Eagle"]cross multiplication and a few simplifications[/quote]
Did you mean $ (a^{r}\minus{}1)(a^{s}\minus{}1)\geq0$ for all positive real number a?  :huh:

No.  :wink:  It wasn't wrong but i was wondering about the word "cross multiplication and a few simplifications".
That isn't used in my country.  :lol:


A state had three letters followed by three digits as the format for license plates. In order to increase the number of plates available, the state changed the format to four letters followed by two digits. What is the positive difference between the number of plates available with the new format and the number of plates available with the old format?

Hello,

FTW (For the Win) expects you to do this question in 45 seconds. Is the a faster way than all that multiplication??

Thank you,

Burunduchok

[quote=unskillfulnoob]@burunduchok, no, and actually, you can have up to 120 secs to solve a question.[/quote]

Isn't the limit 45 seconds?

If you don’t change the settings, default is 45 seconds. However, the highest amount you can change it to is 120 seconds.


function
modular arithmetic
number theory
algebra
binomial theorem

Find the last three digits of $ 3^{352}$

hello, we get $ 3^{352}\equiv 241 \mod 1000$.
Sonnhard.

If someone would please post a solution using modular arithmetic, that would be fantastic (since this is derived from an olympiad problem)

[hide="Alternate Solution"]
Using the carmichael function, we find that $ \lambda(1000) \equal{} 100$ therefore $ 3^{100}\equiv 1\pmod{1000}$.

Now the problem becomes finding $ 3^{52}\pmod{1000}$

Simplifying we get, $ 3^{52}\equiv 9^{26}\equiv 81^{13}\pmod{1000}$

We also know that $ 81^4\equiv 561\cdot 561\equiv 721\equiv \minus{} 279\pmod{1000}$

This means that $ 81^6\equiv 561\cdot ( \minus{} 279)\equiv 481\pmod{1000}$

Now we square $ 481$ and multiply by $ 81$ to get the result,

$ 81^{13}\equiv (81^6)^2\cdot 81\equiv (481)^2\cdot 81\equiv 361\cdot 81\equiv \boxed{241}\pmod{1000}$
[/hide]

[hide="Another alternate solution"] 3^352 is 9^176= (10-1)^176. If we expand using binomial theorem, all terms are automatically divisible by 1000 except the last 3. After we realize that the third-to-last term is also divisble by 3, we sum the last 2 terms and get -1759, which is 241(mod 1000). [/hide]

Aha, thank you very much Countdown King. That was a great solution that I was looking for.

Also, tonypr, what is the Carmichael function? It's a bit difficult to understand from Wikipedia and Wolfram.

It's the order of number a modulo n.

I perfectly understand http://en.wikipedia.org/wiki/Carmichael_function


Find all functions, $ f: (0,\infty)\rightarrow(0,\infty)$, such that
\[ (x\plus{}y)\cdot f\left (f(x)y\right)\equal{}x^2\cdot f\left(f(x)\plus{}f(y)\right)
\]

Edit: Blaaaarg, I thought it was $ [0,\infty)$ for some reason.

Those are not exactly the same problem.
The solution $ 1/x$ does not work here.

grn_trtle's solution is wrong. Remember $ f$ is only defined on positive real, you can not let $ x\equal{}0$ or $ y\equal{}0$.

folks, please be more cautious when posting a solution or a link. The thread seems more "attractive" when nothing, except for a statement, has been posted in it, right?

To [b]grn_trtle[/b]: "plugging" of $ 0$ into the equation, instead of $ x$ or $ y$, is forbidden.(what makes the problem even easier).

To [b]TTsphn[/b]: They are not the same.

If $ f(a)\equal{}f(b)$, we can compare $ P(a,y)$ and $ P(b,y)$ and get
$ \frac{a\plus{}y}{a^2}\equal{}\frac{b\plus{}y}{b^2},\forall y>0\Rightarrow a\equal{}b$.
Hence $ f$ is injective.
Let $ r$ be the positive root to $ x^2\minus{}x\minus{}1\equal{}0$, then
$ P(r,1): (r\plus{}1)f(f(r))\equal{}r^2f(f(r))\equal{}r^2f(f(r)\plus{}f(1))\\\Rightarrow f(r)\equal{}f(f(r)\plus{}f(1))\Rightarrow f(r)\equal{}f(r)\plus{}f(1)\Rightarrow f(1)\equal{}0$.
Contradiction!
There is no such function.

$ (x\plus{}y)f(f(x)y) \equal{} x^ 2f(f(x)\plus{}f(y))$ and $ f : R^\plus{} \longrightarrow R^\plus{}$

You can prove that $ f(x) \equal{} f(y) \longrightarrow x\equal{}y$
then put $ y \equal{} x^2 \minus{}x$ then we get :

$ f(x)[x^2 \minus{}x\minus{} 1]\equal{}f(x^2 \minus{}x)$

if $ x  \epsilon  (1, \frac{1\plus{}\sqrt{5}}{2})$ then we get :
$ f(x) >0$ and $ [x^2 \minus{}x \minus{}1] < 0$ ==> $ f(x^2 \minus{}x) <0$

but $ (x^2 \minus{}x ) > 0$  ==> $ f(x^2 \minus{} x) > 0$ (because $ f: R^\plus{} \longrightarrow R^\plus{}$ ) ==>  There is no function  .
 :)


During a game of $ \textit{Scrabble}$, Alexandra had tiles with the following letters: R, O, O, V, G, N, and O. She rearranged the tiles to form all possible distinct 7-letter arrangements. In how many unique ways could she have arranged the tiles?

Why are you posting "Just to watch." in multiple threads?

The answer is $ \frac{7!}{3!}\equal{}7 \times 6 \times 5 \times 4\equal{}\boxed{840}$.

If all the letters were different, you'd have: 7!= 5040.

Three letters repeat, though, giving you 7!/3!= 840.

Oh, yeah, 7! means 7 x 6 x 5 x 4 x 3 x 2 x 1


algebra
polynomial
binomial theorem
Pascal\u0027s Triangle

Determine the coefficient of the term containing $x^{n}$ in the expansion of $(1+x)^{2n}$.

[hide]
I got $\binom{2n}{n}x^{n}$

is that correct, and if so, is there a way to simplify that?

thanks.
[/hide]

Well, the coefficient is just $\binom{2n}{n}$ (without the $x^{n}$ part).
Yes that's correct, yes that is fully simplified.

I believe that is correct, as the binomial theorem states that:
$(x+y)^{n}=\sum_{k=0}^{n}\dbinom{n}{n-k}x^{n-k}y^{k}$.
Substituting $1$ for $y$, $2n$ for $n$, the term containing $x^{n}$ will be when $k=n$.

I don't see any obvious way to simplify that, though it may interest you to know that:
$\dbinom{2n}{n}=\sum_{k=0}^{n}{\dbinom{n}{k}}^{2}$.

[quote="undefined117"]thanks for the replies guys!

but davidyko, how does $\binom{2n}{n}=\sum_{k=0}^{n}\binom{n}{k}^{2}$ work? could you derive it (or give me a link that does) or is it some type of theorem?[/quote]

Hm, are you familiar with block walking? It's covered in Ch. 15 of AoPS II. But anyway, here goes:
To walk from $\dbinom{0}{0}$ (top of Pascal's triangle) to $\dbinom{2n}{n}$, we must pass through the $n$th row at $\dbinom{n}{k}$, for some $0 \le k \le n$. The number of ways to go from that point down to $\dbinom{2n}{n}$ is $\dbinom{n}{n-k}$, as we must make $n-k$ more right branches out of the $n$ branches we have left. Therefore the ways to get to $\dbinom{2n}{n}$ passing through $\dbinom{n}{k}$ is $\dbinom{n}{k}\dbinom{n}{n-k}$, which is equal to ${\dbinom{n}{k}}^{2}$. As we can pass through any $k$ in the $n$th row, the result follows.

If you are not familiar with block walking, basically all you need to realize is that to walk from the top of Pascal's triangle to some $\dbinom{n}{k}$, we must make $k$ right branches out of $n$ branches.

[quote="undefined117"]thanks for the replies guys!

but davidyko, how does $\binom{2n}{n}=\sum_{k=0}^{n}\binom{n}{k}^{2}$ work? [/quote]

$(x+1)^{2n}= (x+1)^{n}\cdot (x+1)^{n}$

If you are familiar with the rigorous definition of polynomial multiplication, the conclusion follows immediately (compare the $x^{n}$ term in both sides).  Note that ${n \choose k}^{2}={n \choose k}{n \choose n-k}$.

Edit:  A massive generalization.

http://mathworld.wolfram.com/Chu-VandermondeIdentity.html

davidyko, i read what you said and CH. 15 AoPS II and i think i get what it is.. thanks again

in response to t0rajir0u's link: OH S*** :D I'll try to understand that when I have more time.. as of now.. final exams

You could also prove the identity by considering a bag of $n$ blue and $n$ red balls. Then $\binom{2n}{n}$ is the number of ways of picking $n$ balls from this bag. Now, this could be counted in another way: you could pick $k$ red balls and $n-k$ blue balls, where $k$ varies from $0$ to $n$. Therefore, \[\binom{2n}{n}= \sum_{k=0}^{n}\binom{n}{k}\binom{n}{n-k}= \sum_{k=0}^{n}\binom{n}{k}^{2}\ \blacksquare\]


conics
ellipse
geometry
geometry theorems

Butterfly theorem is true for circle and ellipse!
Now which is a curve that butterfly theorem is true, example is it true for hyperbolla?
There is a version of butterfly theorem for space geometry and I see it true for ellipsolid and a similar question: which surface that butterfly theorem is true?

Butterfly Theorem is true for all conics.

I have found a proof involving Projective Geometry. I am still struggling to understand it but I will post it anyway.

[i]In circle $O$. draw any chord $AB$ with midpoint $M$. Through M two other chords $FE$ and $CD$ are drawn. $CE$ and $FD$ intersects $AB$ at $Q$ and $P$ respectively. Prove that $MP=MQ$.[/i]

The above is the theorem for a circle but the proof applies to all conics.

Let $K$ be the intersection of $DF$ and $EC$. Let $I$ be the intersection of $FC$ and $DE$. Let $N$ be the intersection of $AB$ and $KI$.

$KI$ is the polar of $M$ with respect to the conic (in this case the circle). Therefore $M,A,B,N$ form a harmonic range.

Thus $\frac{MB}{MA}= \frac{BN}{NA}$ and since $MB=MA$, $N$ is at infinity.

Hence $AB$ is parallel to $KI$. Now $KE, KM, KD, KI$ is a harmonic pencil. Therefore $Q,M,P,N$ is a harmonic range and $\frac{MQ}{MP}= \frac{QN}{NP}$. Since $N$ is at infinity $MQ=MP$.

Can somone explain some of the terminology?


abstract algebra
trigonometry
integration
real analysis
real analysis unsolved

Define $ P_r(\theta) \equal{} \sum_{n \equal{} \minus{} \infty}^\infty r^{|n|}e^{in\theta}$

(i) Show that $ P_r(\theta) \equal{} \frac {1 \minus{} r^2}{1 \minus{} 2r\cos\theta \plus{} r^2}$

(ii) Let $ f\in L^1(T)$ show that $ \sum_{n \equal{} \minus{} \infty}^\infty r^{|n|}\hat{f}(n)e^{in\theta} \equal{} (P_r*f)(t)$

(iii) show that $ P_r(\theta)_{0\leq r < 1}$ is a summation kernel
 that is show that
a)$ \frac {1}{2\pi}\int_\pi^\pi P_r(\theta)d\theta \equal{} 1$
b) $ P_r(\theta) \geq 0$ for all $ \theta$
c)$ \max_{\delta\leq\theta\leq\pi}P_r(\theta) \rightarrow 0$ when $ r \rightarrow 1$, if $ \delta > 0$

All calculations are straightforward, e.g.,

$ P_r(\theta)$ is absolute convergent for $ |r|<1$. Therefore, $ P_r(\theta)\equal{}1\plus{}\sum_{n\equal{}1}^{\infty}\left[\left(re^{i\theta}\right)^n\plus{}\left(re^{\minus{}i\theta}\right)^n\right]\equal{}1\plus{}\frac{re^{i\theta}}{1\minus{}re^{i\theta}}\plus{}\frac{re^{\minus{}i\theta}}{1\minus{}re^{\minus{}i\theta}}$ from which the claim follows by moving to the common denominator $ \left(1\minus{}re^{i\theta}\right)\left(1\minus{}re^{\minus{}i\theta}\right)\equal{}1\minus{}2r\cos\theta\plus{}r^2$


Prove or prove this ineq :
For all a,b,c>0 we have :
$ \frac{ab\plus{}c^2}{c^2\plus{}ac}\plus{}\frac{bc\plus{}a^2}{a^2\plus{}ab} \plus{}\frac{ca\plus{}b^2}{b^2\plus{}bc} \geq3$
Hope to see many comments from all nice friends soon
 :oops:

Using AM-GM and Muirhead

 ${ LHS\ge 3\sqrt [3]{\prod \frac {a^2 + bc}{a^2 + ab}}}\ge RHS$

 because $ \sum_{cyc} a^3b^3+\sum_{cyc} a^4bc\ge\sum_{sym}a^3b^2c$

[quote="nkht-tk14"]Prove or prove this ineq :
For all a,b,c>0 we have :
$ \frac {ab \plus{} c^2}{c^2 \plus{} ac} \plus{} \frac {bc \plus{} a^2}{a^2 \plus{} ab} \plus{} \frac {ca \plus{} b^2}{b^2 \plus{} bc} \geq3$
Hope to see many comments from all nice friends soon
 :oops:[/quote]
[url]http://www.mathlinks.ro/viewtopic.php?t=281468[/url]
they are one ! :lol:


Το 4ο θέμα του Ευκλείδη της Γ' Λυκείου ζητούσε να δειχθεί ότι η συνάρτηση f που ικανοποιούσε μια δεδομένη συναρτησιακή εξίσωση ήταν περιττή.

Φυσικά σαν μαθηματικοί αμέσως θέσαμε τα ερωτήματα: Τι άλλο μπορούμε να πούμε για την f; Είναι μονότονη; Είναι συνεχής; Μπορούμε να βρούμε όλες τις f που ικανοποιούν την συναρτησιακή εξίσωση;

Μετά από κάποια δουλειά καταλήγουμε στο ερώτημα: Να βρεθούν όλες οι συναρτήσεις $ f: \mathbb{R} \to \mathbb{R}$ που ικανοποιούν
1) f(x+y) = f(x) + f(y)  για κάθε x,y
2) f(f(x)) για κάθε x.

Θα μελετήσουμε πρώτα τις συναρτήσεις f που ικανοποιούν μόνο την (1) και μετά θα πάμε και σε αυτές που ικανοποιούν και την (2).

Η (1) ονομάζεται συναρτησιακή εξίσωση του Cauchy. Εύκολα κάποιος "μαντεύει" τις "κλασσικές" λύσεις $ f(x) \equal{} ax$. Θα δούμε όμως στην συνέχεια ότι υπάρχουν και αρκετές άλλες λύσεις.

Αυτό που μπορούμε να δείξουμε σχετικά εύκολα και αφήνεται σαν άσκηση σε όσουν δεν το έχουν ξαναδεί είναι
3) f(qx) = qf(x) για κάθε ρητό q και για κάθε πραγματικό x.

Αν η (3) ίσχυε για κάθε πραγματικό αριθμό q τότε αυτομάτως θα είχαμε $ f(x) \equal{} ax$ όπου $ a \equal{} f(1)$. Αυτό όμως δεν ισχύει.

Δυστυχώς όμως για να γράψουμε κάτω τις άλλες συναρτήσεις που ικανοποιούν την (1), θα χρειαστεί να ξεφύγουμε λίγο από τα σχολικά/ολυμπιακά θέματα.

Για να βρούμε τις υπόλοιπες συναρτήσεις θα βασιστούμε στο ακόλουθο θεώρημα του Hamel

Υπάρχει υποσύνολο $ A \equal{} \{a_i: i \in I\}$ του R ώστε:

Κάθε πραγματικός αριθμός x μπορεί να γραφτεί [i]με μοναδικό τρόπο[/i] στην μορφή $ x \equal{} \sum_{i \in  I}{q_ia_i}$ όπου τα $ q_i$ είναι ρητοί και [i]όλα εκτός από πεπερασμένο αριθμό[/i] ισούνται με 0.

(Το Ι εδώ είναι απλώς ένας "συνολοδείκτης" που με βοηθάει στο να γράφω πιο εύκολα αυτά που θέλω.)

Δεν θα βάλω απόδειξη αυτού του θεωρήματος. (Για όσους έχουν δει αποδείξεις που χρησιμοποιούν αξίωμα επιλογής/λήμμα του Zorn η απόδειξη είναι σχετική απλή - για τους υπόλοιπους είναι σχεδόν αδύνατη και δεν νομίζω να προσθέτει τίποτα στην κατανόηση αν την βάλω.)

Χρησιμοποιώντας όμως το θεώρημα αυτό μπορώ εύκολα να κατασκευάσω f που να ικανοποιούν την (1)

Ορίζουμε με οποιονδήποτε τρόπο την f στο σύνολο Α, ας είναι f(a_i) = r_i και μετά την επεκτείνουμε στο R ορίζοντας
$ f(x) \equal{} \sum_{i \in  I}{q_ir_i}$, όπου $ x \equal{} \sum_{i \in  I}{q_ia_i}$.

Η ύπαρξη και η μοναδικότητα της γραφής κάθε πραγματικού x με αυτόν τον τρόπο δείχνουν ότι η f είναι όντως καλώς ορισμένη (έχω δώσει ακριβώς μία τιμή για κάθε x). Μέσω της (3) εύκολα ελέγχεται ότι η f ικανοποιεί την (1).

Σαν άσκηση αφήνω να δειχθεί ότι όλες οι f που ικανοποιούν την (1) είναι αυτής της μορφής. (Ουσιαστικά περιορίζουμε την f στο Α και μετά βλέπουμε ότι η (3) μας απαγορεύει να την επεκτείνουμε με οποιονδήποτε άλλο τρόπο.)

Αφήνω επίσης σαν άσκηση να δειχθεί ότι όλες αυτές οι f, εκτός αν είναι μια από τις κλασσικές λύσεις, δεν είναι ούτε συνεχείς, ούτε μονότονες.

Τώρα μπορούμε να πάμε και στην συνθήκη (2). Η f που έγραψα πιό πάνω δεν την ικανοποιεί απαραίτητα. Το κόλπο εδώ είναι να χωρίσω το σύνολο Α σε "ζευγάρια". Πιο συγκεκριμμένα, μπορώ να γράψω $ A \equal{} B \cup C$ με $ B \cap C \equal{} \emptyset$ και $ B \equal{} \{b_j: j \in J\}, C \equal{} \{c_j: j \in J\}$. Λέγοντας ζευγάρια εννοώ τα $ (b_j,c_j)$ .

Μπορώ τώρα να ορίσω $ f(b_j) \equal{} c_j, f(c_j) \equal{} b_j$. Στο σύνολο Α τώρα η f ικανοποιεί και την (1) και την (2). Επίσης δεν είναι δύσκολο να ελεγχθεί ότι αν επεκτείνω την f ακριβώς όπως προηγουμένως, θα εξακολουθεί να ικανοποιεί τις (1) και (2).

Πιο δύσκολο είναι να δειχθεί ότι με αυτήν την ιδέα μπορούμε να πάρουμε όλες τις f που ικανοποιούν τις (1) και (2). (Σε αυτήν την περίπτωση θα χρειαστεί να χωρίσουμε  το Α σε τρία σύνολα B,C,D. Στα B και C ορίζουμε την f όπως προηγουμένως, στο D ορίζουμε είτε $ f(d_k) \equal{} d_k$ είτε $ f(d_k) \equal{} \minus{}d_k$. Μια πρόσθετη δυσκολία αυτής της απόδειξης είναι ότι δεν αρκεί να έχώ μόνο ένα σύνολο Α που μου δίνει τα θεώρημα Hamel αλλά "πολλά" τέτοια Α. Επειδή παρουσιάζεται αυτή η δυσκολία ο μόνος τρόπος που βλεπω για να δείξω ότι όλες οι λύσεις είναι αυτής της μορφής είναι να αποδείξω το θεώρημα του Hamel, αλλά τότε θα ξεφύγουμε πολυ περισσότερο από ότι έχουμε ήδη ξεφύγει.)

Προς moderators: Αν και το θέμα ξεφεύγει από την ύλη των ολυμπιάδων, η συναρτησιακή εξίσωση του Cauchy   (όπως γνωρίζετε) είναι αρκετά σημαντική και οποιαδήποτε περισσότερη κατανόηση ίσως φανεί χρήσιμη. Γι'αυτό και έβαλα το θέμα στις ολυμπιάδες. Αν όμως θέλετε μπορείτε να το μετακινήσετε.

[quote="Athinaios"]Mαλιστα.. πραγματικα ενδιαφερον το θεμα που αγγιζουμε.. thanx παντως για την ολη προσεγγιση που κανεις, ειναι πολυ διδακτικη. Θα κοιταξω να ασχοληθω και λιγο παραπανω με γραμμικη αλγεβρα ωστε να τα καταλαβω πιο καλα.
 Στο ιντερνετ βρηκα κανα δυο ωραιες σημειωσεις,
http://www.math.rutgers.edu/~useminar/cauchy.pdf

http://renyi.hu/~p_erdos/1965-19.pdf[/quote]

Ηλία, το δεύτερο είναι ερευνητικό άρθρο και σαφώς πιο δύσκολο να κατανοηθεί. Το πρώτο εξηγεί ωραία αρκετά πράγματα αλλά όπως και εγώ δεν βάζει την απόδειξη του θεωρήματος του Hamel. Νομίζω όμως πως προς το παρόν δεν χρειάζεται να ανησυχείς για αυτό. Η απόδειξη είναι περισσότερο "τεχνική" και πρέπει πρώτα να έχεις τα κατάλληλα εργαλεία από την θεωρία συνόλων.

Σίγουρα η γραμμική άλγεβρα θα βοηθήσει περισσότερο στην κατανόηση των όσων έχω γράψει. (Αν και ηθελημένα προσπάθησα να αποφύγω την χρησιμοποίηση όρων της γραμμικής άλγεβρας.)


[quote="Utkirstudios"]let x,y,z be real numbers then prove that:
\[ x(x+y)^{3}+y(y+z)^{3}+z(z+x)^{3}\ge \frac{8}{{27}}(x+y+z)^{4}\]
[/quote]
$ y=x+u$ and $ z=x+v$ helps, but it's very ugly.

[quote="kunny"]Let $ x+y=2a,\ y+z=2b,\ z+x=2c\Longleftrightarrow x=c+a-b,\ y=a+b-c,\ z=b+c-a$, where $ a,\ b,\ c$ are the side length of a triangle.
Under this condition, we can rewrite the given inequality as $ a^{3}(c+a-b)+b^{3}(a+b-c)+c^{3}(b+c-a)\geq \frac{1}{27}(a+b+c)^{4}$.[/quote]

How can we talk about side lenghts of a triangle when we have real numbers.. :?:

[quote="Shanku"][quote="arqady"][quote="gold46"]Let $x , y ,z $ be real numbers.Prove that
$x(x+y)^3+y(y+z)^3+z(z+x)^3 \ge \frac{8}{27} (x+y+z)^4$.[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=158931[/quote]
You have not shown your 'ugly' solution in the link, arqady.[/quote]
For you, Shanku! :)
Let $y=x+u$ and $z=x+v.$
Hence, our inequality is equivalent to
$216(u^2-uv+v^2)x^2+3(40u^3+39u^2v-15uv^2+40v^3)x+$
$+19u^4+49u^3v+33u^2v^2-5uv^3+19v^4\geq0.$
Let $u=tv.$
Id est, it remains to prove that
$(40t^3+39t^2-15t+40)^2-96(t^2-t+1)(19t^4+49t^3+33t^2-5t+19)\leq0,$
which is equivalent to
$224t^6-240t^5-33t^4-974t^3+2127t^2-1104t+224\geq0,$ which is true.

[quote="MathUniverse"][quote="arqady"]
which is equivalent to
$224t^6-240t^5-33t^4-974t^3+2127t^2-1104t+224\geq0,$ which is true.[/quote]

But, how can you prove this without computer? :P[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1913072#p1913072

Another way.
$x(x+y)^{3}+y(y+z)^{3}+z(z+x)^{3}\ge \frac{8}{{27}}(x+y+z)^{4}\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}(19x^4+49x^3y-5x^3z+33x^2y^2-96x^2yz)\geq0\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}(19x^4+22x^3y+22x^3z+33x^2y^2-96x^2yz)\geq27\sum_{cyc}(x^3z-x^3y)$.
Let $x+y+z=3u$, $xy+xz+yz=3v^2$, where $v^2$ can be negative, $xyz=w^3$ and $u^2=tv^2$.
Hence, our inequality is equivalent to
$19u^4-18u^2v^2+3v^4-4uw^3\geq u(x-y)(y-z)(z-x)$.
We'll prove that $f(w^3)\geq0$, where $f(w^3)=19u^4-18u^2v^2+3v^4-4uw^3$.
We see that $f$ is linear function. Hence, $f$ gets a minimal value, when $w^3$ gets an extremal value,
which happens, when two numbers from $\{x,y,z\}$ are equal.
Since $f(w^3)\geq0$ is homogeneous inequality of degree $4$, it remains to prove it for $y=z=1$,
which gives $(x-1)^2(19x^2+82x+115)\geq0$, which is true.
Hence, we need to prove that:
$\left(19u^4-18u^2v^2+3v^4-4uw^3\right)^2\geq u^2(x-y)^2(y-z)^2(z-x)^2$.
But $(x-y)^2(y-z)^2(z-x)^2=27(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)$, which gives
$43u^2w^6-2u(22u^4+9u^2v^2+12v^4)w^3+361u^8-684u^6v^2+357u^4v^4+9v^8\geq0$.
Hence, it remains to prove that:
$u^2(22u^4+9u^2v^2+12v^4)^2-43u^2(361u^8-684u^6v^2+357u^4v^4+9v^8)\leq0$,
which is $u^2v^8(t-1)^2(557t^2+10t+9)\geq0$, which is obvious.

i think sos may help here,but not so nice

[quote=Utkirstudios]let x,y,z be real numbers then prove that:
\[ x(x+y)^{3}+y(y+z)^{3}+z(z+x)^{3}\ge \frac{8}{{27}}(x+y+z)^{4}\][/quote]
Because, we have
\[\text{LHS-RHS} = \frac{1}{1026} \sum (19x^2+18xy+13xz-19y^2-31yz)^2+\frac{29}{82878}\sum(18xy+13xz-31yz)^2 \geqslant0.\]


I haven't found anything where you can suggest new forums so I thought I would make one.
I personally would like there to be a forum for marathons and contests so people wouldn't have to find them all if they want to participate.

You can always suggest a new forum here. If it is a good idea, we will definitely support it. However I feel that in many forums marathons are the topics that keep them going, so removing them would not help.

You will have to use the favorites function to keep track of your marathons.


geometry
geometric transformation
reflection
puzzles

Put 8 queens in a chess board such that they can't take each other.

B for Blank, Q for Queen.

BBBBQBBB
BBBBBBQB
QBBBBBBB
BBQBBBBB
BBBBBBBQ
BBBBBQBB
BBBQBBBB
BQBBBBBB

Actually there are 92 ways if you allow for reflections and rotations.


prove thatt 13 divides 10^2p - 10^p + 1    Hint: use 6k+1,6k+5

Please just show me the start of one of these, I have been stuck on this problem for many hours now...

Well, it fails for $p=2$ and $p=3$, i think it wants the problem proven for $p$ prime and greater than or equal to 5.

[hide="hint"]
It is useful to write $10^{2p}-10^{p}+1$ as $\frac{10^{3p}+1}{10^{p}+1}$. Now just try to determine all positive integers $n$ such that $10^{n}+1$ is divisible by 13.
[/hide]

Sorry about the mistake. The question did state that p>3 and It is under the chapter of module. So I think somehow I need to incorporate that when solving it.

I don't think this problem works. If we take 10^2p - 10^p + 1 mod 11, we have:
[(-1)^2p - (-1)^p + 1] mod 11, which equals [2- (-1)^p] mod 11, which can never be 0 mod 11. So... either the book is trying to trick you, or you copied the problem wrong.

Why are you taking it mod 11 if the question is asking you to show that it is divisible by 13?

The problem asks for expression = 0 (mod 13)

i have an idea :D
if $p=6k+1$ , we need to prove that $13|10^{12k+2}-10^{6k+1}+1$
but , we have $10^{12k+2}=10^{12k}.100 \equiv 100 \equiv 9(mod 13)$
and $10^{6k+1}\equiv 10 (mod 13)$
so we obtain  $10^{12k+2}-10^{6k+1}+1 \equiv 9-10+1 \equiv 0 (mod 13)$
similar in case $p=6k+5$ we get the desired result

[hide]why do you consider y in the form $6k+r$ ??? because $10^{6}\equiv 1(mod 13)$ is useful[/hide]


I let Maple v10 chew on it for 3167.0 seconds and still no solution: difficult indeed!

Before posting Polish MO problems, please check the resources section - many of them are already posted.

http://www.mathlinks.ro/Forum/resources.php?c=137&cid=39&year=2006

By clicking the problem numbers at the left of the page, you get to the respective topics, e. g. http://www.mathlinks.ro/Forum/viewtopic.php?t=82894 .

  darij

However, the problem hasn't been solved completely. You should read the topic again.  :D

[quote="ZetaX"]Being solved or not is no reason to repost them...[/quote]
I remember I post the solution right now on Mathlinks ,but I forgot where I post it .


geometry
ratio
angle bisector
geometry unsolved

Draw the nine-point circle $ (N,NM_a)$ intersecting the line $ T_aM_a$ again at the foot $ H_a$ of the A-altitude. Let $ X_a,Y_a$ be the unknown tangency points of  the incircle $ (I)$ and the A-exincircle $ (I_a)$ with $ BC.$ Since cross ratio $ (A,T_a,I,I_a)$ is harmonic, then so is the orthogonal projection on $ BC,$ namely $ (H_a,T_a,X_a,Y_a).$ $ M_a$ is also midpoint of $ X_aY_a,$ thus  by Newton's theorem, we have ${M_aY_a}^2 \equal{} M_aT_a \cdot M_aH_a$ $\Longrightarrow$ segment $ \varrho \equal{} M_aY_a$ is constructible. Circle $ (M_a,\varrho)$ meets $ M_aT_a$ at $ X_a,Y_a,$  such that $ X_a$ lies between $ H_a,T_a.$ Construct the incircle $ (I)$  tangent to $ M_aT_a$ at $ X_a$ and internally tangent to $ (N)$ in the upper halfplane. Draw the angle bisector of $ \angle BAC$ passing though $ I,T_a.$ This cuts the perpendicular line  to $ M_aH_a$ through $ H_a$ at vertex $ A.$  Tangents from $ A$ to $ (I)$  complete $\triangle ABC.$


Note: It seems  not factorable
What is x: $ 5x^2 \plus{} 6x \plus{} 11 \equal{} 0$ (mod 23)

[hide="solution"]
We can complete the square working in $ \pmod{23}$:

$ 5x^2 \plus{} 6x \plus{} 11 \equiv x^2 \plus{} 6*5^{ \minus{} 1}x \plus{} 11*5^{ \minus{} 1}$

Since $ 5*14 \equiv 1$, $ 14 \equiv 5^{ \minus{} 1}$.

Therefore,

$ x^2 \plus{} 6*14x \plus{} 11*14 \equiv 0$
$ x^2 \plus{} 15x \plus{} 16$
$ (x \plus{} 15*2^{ \minus{} 1})^2 \plus{} 16 \minus{} 19^2 \equiv 0$
$ (x \plus{} 19)^2 \equiv 0$
$ x \equiv \minus{} 19 \equiv 4$.

So $ x \equal{} 4$.

[/hide]

Two comments:

1) Your very first line is not true.  What you meant is that $ 5x^2 \plus{} 6x \plus{} 11 \equiv 0$ implies $ x^2 \plus{} 6*5^{ \minus{} 1}x \plus{} 11*5^{ \minus{} 1} \equiv 0$ (by multiplying through by $ 5^{ \minus{} 1}$, since $ (5, 23) \equal{} 1$).

2) Your sig currently reads, "High school cliche groups aren't groups because they don't have an identity!"  I like the joke, but I think you want "[url=http://www.merriam-webster.com/dictionary/clique]clique[/url]" ("a narrow exclusive circle or group of persons"), not "[url=http://dictionary.reference.com/browse/cliche]cliche[/url]" ("a trite, stereotyped expression").


inequalities
induction
inequalities proposed

If $ a_1 ,a_2 ,...,a_n$ are positive numbers and $ k$ is no-negative integer ,prove that

$ \frac{a_1 ^k \plus{}a_2 ^k \plus{}...\plus{}a_n ^k}{n}\leq \frac{a_1 ^{k\plus{}1} \plus{}a_2 ^{k\plus{}1} \plus{}....\plus{}a_n ^{k\plus{}1}}{a_1 \plus{}a_2 \plus{}...\plus{}a_n}$

[quote="Rafikafi"]Just Chebychef[/quote]

Okay.It's Chebychef  :P

Yes,I can see that the proof is very easy by Chebyshev's inequality

I used Cauchy and I completely mess up everything

Thanks for the Hint   :P

Well if you really want to you can give a Cauchy-Schwartz proof too!

Proceed by induction: For $ k\equal{}0,1$ there's nothing to prove.
Cauchy gives
\[ a_1^{k\plus{}1}\plus{}a_2^{k\plus{}1}\plus{}\cdots\plus{}a_n^{k\plus{}1}\geq \frac{\left(a_1^k\plus{}a_2^k\plus{}\cdots\plus{}a_n^k\right)^2}{a_1^{k\minus{}1}\plus{}a_2^{k\minus{}1}\plus{}\cdots\plus{}a_n^{k\minus{}1}}\]
By the inductive hypothesis we know that
\[ \frac{a_1^k\plus{}a_2^k\plus{}\cdots\plus{}a_n^k}{a_1^{k\minus{}1}\plus{}a_2^{k\minus{}1}\plus{}\cdots\plus{}a_n^{k\minus{}1}}\geq \frac{a_1\plus{}a_2\plus{}\cdots\plus{}a_n}{n}\]
so we obtain
\[ a_1^{k\plus{}1}\plus{}a_2^{k\plus{}1}\plus{}\cdots\plus{}a_n^{k\plus{}1}\geq (a_1^k\plus{}a_2^k\plus{}\cdots\plus{}a_n^k)\frac{a_1\plus{}a_2\plus{}\cdots\plus{}a_n}{n}\]
Which is what we wanted.
 :lol:

You can solve this with MUİRHEAD too  :P

If you know how to prove it, yeah  :lol:


[quote="Rust"]1. $y=(5+\sqrt{21})^{x}\to y+\frac{4}{y}=8$[/quote]
You are wrong! Product of terms is equal to $4^{x}$.

[color=purple]I think these above problem are quite hard and i hope you can write clearly  :) [/color]


number theory
least common multiple
modular arithmetic

What is the least positive integer that has a remainder of 0 when divided by 2, a remainder of 1 when divided by 3, and a remainder of 2 when divided by 4?

If it leaves a remainder of 0 when divided by 2, then it must be an even number. Let's try a few numbers that fit the first and last requirement, and see if they fit the 2nd.

6 - No
10 - Yes!

Another solution

We note that in order for a number to satisfy these conditions, it must be 2 less than a multiple of 2, 3, and 4.  So the answer is $ \text{LCM}(2,3,4)\minus{}2\equal{}\boxed{10}$

[hide="Modular Arithmetic"]
We have \[x\equiv 0\mod 2\equiv 1\mod 3\equiv 2\mod 4\implies x+2\equiv 0\mod 2,3,4\implies x+2\equiv 0\mod 12\implies x\equiv -2\mod 12\implies x=\boxed{10}\][/hide]

The last statement implies the first statement. Then we just check numbers that are 2 mod 4. We only need to check a maximum of 3 by CRT and we find that 10 works.

[hide=Sol]The answer should be 2 less than the LCM of 2,3,4. ANs = 3*4-2 = 10[/hide]


Fie a apartine lui [0,1) un nr scris in forma zecimala
construim un sir xn, n>=1 astefl:
-  x1 se obtine permutand in mod arbitrar primele trei zecimale ale lui a
-  x2 se obtine permutand in mod arbitrar zecimalele a doua, a treia, si a patra ale lui x1
- si in general x(n+1) se obtine permutand in mod arbitrar zecimalele n, n+1,n+2 ale lui xn, oricare n>=1;

 a) aratati ca orice sir astfel construi este convergent catre o limita b
 b ) este posibil ca a sa fie rational si b sa fie irational?
 c) gasiti a pentru care orice sir (xn)n>=1 , construit ca mai sus, are limita irationala

dat cu copy paste din arhiva, nu am mai modificat in latex ca oricum nu-i mare lucru de scris

Problema aceasta seamana cu o problema de la un concurs interjudetean...mi se pare ca autorul ei era Dan Schwarz. Exista si o alta solutie in afara de cea oficiala?

seamana cu o problema de la baraj de anul trecut, si s-a mai dat si la bucuresti la sector, anul acesta.


geometry
trigonometry
symmetry
circumcircle
geometric transformation
reflection
geometry unsolved

Let $ ABC$ is a right triangle, $ \angle A\equal{}90^0$, $ AB\equal{}AC$. Let $ D$ is midpoint of $ BC$, $ M$ is a change point on $ AD$. Let $ N, P$ are othogonal projection of point $ M$ on $ AB$ and $ AC$ respectively. Let $ d$ is a line through point $ B$ and perpendicular with $ AB$. Let $ PD$ meet $ d$ at $ E$. Let $ H$ is othogonal projection of point $ N$ on $ PD$.

1) Prove that: $ B, M, H$ are collinear.

2) Find place of point $ M$ for area of $ \triangle ABH$ is maximum.

3) Prove that when point $ M$ change on $ AD$. The line $ HN$ allway pass a fixed point.

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My school site: http://www.chuyenquangtrung.com.vn

[img]http://i004.radikal.ru/0805/55/2457790a036a.jpg[/img]
[color=darkblue][size=150][b]1)[/b][/size]$ ANMP$ is square. Let $ \angle{HNM} \equal{} \alpha$. Then $ \angle{PNH} \equal{} 45^\circ \minus{} \alpha \equal{} \angle{PMH}$ (because $ MNPH$ is cyclic). Also $ \angle{NPM} \equal{} \angle{NHM} \equal{} 45^\circ$. Then $ \angle{ABM} \equal{} 180^\circ \minus{} \angle{BNH} \minus{} \angle{NHM} \equal{} 45^\circ \minus{} \alpha$. So $ \angle{NMB} \equal{} 90^\circ \minus{} \angle{NBM} \equal{} 45^\circ \plus{} \alpha$ and $ \boxed{\angle{BMH} \equal{} 45^\circ \plus{} \alpha \plus{} 90^\circ \plus{} 45^\circ \minus{} \alpha \equal{} 180^\circ}$, so $ B,M,H$ are collinear.

[size=150][b]2)[/b][/size] $ S_{ABH}$ is $ max$, when $ \dfrac{1}{2} AB \cdot BH \cdot \sin{(45 \minus{} \alpha)}$. Because $ \dfrac{1}{2}AB \equal{} const$, $ S_{ABH} \equal{} max$, when $ BH \sin{(45 \minus{} \alpha)} \equal{} max$. But in triangle $ BHD$ we have $ BH \equal{} \dfrac{BD \sin{(45 \plus{} \alpha)}}{\sin{45}}$. Again $ \dfrac{BD}{\sin{45}} \equal{} const$, so we must find $ \sin{(45 \minus{} \alpha)}\sin{(45 \plus{} \alpha)} \equal{} max$ or $ \cos{2\alpha} \minus{} \cos{90} \equal{} \cos{2\alpha} \equal{} max$.  But $ \cos{2\alpha} \equal{} max$, when $ \alpha \equal{} 0$ or $ \boxed{M \equal{} D}$.

[/color]

2) Find place of point $ M$ for area of $ \triangle ABH$ is maximum.

$ S_{ABH}\equal{}\frac{1}{2}AH.HB$

$ \Longrightarrow S_{ABH}^2\equal{}\frac{1}{4}AH^2.HB^2 \le \frac{1}{4}\frac{(AH^2\plus{}HB^2)^2}{4}\equal{}\frac{1}{16}AB^4$

$ \Longrightarrow S_{ABH} \le \frac{1}{4}AB^2\equal{}\frac{a^2}{4}$

$ \Longrightarrow Max S_{ABH}\equal{}\frac{a^2}{4}$, when $ AH^2\equal{}HB^2 \Longleftrightarrow AH\equal{}BH \Longleftrightarrow H \equiv D \Longleftrightarrow M \equiv D$.

Who can slove problem (3): Prove that when point $ M$ change on $ AD$. The line $ HN$ allway pass a fixed point?

Let $ At$ is parallel to $ BC$,$ K$ is the intersection of $ HN$ and $ At$.We will prove that $ K$is a fix point
easy to see that $ AKDH$ is cyclic obtain $ \angle AKD\equal{}\angle AHP\equal{}45$ then K is a fix point

[quote="thanhnam2902"][color=darkred]Let $ ABC$ is a right triangle, $ \angle A \equal{} 90^0$, $ AB \equal{} AC$. Let $ D$ is midpoint of $ BC$, $ M$ is a change point on $ AD$. Let $ N, P$ are othogonal projection of point $ M$ on $ AB$ and $ AC$ respectively. Let $ d$ is a line through point $ B$ and perpendicular with $ AB$. Let $ PD$ meet $ d$ at $ E$. Let $ H$ is othogonal projection of point $ N$ on $ PD$.
1) Prove that $ B, M, H$ are collinear.
2) Find place of point $ M$ for area of $ \triangle ABH$ is maximum.
3) Prove that when point $ M$ change on $ AD$. The line $ HN$ allway pass a fixed point.[/color] [/quote]

[color=darkblue][b][u]Proof.[/u][/b]

$ 1\blacktriangleright\ ANMHP$ is cyclically (the circle $ w$ with diameter $ AM$ ). Hence $ HM\perp HA$ and $ \boxed {\ \widehat {AMH}\equiv\widehat {HPC}\ }\ (1)$ .

Since $ \widehat {HPC}\equiv\widehat {DPC}\equiv\widehat {DNB}$ (by symmetry) obtain $ \boxed {\ \widehat {HPC}\equiv\widehat {DNB}\ }\ (2)$ . Since $ BNMD$ is cyclically obtain

$ \boxed {\ \widehat {DNB}\equiv\widehat {DMB}\ }\ (3)$ . From three upper relations results $ \widehat {AMH}\equiv\widehat {DMB}$ . But $ M\in AD\ \implies\ M\in BH$ .

[b]Remark.[/b] Denote $ R\in AH\cap BC$ . The point $ M$ is the orthocenter of the triangle $ ABR$ and $ R\in MN$ .

The line $ \overline {BDRC}$ is the image by inversion of the circle $ w$ with the pole $ A$ and the power $ p$ because

$ p \equal{} AN\cdot AB \equal{} AM\cdot AD \equal{} AH\cdot AR \equal{} AP\cdot AC$ .

$ 2\blacktriangleright$ Since $ H$ belongs to the circle with the diameter $ [AB]$ , then the area $ [ABH]$ is maximum iff

the distance of the point $ H$ to the diameter $ AB$ is maximum, i.e. $ H: \equal{} D$ what means $ M: \equal{} D$ .

$ 3\blacktriangleright$ The point $ H$ belongs to the circumcircle (fixed) of the triangle $ ABD$ . Since $ m(\widehat {NHA}) \equal{} m(\widehat {NHB}) \equal{} 45^{\circ}$

obtain that the ray $ [HN$ is the bisector of the angle $ \widehat {AHB}$ . Thus, the line $ HN$ passes through the (fixed) point $ F$

which is the reflection of the point $ D$ w.r.t. the line $ AB$ , i.e. the quadrilateral $ ADBF$ is a square.[/color]


algorithm
LaTeX
combinatorics unsolved
combinatorics

Source : Herbert Wilf's - Combinatorial algorithms

Hi Folks,

First of all let me say that i think this problem should have rather been in the Solved problem section...But it was off-limits to me..Anyways here goes the problem.

You are supposed to show that n*p(n) = SUMMATION (for all m < n) {SOD(n-m)*p(m)}...(I dont know how to use latex yet...will learn it soon)....where p(n) is the number of partitions of n and SOD(n) is the sum of divisors of n...

In the book I used, Herb comes up with a combinatorial proof for the problem but he does not give any example.

So all I am asking for is to kindly prove this theorem (so that i can understand it) and offer an example (so that i can better understand it)...

Regards
-Akash

A combinatorial proof should be more straightforward than other techniques.  What don't you understand about the proof?  As for an example, do you mean an example of what I presume is the bijection Wilf gives or an example of the problem statement?  You can compute the latter yourself.


Counterexample: $ n\equal{}7$ and $ a_1\equal{}a_2\equal{}a_3\equal{}1, a_4\equal{}a_5\equal{}a_6\equal{}a_7\equal{}0$

Similarly, take some of the variables $ 1$ consecutively (the amount of proper, about $ \sqrt{n}$), we get that the inequality is not valid for all $ n \ge 7$. But it's right for $ n\equal{}2,3,4$. For $ n\equal{}5,6$ I haven't got any idea for the time being.


function
inequalities
algebra unsolved
algebra

[quote="pvthuan"]Let $ a,b,c,d$ be real numbers and $ p_1 \equal{} a \plus{} b \plus{} c \plus{} d$, $ p_2 \equal{} ab \plus{} bc \plus{} cd \plus{} da \plus{} ac \plus{} bd$, $ p_3 \equal{} abc \plus{} bcd \plus{} cda \plus{} dab$, $ p_4 \equal{} abcd$.

Express the following
\[ p(a,b,c,d) \equal{} (a \minus{} b)(b \minus{} c)(c \minus{} d)(d \minus{} a)(a \minus{} c)(b \minus{} d)
\]
in terms of $ p_1, p_2, p_3$, and $ p_4$.[/quote]
 :rotfl:  $ p(b,a,c,d)\equal{}\minus{}p(a,b,c,d)$ so this is an anti-symmetric function and cannot be rendered by symmetric functions

:?:  $ p(a,b,c,d)$ is not symmetric, it's sign depends on the order of $ a,b,c,d$.
However $ p(a,b,c,d)^2$ is symmetric and homogeneous of degree $ 12$.
It's called the discriminant and you'll find how to calculate it [url=http://en.wikipedia.org/wiki/Discriminant]here[/url], as the determinant of a $ 7\times 7$ matrix.

Thank you, spenferkel and olorin. My initial aim is to start from
\[ (a\minus{}b)^2(b\minus{}c)^2(c\minus{}a)^2(a\minus{}d)^2(d\minus{}c)^2(b\minus{}d)^2\geq0\]
to make an inequality about $ p_1, p_2,p_3, p_4$. I want $ p_4$ to be bounded by some expressions of $ p_1,p_2,p_3$.


i would like training materials too! Anyone up there for exchange? I don't have much though............

You should work very hard since 5 of the mexicans that are going to IMO this year also can go next year.  Before i get 1st place in OMM, i didn´t know that that was the easy part. By the way, you should start whit all the OMM`s Books and the Engel before you start whit other nations problems, since many of them are more dificult than Mexican´s problems. However, good luck.

Yeah the engel book¡¡ i´ve had a copy two years ago  i know all the problems¡¡ And the omm´s books 3 years ago¡¡ so i think it´s time to start with other countrys. OH dont worry i´m prepared¡¡ well see who wins in campeche  :P  Best of luck in the imo ¡¡¡

[quote="marko avila"]HELLO¡¡  Anyway i am fortunate to live about 6 blocks away from the hotels in wich the imo will be held in merida, so i will go visit and admire this  event (wich i hope to participate in next year) . So i was wondering if anyone might have any type material that we could exchange so i could study more  :oops: . Valentin : i heard you were giving booklets of the romanian olympiad last year or 2 Are you giving something now? money is no object :D  anyway i´d also like to meet you. also Darij: maybe you can give me something to help out on my geometry ; it´s not that bad bout it could be so much better¡¡ :lol: i´d appreciate it very much¡¡ finally Pascual: i know you´re going to the imo maybe you could help me out andí´d also like to meet you. anyway i´m going to help out in the imo, i´ll be carrying boxes, giving out exams, and paper etc. my name is marco. i would love to meet you guys¡¡ so if you have anything at all that might help me out please sell it to me or something like that :blush:  Thanks in advance¡¡ :D  :lol:[/quote]
marco please tell me something about this IMO
are we going to visit any nice place there?
is the beach near?
do you have any idea were are we gonna stay?
 ;)

Well it will be very nice¡¡¡ you will be staying either at the hyatt, the fiesta americana or the holiday inn they are all 4 to 5 stars. the beach is about 20 minutes away by car but you would have to go on your own expense cause it´s not part of the trip :? ; but you will go visit Chichen itza and Dzibichaltun wich are ancient cities :lol: . the judges will go to the beach  :lol:  they will go to a nice hotel called the reef club. it will be very nice¡¡¡ :D

[quote="marko avila"] the judges will go to the beach  :lol:  they will go to a nice hotel called the reef club. it will be very nice¡¡¡ :D[/quote]Sounds nice :D

[quote="Albanian Eagle"]Valentin you are a lider or something?are you coming to IMO 2005?[/quote]I will be a coordinator (which is what I think marko ment by "judge").

[quote="Simo_the_Wolf"]so we are not together in the same hotel?[/quote]
so i think it will be like in Japan.
we were in different buildings,
if they sort them alfabetically i hope "we" will be in the best hotel...

[quote="Albanian Eagle"]all right!
can anyone tell me who is the best in math in this forum?[/quote]

They are too numerous too mention. Go and check for yourself in the various problem-solving sections!  ;)

[quote="Albanian Eagle"]and an other question, is there any "IMO problem proposer" in this forum?[/quote]

Valentin Vornicu for instance. Ideahitme as well.  :)

[quote="Albanian Eagle"]what do you think is a god material of NT for preparing for the IMO.I just prepared with A.Engels book, and Paul Zietz's "the art and craft of problem solving"  :?:[/quote]
standard books include introduction to the theory of numbers by ivan niven or the book written by hardy and wright.


function
logarithms
algebra unsolved
algebra

[quote="santosguzella"]Let $ h: R \rightarrow R$ a function such:
1) $ h(xy) \equal{} h(x) \plus{} h(y)$
2) $ h\Big(\max\Big(\frac {m}{2^p3^q},\frac {n}{2^p}\Big) \plus{} \min\Big(\frac {m}{2^p},\frac {n}{2^p3^q}\Big)\Big) \equal{} \log_2 (q \plus{} 1)$, for all $ m,n \in Z$ and $ p,q \in N$
Prove that if $ x \in Q$ then $ h(x) \equal{} 0$[/quote]

This is a crazy problem already posted with a lot of errors in the statement. Stop post it !

It's obviously wrong : Take in "$ 2)$" $ m \equal{} n \equal{} p \equal{} q \equal{} 1$ and you have $ h(\frac {2}{3}) \equal{} 1\neq 0$

See [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=182632[/url]