vector
calculus
real analysis
real analysis unsolved

Two vectors $ \underline{\bold{a}}(\underline{\bold{x}}, t)$ and $ \underline{\bold{b}}(\underline{\bold{x}}, t)$ are related by

${{ \frac{{{\delta}} \underline{\bold{a}}}{{\delta t}}} = \nabla \times \underline{\bold{b}} ,       \frac{{{\delta}} \underline{\bold{b}}}{{\delta t}}} = - \nabla \times \underline{\bold{a}}$

Show that

${{ {\iint_S (\underline{\bold{a}}  \times \underline{\bold{b}}}).\underline{\bold{dS}} = -\frac{{1}} {2}} \frac{{d}} {dt}}{\iiint_V (a^{2} + b^{2})d\tau}$,
where $ a=|{\underline{\bold{a}}|}$, $ b=|{\underline{\bold{b}}|}$ and $ S$ is the surface enclosing the fixed volume $ V$.

If $ \underline{\bold{a}}$ and $ \underline{\bold{b}}$ are both irrotational in $ V$, then deduce that

$ {\iint_S (\underline{\bold{a}}  \times \underline{\bold{b}}}).\underline{\bold{dS}} =0$

Vector Calculus is fun! Start with the divergence theorem. Use the identity $ \nabla\cdot(\vec F\times\vec G)\equal{}\vec G\cdot (\nabla\times \vec F)\minus{}\vec F\cdot (\nabla\times \vec G)$ (prove it by expanding both sides). Then use the given differential equations for $ \vec a$ and $ \vec b$. Finally, $ \frac{d}{dt}|\vec a|^2\equal{}2\langle \vec a, d\vec a/dt\rangle$ by the product rule (which applies to any [url=http://en.wikipedia.org/wiki/Bilinear_form]bilinear form[/url], not just the usual product).

would it be possible for you to go into a little bit more detail? I'm beginning to struggle with this topic, and I think a well worked example would be very beneficial. Thanks.

actually there's no need, what you said was perfectly sufficient. Thanks for your help.


LaTeX
induction
function
logarithms
algebra unsolved
algebra

I don't now the answer of this :
Let $\{a_n\}$ be a sequence of integers such that for $n \geq 1$ we have $(n-1) \cdot a_{n+1} =(n+1)\cdot a_n - 2(n-1)$.
If 2000 divides $a_{1999}$, find the smallest $n\geq 2$ such that 2000 divides $a_n$.

Sorry that I can't type LATEX, I'm the beginner...

[i]Edited by Myth[/i]

Is this what you wanted to write :)?

Let $a_{n}$ be a sequence of integers such that for $n\geq 1$:
$(n-1)\cdot a_{n+1}=(n+1)\cdot  a_{n} - 2(n-1)$
if 2000 divides $a_{1999}$ , find the smallest $n\geq 2$ such that 2000 divides $a_{n}$

You can learn $\LaTeX$ from examples. Just examine you post now.

You can prove by induction that $a_n=(a_2-2)\cdot\frac{n(n-1)}{2}+2(n-1)$. The rest should be easy...  :?

uhm...
how about trying with generating functions? everything coming out is quite easy (and short) to integrate...
i didn't make last calculations, but it doesn't look impossibile at all, just some expanding $ \ln (1-x) $ and $ (1-x)^{-n} $ for a couple of values of $n$, or something like this :)
of course it's not as short and as elegant as your proof, but it's quite a safe way..

As for me, I must solve in such a way which Japanese high school student can understand.


logarithms
inequalities
function
inequalities proposed

Sorry for that I have just edited the problem. Coud you correct your answer, SplashD? :omighty:


Given any triangle, let its circumradius and inradius be R and r respectively, prove that R \geq 2r.

Look at [url]http://www.kalva.demon.co.uk/imo/isoln/isoln626.html[/url] for a rigorous proof.

By Feuerbach's theorem we have the incircle is tangent to the 9-point circle.
Notice that the radius of the 9-point circle is R/2, we have r<R/2, which is the result.

If you look at Mitrinovic's ineq you have that:
3sqrt(3)*r<=p<=3sqrt(3)/2*R and talking the 1st and last relations you get the desired ineq! :)

There is another way to prove it: you know that sum(cosA)=1+r/R and also that sum(cos(A))<=3/2.

sum(cos(A))<=3/2 is easy to prove as the function f: (0,pi/2)-->R, f(x)=cos(x) is concave.

and the identity follows from the fact that:
sum(cos(A))=1+4 prod(sin(A/2)) (you prove this by transforming the sums in products and having in mind that A+B+C=pi)

then we have that:
prod(sin(A/2))=r/(4R) ; you get this by replacing sin(A/2) with sqrt((p-b)(p-c)/(bc))

and you have another solution...


Show that the speed of longitudinal waves along a spring of force constant k is $ \sqrt{\frac{kL}{\mu}}$,where L is the length of the spring and μ is the mass density. You are not allowed to use the derivation of the wave equation.

i beleive $ L$ is the extension of the spring.

[quote="Nihat"]Show that the speed of longitudinal waves along a spring of force constant k is $ \sqrt {\frac {kL}{\mu}}$,where L is the length of the spring and μ is the mass density. You are not allowed to use the derivation of the wave equation.[/quote]

Start with the Work-Energy theorem.

	 \[ W = \int_{0}^{L} \dot{p}\,dx =\int_{0}^{L} -kx\,dx
\]

Differentiating under the integral,

	 \[ \dot{p}(L)\frac{\partial L}{\partial u}= -kL\frac{\partial L}{\partial u}
\]

We can eliminate $ \frac{\partial L}{\partial u}$  from the above,

	 \[ \dot{p}(L)=-kL
\]

From the Hamilton – Jacobi equations: $ \dot{p}= -\frac{\partial \mathcal{H}}{\partial u}$ .

	 \[ -\frac{\partial\mathcal{H}}{\partial u}|_{u=L}=-kL
\]

The wave itself has no potential energy and thus $ \mathcal{H}=\frac{p^2}{2m}$ .

	\[ \frac{\partial \mathcal{H}}{\partial u} =\frac{p}{m}\frac{\partial p}{\partial u}=v\frac{\partial(mv)}{\partial u}
\]

	\[ \frac{\partial(mv)}{\partial u}=m\frac{\partial v}{\partial u}+v\frac{\partial m}{\partial u}
\]

In the above expression, $ \frac{\partial v}{\partial u}\sim \nabla\cdot\vec{v}$ .

$ \nabla\cdot\vec{v} = 0$ , unless the spring is a “compressible fluid.”  Therefore,

	 \[ \frac{\partial\mathcal{H}}{\partial u} = v^2\frac{\partial m}{\partial u} = kL
\]

	\[ \frac{\partial m}{\partial u}\doteq \mu
\]

	\[ \Rightarrow \mu v^2 = kL
\]

	\[ \boxed{\therefore v =\sqrt{\frac{kL}{\mu}}}
\]


number theory
least common multiple
greatest common divisor
percent

LCM: Least Common Multiple
and
GCF: Greatest Common Factor

Not to get these two mixed up, here's some questions:
1.) What is the GCM of any pair of positive integers?
2.) Is there a CM (Common Multiple) for a pair of integers with 0 as one of the integers? If so which integer is it? If not why?

3.) What is the GCM of any pair of negative integers?
4.) Is there a CM to a pair of a positive integer and a negative integer?

5.) What is the LCF of a pair of positive integers?
6.) Is there a CF (Common Factor) for a pair of integers with 0 as one of the integers? If so which integer is it? If not why?

7.) What is the LCF of a pair of negative integers? Does it matter which signs the integers have?
8.) Is there a CF to a pair of a positive integer and a negative integer? What does this tell us about LCFs of all integer pairs?

9.) Show that there is (not) a pair of integers in which the LCM and the GCF of the numbers are the same.

First problem series I typed up to make it look like a problem set from a text book :)

Ur 10 digit problem fryed my brain..im too lazy to try this one.  But is it masters round?

Okay, I take this post back.

--deleted--

It's supposed to be LCF and GCM :lol:

Mmm, fried brains, yum-num :twisted:

All this is one-hundred percent original so I have no idea what the level of difficulty is, but it sure is interesting! (or is it only me? :) )

It's interesting because it actually makes you think instead of using predefined concepts. Yes, they are supposed to be GCM and LCF.

[hide]

1. There's no such thing because they go on infinitely.

2. Yes, it is 0. 0 is a multiple of everything because n*0 = 0.

3. -LCM(|x|,|y|) where x and y are the two integers

4. Yes, you can multiply them together.

5. 1, because 1 is a factor of every number.

6. Yes, it is 0 again. 0 if a factor of everything as well, because

since 0 is a multiple, the only way that n*x = 0 is if x=0 as well.

7. -GCF(|x|,|y|) where x and y are the two integers.

8. Yes, 1 is an example. They would be negative? Because all of the positive factors of an integer have a negative factor corresponding to it as well.

9. The pair (1,1) has LCM of 1 and GCF of 1...?

[/hide]

took a while, but it wasn't too bad


function
integration
real analysis
real analysis unsolved

Now comes my favorite one :lol: .

Let $E_k$, $k=1,2,3,\dots$ be a family of measurable subsets of $\mathbb R$ such that $m(E_k)>\frac12$ for all $k$ and $m(E_k\cap E_\ell)\leq \frac 14$ for $k\neq \ell$. Prove  that

a) $m\Bigl(\bigcup_{k\geq 1}E_k\Bigr)\geq 1$ (not very hard)

b) $m\Bigl(\bigcup_{k\geq 1}E_k\Bigr)> 1$ (hard)

Seems like it's time for a [hide="hint"] Use Cauchy-Schwartz on the sum of the characteristic functions of $E_k$, $k=1,\dots, N$ [/hide]

Since nobody solved it for a month, let me post a solution to a). I'll leave b) as a challenge.

Let $f_k$ be the characteristic function of $E_k$. Let $E=\bigcup_k E_k$. If $m(E)>1$, we are done. Otherwise, take a large $N$ and consider the function $F=\sum_{k=1}^N f_k$. By Cauchy-Schwartz, we have $\left(\int_{E} F\,dm\right)^2\leqslant m(E)\int_E F^2\,dm$. But the LHS is at least $(N/2)^2$ while the RHS is at most $m(E)$ times $N+\frac{N(N-1)}4=\frac{N(N+3)}4$ because $\int f_k^2\,dm=m(E_k)\leqslant 1$ and $\int f_k f_\ell\,dm\leqslant \frac14$. Thus ${m(E)\geqslant \frac{N^2}{N(N+3)}}$. Passing to the limit as $N\to\infty$, we get $m(E)\geqslant 1$.


calculus
integration
logarithms
function
derivative
calculus computations

This function does not have an elementary anti-derivative, it involves the polylogarithm function.  Try it here (for ln(x) use Log[x]):

http://integrals.wolfram.com/index.jsp


geometry
3D geometry
prism
geometry proposed

Is it true that for each $ n$, the regular $ 2n$-gon is a projection of some polyhedron having not greater than $ n \plus{} 2$ faces?

Is a regular prism 2$n$-gonal be or goal?


number theory
least common multiple
number theory unsolved

I have some ideas for this problem but didn't manage to finish it. Could somebody solve this?

Prove that for all $n\in\mathbb{N}$ \[[1,2,\dots,n]=\left[{n\choose 1},{n\choose 2},\dots,{n\choose n}\right]\] iff $n+1$ is a prime number.

very nicebut I posted it before in Mathlink .The orginal problem from AMM 2686:
    Prove that  $(n+1)\left[\binom {n}{0}+\binom {n}{1}+..+\binom {n}{n}\right]=[1,2,..,n+1]$

I found it, thank you.
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=34798


MOEMS is an abbreviation which means Math Olympiads for Elementary and Middle Schools; Elementary and Middle School is defined to be from 1st to 8th grades inclusive. So high schoolers answering the questions do NOT belong here. Please don't answer the questions here unless you have something very important to say. Thanks! :)

[quote="Geometry"]Where are freakin' questions? :rotfl:[/quote]Please behave on the forums. This is not a place for funny jokes. If you want to play with this kind of joke in AoPS, go to [url=http://www.mathlinks.ro/index.php?f=139]Games & Fun Factory[/url].

@ EVERYONE: Please stop spamming or posting non-informative posts. Have some sense of maturity, respect and responsibility for the forums.

 :spam:  <-----This will not be tolerated.

You must first register.
More information may be found [url=http://www.moems.org/]here[/url].


In the tennis championship between players A and B, the first player to win 3 sets is champion. Find the number of different ways for player A to win the championship.

[hide="I dont know"] You can win in 3, 4, or 5 games, since in 6, someone has to have one three already.
With 3 games, there's only one way. The thing is, with 4 games, you know you have to win the fourth game otherwise it's the same with the 3 game way. So maye its

$ \binom{3}{3}+\binom{3}{2}+\binom{4}{2}=10?$[/hide]

Oh yeah... Xpmath is right.

Btw, you can't put apostrophes in hide tags; they don't work for some reason.


I was looking through my old awards from junior high school, and found that back in 7th grade (I'm in 12th now), I had received an award for Third Place on the AMC 8.  What does this mean?  Does it mean third highest score anyone got? Third highest in the State of California?  Third highest in my middle school?

I'd imagine it would mean 3rd highest in your school, or possibly region. There should be a pretty big jam for perfect scores in California.

I'm fairly certain it means third place in your school.  Each school receives (as part of a packet) some standard pins and certificates, among these a first, second, and third place certificate.  Awards are not given by region or state unless your state does something special; the only other awards the AMC gives are honor roll and merit roll of distinction, the former being given to all competitors above a certain score and the latter being given to all 6th graders or below above a certain score.

I think it means third highest in your section. A school has multiple sections, and your teacher decided which section you were in. It might have been by math class, approximate level, grade, or other stuff.

depends, what was your score:
generally it would be third in your school. I dobut that AMC gives awards for region or state. If you got above a twenty (or so) your name is put in a booklet.

It so... definitley means the 3rd place in the school...

My brother got the first place in 8th grade but he didn't make the perfect scores....

there are alot of people in the world who get the full scores...

too bad...I only got 13 right(^.^) not even in the top ten in the whole school...

[quote="AMC Website"]The top three students for each school section will receive respectively a gold, silver, or bronze Certificate for Outstanding Achievement.[/quote]

[quote="Teacher's Manual"]
(4) Section Letter
For each math class participating in the AMC 8, assign a different section letter of the alphabet (A-Z). These letters are assigned at the Contest Manager's discretion. At least 10 students should be assigned to each Section Letter. If you return results with Sections which have fewer than ten people, we will combine these Sections into a single Section. Certificates are given to the top 3 students in each section (see page 6).[/quote]

[quote="Teacher's Manual"]# A list of the top 20 students in your school.
# A report for each section with a minimum of 10 students. This report lists the score of each student.
# One combined report for all sections of size less than 10.[/quote]


Στην ιστοσελίδα του pathfinder ένας μαθηματικός έθεσε το ερώτημα :
 Είναι σωστή η πρόταση : $3 \leq4$ ;
 Παρακαλώ τους φοιτητές να εκφράσουν  την άποψή τους  εδώ στο forum αλλά κυρίως να ρωτήσουν και τους καθηγητές τους των Μαθηματικών να  πουν την άποψή τους.
 Τη δική μου άποψη την έγραψα σε εκείνο το  forum και δεν την εκφράζω και εδώ , πριν διαβάσω τη γνώμη σας.
 Διότι , όταν  λέει ο Αμβρόσιος ότι σε αυτή τη χώρα όλα διαλύονται , τον κοιτάζουμε χαμογελώντας.(Αμβρόσιε κράτα γερά και μαχητικά!)Σε λίγο ακόμα ,  με αυτά που διαβάζω ... εγώ τουλάχιστον σαλτάρω .

Βοηθείστε(όχι να σαλτάρω) !

  Μπάμπης

[color=red] '' Εναντίον της βλακείας και οι Θεοί μάταια μάχονται !
                                                                      Schiller [/color]

Ayto to theoro ena erotima pou exei na kanei kathara me ta themelia
twn mathimatikon, ta simbola 3,4 kai opoiadipote simbola
den ekfrazoun tipota an emeis den exoyme antisitixisei
se ayta ta noimata pou theloume. ipo to prisma ayto an exoume
orisei ti einai oi fisikoi arithmoi kai exoume thesei axiomata pou tha
kathorizoyn to pas tha allilepidroun metaxi yous tote mporoume na  pisteuoume s ayta kai na kanoume mathimatika.

Tora epidi i pisti mou epi toy erotimatos diamorfonete apo ta axiomata toy Peano kanw link to site
http://eclass.uoa.gr/claroline/document/document.php
kai to biblio einai Φυσικοί -- Ακέραιοι -- Ρητοί (Terence Tao- Analysis)

Ως φοιτητης απαντω σε αυτο το ωραιο ερωτημα...

Το εχω ξανασυναντησει και στο παρελθον (μακρινο....3+ χρονια...)

Λοιπον, ας περιοριστουμε στο οτι το συμβολα 3 και 4 ειναι αυτα του σωματος $\mathbb{N}$ ...

Φανταστειτε ενα συνολο, εστω Μ, τετοιο ωστε να περιεχει ολους τους αριθμους που ειναι αυστηρα μικροτεροι του 4:$M= \{x|x<4 \}$

Eστω επισης Μ' το συνολο που περιεχει ολα τα στοιχεια του Μ [b]και[/b]  το 4. Τοτε $3 \in M$ και $M \subset M'$

Ετσι $3 \in M'$ και $3\leq 4$

Eνα παρομοιο προβλημα ειναι το εξης:

Αν $a<1$ τοτε $a\leq 1$ ???

Aυτα...
Θα ηθελα να ακουσω και αλλες αποψεις ([i]και βεβαια του μαθηματικου του pathfinder[/i]....)

 :)

Kata tin apopsi mou fisika kai einai logiki i protasi $3\leq 4$. Prwta-prwta an to paroume me tin logiki, otan enas arithmos $a$ einai mikroteros apo ton $b$ tote den ton empodizei tipote apolitws na einai kai mikroteros 'h isos me auton...

Isxiei diladi i logiki sinepagwgi $a<b \Longrightarrow a\leq b$.

Epeidi tixainei oi arithmoi autoi na einai kai fisikoi as to paroume Sinolothewritika (Simfwna me tin Sinolothewria tou Zermelo - Fraenkel) stin opoio orizetai oti to \mathbb{N} einai i tomi olwn twn epagwgikwn sinolwn kai orizetai i Sinartisi tou amesws epomenou (legetai Successor sta Agglika kai simvolizetai me S) ws eksis:

$0=\varnothing$
$1=\{0\}=\{\varnothing\}$
$2=\{0,1\}=\{\varnothing,\{\varnothing\}\}$
$3=\{0,1,2\}=\{\varnothing,\{\varnothing\},\{\varnothing,\{\varnothing\}\}\}$
$4=\{0,1,2,3\}=\{\varnothing,\{\varnothing\},\{\varnothing,\{\varnothing\}\},\{\varnothing,\{\varnothing\},\{\varnothing,\{\varnothing\}\}\}\}$
........

Opws gnwrizoume i praksi me tin opoia sigrinoume sinola einai i sxesi "iposinolo" kai opws fainetai apo ta parapanw (den iparxei logos na mpleksoume mesa tin domi twn Dedekind-Peano gia to oti einai kala orismeni i parapanw sinartisi S kai plirei kapoia aksiwmata) isxiei oti to $3$ ws sinolo (auto pou egrapsa parapanw) einai gnisio iposinolo tou $4$ (ara mporei na einai gnisio iposinolo 'h ison me auto). Opws kai na to pareis einai thema mathimatikis logikis...

(Einai auto to "h" to diazeuktiko pou kanei oli tin douleia!! Leme: mikrotero 'H ison)

Auti einai i gnwmi mou, (ean fisika ekeinos pou to egrapse milouse gia tous arithmous $3, 4$ ws fisikous kai apodexomenos tin Sinolothewria me ta gnwsta tis aksiwmata. Ean oxi, tote zise Mai mou na fas trifilli... Den kserw ti na pw!).

Alexandros

Y.G. (Akomi kai oi mathites tis 3hs Likeiou to antimetwpizoun polles fores stis askiseis tous, p.x. sto Thewrima Bolzano theleis na deikseis oti mia riza vrisketai sto diastima $(0,\pi)$ kai anti autou apodeiknieis oti vrisketai sto $(1,3)$ (pou apla einai isxirotero) kai META epanerxesai kai pali (gia na ikanopoiiseis ton siggrafea tou vivliou pou sou zitaei oti "Na deikseis oti iparxei riza sto diastima $(0,\pi)$") kai les oti telika i riza anikei sto $(0,\pi)$. Einai akrivws to idio pragma! Otan egw exw sto portofoli mou 2 eurw tote FISIKA exw ligotera 'H isa apo 1.000.000 eurw!!)

στο N δηλ στους φυσικους εχουμε τον ορισμο ''ολικως διατεταγμενο συνολο'' το οποιο μας οδηγει στην αρχη της καλης διαταξης η οποια με βαση της αποδειξης του Θαναση μας δινει οτι ειναι σωστο. μηπως χανω κατι?

Καλημέρα σας
αφού καλημερίσω τον φίλτατο Κύριο Στεργίου κι του ευχηθώ υγεία θα τον ρωτήσω γιατί χαμογελάτε όταν το λέω αυτό ?? μήπως επειδή είναι υπερβολή? ισώς κι ναι αλλά εστω κι υπερβολικά εκφράζει την αλήθεια
κι τι ακριβώς σας κάνει να σαλτάρετε?? μήπωως αυτά που λέω ? αν πραγμαατικά είναι έτσι δυστυχώς φταίει η κατάσταση που με αναγκάζει να λέω τέτοια πράγματα κι αφού ευχαριστήσω θέρμα την προτροπή του (εύχομαι κι εγώ το ίδιο σε αυτον) περιμένωτην απάντσή του
όσον αφορά τώρα την ερώτηση αν εφαρμόσουμε τα θεμέλιωδη θεωρήματ της διάταξης 'Η  την μαθηματική λογική για το ανισοτικό σύμβολο μεγαλύτερο ή ίσον θα πάρουμε την απάντηση που θέλουμε
χαιρετώ

Η άρνηση της πρότασης  $p : 3 \leq 4$ είναι η $q: 3 > 4$ που είναι ψευδής.Αλλά δε μπορεί να είναι συγρόνως σωστές μια πρόταση και η άρνησή της.Άρα η $p$ είναι σωστή.
  Συνεχίστε ! Τελικά μόνο η ανταλλαγή απόψεων φέρνει πρόοδο.Μόνο που πρέπει να έχουμε το θάρρος να παραδεχόμαστε και το λάθος μας.Διαφορετικά γιατί βάζουμε ερωτήματα στα fora ? Αν είμαστε σίγουροι ότι η άποψή μας είναι σωστή , ας την κρατήσουμε μόνο για τον εαυτό μας.Αν έχουμε αμφιβολία , τότε μας τιμά να αλλάξουμε γνώμη και να γίνουμε καλύτεροι.Αυτονόητο!
 Τώρα καταλαβαίνω πόσο μπροστά είναι τα μέλη αυτού του forum που αγαπάνε να μαθαίνουν από τους άλλους , να κάνουν και λάθη , να ζητάνε ευγενικά συγνώμη και μάλιστα με την υπογραφή τους  και να προχωράνε !

[u] Μπάμπης[/u]


function
algebra
domain
college contests

Let f,g:[a,b]->[a,b] be continuous functions s.t. f(g(x))=g(f(x)) for all x in the domain and f is monotonous. Prove that there is a c in [a,b] s.t. f(c)=g(c)=c.

nice one. vietnam undergrad 2003.

first of all each of f and g have a fixed point.  if f is increasing then let g(d)=d be a fixed point of g. Then f(g(d))=g(f(d)) => f(d) = g(f(d)) which means that also f(d) is a fixed point. In this way we create a sequence (f(n) (d) )n which is a bounded sequence, and also monotonous (since f is increasing), thus convergent. it's limit is the point c we are looking for.

the case when f is decreasing is even easier: consider a random fixed point of f, f(c)=c. Then g(c)=f(g(c)) then g(c) is also a fixed point, and if g(c)<c then g(c) = f(g(c)) > f(c) =c which is a contradiction! again a contradiction arises when we try g(c)>c. Thus only one option remains valid: g(c)=c=f(c) and we are done.

hope all of you know why a continuous function f:[a,b]->[a,b] has a fixed point :D:D:D

well consider the auxiliar function g(x)=f(x)-x. Obviously it's continous, and g(a)=f(a)-a \geq 0 and g(b)=f(b)-b \leq 0 thus it must also take the intermediate value 0 ...

maybe your "how" was directed to another thing?

[quote="Valentin Vornicu"]
the case when f is decreasing is even easier: consider a random fixed point of f, f(c)=c. .....and if g(c)<c then [b]g(c) = f(g(c)) > f(c) =c[/b] which is a contradiction! again a contradiction arises when we try g(c)>c. Thus only one option remains valid: g(c)=c=f(c) and we are done.
[/quote]

I think the > in your proof should be replaced by  \geq. I don't see why should f be injective. This doesn't change anything. The proof is still valid, but a nicer contradiction is obtained.

f is decreasing ( that is f(x)<f(y) when x>y) and g(c) I supposed it is smaller than c which means that f(g(c))>f(c). nu equality involved.

So what you mean is that you only consider the situations f is increasing (x>y => f(x)>f(y))and f is decreasing (x>y =>f(x)<f(y)).
Then I suppose that you consider that the monotonous functions are injective. In Romania we use [b]strictly monotonous[/b] for this. So what does the monotonous in the [b]problem text[/b] stand for?

in Romanian you use the word "monoton". In English monotonous means either increasing or decreasing, that is (f(x)-f(y))(x-y) > 0 or (f(x)-f(y))(x-y) <0 for any x,y. I belive that you have an English problem here :D

I don't think monotonous means strict inequalities. I also think that we should use "strictly monotonous" if we're referring to monotonous and injective functions.

Well, if you see [url=http://mathworld.wolfram.com/MonotonicFunction.html]this link[/url] and [url=http://mathworld.wolfram.com/MonotoneDecreasing.html]that link[/url], you'll understand where the confusion comes from.

Personally, I can't even remember what I learnt monotone to be during training sessions... (this is what query forms are for...)

In high school books, 'increasing' and 'decreasing' mean strict inequalities.
But in the most math analysis books, it allows equality. So we should use 'strictly increasing(decreasing)'


[quote="tµtµ"]By rearrangement we have

$\left[\frac{(a+1)^3}{b}+\frac{(b+1)^3}{c}+\frac{(c+1)^3}{a}\right] \geq \sum (1+a)^3/a \geq 81/4$ with min for $a=b=c=1/2$[/quote]
I know but how to prove the minimum is 81/4?

What is the minimum of $(1+a)^3/a$ for $a > 0$ ?

What I mean is that why it get the minimum when a=b=c?

[quote="Arne"]Do you know when we have equality in the rearrangement inequality? :)[/quote]
i've forgot it.......  :oops:
and it's really a easy problem but i can't do it.  :blush:


trigonometry
geometry
geometric transformation
dilation
vector
function
complex numbers

Let $ z$ be a complex number. What is the locus of points on the complex plane such that $ \arg \left( \frac {z \minus{} 3}{z} \right) \equal{} \frac {\pi}{4}$?

Assume that there does exist some complex number $ z$ that satisfies the given equation. First, note that $ \frac {z \minus{} 3}{z}$ must be in either quadrant 1 (as its argument is in quadrant 1).

We hence have that $ \arctan(\frac {z \minus{} 3}{z}) \equal{} \frac {\pi}{4}$. Hence, $ \frac {z \minus{} 3}{z} \equal{} \tan(\frac{\pi}{4}) \equal{} 1$. Hence, $ z \minus{} 3 \equal{} z$. Since there are no complex numbers that satisfy the equation, we see that there are points that satisfy the given equation.

I'm not so sure, but I think $ z \equal{} 3 \plus{} 3i$ works, for example.

Edit: In general, I have $ z \equal{} 3\left(\frac { \minus{} x \plus{} 1}{2x^2 \minus{} 2x \plus{} 1} \plus{} \frac {xi}{2x^2 \minus{} 2x \plus{} 1}i\right)$ for all real $ x$ .. which might be a circle?

Hmm .. I don't know anything about inversions, but it you take the circle centered at $ \left( \frac 32, \frac 32 \right)$ and goes through the origin, and invert it with respect to the unit circle, dilate by $ 3$, and subtract from $ 1$, I think you get the line $ y\equal{}x$ ..

Excuse me for the incorrect proof. My mistake was stating that the argument of $ \frac{z \minus{} 3}{z}$ was $ \arctan{\frac{z \minus{} 3}{z}}$.

Let $ z \equal{} a \plus{} bi$. Note that $ \frac{a \plus{} bi \minus{} 3}{a \plus{} bi} \equal{} \frac{(a \plus{} bi \minus{} 3)(a \minus{} bi)}{a^2 \plus{} b^2} \equal{} \frac{(a^2 \plus{} b^2) \minus{} 3a \plus{} 3bi}{a^2 \plus{} b^2} \equal{} 1 \minus{} \frac{3a}{a^2 \plus{} b^2} \plus{} \frac{3b}{a^2 \plus{} b^2}i$. Since its argument is 1, we have that:

$ \frac{1 \minus{} \frac{3a}{a^2 \plus{} b^2}}{\frac{3b}{a^2 \plus{} b^2}} \equal{} 1$, i.e., $ 1 \minus{} \frac{3a}{a^2 \plus{} b^2} \equal{} \frac{3b}{a^2 \plus{} b^2}$, i.e., $ a^2 \plus{} b^2 \minus{} 3a \equal{} 3b$. Hence, $ a^2 \minus{} 3a \plus{} \frac{9}{4} \plus{} b^2 \minus{} 3b \plus{} \frac{9}{4} \equal{} \frac{9}{2}$, so $ (a \minus{} \frac{3}{2})^2 \plus{} (b \minus{} \frac{3}{2})^2 \equal{} (\frac{3}{\sqrt{2}})^2$. Hence, the locus of points that satisfy the given equation is a circle with radius $ \frac{3\sqrt{2}}{2}$, centered at $ (\frac{3}{2}, \frac{3}{2})$.

Excuse me if I've made any more mistakes, be it in my reasoning or algebra.

Yep :) Just make sure you show that it can be the whole circle, because points of the form $ a(1\plus{}i)$ can either have argument $ \frac{\pi}{4}$ or $ \pi \plus{} \frac{\pi}{4}$

I approached it by saying $ \frac {z \minus{} 3}{z} \equal{} t \plus{} ti$ for some $ t > 0$. Solving for $ z$, we get $ z \equal{} \frac {3}{(1 \minus{} t) \minus{} ti}$, which yields the same region as above if you take the real and imaginary parts.

If $ z \equal{} x \plus{} iy$, then $ z' \equal{} z \minus{} 3 \equal{} (x \minus{} 3) \plus{} iy$

Hence if the angles made by the respective radius vectors are $ \theta_1$ and $ \theta_2$ then

$ \tan \theta_1 \equal{} \frac {y}{x}$ and $ \tan \theta_2 \equal{} \frac {y}{x \minus{} 3}$

The given condition implies $ \tan (\theta_2 \minus{} \theta_1) \equal{} \frac {\tan \theta_2 \minus{} \tan \theta_1} {1 \plus{} \tan \theta_1 \tan \theta_2} \equal{} 1$

which is the circle $ x^2 \plus{} y^2 \minus{} 3x \minus{} 3y \equal{} 0$ with centre $ \left( \frac {3}{2} \ , \frac {3}{2} \right )$ and radius $ \frac {3 \sqrt 2}{2}$

[hide="A purely geometric solution"] What the problem is asking for is the set of points $ z$ such that the angle between the vectors $ z$ and $ z \minus{} 3$ is constant (in particular, $ 45^{\circ}$).

Here's an equivalent formulation:  the origin is $ O$, the point $ z$ is $ Z$, and the point $ 3$ is $ P$.  Then we want the points $ Z$ such that $ \angle OZP \equal{} 45^{\circ}$.

By inscribed angles, this is clearly a circle.  Call its center $ C$; then $ \angle OCP \equal{} 90^{\circ}$, so $ \triangle OCP$ is a right isosceles triangle, so the radius of the circle is $ \frac{3 \sqrt{2}}{2}$.  Its center is $ \frac{3}{2} \plus{} \frac{3}{2} i$ by inspection.  Note that we actually do not have every point on the circle: the points $ 0$ and $ 3$ are missing.  [/hide]

I'm not sure how that works, but doesn't the arg function output negative values? So the angle between z and z-3 could be negative -45 or whatever, and that would not be in the locus?

Point taken.  The correct locus is the part of the circle strictly above the $ x$-axis, then, I believe.


trigonometry
inequalities proposed
inequalities

Prove that $ (\cos x)^{\cos^2x}<(\sin x)^{\sin^2x}$ for every $ x\in(0,\frac{\pi}{4})$

[quote="jerry_smith"]Prove that $ (\cos x)^{\cos^2x} < (\sin x)^{\sin^2x}$ for every $ x\in(0,\frac {\pi}{4})$[/quote]
Must be :  $ (\cos x)^{\cos^2x} > (\sin x)^{\sin^2x}$ for every $ x\in(0,\frac {\pi}{4})$
Let $ \sin^2x\equal{}t$ , then we must prove : $ (1\minus{}t)^{\frac{1\minus{}t}{t}} > t$ for every $ x\in(0,\frac {1}{2})$ .Which is true by Bernoulli inequality.


What is the formula for derangement and how do you prove it?

To make the problem clear:

Let $a_{1}, a_{2}, ...a_{n}$ be a permutation of the set $S_{n}={1,2,...,n}$. An element $i$ in $S_{n}$ is called a fixed point of this permutation if $a_{i}=i$.

A derangement of $S_{n}$ is a permutation of $S_{n}$ having no fixed points.

A common form of this problem:

There are $n$ envelopes addressed to $n$ different people. In how many ways can nobody get his/her own envelope?


I have few questions regarding critical point.

If y = x^2 + 2x
, then y` = 2x + 2

If the domain of x is [-2,1], is -2 is critical point of this equation?

In general, can critical points be at the end point of an closed interval?
How about on the open interval?  Can the critical points be at the end point of an open interval?

[quote="bobbym"]Hi aswdaniel;

You mean at -1. don't you?[/quote]

yes sorry.

Let me put the statement again with correction

x= -1 and domain of x is [-1,1]

Can you give me an advice for my question?

Hi;

[quote]x= -1 and domain of x is [-1,1] [/quote]

I read the above as the closed interval -1 to 1 inclusive, so I do not see any reason why the critical point -1 cannot be in that interval.

This isn't HSB material. There's a calculus forum.

Anyway, iff the domain is inclusive and an endpoint is a critical value, it's a valid critical value.


Let $n$ and $k$ be positive integers, and let set $S = \{1,2,\dots,n\}$. A subset of $S$ is called 'skipping' if it doesn't contain consecutive integers. How many $k-\mbox{element}$ subsets of $S$ are there? Also, how many skipping subsets of $S$ are there total?

[hide]The amount of total skipping subsets is equal to $\sum_{i=0}^{n}F_{n}$ where $F_{n}$ are the Fibonacci numbers.
This is easy to prove by induction.  Show the base case.  THe skipping subsets of {1}=1 which is equal to the number we got in our sum.
Assume it true for all $F_{n}$ up to n.  To prove it true for $F_{n+1}$, let's analyze our subsets.  On top of our subsets for that doesn't include n+1, we have $(\sum_{i=0}^{n-1}F_{n})+1$ that does include n+1.  Listing out the sums,
1,1,2,3,...,$F_{n}$
1,1,2,3,...,$F_{n-1}$,1.  By adding diagonally and putting the end one at the beginning, one gets 1,1,2,3,...,$F_{n+1}$=$\sum_{i=0}^{n+1}F_{n}$[/hide]


Prove that, given any string of natural numbers you can append some more natural numbers to the right to make it a power of 2.


In the beginning of the section in my algebra 2 textbook, a parabola was suppoed to have the equation $ (y\minus{}k)^2\equal{}4p(x\minus{}h)$ for a horizontal axis (what does that mean?) and $ (x\minus{}h)^2\equal{}4p(y\minus{}k)$ for a vertical axis. I guess it says horizontal by meaning the parabola goes right or left, or something.

In the end of the section, it listed the parabola equation as $ y\equal{}a(x\minus{}h)^2\plus{}k$ and $ x\equal{}a(y\minus{}k)^2\plus{}h$

I know p in the first part is the distance from the focus to the vertex, but I don't know what a is.

Also, which is standard? A little confusing ^_^  :blush:

[quote="Bicameral"]...a parabola was suppoed to have the equation $ (y \minus{} k)^2 \equal{} 4p(x \minus{} h)$ for a horizontal axis...it listed the parabola equation as $ x \equal{} a(y \minus{} k)^2 \plus{} h$...which is standard?[/quote]

[hide]Both are the same anyway. We can write the first one as $ x \equal{} \frac {(y \minus{} k)^2}{4p} \plus{} h$ or $ x \equal{} a(y \minus{} k)^2 \plus{} h$ with $ a \equal{} \frac {1}{4p}$.[/hide]


inequalities
geometry
cyclic quadrilateral
IMO
IMO 2003

First Day - 44th IMO 2003 Japan

1. Let A be a 101-element subset of the set S={1,2,3,...,1000000}. Prove that there exist numbers t1, t2, ..., t100 in S such that the sets

Aj = { x + tj | x is in A }  for each j = 1, 2, ..., 100

are pairwise disjoint.

2. Find all pairs of positive integers (a,b) such that the number

a2 / ( 2ab2-b3+1) is also a positive integer.

3. Given is a convex hexagon with the property that the segment connecting the middle points of each pair of opposite sides in the hexagon is \sqrt (3) / 2 times the sum of those sides' sum.

Prove that the hexagon has all its angles equal to 120.

Second Day - 44th IMO 2003 Japan

4. Given is a cyclic quadrilateral ABCD and let P, Q, R be feet of the altitudes from D to AB, BC and CA respectively. Prove that if PR = RQ then the interior angle bisectors of the angles < ABC and < ADC are concurrent on AC.

5. Let x1 <= x2 <= ... <= xn be real numbers, n>2.

a) Prove the following inequality:

( \sum ni,j=1 | xi - xj | ) 2 <= 2/3 ( n2 - 1  )  \sum ni,j=1 ( xi - xj  )2

b) Prove that the equality in the inequality above is obtained if and only if the sequence (xk) is an arithemetical progression.

6. Prove that for each given prime p there exists a prime q such that np - p is not divisible by q for each positive integer n.

Gluck.

if you say "topics below", you should sticky these, Valentin :)


On a plane, there is a triangle ABC with center G, and an arbitrary point P. Prove that
$\displaystyle \sum_{cyclic} PA \cdot PB \cdot (PA+PB) \cdot AB \geq 3AB \cdot BC \cdot CA \cdot PG$.


Yeah, well-- I THINK this problem is from the old Art of Problem Solving Introduction to Algebra book, and I can't seem to setup the right equations. Anyways, your help is greatly appreciated (please include the steps in solving it, too!) :

A train traveling from Aytwon to Beetwon meets with an accident after 1 hour. The train is stopped for 30 minutes, after which it proceeds at four-fifths of its usual rate, arriving at Beetown 2 hours late. If the train had covered 80 miles more before the accident, it would have been just one hour late. What is the usual rate of the train?

IT BOGGLES MY MIND!!! :what?:  :wallbash_red:  :cursing:

The mathlover, MATTH.

In this problem, we are given the most information about the times. So, our goal should somehow be to make the equations based on the information given in terms of the time. In other words, we have $ d\equal{}rt,$ but we must make our equations such that it is in terms of the time. So, we express them as $ t\equal{}\frac{d}{r}.$
This was problem #40 from AoPS Vol. 1. And I couldn't exactly find another effective way to do it, so it is from the Solutions Manual.

So we notice that, if we break this problem up into parts, we see that we have the first hour same for both the cases (the cases being: the accident after an hour, and the accident after going for 80 more miles). For the first hour, both the trains went for one hour (of course) and traveled at $ x$ miles per hour. So that appears to be the same. Next, we bring into consideration the next part; in the first case, the train goes to traveling at $ \frac{4x}{5},$ in the second case, however, after going the 80 miles, the train reverts to going at $ \frac{4x}{5}.$ So we see that at the very beginning, both the trains were going at $ x,$ and at the very end, both the trains were going at $ \frac{4x}{5}.$ Thus, the only difference in the arrival timings was in the 80 miles that one missed and the other gained. In the 80 miles, the first case took (we are representing this in terms of time. So, $ t\equal{}\frac{d}{r}$) $ \frac{80}{\frac{4x}{5}},$ where as the train in the second case took $ \frac{80}{x}.$ Since, the first case took two more hours and the second case took one more hour, we have
$ \text{First Case}\minus{}\text{Second Case} \equal{} 1.$ Applying this concept to this problem, we get:
$ \frac{80}{\frac{4x}{5}} \implies \frac{100}{x} \minus{} \frac{80}{x} \equal{} 1 \implies \frac{20}{x} \equal{} 1 \implies \boxed{x \equal{} 20}.$
Notice how the usual rate was represented by $ x.$ It is usually a good technique to represent what the problem is asking for with a variable.

Just as a side note, I didn't get this problem myself the first time. This is a really good problem. This was from the AHSME, so I think it belongs in the High School Basics section more maybe.


I need a brilliant idea on th following problem.

Let $ABC$ be a triangle, $P$ is a point inside of it. Suppose $AP$ meets $BC$ at $A'$, $BP$ meets $AC$ at $B'$ and $CP$ meets $AB$ at $C'$. Let $H$ be the foot of perpendiculat from $A'$ to $B'C'$. Prove that feet of perpendiculars from $H$ to $BB'$, $CC'$, $AB$ and $AC$ lie on the circle.

[quote="hunter_cvp_ic"]it here
http://www.mathlinks.ro/Forum/viewtopic.php?t=110488[/quote]
Yet  another topic without solution.

To myth,

It is nice to see you back here. Anyway, this topic has been discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=16824


How can I use Lucas Theorem to find C( 156, 120 ) modulo 7 ?

Here is the wikipedia link: http://en.wikipedia.org/wiki/Lucas%27_theorem

If we go by the theorem as given on wikipedia, we will encounter C( 2, 3 )  as an intermediate which I don't understand how should be dealt with. Could someone help me out ?

Okay! Found it myself :)

Do let me know if you want to find this thing out.


If a, b, and c are the three sides of a triangle, prove the following inequality:

[size=150]

(c+a-b)^4 /[a(a+b-c)]  +  (a+b-c)^4 /[b(b+c-a)]  +  (b+c-a)^4 /[c(c+a-b)]  >=  ab + bc + ca

[/size]

>= means greater than or equal to

how do you type those special characters on this forum?

thanks a lot  :)

EDIT: also when will equality take place?

See here:
 http://www.artofproblemsolving.com/Forum/viewtopic.php?t=135712


Let  $a,b,c>0$ such that $ abc=1$. Prove the following inequality:

                   $\sum\frac{a+b}{c}\geq2(\sum{a}+\sum{\frac{1}{a}}-3)$ ;)

And when I think it appeared in the romanian shortlist with my name under it...  :blush:  I'm ashamed. Anyway, as I said before this problem is absolutely trivial and should not appear on a shortlist. But it was not the only one.  ;)


algebra
polynomial
inequalities
algebra unsolved

Find all real polynomials $ f$ with $ x,y \in \mathbb{R}$ such that
\[ 2 y f(x \plus{} y) \plus{} (x \minus{} y)(f(x) \plus{} f(y)) \geq 0.
\]

Let $ b\equal{}f(a)$, we have
$ P(a\minus{}y,y): 2by\plus{}(a\minus{}2y)(f(a\minus{}y)\plus{}f(y))\ge0$ and
$ P(y,a\minus{}y): 2b(a\minus{}y)\minus{}(a\minus{}2y)(f(a\minus{}y)\plus{}f(y))\ge0$.
Divide both inequalities by $ a\minus{}2y$ and let $ y\rightarrow\infty$, we have
$ f(a\minus{}y)\plus{}f(y)\rightarrow b\equal{}f(a)$.
Since $ f$ is polynomial, it is not hard to see
$ f(a\minus{}y)\plus{}f(y)\equal{}f(a)$.
Since $ a$ is arbitrary, Cauchy equation holds, so $ f$ has to be linear.
It is easy to see $ f(x)\equal{}cx$ with $ c\ge0$.

[quote="xxp2000"]Let $ b \equal{} f(a)$, we have
$ P(a \minus{} y,y): 2by \plus{} (a \minus{} 2y)(f(a \minus{} y) \plus{} f(y))\ge0$ and
$ P(y,a \minus{} y): 2b(a \minus{} y) \minus{} (a \minus{} 2y)(f(a \minus{} y) \plus{} f(y))\ge0$.
Divide both inequalities by $ a \minus{} 2y$ and let $ y\rightarrow\infty$, we have
$ f(a \minus{} y) \plus{} f(y)\rightarrow b \equal{} f(a)$.
Since $ f$ is polynomial, it is not hard to see
$ f(a \minus{} y) \plus{} f(y) \equal{} f(a)$.
Since $ a$ is arbitrary, Cauchy equation holds, so $ f$ has to be linear.
It is easy to see $ f(x) \equal{} cx$ with $ c\ge0$.[/quote]
Is $ a\minus{}2y$ necessaraly possitive? I think not

We do not need $ a \minus{} 2y > 0$. Actually $ a \minus{} 2y < 0$ if $ y\rightarrow\infty$.

As long as the upper bound and lower bound for $ f(a \minus{} y) \plus{} f(y)$ are converging to same limit, it should be fine.

ok that's right. Is there some sort of problem if a-2y=0

Let $ y>\frac a2$, divide with $ a\minus{}2y$, use $ y\rightarrow\infty$.


algebra
polynomial
linear algebra
matrix
linear algebra unsolved

It's easy to prove that if $1-AB$ is invertible then $1-BA$ is invertible. Also $I-AB=0$ iff $I-BA=0$. My conjecture is :
If $A,B\in M_n(\mathbb F)$  then \[ \mbox{Rank}(1-AB)=\mbox{Rank}(1-BA) \]

Good Luck  ;)  ;)

Since $I_n-AB$ and $I_n-BA$ have the same characteristic polynomial, let`s denote it by $P$.( Then $P=a_nX^n+...+a_0$ where $a_i$ represents the sum of all minors of order $i$.) So $0$ has the same geometric multiplicity for both $I_n-BA$ and $I_n-AB$ thus $\mbox{Rank}(I_n-AB)=\mbox{Rank}(I_n-BA)$.

[quote="spix"]So $0$ has the same geometric multiplicity for both $I_n-BA$ and $I_n-AB$ thus $\mbox{Rank} (I_n-AB) = \mbox{Rank} (I_n-BA)$.[/quote]

I think that's [b]algebraic[/b] multiplicity. Two matrices with the same characteristic polynomial can have different ranks. If you see something I don't, please let me know.

It's not obvious for me that characteristic polynomials of $I-AB$ and $I-BA$ are equal.  Maybe I'm stupid now. :?  :?

perfect_radio, I don`t know for sure I kinda just said that as facts hoping it`s true  :oops:
Nevertheless, I found the solution, and it seems quite simple .
 It`s the same as proving that $\det(I_n-AB)=\det(I_n-BA)$ .. I guess you know what I`m talking about... ($A_z=A+zI_n$)

Let $ p_{A}$ denote the characteristic polynomial of $ A$. Then

$ p_{I-AB}(x) = det(xI-(I-AB)) = det((x-1)I+AB) = p_{-AB}(x-1) = p_{-BA}(x-1) = det(xI-(I-BA)) = p_{I-BA}(x)$

I don't know how to prove Omid's conjecture.

Let $ S=\{x: x=BAx\}$ and $ T=\{y: y=ABy\}$

$ S$ is the null space of $ I-BA$ and $ T$ is the null space of $ I-AB.$

Suppose $ x\in S.$ What can you say about $ Ax?$ If $ y=Ax,$ then $ By=BAx=x,$ and it follows that $ ABy=Ax=y.$ Hence, $ y=Ax\in T.$

So $ AS\subseteq T$ and $ BT\subseteq S.$

Now consider the mapping $ A$ restricted to $ S$ and $ B$ restricted to $ T.$ Since $ BAx=x$ for all $ x\in S,$ we must have that $ A$ is injective on $ S.$ Similarly, $ B$ is injective on $ T.$ Then $ \text{dim }S\le\text{dim }T$ and $ \text{dim }T\le\text{dim }S.$

Thus, $ \text{dim }S=\text{dim }T.$ $ (I-BA)$ and $ (I-AB)$ have the same nullity and hence the same rank.

Throughout this problem, we've been assuming that $ A$ and $ B$ were square matrices. What if they're not?

Suppose $ A$ is an $ m\times n$ matrix and $ B$ is an $ n\times m$ matrix. I do think my proof still works, but the result is that $ (I_{m}-AB)$ and $ (I_{n}-BA)$ have the same nullity, which is not the same thing as having the same rank.


logarithms
real analysis
real analysis solved

$(u_n)  $ is a sequence of positive real numbers such that

$ \frac{u_1+...+u_n}{nu_n} \longrightarrow  \alpha $
Prove that
$ \frac{u_1+2u_2+...+nu_n}{n^2u_n} \longrightarrow
\frac{\alpha}{1+ \alpha} $

$\alpha>0$
$A_n=u_1+...+u_n$

Hints:

1/ Prove $A_n$ does not converge.If it does get a contradiction.
2/ Find a relation between $\sum_{k=1}^nA_k$ and $\sum_{k=1}^{n}ku_k$
Using the fact that $A_n \sim \alpha nu_n$
3/ $\frac{\limits\sum_{k=1}^{n}ku_k}{n^2u_n}
\sim \frac{(n+1)A_n}{n(A_n/\alpha)(\alpha +1)}$

Suppose $A_n$ is convergent. Let $\beta\in\mathbb{R}$ be its limit.
Then $u_n$~$\displaystyle \frac 1n\frac\alpha\beta$, but from this results that $A_n$~$\ln n\displaystyle\frac\alpha\beta$ which is divergent, contradiction.
So $A_n\to\infty$.
$\displaystyle\sum_{k=1}^nku_k=(n+1)A_n-\sum_{k=1}^nA_k$ and $A_n$~$\alpha nu_n$ so $\displaystyle\sum_{k=1}^nku_k$~$\displaystyle(n+1)A_n-\alpha\sum_{k=1}^nku_k\Rightarrow \sum_{k=1}^nku_k$~$\displaystyle\frac1\alpha(n+1)A_n$.
Now $\displaystyle\frac{\sum_{k=1}^nku_k}{n^2u_n}$~$\displaystyle\frac{(n+1)A_n\cdot1/(\alpha+1)}{nA_n/\alpha}~\frac\alpha{\alpha+1}$. q.e.d.

Next time please post in the proper forums, in the College Playground category.


You and a friend are stranded on a desert island.  You have one large bottle of water (of arbitrary shape) between you, and you wish to split it exactly evenly - that is, one of you drinks exactly half its volume, and the other exactly the other half.

How can you do it?

(If you can't do it, what restrictions do you need on the shape of the water bottle to be able to do it?)

The bottle has to be symmetric so that axis of symmetry passes through a pt on the edge of the opening of the bottle and its bottom. the bottle has to have circular symmetry from top view. Now tilt bottle so that edge of water touchees both the edge of bottome and edge of opening. Exactly half of water will pour out. :)

This question looks like it is related to the discussion of a question gone awry on the Mock NYSML contest [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=79871&postorder=asc&start=20]here.[/url]

Meh, why bother, just take all the water for yourself. Then you can eat your friend if you need to.


trigonometry
calculus
calculus computations

(i)Find the Maclaurin series for cosx as far as the term in x^4
(ii)Deduce the the Maclaurin series for f(x) = (1/2)cos(2x) as far as the term in x^4
(iii)Hence write down the first three non-zero terms of the series for f(3x)


The combined circumfrence of 2 different-sized circles is 603. If the area of the larger cirlce is 121 times the smaller, what is the area of the smaller circle? Round your answer to the nearest whole number.

[hide="solution"]Since the area of the larger circle is 121 times larger, the radius of the larger circle is 11 times longer.  If the radius of the smaller circle is $r$, then $12r=603$ so $r=\frac{201}{4}$.

Therefore, the area of the smaller circle is $\pi\frac{40401}{16}$[/hide]

[quote="cincodemayo5590"][quote="jli"][hide="quoted"] If the radius of the smaller circle is $r$, then $12r=603$ so $r=\frac{201}{4}$.[/hide][/quote]

Wait, what?
[hide="It should be:"]
\[2 \pi (r+11r) = 603 \implies r = \frac{201}{8 \pi}\implies \pi r^{2}= \frac{ 201^{2}}{8^{2}\pi}\approx \boxed{201}\]
[/hide][/quote]

oops...i suck

youre right


inequalities
trigonometry
inequalities unsolved

Zhaobin, but isn't your inequality equivalent with $\sum cos^{2} A \geq\sum cos^{3} B$.
But since $ABC$ is acute then $cos^{2} A\geq cos^{3} A$. :D
Am i missing something? :?

yes,i think you have miss something :lol:
my inequality is :    $2 (\sum \cos^2 A)^2 \geq \sum (\cos^2 A)+\sum (\cos^3 A)$.

Ok, sorry. It's kind of late and and i haven't seen the power. :oops:

[quote="zhaobin"]$A,B,C$ are three angles in an acute-triangle.prove:
$2(\cos^2 A+\cos^2 B+\cos^2 C)^2 \geq \cos^2 A+\cos^2 B+\cos^2 C+\cos^3 A+\cos^3 B+\cos^3 C$[/quote]

Wrong. Try, for instance, A = 32°, B = 73° and C = 75°.

  darij

The inequality is true for obtuse angle triangles
First note that
(CosA)^2 + (CosA)^3  = 2.CosA)^2 .(CosA/2)^2

Assuming the triangle is obtuse

(CosA)^2  >=  [(CosA)^2. CosA/2)^2 ]/{ (CosA)^2  + (CosB)^2  + (CosC)^2 }

because the denominator on the right side which  can be expressed as 1 - 2( CosA )(CosB  )(CosC) is greater than 1.

Adding this inequality to the two similar ones where A is replaced by B and C respectively we obtain the inequality to be proved.


Which are there more of among the natural numbers between $1$ and $1,000,000:$ numbers that can be represented as a sum of a perfect square and a positive perfect cube, or numbers that cannot be?

Did I misread something? This seems too easy...  :huh:
[hide="Solution"]We're looking for the number of numbers in the form $a^{2}+b^{3}.$ Note that $0\le a \le 999$ and $1 \le b \le 100.$

Hence, there are $(100)(1000)=10^{5}$ numbers in the form $a^{2}+b^{3}$ formed by these values of $a$ and $b.$ (We overcounted, since several of these sums are greater than $10^{6}.$)

However, $10^{6}-10^{5}=9\cdot 10^{5},$ which is much greater than $10^{5},$ so....

 :huh:
[/hide]

Yeah... that's what I got. That doesn't even account for overlaps and it's still way too small.

Edit: What about negative cubes?

Are you sure that the problem is correct?

According to Kalva, that is indeed the correct problem statement (or has the right meaning, in any case).


Henry's Hamburger Heaven orders its hamburgers with the following condiments: ketchup, mustard, mayonnaise, tomato, lettuce, pickles, cheese, and onions. A customer can choose one, two, or three meat patties, and any collection of condiments. How many different kinds of hamburgers can be ordered?

(A) 24 (B) 256 (C) 768 (D) 40; 320 (E) 120; 960

[hide]First, the customer has $ 3$ choices of meat patties. Second, for each condiment, the customer can either choose to have the condiment or not, so there are $ 2^8$ choices for condiments.

In total, $ 3*2^8\equal{}768$ different kinds of hamburgers can be ordered $ \Rightarrow \boxed{C}$.[/hide]

[quote="mathemonster"]how do you do the problem

8 objects how many distinct groups of any number of objects can be made?[/quote]

Well, for each condiment, you can either choose to (1) have it, or (2) not have it. Therefore, since there are 8 objects, each with two choices, we have $ 2^8\equal{}256$.


8. In professional football it is possible to score 2 points (for a safety), 3 points (for a field goal), or 6 points (for a touchdown.) If a touchdown is scored it is possible to score one additional point. The score of a football game is 21 to 17. Find the total number of possible combinations of points to achieve this final score.

How do you guys do this?
[hide="Answer"]180 ways[/hide]


I tried to solve the system:

x+4y-z-6=0
2x-y+z-3=0
3x+2y+3z-16=0

with Gauss-Jordan Elimination.  I got (14/5,1/5,-12/5).  However, that does not check correctly.  Can someone post how to solve this correctly, with Gauss-Jordan Elimination?

$ (2x\minus{}y\plus{}z\minus{}3)\minus{}2(x\plus{}4y\minus{}z\minus{}6)\equal{}\minus{}9y\plus{}3z\plus{}9\equal{}0 \iff y\minus{}\frac{1}{3}z\minus{}1\equal{}0$
$ (3x\plus{}2y\plus{}3z\minus{}16)\minus{}3(x\plus{}4y\minus{}z\minus{}6)\equal{}\minus{}10y\plus{}6z\plus{}2\equal{}0 \iff 5y\minus{}3z\minus{}1\equal{}0$

$ (x\plus{}4y\minus{}z\minus{}6)\minus{}4\left(y\minus{}\frac{1}{3}z\minus{}1\right) \equal{} x\plus{}\frac{1}{3}z\minus{}2\equal{}0$
$ (5y\minus{}3z\minus{}1)\minus{}5\left(y\minus{}\frac{1}{3}z\minus{}1\right) \equal{} \frac{4}{3}z\minus{}4\equal{}0 \iff z\minus{}3\equal{}0$

$ \left(x\plus{}\frac{1}{3}z\minus{}2\right)\minus{}\frac{1}{3}(z\minus{}3)\equal{}x\minus{}1\equal{}0$
$ \left(y\minus{}\frac{1}{3}z\minus{}1\right)\plus{}\frac{1}{3}(z\minus{}3)\equal{}y\minus{}2\equal{}0$

Thus, $ (x, y, z)\equal{}(1, 2, 3)$.

hello, the Gauss-Eliminaton gives us a shorter way,multiplying the first equation by -2 and adding to the seconde we get
$ \minus{}9y\plus{}3z\equal{}\minus{}9$
or
$ \minus{}3y\plus{}z\equal{}\minus{}3$ $ (1)$
multiplying the first equation and adding to the third equation we get
$ \minus{}5y\plus{}3z\equal{}\minus{}1$  $ (2)$ (after dividing by -2)
multiplying $ (1)$ by $ \minus{}3$ and adding to $ (2)$ we get
$ y\equal{}2$ and so on.
Sonnhard.


$9)$ Simplify

              $\sqrt[3]{x\sqrt[3]{x\sqrt[3]{x\sqrt{x}}}}$

[hide]
Just put it in fractional exponents, then a pattern occurs where it switches between $x^{\frac12}$ and $x^{\frac32}$, and the expression after simplification is $\boxed{\sqrt x}$.[/hide]


superior algebra
superior algebra solved

$\mathbb Q[i]$ has exactly two endomorphisms
$1_{\mathbb Q}$ and $f(z)=\overline z$ right?

As field: yes

Proof:
You have $f(1)=1 \implies f(n)=n \implies f(\frac{a}{b})=\frac{a}{b} , a,b,n \in \mathbb{Z}$
Now all is given by $f(i)$, but $f(i)^2=f(-1)=-1$, so there are only $2$ possibilities, giving your endomorphisms


probability
symmetry
MATHCOUNTS
articles
AoPSwiki
Pascal\u0027s Triangle
number theory

[quote="jhredsox"]If they were to ever ask you something about pascal's triangle in mcounts, what would you need to know?
-jorian

p.s. i didn't really know that much about it until we discussed it in algebra 1 today[/quote]
If you just want to learn more about Pascal's Triangle, take a look here:
[url]http://ptri1.tripod.com/[/url].
However, there probably won't be too many problems that involve Pascal's Triangle, but if there are, they can probably be solved with just combinatorics.
(It's still a good tool to know though.  :) )

[quote="perfect628"]You can have that one, but I meant questions that don't ask ABOUT Pascal's trianghle.[/quote]
No, I agree with you, the triangle itself is relatively useless (the content is what counts). I was just giving an example of the only problem I can think of that uses pascal's.

I have to disagree on this one.  I find myself using Pascal's triangle all the time doing MATHCOUNTS.  Since P.T. is just a collection of combinations, you can do most problems other wasy, but really knowing P.T. can make things quick.

For example the question, what is the probability that you get 3 heads when you flip a coin 4 times looks at the 4th row of Pascal :

1   4   6   4   1

and the answer is 4(the second 4) out of the 16 total in the row

Occasionally you have problems that ask for the sum of several combinations and you can use the properties of PT to add them.

The sum of the traingular numbers is seen quickly in Pascal.

You can go on 4ever.  Again, these things can be answered by combinations for the most part, but exploring PT for it's patterns canbe very useful.

Asking "what do I need to know" though is not the best approach.  Explore for patterns, and understand how PT uses combinations.  Looking at the sum of any row, and WHY you get what you get for the sum of any row has come in handy regularly too.

I understand what you are saying, and I didn't even know the formula for $\binom{n}{k}$ until the end of fifth grade when I just knew it as the kth term in row n of the triangle for a long time. Pascals is really useful, but when you see it as combinations than it becomes a lot less important. I have starting seeing such problems as combinations not pascals. So pascals isn't important for me. That is what I and others (I think) were saying. Also, I know all combinations up to $\binom{6}{x}$ by heart.

here's an example of a problem, how many paths are there to spell out MTHOMBA from top to bottom?

   M
  TT
  HHH
OOOO
 MMM
  BB
  A

yeah, the orientation is bad, deal with it.

[quote="frost13"]Bpms - I hope you don't misunderstand me, I agree that most problems that might use PT might be done more easily just using combinations, and for the more advanced student PT may seem like more or a trick or memorized cheat.  I would just warn that if you find yourself not valueing PT anymore because you have advanced to see things as combinations, you may be hurting yourself.  It's a took to at least keep handy.  My guess is that if I asked you what (x+y)^5 is you'd know immediately to use the binomial expansion and the combinations that go with it.  Wouldn't it be easier to recognize that the 5throw of PT contains the combinations you need and therefor the coefficients in the expansion are 1  5  10  10  5  1.  Would you really do 6 combinations(or just the 3 and use symmetry) instead of just using the 5th row of PT?  I'm not saying memorize 20 rows or anything, just have a few on hand for quick problems.  Knowing PT for me now is like knowing my multiplication tables.  Quick facts that come in handy, but I still understand where they come from.

Another example might be finding the sum of the first 20 triangular numbers.  I suppose you could memorize a formula, or derive one on the spot, but knowing the "hockey stick" technique in Pascal's triangle can help you get it quick without actually drawing it out.  You can quickly see the sum of the first n triangular numbers is (n+2)_C_3.  I have too many other things memorized to keep that one in my head, but I know PT can get it for me:-)

Again, I think you "kids" on this site are amazing and I am one of the few adults who come in here regularly because I respect your guys(and girls) work and am always looking for things to use/discuss with my own students(some of who are in and out of here).   Every once in awhile I just like to throw in a bit of advice.  After sending 10 kids to Nats in 8 years, and 6 trips myself I've seen a lot of problems and a lot of different ways to do them.  At this point I'm not discarding ANY strategies!!!

Man that was long winded....  I guess the moral is : if you are newer to the problem solvig world I suggest you take a long look at Pascals triangle.  Once you are more advanced get a real strong feel for the combinations and keep PT in your back pocket for when you need a little extra help.

Keep up the good work everyone and hopefully we can all meet up in Texas come May:-)[/quote]
You are completely right, I learned pascals triangle in my math class in fifth grade (along with modular arithmetic; it was quite the class). There the only identity we learned was pascal's identity which we all thought was obvious since all we had seen was the triangle. None of us knew the formula for combinations, and I personally had generated the triangle so many times I know it by heart up to 1 6 15 20 15 6 1. You are right about quickness as well. Also, about what you are saying about Texas, I hope so too, I saw you and Mr. Boyd talking at nats last year, but you probably had no clue who I was. This year you moved to MA? Well, I hope that you make it to nats again although it is obvious that you (and Ravi B) are great coaches (I hope I make it too...).

In my opinion*, Pascal's isn't as useful as understanding why it corresponds the way it does. I.e. don't get so accustomed to problem-like-above -> Pascals' that when you come across this problem:
How many ways can you go from the upper right square on an 8x8 chessboard to the lower left if you can only move one square left or down on the squares and the upper left and lower right corners are cut out of the chessboard?

you are befuddled.

*I only went to Nats once, and didn't do wonderfully then. So my opinion isn't as important as Neal's.

[quote="bpms"]Here is one nationals sprint problem that is very important with pascals. How many rows out of the first 20 of pascals have all the numbers except the first and last ones being even.[/quote]

how would u do something like this

i know pascals isn't super important, i was just asking what was the maximum question they would test u on this

-jorian

let me change what i said.....how would u solve this?
-jorian

Just think about the parity of the numbers, I guess a possible brute force way would be to just develop the entire triangle mod 2.

im sorry again, but i'm horrible at math so i didn't understand much about what u said

.....as for mods, i searched, but it was way to general....anyone want to im w/ me to explain it fully?
-jorian

ty

The mod m value of a number is the remainder when that number is divided by m.

[quote="frost13"]jorian, have you tried making a Pascal's Triangle?

If you can make it quickly until you notice a few rows where all the interior numbers are even you may find a pattern.  That's what much of MATHCOUNTS is...looking for patterns.  If you still can't find it ask again and I'll GIVE you the answer.

You can't just jump right into modular arithmetic until you have the other basics down.  Other people said Pascal's Triangle isn't important to them since they have combinatorics down pat.  If you don't have combinatorics down, then Pascal's Triangle can be very important![/quote]

What basics do you think are required to fully understand these?

[quote="nebula42"][quote="jhredsox"]so.........not everything gos under a certain mod

wait, ..............................o its harder than when i just saw ur explanation

idont get it

u guys are too good for me

 :ninja:
-jorian :([/quote]
Take a look at the [url=http://www.artofproblemsolving.com/Wiki/index.php/Modular_arithmetic/Introduction]article in the AoPS Wiki[/url].

And try reading over it, so if you need to ask another question again, you can ask a more specific one that someone could help you with more easily.  :wink:

You've also posted this question [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=116127]here[/url], and you received several good replies. Try looking over them at least once or twice before you ask the exact same question again.  :wink:[/quote]

first off, stop winking at me nebula :huh:

ok, so i think i know what mod means, just not how to apply it

so....

8 is 0 in mod 4
9 is 3 in mod 6
343 is 0 in mod 7

the mod thing on aopswiki was acceptional, except there were many crucial parts left out, so im confused..........a lot..still

the big question is, how do u apply it to questions?
-jorian

@nebula  :wink:  :wink:

there are many ways to use mods, here i sone example from a question on a test my team did the other day :

The sum of each pair of positive integers x, y that satisfy 3x + 4y = 50 is computed.  What is the largest such sum?

Since there are two variables there will be an infinite # of solutions, but not an infinite # of solutions with x and y both being positive integers.  You could probably do this with just guess and check, but here is how to use mods.

3x + 4y = 50

taking the remainder when dividing by 3  (mod 3) you get

0x + 1y = 2 (mod 3)

so y = 2 (mod 3)

so y can equal 2, 5, 8, 11, 14, 17, .......

now you can plug those values into the equation to infd corresponding values of x

3x+4(2) = 50 --> x = 14  -->  x + y = 16
3x+4(5) = 50 --> x = 10  -->  x + y = 15
3x+4(8) = 50 --> x =  6   -->  x + y = 14

It appears 16 is the largest sum of x + y!

there you go, one of MANY possible ways to use modular arithmetic.  If the equation uses less friendly coefficients like 19x + 23y = 386 you are less likely to guess and check, but mod 19 would be helpful

Ok............I formally GIVE UP on trying to know more than 1 thing about mods before Math Counts this year. :(
-jorian

[quote="solafidefarms"]How many ways can you go from the upper right square on an 8x8 chessboard to the lower left if you can only move one square left or down on the squares and the upper left and lower right corners are cut out of the chessboard?[/quote]

How can you solve this problem using Pascal's triangle?

The point is, you can't get so locked into Pascal and not why it works that you can't figure out that problem. The problem in question you can't use Pascal's on, but the ways to get to any specific point is the sum of the ways to get to the point above it plus the ways to get to the point immediately left of it. (Or, of course, you can count how many paths go through those corner squares that are cut out - 2 - and subtract from the total.)


please explain with a pie chart.

Data:      Subjects

                  science=45
                 english = 35

                 hindi     =15
                 maths   =30
                 history=25
                  geography =50

Draw a pie chart n explain. number of students is given by the number next to the = sign.

please reply soon .i know it is a dumb question but i've forgotten.

Well, a pie chart is used for comparing the values of different things. A pie chart is basically with a circle divided into sectors for each quantity. The smaller sectors represent the smaller quantities, and the larger sectors represent the larger quantities. You find the size of every sector by finding the angle measure. Of course there are 360 degrees at the center of a circle. The first step to finding the degree measure of every sector is to find the total quantity. The total quantity will be the total number of students. This total will represent 360 degrees of the circle. The total number of students is 200. Now lets see what the degree measure is for just one student. We find this by dividing 360 by 200. See why? Dividing 360 by 200 we get 9/5 of a degree measure. Now we just multiply this number by the number of students in each subject to find the degree measure for this subject. Once we have the degree measures for each subject, we can use a protractor to draw a circle and divide the circle into sectors with these degree measures.


If $ a,b,c$ are positive integers less than $ 10$, then $ (10a \plus{} b)(10a \plus{} c) \equal{} 100a(a \plus{} 1) \plus{} bc$ if:
$ \textbf{(A)}\ b \plus{} c \equal{} 10 \qquad\textbf{(B)}\ b \equal{} c \qquad\textbf{(C)}\ a \plus{} b \equal{} 10 \qquad\textbf{(D)}\ a \equal{} b \\
\textbf{(E)}\ a \plus{} b \plus{} c \equal{} 10$

[hide="Whee"]
Expand.
$ 100a^2 \plus{} 10ac \plus{} 10ab \plus{} bc \equal{} 100a^2 \plus{} 100a \plus{} bc$
Simplify.
$ 10ac \plus{} 10ab \equal{} 100a$
Divide by 10a.
$ c \plus{} b \equal{} 10$

$ \boxed{\text{A}}$[/hide]


Q.) Four distinct integers are in AP. If one of these integers is the sum of squares of  the remaining three, find the numbers.

[quote="ocha"]we have $ a \minus{} d,a,a \plus{} d,a \plus{} 2d$

now, $ a \plus{} 2d \equal{} (a \minus{} d)^2 \plus{} a^2 \plus{} (a \plus{} d)^2 \equal{} 3a^2 \plus{} 2d^2$

$ \therefore \frac { \minus{} 3a}{2d} \equal{} \frac {d \minus{} 1}{a \minus{} 1}$

Note that $ a,d$ are integers $ (\because a$ is a term of the AP, which are all integers..$ )$

Observing the negative sign on the LHS, we see that $ a \equal{} 0$, hence $ d \equal{} 1$

The AP is $ \boxed{ \minus{} 1,0,1,2}$, checking this; $ ( \minus{} 1)^2 \plus{} 0^2 \plus{} 1^2 \equal{} 2$ which is true[/quote]
I dont get the judgement. Its just as if assuming them to be consecutive.


geometry
geometric transformation
reflection
inequalities open
inequalities

Let $ x,y,z$ positive real numbers such that $ x\plus{}y\plus{}z\equal{}1$, prove that
\[ \frac{1}{1\minus{}xy}\plus{}\frac{1}{1\minus{}yz}\plus{}\frac{1}{1\minus{}zx}\leq\frac{27}{8}\]

[quote="Sunjee"]Let $ x,y,z$ positive real numbers such that $ x\plus{}y\plus{}z \equal{} 1$, prove that
\[ \frac{1}{1\minus{}xy}\plus{}\frac{1}{1\minus{}yz}\plus{}\frac{1}{1\minus{}zx}\leq\frac{27}{8}\]
[/quote]
We need prove that:
$ \sum\frac{1\minus{}9x}{1\minus{}x}\geq 0$
or $ \sum\frac{(1\minus{}9x)(\frac{5}{3}\plus{}3x)}{(1\minus{}x)(\frac{5}{3}\plus{}3x )}\geq 0$
Suppose that $ a\geq b\geq c$
By Chebyshep we have:
$ \sum\frac{(1\minus{}9x)(\frac{5}{3}\plus{}3x)}{(1\minus{}x)(\frac{5}{3}\plus{}3x)}$ $ \geq (\sum_{x;y;z}(1\minus{}9x)(\frac{5}{3}\plus{}3x) )(\sum_{x;y;z}\frac{1}{(1\minus{}x)(\frac{5}{3}\plus{}3x)}$
We need prove that:
$ \sum_{x;y;z}(1\minus{}9x)(\frac{5}{3}\plus{}3x)\geq 0 \equal{} 5\minus{}12(x\plus{}y\plus{}z)\minus{}27(x^{2}\plus{}y^{2}\plus{}z^{2})\geq 0$
or $ 27(a^{2}b^{2}\plus{}b^{2}c^{2}\plus{}c^{2}a^{2})\plus{}12(ab\plus{}bc\plus{}ca)\leq 5$
Put $ p \equal{} a\plus{}b\plus{}c;q \equal{} ab\plus{}bc\plus{}ca;r \equal{} abc$.We have:
$ 27q^{2}\minus{}54r\plus{}12q\leq 5$
Done


calculus AMC AIME USA(J)MO USAMO Vieta inequalities

[quote="gauss1181"]Study past AMC 12 and AHSME exams.
As in books, you should have all the Intro and Intermediate books, plus the AoPS Volumes 1 and 2.
In addition, you should buy some resources made by the AMC makers too, including the Contest Books on the website (and on this site) plus the 21st Century Contests AMC CD, which has all of the AMC, AIME, and USAMO tests from the 21st century. You should try to memorize all the theorems/basic stuff that's in those AoPS books, like Vieta's Formulas, AM-GM Inequality, DeMoivre's Theorem, etc. Also, some trig formulas are good for memorization.
I'm not exactly sure if calculus can help...though you can still do well without using it...(I did so and got like a 120s-130s score without calculus methods, that was unofficial for me)

But good luck![/quote]

Calc will not help at all.

Dont JUST memorize the formulas. uindetsand where they come from, how to prove them, ect. That will help a lot.

Going through AoPS Volume 2 gives you enough tools to ace the AMC 12 - actually doing it is more a matter of being quick and having experience, which you can learn to do by practice. And by "go through" I mean actually do most to all of the problems. You should be able to do that pretty easily between now and February. Calculus will not help one bet in all probability. The questions are designed so that calculus will be unnecessary.

abacadaea is right about the memorization - you shouldn't memorize formulas. Learn how to prove them and you'll be able to use them a lot better.

[quote="math92"]So, I want to do well in the AMC 12 this year.

What can I do to prepare?
What books should I get if any?
What theorems should I know? (I barely know any, so please state even the obvious ones)
Will knowing calculus help me?

Thanks.[/quote][hide="So, I want to do well in the AMC 12 this year."]Good for you![/hide][hide="What can I do to prepare?"]You can do old tests, look around the high-school basics and intermediate forums for problems that you can do, finish the introductory books series from AoPS, do AoPS Vol. 1, and if you can, Vol. 2, and DO NOT cram![/hide][hide="What books should I get if any?"]All of the above mentioned if you don't have them already[/hide][hide="What theorems should I know?"]Pythagorean theorem, Chicken McNugget theorem, and your similar triangles. If you are aiming for a perfect, you should know Ptolemy's theorem, Stewart's theorem, all about complex numbers, Vieta's Formula, Principle of Inclusion-Exclusion, there's lots.[/hide][hide="Will knowing calculus help me?"]Only a little. If(for any odd reason(this was a MathCounts problem once)) there is something like a problem involving the area under a random curve, it's nice to know basics of integral calculus, but the probability of getting that kind of problem is so slim, you'd win the lottery before that happens(learn it quick, the test writers might see this! :P )[/hide][hide="Thanks."]You're welcome![/hide]
[quote="mathemonster"]for aime AcoPs is good after volume 2 right?[/quote]Yes.
[quote="worthawholebean"]abacadaea is right about the memorization - you shouldn't memorize formulas. Learn how to prove them and you'll be able to use them a lot better.[/quote]
This is the entire AoPS Religion. *chants and walks slowly away*

[quote="1=2"]You can do old tests, look around the high-school basics and intermediate forums for problems that you can do, finish the introductory books series from AoPS, do AoPS Vol. 1, and if you can, Vol. 2, and DO NOT cram![/hide][hide=What books should I get if any?]All of the above mentioned if you don't have them already.[/quote]

I would emphasize doing the old tests here, since it's unlikely that you'll have enough time to get through all of these before the AMC12.

Approximately 500 students qualify for USAMO.

This is my 1000th post!!!

It's also easier to get in as high school sophomores and freshmen (and under)

Do u guys read the vecctor and matrices chapters of v2 because i dont think they will help for aime..i might skip them
wait how is math92 a birch and swinnerton but so inexpierienced???

[quote="gauss1181"][quote="mathemonster"]aops v1 + v2 will easily be enough to qualify for aime. The intro's would be overkill if youve read volume 2 correct? Gauss have you really done all those book thoroughly? how long did it take??!![/quote]
Yeah I just overkill myself reading all the intro books thoroughly. This helps me to keep the Mathcounts tricks with me when I do AMCs.

BTW, national mathcounts is kinda like AMC 12, only a teeny bit easier. If you can ace any middle school contest or AMC 10, AMC 12 should be the perfect level for you.[/quote]

No, national mathcoutns is (much) easier than amc 10. AMC 10 is easier than 12.

I think he's just wanting tips for getting a perfect on AMC 12.

Hey, don't skip [b]anything[/b] in v2. Read it all with brute force thinking. It should [b]prepare you well.[/b]

[quote="mathemonster"]have you read it thoroughly?[/quote]
Yeah of course! :D

[quote="gauss1181"]BTW, national mathcounts is kinda like AMC 12, only a teeny bit easier.[/quote]

I disagree; I can get almost perfect AMC 12, but not so close to perfect nats. I actually think that is easier than the Nats MC, it just needs a couple more theorems.

AMC 12 has much more difficult problems than MC Nats but much less time pressure.

I don't recommend buying all the intro books. You don't have nearly enough time to get through them all thoroughly and that's pretty expensive. Just do the classic vol. 1 and 2. They're both very good. Read through the sample material on vol. 1 to see if it's too easy - if you can do 95% of the problems, you should just get vol. 2.

what do u mean "brute force thinking" gauss1181?

I think he means that if you get stuck, keeep going.

calculus is not as useless as some people here make it sound. i suggest you learn up to dervatives, (which includes limits because that is how they are defined) and partial derivatives. also you should learn lagrange multipliers as they allow you to maximize or minimize a function under a certain constraint.

for example: minimize x+1/x if x is a real number different from 0.

solution: differentiating-:1-1/x^2, and set equal to 0. 1-1/x^2=0 x=1 is either a minimum or maximum(or neither but this is rare) ok so it is a minimum we are done.

or:maximize 3x+4y if x^2+8xy+y+3x+y^2=14    or whatever.
one typical problem of this kind is maximizing area given perimeter, maximizing volume given surface area, minimizing surface area given volume, etc.

Calculus definately should not be nessacary. For the questions in the post above this, you should become familiar with RMS-AM-GM-HM, Cauchy-Scwartz.

As an example, take the first question in that post.

By AM-GM, $ \frac{x\plus{}\frac{1}{x}}{2} \geq \sqrt{1} \implies x\plus{}\frac{1}{x} \geq 2$.

Therefore, the minimum is $ 1$.

one typical problem of this kind is maximizing area given perimeter,
This screams AM-GM

 maximizing volume given surface area, minimizing surface area given volume, etc.
Again, AM-GM

what if i told you to minimize x^2+2y^2 subject to the constraint xy=1? how would you do that without some messy algebra?

AM-GM also works fine, we immediately get that $ \frac{x^2\plus{}2y^2}{2} \ge \sqrt{2}xy \equal{} \sqrt{2}.$

Lagrange Multipliers tend to be much more messy then other methods. Also, you need to prove that there is a minimum.

when people say lagrange multipliers do they mean taking the derivative and setting equal to zero??

No, I think Lagrange Multipliers belong to multivariable calculus.

AP Calculus AB and BC only cover single-variable.

[quote="worthawholebean"]Going through AoPS Volume 2 gives you enough tools to ace the AMC 12 - actually doing it is more a matter of being quick and having experience, which you can learn to do by practice.
[/quote]

I realize that the original poster has probably already gotten all the information he needs, but I'll just give my opinion.

I [b]highly[/b] agree with this part. Volume 1 is most likely enough to get you most of the first 15 or so without a problem, as long as they don't include topics that don't appear in the book. However, doing most of Volume 2 (by most, I mean all of the chapters that actually relate to topics that appear in the AMC 12) pretty much gives you all the knowledge you need for any problem. It gives you a bit of experience, if you do the problems, but many of them aren't AMC 12 style and you also don't have the time restraints. To get the other experience you need, you should do many problems.

For example, since the time I took the AMC 12, I've actually gone through most of volume 2 that applies to the AMC 12 (not including the matrices, limits, etc) thoroughly. The only ones that I haven't really done thoroughly (once again, besides the construction, matrices, limits, vectors, etc) are chapters 6, 13, 16, and 17. However, I've gone through those chapters once, earlier, so I have most of the ideas down, not just everything. With that much, I've improved from getting around 100s to a 127.5 on my last practice test, which was with two very stupid mistakes. However, I think doing lots of problems helps just as much as having the knowledge required.

So, basically, you should make sure you've done the following:
1) [b]Have the knowledge to solve the problem. [/b]By this I mean, know the tools that are used to solve the problem, be able to actually understand what the problem asks, etc, like knowing how to use and manipulate basic complex numbers, several theorems like the Pythagorean (must), areas, etc. This can be done by reading books, like volume 2.

2) [b]Make sure you've done lots of problems[/b]. They don't just have to be AMC problems; they definitely can be, but you should also do some harder problems along with a few easier ones. Chances are, if you've done a lot of problems, you'll be able to employ a technique that you've already used on a previously solved problem. There are so many problems that I've seen for the first time that I already know how to solve, because I've seen problems similar to them.

3) [b]Get used to the conditions of taking the test.[/b] Okay, maybe this is more important at the AIME and USAMO (especially USAMO), but you should still be used to the time available, and know what kind of a pace you have to set. The AMC 10 and 12 have a much bigger speed factor than the AIME or USAMO, so make sure you don't work too slowly and miss the chance to solve easy problems on the real test. For this, you should try to take practice tests in conditions as close as possible to the real test. I admit, that's impossible, but you can still give yourself the real time limit, make sure your surroundings are quiet and take the test around the time you'll have to take it for real.

Much of that probably isn't true, since I've only taken the AMCs for two years, so I don't have as much experience as most people do.

That is some very good advice, I have the AoPS Volume I which I'm almost done going through. Do you think I should get the AoPS Volume II and go through that December, January, February? My goal is to get past 100, hopefully 110 or so (as high as I can go) and get into AIME...thanks!

Well, what else would you use to prepare, besides old competitions? I'd recommend getting it, so you can learn some of the more important techniques and concepts that are useful. Also, some things that come up on the AMC 12 aren't in volume 1, such as more advanced complex numbers, trig, etc, and they are covered very well in volume 2. However, make sure you're comfortable with *everything* in volume 1, since I find you can solve a lot of problems only using stuff from volume 1. Also, as I said before, you should also take lots of practice tests and do competitions. In my opinion, that's just as important as getting volume 2, because you already have the tools required to handle most AMC problems.

I remember there's a topic talking about what percentage of chapter that could involve in AMC 12. Like algebra 10% ............. But I can't search them, Anyone help me.


Gauss
number theory theorems
number theory

hey
i'm new in this forum and i try to gather many identities of factorisation in a document.
i open this discussion so you give me identities remarkable that you know in order to enrich my document and if you're interested I can send you.

(sorry for my bad english, i am french)

[quote="ZetaX"]See the formulary etc. in this subforum, it already contains a lot. Another thing to mention are the Gauss' and Lucas' identities which I didn't add untill now...[/quote]

oh yes ok, thank you very much  :lol:


[hide]$ x \equal{} 9 \minus{} 4\sqrt 5\implies \frac 1x \equal{} \frac {9 \plus{} 4\sqrt 5}{9^2 \minus{} 5\cdot 4^2} \equal{} 9 \plus{} 4\sqrt 5$ therefore $ x \plus{} \frac 1x \equal{} 18$.[/hide]

He was actually after the quantity $ y \equal{} \sqrt {x} \plus{} \frac {1}{\sqrt {x}}$.

[hide="Answer"]Note that $ y > 0$, then square to find $ y^2 \equal{} x \plus{} \frac {1}{x} \plus{} 2$.

[b]nayel[/b] calculated $ x \plus{} \frac {1}{x} \equal{} 18$ in the previous post, so we have $ y^2 \equal{} 20 \implies \boxed{y \equal{} 2 \sqrt {5}}$[/hide]


trigonometry
inequalities
function
inequalities proposed

Let $ 0\le \alpha,\beta,\gamma\le \frac{\pi}{2}$ and $ \alpha\plus{}\beta\plus{}\gamma\equal{}\pi.$ Prove the inequality $ \sin\alpha\plus{}\sin\beta\plus{}\sin\gamma\le2(\sin\alpha\sin\beta\plus{}\sin\beta\sin\gamma\plus{}\sin\gamma\sin\alpha).$

We can show that
$ \sin \alpha \plus{} \sin \beta \geq 1$.
This is the case because $ \beta \geq \frac {\pi}{2} \minus{} \alpha$ (as $ \gamma \leq \frac {\pi}{2}$), and so, since both $ \beta$ and $ \frac {\pi}{2} \minus{} \alpha$ are in the range $ [0,\frac {\pi}{2}]$ where the sine function is increasing, $ \sin \beta \geq \sin(\frac {\pi}{2} \minus{} \alpha)$.  Therefore,
$ \sin \alpha \plus{} \sin \beta \geq \sin \alpha \plus{} \sin(\frac {\pi}{2} \minus{} \alpha) \equal{} \sin \alpha \plus{} \cos \alpha \equal{} \sqrt{1 \plus{} \sin 2 \alpha} \geq 1$.
Now, we can write the inequality as
$ \frac {1}{\sin \alpha} \plus{} \frac {1}{\sin \beta} \geq \frac {1}{\sin \alpha \sin \beta}$.
The same can be written for other pairs of angles, so adding,
$ \frac {2}{\sin \alpha} \plus{} \frac {2}{\sin \beta} \plus{} \frac {2}{\sin \gamma} \geq \frac {1}{\sin \alpha \sin \beta} \plus{} \frac {1}{\sin \beta \sin \gamma} \plus{} \frac {1}{\sin \gamma \sin \alpha}$,
which is equivalent to what needs to be shown.

[quote="rogue"]Let $ 0\le \alpha,\beta,\gamma\le \frac {\pi}{2}$ and $ \alpha \plus{} \beta \plus{} \gamma \equal{} \pi.$ Prove the inequality $ \sin\alpha \plus{} \sin\beta \plus{} \sin\gamma\le2(\sin\alpha\sin\beta \plus{} \sin\beta\sin\gamma \plus{} \sin\gamma\sin\alpha).$[/quote]
$ \sin v \equal{} \frac {2\tan \frac {v}{2}}{1 \plus{} \tan^2 \frac {v}{2}}$.
Let $ x \equal{} \tan \frac {\alpha}{2}, y \equal{} \tan \frac {\beta}{2}, z \equal{} \tan \frac {\gamma}{2}$. Then $ xy \plus{} yz \plus{} zx \equal{} 1$.
And then $ \sin \alpha \equal{} \frac {2x}{1 \plus{} x^2} \equal{} \frac {2x}{xy \plus{} yz \plus{} zx \plus{} x^2} \equal{} \frac {2x}{(x \plus{} y)(x \plus{} z)}$
So we have to prove:
$ \sum_{cyc} \frac {x}{(x \plus{} y)(x \plus{} z)} \le \sum_{cyc} \frac {4xy}{(x \plus{} y)^2(y \plus{} z)(x \plus{} z)} \iff$
$ \sum_{cyc} x(y \plus{} z) \le \sum_{cyc} \frac {4xy}{x \plus{} y} \iff$
$ \sum_{cyc} \frac {2xy}{x \plus{} y} \ge 1$
$ 0 < \alpha,\beta,\gamma \le \frac {\pi}{2} \Rightarrow x,y,z \le \tan \frac {\pi}{4} \equal{} 1$
Multiplying with $ (x \plus{} y)(y \plus{} z)(z \plus{} x)$ we have:
$ 2 \plus{} 2(x \plus{} y \plus{} z)xyz \ge x \plus{} y \plus{} z \minus{} xyz$ so it suffices to prove:
$ 2 \ge x \plus{} y \plus{} z \iff$
$ 2 \ge x^2 \plus{} y^2 \plus{} z^2$
Wlog $ x \ge y,z$. If $ y \plus{} z \ge x$: $ 2 \ge x^2 \plus{} y^2 \plus{} z^2 \iff x(y \plus{} z \minus{} x) \plus{} y(z \plus{} x \minus{} y) \plus{} z(x \plus{} y \minus{} z) \ge 0$
If $ x > y \plus{} z$ then $ x^2 \le 1$ and $ y^2 \plus{} z^2 \le (y \plus{} z)^2 < x^2 \le 1$ and hence $ 2 \ge x^2 \plus{} y^2 \plus{} z^2$
And we are done :)

(Equality requires $ xyz\equal{}0$. Wlog $ z\equal{}0$: Then we have $ xy\equal{}1, x^2\plus{}y^2 \le 2$ so $ x\equal{}y\equal{}1$ and hence the only equality case is with the degenerated triangle $ (0,\frac{\pi}{2},\frac{\pi}{2})$)


Prove $ \{\begin{array}{c}\{x,y,z\}\subset (0,\infty )\\\ x+y+z=1\end{array}\|$ $ \implies$ $ (1+x)(1+y)(1+z)\ge 8\cdot (1-x)(1-y)(1-z)$

[hide="An elementary proof."]
[color=darkblue]We"ll use the well-known relations :

$ \{\begin{array}{c}\prod (a\pm x)=a^{3}\pm a^{2}\cdot \sum x+a\cdot\sum yz\pm xyz\\\\ \sum x^{3}-3xyz=\sum x\cdot[\sum x^{2}-\sum yz]\end{array}$.

Therefore, for $ \|\begin{array}{c}\{x,y,z\}\subset (0,\infty )\\\ x+y+z=1\end{array}$ we have the following chain of the equivalencies :

$ \prod (1+x)\ge 8\cdot\prod (1-x)$ $ \Longleftrightarrow$

$ 1+\sum x+\sum yz+xyz\ge 8\cdot(1-\sum x+\sum yz-xyz)$ $ \Longleftrightarrow$

$ 2+9xyz\ge 7\sum yz$ $ \Longleftrightarrow$

$ 2(\sum x)^{2}+9xyz\ge 7\sum yz$ $ \Longleftrightarrow$

$ 2(\sum x^{2}+2\sum yz)+9xyz\ge 7\sum yz$ $ \Longleftrightarrow$

$ 2\sum x^{2}+9xyz\ge 3\sum yz$ $ \Longleftrightarrow$

$ 2\sum x^{2}+6xyz-2\sum yz\ge \sum yz-3xyz$ $ \Longleftrightarrow$

$ 2(\sum x^{2}-\sum yz)+6xyz\ge \sum yz(1-x)$ $ \Longleftrightarrow$

$ 2\sum x(\sum x^{2}-\sum yz)+6xyz\ge \sum yz(y+z)$ $ \Longleftrightarrow$

$ 2(\sum x^{3}-3xyz)+6xyz\ge \sum yz(y+z)$ $ \Longleftrightarrow$

$ 2\sum x^{3}\ge \sum yz(y+z)$ $ \Longleftrightarrow$

$ \sum (y^{3}+z^{3})\ge \sum yz(y+z)$ $ \Longleftrightarrow$

$ \sum (y+z)(y-z)^{2}\ge 0$,

what is truly ![/color][/hide]
[hide="A proof with a strong instrument - Jensen inequality."][color=darkblue]Apply the [b]Jensen's inequality[/b] to the function $ f: (0,1)\rightarrow\mathcal R$, where  : $ f(x)=\ln\frac{1+x}{1-x}$.[/color][/hide]

Why is your proof so long?  Here is a proof I gave in WOOT, and in a private message today:

[hide="Short"] Let $ a = x+y, b = y+z, c = z+x$.  We have the inequality

$ (a+b)(b+c)(c+a) \ge 8abc$

By AM-GM (say $ a+b \ge 2 \sqrt{ab}$ cyclically, or expanding first).  In terms of $ x, y, z$ this is

$ (2x+y+z)(x+2y+z)(x+y+2z) = (1+x)(1+y)(1+z) \ge 8(x+y)(y+z)(z+x) = 8(1-x)(1-y)(1-z)$ [/hide]


Hey guys,

I'm going through a lot of problems within next few weeks before AMC-12 A and decided to propose an idea. If there are large number of people interested ([b]large[/b] means at least 20 people), I will pick out some problems from the ones I go through and make a mock AMC-12 A.

[b]There will not be a original problem made by myself, however [/b].

Does this sound good to anyone? I think it'll be a good idea since that starting this year, calculator is prohibited. Post here if you are interested and when we have at least 20 people, it'll become official.

I'm interested.

Btw, this is my first year taking the AMC 12. :)

Same with me.
I hope it's a good practice.

Sure

The message is too small. Please make the message longer before submitting.

I'll lick your test!  :P ....

OK yeah I'm in.

Hey I'll do it, I'll probably need the practice too.

Ill do it

But when you say you are picking the problems, these problems can come from anywhere right?  They may not be original, but are they all from old AMC contest ?

[quote="mihail911"][quote="Silverfalcon"]I can't guarantee you anything but I'm thinking about sometime between 4:00 to 6:00 EST. This way, people can talk about the problems that day (I won't make a guarantee that you'll get your score that day though).[/quote]
grrr. that's like 1:00-3:00 Pacific time. and i assume that you can only take it in the time allotted. so you can't like post the test and i take it earlier? sorry, i'm new to this.  :blush:[/quote]

Probably not. I might post the test earlier the week though (Saturday instead of Sunday who knows) since I only have like 3-4 problems more to choose (and yes, they are #22-25).

I vote Sunday. Snowboarding is Saturday.

I can do it on Saturday if it's after 4.

can i join? i am available either saturday or sunday

[quote="tenniskidperson3"]I'd like to join also.
My first mock AMC 12, and I'm only in 9th grade![/quote]

I took the AMC 12 last year as a sixth grader, so so what? And I got a 132 :lol:

[quote="abacadaea"]im new to this mock AMC thing, are you just going to post the questions here?[/quote]

Yes, Silverfalcon is going to post on this topic.

[quote="1=2"][quote="fibointepi"]I took the AMC 12 last year as a sixth grader, so so what? And I got a 132 :lol:[/quote]

Let me guess: Bobby Shen. :wink:

When is this starting?[/quote]

are you fo' cereal?
wow!

[quote="fibointepi"]
[quote="abacadaea"]im new to this mock AMC thing, are you just going to post the questions here?[/quote]

Yes, Silverfalcon is going to post on this topic.[/quote]

This is incorrect. This is only discussion thread for starting competition. There will be three more threads in the future regarding this.

1. Problem Thread
This is where the problems will be posted. I'll post the thread 10 minutes prior to the test, and when the time comes up, I'll attach the document so you'll just have to refresh your window.

2. Discussion Thread
This is where you'll discuss your feelings about exams and just talk about the problems in general. The results will also be posted in this thread. DO NOT ASK FOR SOLUTIONS OR ANSWER KEYS. See #3

3. Solution Thread
This is thread solely for solutions. Over the past AMCs, it became very difficult to have solutions to all problems so in this one, this thread will serve to have solutions to ALL problems (they won't be posted at all once though).  They will be posted in the order of request so this thread serves as a request for solution as well.  If you have an alternate solution to the problem(s), this is also the place where you can post it.

Please follow this guidelines. Otherwise, I'll have to delete your posts.  This will make the test easier to navigate for others in future.

Since I want everyone to get their scores as soon as possible, the test will be held TODAY (Saturday, 1/26/08) on 7:00 EST. You will be able to send your answers until 8:20-8:25 but the test itself MUST end on 8:15 (thus, 75 minutes).

There will be a separate thread posted 30 minutes prior to beginning of test (so approximately 6:30 EST). So, spread the word.

For those who signed up, you'll receive private message from me (I'm going to do that now) as well. Now, the thread is locked since the test date has been set.


find all functions $f:\mathbb R\rightarrow\mathbb R$ such that:
                $f(x+f(y))=3*f(x)+f(y)-2*x$   for all x,y in real
i think this problem isn't hard because i have the solution
it's similar with the problem in IMO 1999.

Put $x=y=0$, so $f(f(0))=4f(0)$. Now put $x=-f(0), y=0$ to get $f(-f(0))=-\frac{2f(0)}{3}$. Put $x=f(0),y=t\Rightarrow{f(f(0)+f(t))=10f(0)+f(t)}$ (*) for all $t$.Put $x=f(t)-3f(0), y=f(0)$ to obtain: $f(f(0)+f(t))=3f(f(t)-3f(0))+10f(0)-2f(t)$ (**).From (*) and (**) we have:$f(t)=f(f(t)-3f(0))$ for all $t$.Put $x=f(y)-3f(0)\Rightarrow{f(2f(y)-3f(0))=2f(y)+6f(0)}$, now put $y=0$ in the last equality$\Rightarrow{f(-f(0))=8f(0)}$ but we have that $f(-f(0))=-\frac{2f(0)}{3}$ thus $8f(0)=-\frac{2f(0)}{3}$, so $f(0)=0$.Now just put $y=0$ in the original relation to get $f(x)=x$.
QED
I know this solution is short and seems easy, but believe me, it was not so easy to find it.

P.S. Please somebody check it, it is 2:18 here and I am too tired to do it now.


vector
superior algebra
superior algebra solved

there is a set of points in $\mathbb {R}^n$ such that any ball center at origin contain finitly many points.prove that this set is lattice

What exactly do you mean by lattice? Sorry for this somewhat stupid-sounding question (:)), but I don't get it.

assume that $a_1,\dots a_n$ are independence vector then $x_1.a_1+\dots +x_n.a_n$ when $(x_1,\dots,x_n)\in \mathbb{Z}^n$ is lattice which its basis is $a_1,\dots a_n$

That's what I thought, but I'm not at all sure it's true. What if the set of points is obtained from a lattice by deleting some vertices? Is that still a lattice? Or maybe we have to show that the set is part of a lattice?

What's the source of this problem?

Without some additional conditions, this question is nonsense- consider an arbitrary finite set.

Do we also know that the set is a group under vector addition? That question is fairly standard algebra.

Is it even true in $\mathbb{R}^1$? What if we take our set of points to be $\{1,\sqrt{2}\}$, for example?

I think it's clear by now that it's not true. Let's just wait and see what sam-n has to say :).

oops :blush: I always forgot to mention key word :D  again i write the problem: assume that we have set of points in $\mathbb{R}^n$ suchthat they are additive group if any ball aroun the origin contains finite points then prove that they are lattice. :blush:


Given the convex quadrilateral ABCD, one constructs the semicircles having AB, BC, CD and AD as diameters, all of them having interior points common to the interior of ABCD, and note M, N, P and Q their midpoints respectively.
Then MP = NQ and they are perpendicular.

Best regards,
sunken rock

I hope I understand this correctly. This is a variation of Van Aubel's Theorem right?

Note that $ M$ is the center of the square constructed inward with $ AB$ as its side. Let $ ABXY$ be the square whose center is $ M$.

We'll use complex number. We have $ x\equal{}b\minus{}(a\minus{}b)i$. So $ m\equal{}\frac{a\plus{}x}{2}\equal{}\frac{a\plus{}b\minus{}(a\minus{}b)i}2$. Likewise, $ n\equal{}\frac{b\plus{}c\minus{}(b\minus{}c)i}2$, $ p\equal{}\frac{c\plus{}d\plus{}(c\minus{}d)i}2$, $ q\equal{}\frac{d\plus{}a\minus{}(d\minus{}a)i}2$. Then $ \frac{n\minus{}q}{m\minus{}p}\equal{}\frac{b\plus{}c\minus{}d\minus{}a\minus{}(b\minus{}c\minus{}d\plus{}a)i}{a\plus{}b\minus{}c\minus{}d\minus{}(a\minus{}b\minus{}c\plus{}d)i}\equal{}\minus{}i$. So $ NQ\perp MP$. Also, $ |\frac{n\minus{}q}{m\minus{}p}|\equal{}1$, thus $ NQ\equal{}MP$.


search inequalities LaTeX calculus trigonometry

Have you tried to search forum for it?

If you want to get a quick reply then it is better to place your problem into the Inequalities Section.

[quote="_NightWish_"]Sa se demonstreze intr-un triunghi urmatoare inegalitate

       :sqrt: (bc) /a cos(A/2) +  :sqrt: (ca) /b cos(B/2) +  :sqrt: (ab) /c cos(C/2) >= (3 :sqrt: 3)/2[/quote]

Deoarece cos (A/2) = sqrt ( (p (p-a)) / (b c) ), inegalitatea este echivalenta cu ( f(a) + f(b) + f(c) ) / 3 >= sqrt (3) / 2, unde f: (0,p) -> R, f(x) = sqrt (p) sqrt (p-x) / x. Daca se poate studia convexitatea lui f (prin studiul semnului celei de-a doua derivate), mai departe ramane de observat doar de aplicat inegalitatea lui Jensen pentru ca f( (a+b+c) / 3 ) = f( 2p / 3 ) = sqrt (3) / 2. qed

It is trespass ! You can learn the LaTeX at the most in a week. You try, success !
Here is a solution without the calculus computationals (Jensen,a.s.o.):

$\exists\ x,y,z>0,\ a=y^2+z^2,\ b=z^2+x^2,\ c=x^2+$ $y^2$

$(s=x^2+y^2+z^2,\ s-a=x^2,\ \ldots\ ).$

$S=\sum \frac{\sqrt{bc}}{a}\cdot \cos \frac A2 =\sum \frac{\sqrt{s(s-a)}}{a}=\sqrt{x^2 +y^2 +z^2 }\cdot \sum \frac{x}{y^2 +z^2}\ge \ ^{*)}$

$\frac{1}{\sqrt 3}\sum x\cdot \sum \frac{x}{y^2+z^2}\ge ^{*)}\sqrt 3\cdot \sum \frac{x^2}{y^2+z^2}\ge \frac{3\sqrt 3}{2}.$

$^{*)}$ [u]Remark.[/u] For used Tchebaschev's inequality: $x\le y\le z\Longleftrightarrow \frac{x}{y^2+z^2}\le \frac{y^2}{z^2+x^2}\le \frac{z^2}{x^2+y^2}.$