inequalities
quadratics
inequalities proposed

Let $ x,y,z$ be real numbers.
(a) If $ A \equal{} \sum x^4 \minus{} \sum x^2y^2$, $ B \equal{} \sum x^2y^2 \minus{} xyz\sum x$, $ C \equal{} \sum x^3y \minus{} xyz\sum x$, $ D \equal{} \sum xy^3 \minus{} xyz\sum x$, then
$ (1) \ \ \ \ AB \ge CD$;
$ (2) \ \ \ \ AB \ge (\frac {C \plus{} D}{2})^2$;
(b) if $ S_1 \equal{} \frac 1{2}\sum (y^2 \plus{} z^2)(y \minus{} z)^2$, $ S_2 \equal{} \sum yz(y \minus{} z)^2$ and $ S_3 \equal{} \sum x^2(y \minus{} z)^2$, then
$ (3) \ \ \ \ 2S_1S_3 \ge S_2^2 \plus{} S_3^2$.

A hint: In the inequality $ AB \minus{} CD\ge 0$, equality holds for $ x \plus{} y \plus{} z \equal{} 0$.

On the other hand, the following inequalities holds, too: $ AB\ge \frac 1{2}(C^2 \plus{} D^2)$ and $ AB \ge C^2 \minus{} CD \plus{} D^2$.
The latter is the sharpest inequality I found ever.  :P  It is equality for $ x \equal{} 1$, $ y \equal{} 2$, $ z \equal{} 3$.

Nice, VasC. :)
I found (1),(2),(3) are equivalent to $ (x\plus{}y\plus{}z)^2(x\minus{}y)^2(y\minus{}z)^2(z\minus{}x)^2 \ge 0$ which is obviously true. :)

[quote="can_hang2007"]Nice, VasC. :)
I found (1),(2),(3) are equivalent to $ (x \plus{} y \plus{} z)^2(x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2 \ge 0$ which is obviously true. :)[/quote]
This is true.
And proves also "the sharpest" inequality $ AB \ge C^2\minus{}CD\plus{}D^2$. :wink:

But how shows a proof?

[quote="can_hang2007"]The last inequality is the sharpest, since it's equality! :)
To prove it, I use the $ pqr$ technique :).[/quote]
I found this inequality from
$ \sum [2x^2 \minus{} y^2 \minus{} z^2 \minus{} pxy \plus{} (p \plus{} q)yz \minus{} qzx]^2 \ge 0$,
that is
$ \sum x^4 \plus{} (p \plus{} q \plus{} r \minus{} 1)xyz\sum x \ge p\sum x^3y \plus{} q\sum xy^3 \plus{} r\sum x^2y^2$,
for $ 3(1 \minus{} r) \equal{} p^2 \plus{} pq \plus{} q^2$.

The discriminant $ D1(q,x,y,z)$ of the quadratic of $ p$ is nonpositive, and the discriminant $ D2(x,y,z)$ of the discriminant $ D1$, which is a quadratic of $ q$, is also nonpositve: $ D2 \equal{} C^2 \minus{} CD \plus{} D^2 \minus{} AB\le 0$.


What would happen if you add copper(II) chloride and silver nitrate together? Would you get copper(II) nitrate and silver chloride?

[hide="Opinion"]Yes, you would get a white precipitate of silver chloride since this salt has a low water solubility. However, cupric nitrate is water soluble, so we would end up with a solution containing solvated copper(II) ions and nitrate ions, with most of the silver and chloride ions present as solid AgCl. Upon standing, the precipitate would very likely turn grey and the solution would get a yellowish-green look.[/hide]


Today, I'm being political because I actually talked to my Mom today.  Ahh..Anyway, please, everyone go to http://www.gtinitiative.org/  and click on the GTI:  The Slide Show

I gotta give you some "inside information" on the GTI...THEY ARE JERKS. THEY ARE A BUNCH OF PROFESSORS TRYING TO SOUND SMART BY USING BIG WORDS AND HYPOTHETICAL GRAPHS USING LOGARITHMIC FUNCTIONS.  Who does that?  They go:  this is what society is going to be like if we don't implement this...[imagine a street full of bums] and if we do implement this...[imagine living in a mansion]...Yeah, that's in really informal language...but I think that's basically the gist of GTI.

My mom really wanted to get their job (she's really scary trying to make the world "a better place" and providing everyone with their basic needs) and she was so intent on getting it.  However, GTI, a supposed non-profit organization whose mission statement is like:  We help people answer the question:  What can I do?, rejected her based on her location.  Yeah, your organization is really based on getting a global action to make the world a better place when your company won't even interview someone that is 45 minutes away.  Oh yeah, definitely.   Anyhow, she make a powerpoint just for their company to improve their system and stuff, and they decided to see it, and allowed her 15 minutes, not with the president of the company (he refused to talk to her), but with a low level employee that was 25 years old.

So, she parked her car near their building, and the guy tells her to park her car in their parking lot because the parking around that neighborhood is basically just for local.  Also, my mom thought that if she was invited to park in their parking lot that the guy is inviting her to talk with him for a while...WRONG.  Anyway, she parks...and apparently she blocked in this professor or somethinkg, so she had to move.   Then, she goes and parks back in.  Yet again,  she needs to park somewhere else...And then after that she goes back in his office and he says, "Oh we're not interested in looking at your powerpoint...Get away."   Simplified, of course.  Excuse some people, but people don't take three hours of their life to be told "park here, no park here, no! park here, PARK HERE!"

Anyway, can some people post some of their comments on GTI and their like complete vagueness?

If you want to see my mom's powerpoint, IM me at KillingInSLeep :-P

Haha, I wonder if anyone is going to respond..it's kind of long and monotonous.

That post made very little sense. I've no idea what GTI is.

Well, uh, you might want to click on the link that he posted, and then go to the slide show (like he said), and then you will understand GTI and their mission.  He is basically saying that the world is fully of hypocrisy.  His mom who wants to make the world a better place tried to get a job with GTI which supposedly shares her point of view.  However when she tried to get a job they would not give her any time and were rude to her.  I mean, if you are going to save the world, why not start with yourself?!

This is obviously something serious in his life you might want to be more sympathetic.  He did leave contact info or you could PM him if you were confused.

:P  Well... GTI stands for the "Great Transition Initiative"  and basically it's a bunch of little, pompous, hippy professors that just talk about making the world "a better place."  The reason why this post makes little sense, is because GTI makes LITTLE SENSE.

[quote="JS1527"]Well, uh, you might want to click on the link that he posted, and then go to the slide show (like he said), and then you will understand GTI and their mission.[/quote]
All well and good, if I had powerpoint!

[quote="confuted"]
All well and good, if I had powerpoint![/quote]Not to worry, I'm sure PowerPoint would give you plenty of trouble.  I don't have it either, but I generally don't read MS Office files from the web.  If something is worth saying, it should be presented in an open file format, such as HTML or PDF.

If I had to guess, I would say that this "Great Transition Initiative" is a plot by some high school English teachers to improve the flow of language.

You can download a powerpoint viewer here: http://www.microsoft.com/downloads/details.aspx?FamilyID=428d5727-43ab-4f24-90b7-a94784af71a4&displaylang=en  It's only like 2mg.

[quote="confuted"][quote="failure"] If something is worth saying, it should be presented in an open file format, such as HTML or PDF.[/quote]
PDF is not an open file format; Adobe has just made the reader available for many operating systems. It is not open source.[/quote] Notice that I said "open" and not "open source," meaning that the file specifications are freely available, allowing anyone to write a writer or reader for the format without a lot of guesswork.  Ghostscript and PDFLaTeX are examples of third-party programs that can read and write (respectively) PDF files.  Take a look here: http://partners.adobe.com/asn/tech/pdf/specifications.jsp .  PowerPoint, on the other hand, is a proprietary and undocumented format.  The main point is that I shouldn't have to rely on a specific piece of software to view web content.  If I do, I probably won't bother viewing it.

I could be wrong on this, but I don't believe that the words "open source" are even applicable to file formats.

[quote="failure"]Notice that I said "open" and not "open source," meaning that the file specifications are freely available, allowing anyone to write a writer or reader for the format without a lot of guesswork.  Ghostscript and PDFLaTeX are examples of third-party programs that can read and write (respectively) PDF files.  Take a look here: http://partners.adobe.com/asn/tech/pdf/specifications.jsp[/quote]
My deepest apologies; I honestly was under the impression that the specifications for the PDF format were not public, but rather had been "reverse engineered" by a number of programmers. Your link, however, conclusively demonstrates that I was incorrect.

Just curious, is GTI truly non-profit? And they won't benefit more than others from this "global citiens movement", right?


inequalities
geometry
algebra
polynomial
area of a triangle
Heron\u0027s formula
algebra unsolved

A sequence u_n, n \in N, is defined by u_n = 2*u_(n-1) + 2*u_(n-2) - u_(n-3). Assume that the square roots of the start values u_0, u_1 and u_2 satisfy the triangle inequalities, then show:

a.) All triangles with side lengths (u_n)^0.5, (u_(n+1))^0.5 and (u_(n+2))^0.5 have the same area.

b.) lim_{n->infinity} ((u_n)^0.5 + (u_(n+1))^0.5)/(u_(n+2))^0.5 = 1.

There's also a completely geometric, non-computational argument for part b).

We see easily that the u_i's are increasing and in fact approach infinity.  But we also derived, in part a), that any 3 consecutive u_i^.5 form edges in some triangle.  As n-->infinity, since the area is constant, the distance from the vertex opposite u_(n+2)^.5 must approach 0 implying u_n^.5+u_(n+1)^.5-->u_(n+2)^.5 and hence lim ((u_n^.5+u_(n+1)^.5)/u_(n+2)^.5)=1, as required.


using the product and quotient laws of logarithms to prove the power law of logarithms

${log}_a\, p^c\,$= c ${log}_a\, p\,$
where a element (1,infinity),
p, q element (0, infinity),
c element (-infinity, infinity)

:huuh:

the proof is obvious and comes strieght from the Defenitions

$p^c=p.p.p....p$ (c times).

$\log_{a}{p^c}=\log_{a}{p.p.p...p}=log_{a}p+\log_{a}{p}+...+\log_{a}{p}$ (c times) $=c\cdot \log_{a}{p}$

$\mathbb{QED}$

Hmm, are there such obvious proofs for any other logarithmic identities? Besides the logarithm addition and subtraction ones, that is.

yeah, in my opinion all the proofs dealing with Logs come streight from defenitions. and are easily proven.

[quote="rrusczyk"]What if c is not an integer?[/quote]

again if you think about how we define the Log of a number in some base of some other number, it will be trivial.


How do you take to the $ n$th power an infinite series? For example, $ (1\plus{}x\plus{}x^2\plus{}x^3\plus{}...)^5$?

Thanks.

generating functions eh
well 1+x+x^2...=1/(1-x) so if you want to raise that to the fifth power then do 1/(1-x)^5 and have fun
btw that reduces to 1/(1-{x-x^2+x^3-x^4+x^5}) which is another infinite geometric series
1+(x-x^2+x^3-x^4+x^5)+(x-x^2+x^3-x^4+x^5)^2+...
alright maybe that does not help much but it's the best that i can do...
EDIT: oh haha,i lose. it's n-1 C n-1 + (n C n-1)x + (n+1 C n-1)x^2+...
taken right out of a book

@funtwo: 1/(1-x) implies -1<x<1 which is not necessarily the case

Anyways, you can use induction to prove that $ \displaystyle (\sum^{\infty}_{k\equal{}0} x^k)^n \equal{} \displaystyle\sum^{\infty}_{k\equal{}0} \binom{n\plus{}k\minus{}1}{n\minus{}1}$.

[hide="Outline"]

$ (1\plus{}x\plus{}x^2\plus{}\cdots)^1\equal{}1\plus{}x\plus{}x^2\plus{}\cdots$ so the base case n=1 works.

$ (\displaystyle\sum^{\infty}_{k\equal{}0} x^k)^{n\plus{}1}\equal{}(\sum^{\infty}_{k\equal{}0} x^k)^n (\sum^{\infty}_{k\equal{}0} x^k)\equal{}(\sum^{\infty}_{k\equal{}0} \binom{n\minus{}1\plus{}k}{n\minus{}1}x^k)(\sum^{\infty}_{k\equal{}0} x^k)$ and you can combine that witht he Hockey Stick Identity to finish the proof.

[/hide]

ah yes of course this isn't generating functions


category theory
superior algebra
superior algebra unsolved

1．	Give an example in TOP of a morphism that is monic and epic but does not have a retraction.
2．	Let Rng be the category of rings without unit, F the functor from Ring to Rng that forgets the unit.Show that any object in Rng possesses a universal from it to F.
    About 2 I have already know that every object of Rng denoted R may be embedded in a ring S. The ring S =R+Z where “+” means direct sum. And define multiplication in S by (r1,k1)(r2,k2)=(r1r2+k2r1+k1r2,k1k2). Verify that S is a ring with identity (0,1). u: R--S define by u(r)=(r,0) is a monomorphism of Rng. But I cannot show that (S,u) is the universal from R to F. Who can help me? Thanks very much!

1. remark that monic / epic morphisms coincide with injective / surjective continuous maps in the category TOP (this is a very rare property of concrete categories!). consider the standard example $[0,2 \pi ) \to S^{1}, t \to \exp(it)$.

2. you just have to check that every homomorphism $f : R \to T$ in Rng, where $T \in$ Ring, extends via $R \to S$ uniquely to a homomorphism $\overline{f}: S \to T$ in Ring. obviously this has to be $\overline{f}(r,z) = f(r)+z*1$, which makes it unique and suggests a proof for existence. it's very simple.


integration
function
calculus
calculus computations

Could someone help me with this partial differential equation?
$ u_t \equal{} \alpha u_{xx}$

I did some work, and the solution is :
$ u(t,x) \equal{} \frac{1}{\sqrt{2\pi}} \int_{\minus{}\infty}^{\infty} F(\xi) e^{\minus{}\alpha \xi^2 t} e^{i \xi x} d\xi$

Thanks. Is there a way to simplify that?

Let $ F(\xi) \equal{} \frac{1}{\sqrt{2\pi}} \int_{\minus{}\infty}^{\infty} f(x) e^{\minus{}i \xi x}\, dx$.
If f is very small, $ u(t,x) \equal{} \frac{1}{2\sqrt{\pi \alpha t}} \exp\left(\minus{}\frac{x^2}{4 \alpha t} \right)$.

lol yes that does make sense... *rubs chin*

[quote="TeXer"]Let $ F(\xi) \equal{} \frac {1}{\sqrt {2\pi}} \int_{ \minus{} \infty}^{\infty} f(x) e^{ \minus{} i \xi x}\, dx$.
If f is very small, $ u(t,x) \equal{} \frac {1}{2\sqrt {\pi \alpha t}} \exp\left( \minus{} \frac {x^2}{4 \alpha t} \right)$.[/quote]

How'd you get that? I'm looking to learning how to solve PDQs and I can't find anywhere legit from where I can learn =/

Need to take the Fourier Transform of the PDE.  In my opinion, the best intro to PDEs is "Basic Partial Differential Equations" by Bleecher and Csordas.  You'll like it: it's a smooth read.  Get the book, take an intro class, use both books, and in one year, you'll be good at them. :)

[quote]How'd you get that? I'm looking to learning how to solve PDQs and I can't find anywhere legit from where I can learn =/[/quote]
Take the Fourier Transform of the PDE.

Key words: heat equation, diffusion equation, Gaussian.

The result can be expressed as a convolution. Let $ G(x,t)$ be a function that you see above, namely
\[ G(x,t) \equal{} \frac1{\sqrt {4\pi \alpha t}}\exp\left( \minus{} \frac {x^2}{4\alpha t}\right)\]
Note that considered as a function of $ x,$ $ G(x,t)$ is the density of a normal distribution with variance $ 2\alpha t.$

The solution to the PDE is then given by
\[ G(\cdot,t)*f(x) \equal{} \int_{ \minus{} \infty}^{\infty}G(x \minus{} y,t)f(y)\,dy \equal{} \int_{ \minus{} \infty}^{\infty}f(x \minus{} y)G(y,t)\,dy\]
I did include a chapter on the heat equation in my [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=249955]Fourier analysis class notes.[/url]


I wanted to resurrect the topic. I happen to find the first one when I went through the search option...

This is basically what you need to do:
In order from strongest to weakest, choose five "categories" that best fit your self-identity. Not character traits, like "nice" or "inquisitive", but categories - in the sense of belonging to a group.

1. I am an AoPSer :lol:
2. I am a Roman Catholic
3. I am a Korean
4. I am a human being.... :huh:
5. I am a free thinker

[quote="djimenez"]Why do you want to cathegorize people... C'mon, most of the people is quite complex to fit exactly in a cathegory (or intersection of cathegories), except if you are talking about things like gender (still there are people who don't feet well in neither of the two), nationality (there are lots of people with double nationality) or something like that, that really don't say that much about how a person is really.[/quote]

Absolutely agree!

[quote="Dark_Shadow"]Here, we have the "part of no set" question.  Either you are part of some group, or you aren't.  If you aren't, you belong in a group with those who aren't.  Contradiction.  Therefore everybody is part of some group.

Everybody fits in some categories.  Even the category of "those who oppose categories."[/quote]
(a) It is, I think, a profundly simplist way of seen things, that is may be mathematically correct, but otherwise it has no real sence.
(b) Why do you want to cathegorize everything.
(c) You are using Aristotelic Logic, hence you are not considering the third excluded :D
(d) You wold have to be too specific with respect to your definition of the conditions to be part of one or other group, and you would probably will be defining either a too general set of conditions that will partially describe a huge population, or too specific that you are defining a too small group. Unless, of course, you want a "market oriented" type of analysis.

I am an organism
I am a vertabrate
I am a mammal
I am a primate
I am a homo sapien.

I'm also hot.  ;)

I'm a guy (w00t).
I'm Chinese (azn pride).
I'm atheist (don't want to get involved in religious debate topic).
I'm a mathy person (no way).
I'm a liberal (yup).

How to recognize me in the crowd. I have ...

Average height
Average weight
A common face
No distinct facial signs
Jeans & Nike shoes

:D :D :D

[quote="MathFiend"]I am a Christian
I am a math nerd
I try not to care what other people think about me
I avoid confrontation
I am constantly singing[/quote]

Well, that in fact says a little more about you than most of the other posted, but according to the "original idea"

[quote="xxreddevilzxx"]Not character traits, like "nice" or "inquisitive", but categories - in the sense of belonging to a group.[/quote]

fine then  :P

I am a Christian
I am a math nerd
I am a leader, not a follower
I am an AoPSer
I am an insomniac :)

better?  ;)

Im a Sexy Beast
Im Indian
Im tall
Im Smart

I'm super smart.

I'm very athletic.

I'm extremely knowledgeable about everything.

Just kidding :P.

I've been trying to convince people that I'm in category #4 for over a year now. Earlier I was trying to prove that I
[hide][/hide]

[quote="joml88"]...you do not have to participate in every single thread in the Fun and Games forum  :P[/quote]
this is the round table, but anywho:

5. i participate in the aops trivia challenge (suck at that)
4. videogame-lete
3. super-smart  :D
2. math nerd (who isn't around here)
1. very good looking (hey, ladies  ;) )

[quote="joml88"][quote="djimenez"][quote="furious"]I am an organism
I am a vertabrate
I am a mammal
I am a private
I am a homo sapien.

I'm also hot.  ;)[/quote]

I rest my point, that says basically nothing about you, or at least nothing different from what a person can say if they see you for half a second![/quote]

You cannot rest your point on furious' facetious remarks...come on that's like equivalent to circular reasoning.

Look, the basic idea of this discussion is to identify 5 groups/categoies that you belong to/fit in that would make you distinguishable from most other people.  If you disagree with this thread just remember that you do not have to participate in every single thread in the Fun and Games forum  :P[/quote]No, I don't have to participate in every single thread on F&G or any other forum in the site (and actually I don't, otherwise, I would have more posts than Grobber :D ), but any way, this is not F&G, it is The Round Table forum, a forum [i]for serious discussion of other topics[/i], and as part of any toppic is the dissidence with someone else's opinion.

By the way, you say that [b]the basic idea of this discussion is to identify 5 groups/categoies that you belong to/fit in that would make you distinguishable from most other people.[/b] Then, could you explain me what is your definition of people, person, and why does the caracteristics organism, vertabrate, mammal, primate, homo sapien makes fourius different from most of the other people? Those same five characteristics apply to every single person I know!

[quote]
Then, could you explain me what is your definition of people, person, and why does the caracteristics organism, vertabrate, mammal, primate, homo sapien makes fourius different from most of the other people? Those same five characteristics apply to every single person I know![/quote]

I quote myself:

[quote]You cannot rest your point on furious' facetious remarks...come on that's like equivalent to circular reasoning. [/quote]

Perhaps a moderator would like to move this thread to F & G ... it seems like it may be better suited there.

A shame - it was one of the most thought-provoking threads the first time around.


[quote="linzk3"]Given that sin(x)*cos(x)=60/69 find sin(x)+cos(x).[/quote]
[hide="Solution"]
We call $ \sin x\equal{}a$ and $ \cos x\equal{}b$.  Thus, $ ab\equal{}\frac{60}{69}\implies 2ab\equal{}\frac{40}{23}$.  Also, we notice that $ a^2\plus{}b^2\equal{}\sin^2 x\plus{}\cos^2 x\equal{}1\implies a^2\plus{}b^2\equal{}1$.  Thus \[ (a\plus{}b)^2\equal{}a^2\plus{}b^2\plus{}2ab\equal{}1\plus{}\frac{40}{23}\equal{}\frac{63}{23}\implies a\plus{}b\equal{}\pm(\sqrt{\frac{63}{23}})\equal{}\boxed{\pm(\frac{\sqrt{1449}}{23})}\]
[/hide]


Prove that
\[ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \le \frac {a^8 \plus{} b^8 \plus{} c^8}{a^3b^3c^3}
\]

[hide]I think we must have a,b,c>0. If not, the counter example is:a=b=-2,c=-1.
With this additional restriction, here's my solution:
Applying AM-GM we have: (a^8+b^8+c^8)/a^3b^3c^3>=ab/c^3+bc/a^3+ac/b^3>=b/ac+c/ab+a/bc
Using AM-GM again, we have: a/bc+b/ac+c/ab>=1/a+1/b+1/c
Therefore, the original inequality is true with [b]a,b,c>0[/b].
The equality occurs when a=b=c
[b]Generalization[/b]: We can solve the more generalized problem by using the same method as above:
Given a,b,c>0. Prove that:1/a+1/b+1/c=<a^(2^k)+b^(2^k)+c^(2^k)/(a^(k).b^(k).c^(k)) with k is any positive integers(you can try proving it, Johnny, since it just pops up from my head and I haven't tried proving it yet.I think it's still true though). :) .By the way, does CGMO stand for "China Girl Mathematics Olympiad"? If it does, I'm surprised by the difficulty of this problem to be chosen.[/hide]

[quote="TZF"][hide="This is a direct application of"]Multiply through by $ (abc)^3$ to get $ \sum_{sym} a^8 b^0 c^0 \geq \sum_{sym} a^3 b^3 c^2$

Because $ ( 8,0,0 )$ majorizes $ (3,3,2)$, this is true by [url=http://en.wikipedia.org/wiki/Muirhead%27s_inequality]Muirhead's Inequality[/url].[/hide][/quote]
Dumbassing, eh? (Well, that's how I'd solve it anyway...)
Since it's homogenous and symmetric, that option seems pretty obvious. :/

Just a brief remark: this one was  Crux (or it was Eureka those days?) 413 problem, proposed by G. C. Giri.

[quote="joh"][quote="ghjk"]By the way, does CGMO stand for "China Girl Mathematics Olympiad"? If it does, I'm surprised by the difficulty of this problem to be chosen.[/quote] is it too easy or too hard?

i think some problems from cgmo are quite easy. so it does not surprise me that this is a cgmo problem.[/quote]
I meant the first side :) . Really? Can you send me some latest problems of this contest(i.e, 2008 or 2007)?

alternatively, we can

[hide]
Use AM-GM on $ a^8, a^8, a^8, b^8, b^8, b^8, c^8,  c^8$ to get $ \frac{3a^8 \plus{} 3b^8 \plus{} 2c^8}{8} \ge a^3b^3c^3$

and add cyclicly and divide by $ a^3b^3c^3$ to get the desired result. Of course, this is essentially Muirhead. [/hide]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=21137]IMO LongList 1967, Poland 3[/url]


calculus
integration
calculus computations

I tried to solve this one by separating the variables, but i got one integral who i do not know how to solve.

$x(y+1)^{2}+(x^{2}+1)y\text{e}^{y}y'=0$

$\int \frac{y\tex{e}^{y}}{(y+1)^{2}}\, \text{d}y$

Thanks for any help.

That integral is rigged to work- try integrating by parts.


$ \csc \alpha \equal{} \csc 2\alpha \plus{} \csc 3\alpha$, where $ \alpha\equal{}180^{\circ}/7$.

i didnt understand ur question. its more of a statement. i dont know what to find or derive or proof

Prove it.

Please justify each statement clearly as I am an amateur "trigonometer".

i hant solved this yet :maybe: , but u can try applying double and triple angle formula here

i think u have titu's book now, u can read it to understand wut i am saying.(pg 5,6)


geometry
algebra
polynomial
Support
MATHCOUNTS
AMC
AMC 8
\/closed

I have already taken Alg II and trig/pre-calc, but some of my algebra skills are lacking.  I thought about taking a class like Alg III on AoPS, but I don't have [i]all[/i] of the skills necessary to begin the class.  I would also like to improve my problem-solving skills.  Are the competition-prep classes worth taking even if I don't plan on participating in any of those competitions?  Will they help improve my problem solving skills?  Should I take a class even though it might be a little too hard for me?  Thanks!  Any help is appreciated!

I think you're better off with one of the specific subject classes.  They're more in-depth, and will patch the holes you think you have more thoroughly than the contest classes.  Which algebra skills are you missing for starting the class?

(Also, it's OK to take a class that's a little too hard as long as it's not way too hard and you're willing to ask questions when you don't understand something.)

Wait if you want to improve your problem-solving skills,  why don't you take part in competitions?

S/he might be too far away, or too old for some competitions, or just doesn't want to put in the effort.

I have a question motivated by the above posts: are there/have there ever been students enrolled in AoPS classes that were in college or beyond at the time they took the class? This doesn't seem too much of a stretch, though it would be interesting to know about what percentage of AoPS students fit this particular mold.

I have a PhD in Computer Science, but I am taking the Olympiad Geometry course...

Thanks for the information!  Some things that I learned in alg2, like some polynomial factoring and sequences, I don't remember well and may need to refresh myself in.  (Sorry that was an awkward sentence.)

I guess the reason why I don't do the particular competitions that AoPS has classes for is that my school doesn't support them.  I did MATHCOUNTS in junior high, and my school put on an AMC8 once, but they never did it again and I didn't know there was another level of competition.

Oh so are you going to start participating now?

I'm not sure what you mean.  I can't participate in competitions right now.  I think I will take an AoPS class, so I may participate in the forums more later.


Given the parabola y=x^2, and the line y=2x-4, find the shortest distance between the line and the parabola (As usual, no calculus).

PS: The question's probably easier than it looks

all you are looking for is the altitud of a triangle fomed by the x,y intersections of the line.
(the altitude formed from (0,0)
[hide="ans"]
$\frac{3\sqrt{5}}{5}$[/hide]

why does that work? i'm a bit confused. isn't the triangle's altitude the distance between the line and (0,0)?

Right triangle, line, parabola, legs parallel to the axes, kthxbai.

i'm [i]still[/i] confused, sorry to bother u all, but first off, i got [hide]$\frac{3\sqrt{2}}{2}$[/hide] instead of amirhtlusa's answer. secondly, i tried the right triangle way, but i don't see how u get the answer without calculus (because u have to find out when the tangent is parallel to the line before making the triangle)


Does there exist a set of $n > 2, n < \infty$ points in the plane such that no three are collinear and the circumcenter of any three points of the set is also in the set?

No. If there is a finite set of such points, then there is a smallest circumcircle, C, among all circumcircles. If there are three points (of the set) on C that don't form an equilateral triangle, then one of the sides of the triangle they form is less than the side of an equilateral triangle inscribed in C, and the circumcenter of C together with the endpoints of that side have a circumcircle smaller than C--a contradiction. Otherwise, there are exactly three points of the set on C and they form an equilateral triangle. By an easy argument, we see that the trisection points of the sides of this triangle must also be in the set. These points are concyclic and they lie on a circle smaller than C--again a contradiction.


Do this quickly!

The density of water is 1.0g/cm^3. Convert this density to kg/m^3.

WHat do you get?

Use factor-label:
$ \frac{1.0\mathrm{g}}{\mbox{cm}^{3}}\cdot\frac{1\mbox{ kg}}{1000\mbox{ g}}\cdot\frac{1000\mbox{ cm^{3}}}{1\mbox{ m^{3}}}}$

JRav, there are $ 1,000,000$ cubic centimeters in a cubic meter.

[quote="JRav"]Use factor-label:
${ \frac {1.0\mathrm{g}}{\mbox{cm}^{3}}\cdot\frac {1\mbox{ kg}}{1000\mbox{ g}}\cdot\frac {1000\mbox{ cm^{3}}}{1\mbox{ m^{3}}}}$[/quote]Correction:

${ \frac{1.0\text{ g}}{\text{cm}^{3}}\cdot\frac{1\text{ kg}}{10^3\text{ g}}\cdot\frac{100^3\text{ cm}^3}{1\text{ m}^3}}$


For each positive integer $ m>1$, let $ P(m)$ denote the greatest prime factor of m.  For how many positive integers n is it true that both $ P(n)\equal{}\sqrt{n}$ and $ P(n\plus{}48)\equal{}\sqrt{n\plus{}48}$?

$ P(n) \equal{} \sqrt {n} \rightarrow n \equal{} k^{2}, k \in Z$
Since $ 25^{2} \minus{} 24^{2} > 48$, we know $ k < 25$.
Using trial and error, $ \boxed{n \equal{} 121}$.

EDIT: removed extraneous solution

I don't think $ n\equal{}16$ works. $ P(16)\equal{}2\ne\sqrt{16}$.

$ n\equal{}121$ seems to work though.

[quote="grn_trtle"]I don't think $ n \equal{} 16$ works. $ P(16) \equal{} 2\ne\sqrt {16}$.

$ n \equal{} 121$ seems to work though.[/quote]

Thanks. $ P(k)\equal{}\sqrt{n} \Rightarrow n\equal{}k^{2}, k \in Z$ AND $ k$ is prime. I forgot the second implication earlier. Thus $ k \in \left \{ 2,3,5,7,11,13,17,19,23 \right \}$, leaving only $ 36$ possibilities to check.


Could we have a section on using Mathematica.
I guess so many of us have encounter problems in writing mathematica programme   :)

YESS!!!
maybe a forum for mathematics programs like that in general


Let f be a real function satisfying f(x)+y = f(x+y) for all real x and y. Assume that f(0) is a positive integer, and that f(2)|f(5). Find f(2008).

[hide]$ x \equal{} 0 \rightarrow f(y) \equal{} y \plus{} c$ for some constant $ c$. $ f(0) \equal{} 0 \rightarrow c$ is a positive integer. $ f(2) | f(5) \rightarrow 2 \plus{} c | 5 \plus{} c \rightarrow 2 \plus{} c | 3 \rightarrow c \equal{} 1$. Thus, $ f(2008) \equal{} 2009$. [/hide]


Να υπολογισετε το οριο:

$ \lim_{n \rightarrow \plus{} \infty}{\frac {0^1 \plus{} 1^2 \plus{} 2^3 \plus{} ... \plus{} (n \minus{} 1)^n}{2^1 \plus{} 3^2 \plus{} 4^3 \plus{} ... \plus{} (n \plus{} 1)^n}}$

Λήμμα Stolz! Να ναι καλά ο κος Λουλάκης που μας το μαθε  :P

Αν ονομάσουμε με a την ακολουθία του αριθμητή και με b του παρονομαστή, έχουμε οτι το ζητούμενο όριο θα συγκλίνει όπου συγκλίνει και το όριο:

$ \lim_{n \to \plus{} \infty} \frac {a_{n \plus{} 1} \minus{} a_n} {b_{n \plus{} 1} \minus{} b_n} \equal{} \lim_{n \to \plus{} \infty} \frac {(n \minus{} 1)^n} {(n \plus{} 1)^n}$
$ \equal{} \lim_{n \to \plus{} \infty} \frac {1} {(1 \plus{} \frac {2} {n \minus{} 1})^n }$

Το οποίο τείνει στο $ \frac {1} {e^2}$

Ακριβώς, well done :thumbup:

Ωστόσο, δεν είμαι σίγουρος για το χαρακτηρισμό "εύκολο", μάλλον θέμα γνώσεων θα το χαρακτήριζα: αν ξέρεις να κάνεις de l'Hôpital σε ακολουθίες, το πρόβλημα είναι εντελώς τετριμμένο... αν όχι, είναι κάπως πιο δύσκολο ;)

Cheerio,

Durandal 1707

[quote="Durandal"]Μάλλον θέμα γνώσεων θα το χαρακτήριζα: αν ξέρεις να κάνεις de l'Hôpital σε ακολουθίες, το πρόβλημα είναι εντελώς τετριμμένο... αν όχι, είναι κάπως πιο δύσκολο ;)

Cheerio,

Durandal 1707[/quote]

Αυτο ακριβως ηθελα να πω με το ευκολο, τετριμενο αν ξερει κανεις Stolz

Ωραία άσκηση και μπορώ να πω ότι έμαθα και το θεώρημα αυτό που δεν το ήξερα!!!! :thumbup:
Bάζω και μία απόδειξη του θεωρήματος για όποιον δεν το έχει ξαναδεί που βρήκα στο ίντερνετ...


Euler
geometry
3D geometry
tetrahedron
LaTeX
linear algebra
matrix

Hi.
I think i once saw a formula somewhere relating the three sides of a triangle ABC to the distances PA, PB and PC from a point P. i cant find it now tho. Does anyone know it? (i think it was very messy but still maybe useful).

Here is the formula in one of its variants:

If P is a point in the plane of a triangle ABC, and we denote BC = a, CA = b, AB = c, PA = x, PB = y and PC = z, then

$\left( y^{2}+z^{2}-a^{2}\right) ^{2}x^{2}+\left( z^{2}+x^{2}-b^{2}\right) ^{2}y^{2}+\left( x^{2}+y^{2}-c^{2}\right) ^{2}z^{2}$
$-\left( y^{2}+z^{2}-a^{2}\right) \left( z^{2}+x^{2}-b^{2}\right) \left( x^{2}+y^{2}-c^{2}\right) -4x^{2}y^{2}z^{2}=0$.

You can also rewrite it as follows:

$\left| 0\;\;\;\;\;c^{2}\;\;\;\;b^{2}\;\;\;x^{2}\;\;\;\;1\right| $
$\left| c^{2}\;\;\;\;0\;\;\;\;a^{2}\;\;\;\;y^{2}\;\;\;\;1\right| $
$\left| b^{2}\;\;\;\;a^{2}\;\;\;\;0\;\;\;\;\;z^{2}\;\;\;1\right| =0$.
$\left| x^{2}\;\;\;\;y^{2}\;\;\;\;z^{2}\;\;\;\;0\;\;\;\;1\right| $
$\left| 1\;\;\;\;\;1\;\;\;\;\;1\;\;\;\;\;\;1\;\;\;\;\;0\right| $

In fact, this formula originates from Euler, and it actually follows from a formula for the volume of the tetrahedron ABCP from the lengths of its edges in the special case when the tetrahedron is flat and the volume is 0. You can also prove it by trigonometry.

BTW, I quoted this formula in [url=http://mathforum.org/epigone/geometry-puzzles/fongpholjan/480ptuvc9pje@legacy]a geometry-puzzles thread[/url], but the formula I gave had a little typo - sorry!

  Darij

you made a slight typo in the latex code of the second equation ... I don't think that \matrix command is a valid latex command. try \begin{array} \end{array} :P

Yes, Valentin, I know, \matrix is a Scientific WorkPlace command, but I was too lazy to retype the SWP matrix in the LaTeX form, hence I now made a little workaround... you see.

  Darij

thanks! (that was the site where i saw it b4)
what is the formula for the volume of a tetrahedron?

It's $\left | \frac 16 \left | \begin{array} {cccc} x_1&y_1&z_1&1\\x_2&y_2&z_2&1\\x_3&y_3&z_3&1\\x_4&y_4&z_4&1\end{array}\right | \right |$, where $(x_i,\ y_i,\ z_i)$ are the vertices of the tetrahedron.


[hide="hint"]first find the difference. and find the decrease of the percentage[/hide]

what math class are you taking?(algebra, geomety, etc.) This is easy stuff.

I took pre-algebra this year (i know... we don't have anything else) and it was on our test... i guess it would be using shinwoo's hints, but i solved it incase people want to know the answer (yes im bored)

[hide]200-180 = 20
20/200 = 1/10 = 10%[/hide]

[hide]to solve for percent of change, you always do:
 amount of change/original value[/hide]

This wasn't meant to be hard..., I knew this already.

i knew that you knew this and this wasn't hard but i did this a long time ago

Percent change-

Change
-------
Original

Edit: Sorry for bumping, I was looking through some old stuff and forgot.  :ninja:

[quote="236factorial"]Source: From my seventh grade final exam review

What is the percent decrease from 200 to 180?[/quote]

[hide]That's easy: $ \%  \equal{} \frac{{200 \minus{} 180}}{{200}}.100\%  \equal{} 10\%$[/hide]

$ \text{percent change }\equal{} \frac{\text {original value} \minus{} \text{ final value}}{\text{ original value}}$

This is useful stuff for physics, many formulas are based on this.


[quote="lasha"]$ 30$ students participated in the mathematical Olympiad. Each of them was given $ 8$ problems to solve. Jury estimated their work with the following rule:
1) Each problem was worth $ k$ points, if it wasn't solved by exactly $ k$ students;
2) Each student received the maximum possible points in each problem or got $ 0$ in it;
Lasha got the least number of points. What's the maximal number of points he could have?
Remark: 1) means that if the problem was solved by exactly $ k$ students, than each of them got $ 30 \minus{} k$ points in it.[/quote]

Maximum number of points is $ 60$. Example: if each of first $ 15$ students will solve only first $ 4$ problems, and each of left $ 15$ students will solve onle the last $ 4$ problems, then each of them will receive exactly $ 60$ points.
Suppose that $ 60$ points isn't maximum then Lasha got at least $ 61$ points(therefore everyone got at least $ 61$ points)
So therefore sum of all students score is at least $ 61 \cdot 30\equal{}1830$ points. There is at least one problem such the sum of scores from this problem is $ [\frac{1830}{8}]\plus{}1\equal{}229$ points. The sum of scores of the students for $ 1$ problem is $ k(30\minus{}k)$ Maximum of such sum is $ 225$. Therefore we have contradiction.
So we have that maximal points that Lasha could have is $ 60$.

[quote=socrates]What if we demand Lasha to be the only student with the  least number of points? What's the maximal number of points he could then have?[/quote]


We call a set $ A$ a good set if it has the following properties:
1. $ A$ consists circles in plane.
2. No two element of $ A$ intersect.
Let $ A,B$ be two good sets. We say $ A,B$ are equivalent if we can reach from $ A$ to $ B$ by moving circles in $ A$, making them bigger or smaller in such a way that during these operations  each circle does not intersect with other circles.
Let $ a_{n}$ be the number of inequivalent good subsets with $ n$ elements. For example $ a_{1}\equal{} 1,a_{2}\equal{} 2,a_{3}\equal{} 4,a_{4}\equal{} 9$.
[img]http://i5.tinypic.com/4r0x81v.png[/img]
If there exist $ a,b$ such that $ Aa^{n}\leq a_{n}\leq Bb^{n}$, we say growth ratio of $ a_{n}$ is larger than $ a$ and is smaller than $ b$.
a) Prove that growth ratio of $ a_{n}$ is larger than 2 and is smaller than 4.
b) Find better bounds for upper and lower growth ratio of $ a_{n}$.

Official Solution:
For the upper bound, to each arrangement assign a string of parentheses. First arrange the circles in such a way that each circle's center is on the x axis, and then cut the tops and bottoms of every circle.
The remaining of each circle consists of two arcs shaped either like $ ($ or like $ )$. Assign this string to this arrangement. (Of course there may be several ways to assign a string to a single arrangement)
This string is correctly parenthesized, because each $ ($ is matched to a corresponding $ )$, and the substring between these two, corresponds to a subarrangement with fewer circles (that is those circles contained in one of the circles in the original arrangement)
It is straightforward to construct back the arrangement from the string of parentheses (Complete the circles by joining each $ ($ with its matching $ )$)
Therefor $ a_n$ is smaller than or equal to the Catalan number of order $ n$ which is $ \frac 1{n\plus{}1}{2n\choose n}\leq 2^{2n}\equal{}4^n$.
For the lower bound, select a string of $ 0$'s and $ 1$'s of length $ n$ and assign an arrangement to it. Reading the string from left to right, replace each $ 0$ by a circle containing no other circles, and each $ 1$ by a circle containing all the other previous circles.
Of course some arrangements can not be achieved by this method. But	if an arrangement can be achieved, it can be achieved from at most one string. This is because, we can easily reverse the procedure. first consider the outermost circles; at most one can contain other circles. Put as many $ 0$'s as there are empty circles at the end of the string. If no circle is left, then we are done. If there is	one left, put a $ 1$ before these zeros, and then put the string representing the contents of this circle at the beginning of our string.
This shows that $ a_n\geq 2^n$.

Here's a different way to get the upper bound for (a).

Set $a_0 = 1$ for convenience. Call a circle in a good set [b] tasty [/b] if it is not contained by any other circles. Define the [b] size [/b] of a tasty circle as the number of circles which are inside it (including itself).

By doing casework on the sizes of the tasty circles, we derive the following recurrence relation for all $n \ge 1$:

$$a_n = \sum_{x_1, x_2, \cdots, x_k \in \mathbb{N}, x_1 + x_2 + \cdots + x_k = n} a_{x_1 - 1} a_{x_2 - 1} \cdots a_{x_k - 1},$$

where the sum is over all unordered partitions $(x_1, x_2, \cdots, x_k)$ of $n$.

It's clear that $a_n$ only increases for all $n$ if we instead sum over ordered partitions $(x_1, x_2, \cdots, x_k)$ of $n$, so let $b_n$ be the set which follows this recurrence relation instead.

In other words, we have $b_0 = 1, b_1 = 1, b_2 = 2, b_3 = 5, \cdots$ and

$$b_n = \sum_{x_1, x_2, \cdots, x_k \in \mathbb{N}, x_1 + x_2 + \cdots + x_k = n} b_{x_1 - 1} b_{x_2 - 1} \cdots b_{x_k - 1},$$

where the sum is over all ordered partitions $(x_1,x_2, \cdots, x_k)$ of $n$.

Let $M(x) = \sum_{k = 1}^{\infty} b_k x^k.$ Then the above recurrence relation implies that:

$$M(x) = \frac{x + xM(x)}{1 - x - xM(x)}.$$

Expanding, we get $xM(x)^2 + (2x-1) M(x) + x = 0$, and solving this quadratic gives:

$$M(x) = \frac{-(2x-1) \pm \sqrt{1 - 4x}}{2x}.$$

By plugging in $x = 0$, it's clear that because $M(0) = 1$, we must have a $-$ instead of a $+$, so that:

$$M(x) = \frac{-(2x-1) - \sqrt{1-4x}}{2x}.$$

By the Binomial Theorem, we then have that the $x^n-$coefficient in $M(x)$ is $\frac{\binom{2n}{n}}{2(2n-1)}$ for $n \ge 1.$

This means that $b_n \le \frac{\binom{2n}{n}}{2(2n-1)}$ for $n \ge 1$, and so clearly the growth ratio of $\{b_n\}$ is smaller than $4.$ Since $a_n \le b_n$ for all $n \in \mathbb{N}$, so is the growth ratio of $\{a_n\}.$

$\square$

P.S. Would like to see a way to improve the upper bound on the growth rate, because this solution bounds $\{a_n\}$ very loosely.


inequalities
IMO Shortlist
inequalities unsolved

For $a,b,c,d$ positive reals such that $abcd=1$ prove that

$\frac{1}{(1+a)^2}+\frac{1}{(1+b)^2}+\frac{1}{(1+c)^2}+\frac{1}{(1+d)^2}\geq 1$

Manlio, I have an advice: instead of posting 50 wrong inequalities, you better post 5 corect ones and also not discussed before. This one is incorrect and also the correct form is discussed on the forum.

Could you please post a link or the correct form? I thought it was wrong too, but then I realized that the LHS can be made as close to $1$ as we want ($a\to 0, b,c,d\to\infty$), so it might actually be a correct and unorthodox inequality.

Sorry if I'm wrong..

Grobber, the link: http://www.mathlinks.ro/Forum/viewtopic.php?t=16027

Harazi, hey... don't we all make mistakes? It seems you're a bit angry on manlio for posting all these problems... :P

I guess it's not bad to have classical ones posted back here... ok for our mathlinks elite it may feel like "again...  :?", but newer people do like having all those "well-known/classical" questions a bit together...

Besides, can't exactly the same be said about the IMO shortlist project... many questions from it have also been posted before... and sometimes a wrong question number is posted, or whatever... still I don't see anyone complaining there ;)


limit
trigonometry
parameterization
parametric equation
real analysis
real analysis unsolved

how do you find the point where a parametric equation crosses it self?

also

1^infiniti is not defined right?

1.  The parametric equation can cross itself at two points $t_1, t_2$.  So you need to solve the following system of simultaneous equations:

$x(t_1) = x(t_2)$
$y(t_1) = y(t_2)$

For $t_1, t_2$.

2.  $\lim_{x \to \infty} 1^x = 1$.  :)

[quote="t0rajir0u"][quote="Gibbenergy"][quote="t0rajir0u"]

2.  $\lim_{x \to \infty} 1^x = 1$.  :)[/quote]
$1^\infty$  is  undeterminate form[/quote]

I never said it wasn't.  :)[/quote]

ummm... ok

one more question.  how do you use power series to find a limit?

example:

$\lim_{x \to 0}{\frac{7\sin x}{e^{2x}-1}}$

[quote="jli"][/quote][quote="t0rajir0u"][quote="Gibbenergy"][quote="t0rajir0u"]

2.  $\lim_{x \to \infty} 1^x = 1$.  :)[/quote]
$1^\infty$  is  undeterminate form[/quote]

I never said it wasn't.  :)[/quote][quote="jli"]

ummm... ok

one more question.  how do you use power series to find a limit?

example:

$\lim_{x \to 0}{\frac{7\sin x}{e^{2x}-1}}$[/quote]

Expand the Taylor Series of the numerator and denominator until you have enough terms to find the limit as a rational function.  The other terms won't matter.

As far as your example:

$\frac{7 \sin x}{e^{2x} - 1} = \frac{7 \left( x - \frac{x^3}{3!} + ... \right) }{ \left( 1 + (2x) + \frac{(2x)^2}{2!} + ... \right) - 1} = \frac{7x - \frac{7x^3}{6} + ... }{ 2x + 2x^2 + ... } = \frac{7 - \frac{7x^2}{6} + ... }{2 + 2x + ...}$

The limit is $\frac{7}{2}$.  (This is the principle behind l'Hopital's rule.)


i think it ia easy and posted before and my solution  here
 (a,b)=$(f_{2n-1},f_{2n+1})$ for f_n is Fibonaci number

[quote="vardanch"]Prove that there exist infinitely many pairs of integers $a, b$ such that $a^{2006}+1$ is divisible by $b$  and $b^{2006}+1$ is divisible by $a$.[/quote]
It is sufficient to prove that there are infinitely many pairs of integers $a, b$ such that $a^{2}+b^{2}+1$ is divisible by $ab$.
But this problem has been disscused many-many times in this forum.
Here is one of many links about that problem.
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=36687[/url]
 :wink:


limit
function
induction
inequalities
calculus
calculus computations

Find the value of

$\sqrt{7-\sqrt{7+\sqrt{7-\sqrt{7+\sqrt{7-\sqrt{7+\ldots }}}}}}$

Define a sequence by
$a_{n+1}= \sqrt{7-\sqrt{7+a_{n}}}\quad a_{0}= 0$.
$a_{n}$ will never be bigger than $\sqrt{7-\sqrt{7}}< 5 < 42$, so we don't have any definitional worries.
We are asked to calculate $\lim_{n\to\infty}a_{n}$.

$g(x) = \sqrt{7-\sqrt{7+x}}$ is a decreasing function, which is annoying to deal with. So instead we consider the sequences $(a_{2n})_{n=0}^\infty$ and $(a_{2n+1})_{n=0}^\infty$, given by $a_{2(n+1)}= h(a_{2n})$ and $a_{2(n+1)+1}= h(a_{2n+1})$, where
$h(x) = \sqrt{7-\sqrt{7+\sqrt{7-\sqrt{7+x}}}}$.

We have that $a_{0}< a_{2}$. By applying the decreasing $g$ on both sides we get $a_{1}> a_{3}$. $h(x)$ is increasing, so obvious induction arguments now give that $a_{2n}$ is an increasing sequence, while $a_{2n+1}$ is a decreasing sequence. Since $x=2$ is a fixed point of $h$ two more trivial inductions show that $a_{2n}< 2$ and $a_{2n+1}> 2$. Therefore, $a_{2n}$ and $a_{2n+1}$ both converge.

It is my claim now that $\frac{d}{dx}\left(h(x)-x\right) < 0$ for $x\in [0,5]$. It is a very not tight inequality, but it is kind of messy to write it out, so I won't. Therefore, we understand that the only fixed point $h$ has in $[0,5]$ is $x=2$. Since both sequences considered at all times lie within $[0,5]$ then this means that $2$ is the only possible value they can converge to (they must converge to a fixed point). Therefore, we have that $\lim_{n\to\infty}a_{2n}= \lim_{n\to\infty}a_{2n+1}= 2$

Nice, but too much paper work for the original question, Kalle.
Solve this equation $x= \sqrt{7-\sqrt{7+x}}$ instead.

It's not quite enough to solve the equation.

Solving the equation will tell you that if the limit exists, it must equal 2.  However, I think you'd still need to show that the limit exists in the first place to use that method.

Yeah, solving $g(x) = x$, i.e. finding the fixed points of $x$, is of course the first thing I did (and that's how I found the fixed point of $h$). But all the work is really to show that the sequence actually converges.

I suppose you may have learned the method of just finding fixed points in the Intermediate forum. I am often there to point out that it's not enough to just find fixed points! If the method you suggest worked, then for any sequence defined by $a_{n+1}= f(a_{n})$ we could just solve the equation $A = f(A)$ to say what $a_{n}$ converges to. Clearly we can't do this, as many such sequences diverge (and if $f$ isn't continuous, then the argument doesn't work at all).

Instead, we need to first prove that the sequence converges. Then, if $f$ is continuous, we can pass the limit and get that the sequence must converge to an $A$ s.t. $A = f(A)$.

As for Moubinool's question, I don't understand what you mean.

[quote="kevinatcausa"]Solving the equation will tell you that if the limit exists, it must equal 2.[/quote]

If you look carefully, 2 is a solution of the equation.
BTW, OP wasn't looking for a limit.

[quote="Moubinool"]Call $a(n)$ the sequence with n-root

Can we find an equivalent of $a(n)-2$ ?

or an asymptotic expansion with two terms ?[/quote]
$a(n)-2\sim c\left(\frac{-1}{24}\right)^{n}+\frac{c^{2}}{60}\left(\frac{1}{24}\right)^{2n}-\frac{c^{3}}{1656}\left(\frac{-1}{24}\right)^{3n}+\cdots$


If a,b,c are real numbers for which: $ a\plus{}b\plus{}c\equal{}0$
Find the minimum and maximum of the expression:
$ D\equal{}\frac{ab}{(c^3\minus{}a^3)(c^3\minus{}b^3)} \plus{} \frac{bc}{(a^3\minus{}b^3)(a^3\minus{}c^3)} \plus{} \frac{ca}{(b^3\minus{}c^3)(b^3\minus{}a^3)}$

[quote="borislav_mirchev"]If a,b,c are real numbers for which: $ a \plus{} b \plus{} c \equal{} 0$
Find the minimum and maximum of the expression:
$ D \equal{} \frac {ab}{(c^3 \minus{} a^3)(c^3 \minus{} b^3)} \plus{} \frac {bc}{(a^3 \minus{} b^3)(a^3 \minus{} c^3)} \plus{} \frac {ca}{(b^3 \minus{} c^3)(b^3 \minus{} a^3)}$[/quote]
$ D\equal{}\frac{2}{(a^2\plus{}ab\plus{}b^2)^2}.$ :wink:

[quote="borislav_mirchev"]It is a very interesting result ... can you post a complete solution or at least some steps ... it is an old Spanish MO problem...

I had no idea how to solve the problem ... but it seems to be very interesting.[/quote]
It's easy!
$ D \equal{} \frac {ab}{(c^3 \minus{} a^3)(c^3 \minus{} b^3)} \plus{} \frac {bc}{(a^3 \minus{} b^3)(a^3 \minus{} c^3)} \plus{} \frac {ca}{(b^3 \minus{} c^3)(b^3 \minus{} a^3)}$ $ \equal{} \frac {a^2 \plus{} b^2 \plus{} c^2 \plus{} ab \plus{} bc \plus{} ca}{(a^2 \plus{} b^2 \plus{} ab)(b^2 \plus{} c^2 \plus{} bc)(c^2 \plus{} a^2 \plus{} ca)}$. But $ a^2 \plus{} b^2 \plus{} c^2 \plus{} ab \plus{} bc \plus{} ca \equal{} a^2 \plus{} b^2 \plus{} ab$ and $ a^2 \plus{} b^2 \plus{} ab \equal{} b^2 \plus{} c^2 \plus{} bc \equal{} c^2 \plus{} a^2 \plus{} ca$. The rest follows.


Do there exist positive integers $n>1, a_{1}, a_{2}, ..., a_{n}$ so that $f(i) = 25a_{i}^{2}-2a_{i+1}$ is a perfect square, for all $i=1,2,...n$ and $a_{n+1}=a_{1}$?

[color=darkred][size=150]CORRECTED![/size][/color]

Well, I have corrected the problem!  :blush: Sorry

$b_{i}=(5a_{i})^{2}-2a_{i+1}$ is a perfect square, implying that $b_{i}\leq (5a_{i}-1)^{2}$. We conclude that $-2a_{i+1}\leq-10a_{i}+1$, or $5a_{i}-\frac{1}{5}\leq a_{i+1}$. Since $a_{i+1}$ is an integer, it results that $5a_{i}\leq a_{i+1}$, Now it is clear that the sequence $a_{i}$ is strictly increasing, contradicting $a_{n+1}=a_{n}$. So the answer is negative.


If two distinct members of the set $ \{ 2, 4, 12, 14, 21, 28, 98\}$ are randomly selected and multiplied, what is the probability that the product is a multiple of 196? Express your answer as a common fraction.

[hide="Solution"]
$ 196\equal{}2^2\times7^2$ The only way we can have two $ 7$'s in the product is to have two multiples of $ 7$. The only four multiples of $ 7$ are (in the set) $ 14, 21, 28, 98$. So far, we have $ \binom42\equal{}6$ possibilities.

Now, we have to check if the products are divisible by $ 4$. If we have two even numbers, we have two $ 2$'s, and we have a $ 4$. The only odd multiple of seven in the set is 21. So we subtract $ 2$ more cases because $ 21\times14$ and $ 21\times98$ don't work ($ 28\times21$ is divisible by $ 4$). That is a total of $ 4$ successful cases so far.

But $ 98\equal{}7^2\times2$. So if we multiply it by any even number, we get a multiple of $ 196$. We have already counted the multiples of $ 7$, so the remaining even numbers are $ 2, 4, 12$. That is three more cases, bringing our total of successful cases up to $ 7$.

Next, we have to find the total cases. That is just $ \binom72\equal{}21$ because there are $ 7$ numbers in the set.

So our final answer is $ \frac7{21}\equal{}\frac13$

$ Answer\equal{}\frac13$
[/hide]


algebra
polynomial
superior algebra
superior algebra unsolved

1. Let K be the splitting field field of a polynomial f(x) in F[x] and supper that K is separable over F.
So exist c in K such that K = F(c)
Let g(x) in F[x] be the minimum polynomial of c.
Show that K is the splitting field of g(x)

2. Let E=F(u) where u is transcendental over F. Let K bea field with F <  K < E. (where < means a subfield of)
and F=/=K.
SHow that u is algebraic over K.

I have no idea about the first one.

For the second one, I said: as K =/= F. So there exists an element q such that K=F(q). iIf q=u, the K=E and q is algebraic over K. If q=/=u, then q is transcendental over F as well. Since if q is algebraic over F, then E=F(q,u) =/= F(u) which leads to a contradiction.
ANd then I can't continue from here...

They are both very easy.

[b]1.[/b]
Take $ c'$ some other root of $ g(x)$ in an algebraic closure $ \bar{K}$ of $ K$
and $ \sigma: K \equal{} F(c)\to F(c')$ the canonical $ F$-isomorphism.
If $ \alpha\in K$ is a root of $ f$, then $ \sigma(\alpha)\in\bar{K}$ is also root of $ f$.
But since $ f$ splits in linear factors over $ K$, we must have $ \sigma(\alpha)\in K$.
$ K$ is generated by such $ \alpha$, so $ \sigma(K)\subseteq K$ and $ c' \equal{} \sigma(c)\in K$.
So all roots of $ g$ in $ \bar{K}$ must lie in $ K$, which gives $ K \equal{} F(c)$ splitting field of $ g$ over $ F$.
($ K$ is generated by roots of $ g$ over $ F$ and $ g$ splits in linear factors over $ K$.)
[b]2.[/b]
Take $ v\in K\setminus F$.
Then $ v\in E \equal{} F(u)$ gives $ v \equal{} {f(u)\over g(u)}$ for some $ f,g\in F[X]$ coprime, $ g(u)\not \equal{} 0$.
But then $ f \minus{} vg\in K[X]$ has root $ u$ in $ E$, and if $ u$ were not algebraic over K, then $ f \minus{} vg \equal{} 0$.
But this gives $ f(0) \minus{} vg(0) \equal{} 0$.
If $ g(0)\not \equal{} 0$, then $ v \equal{} {f(0)\over g(0)}\in F$. Contradiction!
If $ g(0) \equal{} 0$, then also $ f(0) \equal{} 0$ and $ f,g$ have common divisor $ X$ in $ F[X]$. Contradiction!

Yes Olorin. My question was about 1); it's easy but I didn't find.   :blush:


There is no closed form. The value is approximately 0.288788

How do you prove that no closed-form exists, and how do you determine the approximation?

Here's how I know it. I've put it in Mathematica, no symbolical result was found. Using N[] gave me 0.2887880950866024. Then, I've put it in Plouffe's inverter:

[quote="Plouffe"]

Product(1-1/(2^n*P(n)),n=1..infinity), P(n) = up to 3rd deg. polynomials
2887880950866024 = Prod(1-1/(2^n),n=1..inf)

Product(1 (+/-) P(n)/2^n,n=1..infinity), P(n) is a polynomial
2887880950866024 = Prod(1-(1)/2^n,n=1..inf)

[/quote]

That's why I guess there's no closed form.

EDIT:

Few MathWorld references:
[url]http://mathworld.wolfram.com/TreeSearching.html[/url]

[url]http://mathworld.wolfram.com/InfiniteProduct.html[/url]

showing that only way of rewriting this is made through Jacobi elliptic function (not quite a "pure closed form", though).

BTW, this is Sloane [url=http://www.research.att.com/~njas/sequences/A048651]A048651[/url] [/quote]


In a triangle $ ABC$ the ratio between the height and the median relative to the hypothenuse $ \overline{BC}$ is equal to $ \frac {40}{41}$ . Find the ratio between the two legs.

let the foot of the altitude from A be M and the midpoint of BC be N, then MN=9
since the median is half the hypotenuse, BN= 41, BM=50, and AC/BA=AM/BM=4/5

Which segment did you arbitrarily give a length? Could you explain more precisely what you have done?

Since the problem is all about ratios, he simply took $ AM$ EQUALS $ 40$ and $ AN$ EQUALS $ 41$. Then $ MN$ is calculated via Pythagoras, and the rest goes by itself.


Show that there are infinitely many pairs $(x, y)$ of rational numbers such that $x^3 +y^3 =9$.

[quote=SoumikNayak][quote=juna]Let $N=a^3+b^3$. We take $y=t+b$, $x= a-\frac{b^2}{a^2}\cdot {t}$, [/quote]
What is the motivation behind doing this?[/quote]

I can't find the exact source but there is some discussion here.
https://artofproblemsolving.com/community/q2h54186p9111194

Nice problem.
Let $(a,b)$ be a rational solution to $x^3+y^3=9$.
Consider the curve $x^3+y^3=9$ and the point $(a,b)$ on it. The tangent from $(a,b)$ to the curve is the line $a^2x+b^2y=9$. Solving for the coordinates the other intersection of this tangent with the curve leads to the third degree equation $(x-a)^2((b^3-a^3)x+a(9+b^3))=0$ for the $x$ coordinate. This gives another rational point $(\frac{-a(9+b^3)}{b^3-a^3}, \frac{b(9+a^3)}{b^3-a^3})$. Starting from the obvious solution $(1,2)$, we get infinitely many solutions this way.


Determine all function $f : [1,$ ∞ $)$ → $[1,$∞ $)$ satisfying the following two conditions :
   (i)  $\large f(x+1) = \frac{(f(x))^{2}-1}{x}$ for $\large x \geq 1$
   (ii) The function $\large g(x) = \frac{f(x)}{x}$ is bounded

must be $f: R_{+}\to R_{+}$.
(i) $f(x+1)=\frac{(f(x)^{2}-1}{x}\ \forall x\in R_{+}$,
(ii) Exist $A$, suth that $f(x)<Ax$ for $x\ge 1$.
Then only one solution $f(x)=1+x$.

Thanks for your help, Rust.
Could you please write your solution in detail???

[quote="Dtrung"]Could you please write your solution in detail???[/quote]
Assume we have some $A,B\in\mathbb{R},A,B\geq 1$ with
$f(x)\leq A(x+1)$ for all $x\geq B$.
Then a trivial generalization of the argument used [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=131270]here[/url] will give
$f(x)\leq A^{2^{-n}}(x+1)$ for all $x\geq \max\{1,B-n\}$ and $n\in\mathbb{N}_{0}$ using induction.

The induction basis $n=0$ is clear.
And assuming the induction hypothesis,
then for all $x\geq \max\{1,B-(n+1)\}$ we have $x+1\geq \max\{1,B-n\}$,
and so
$f(x)=\sqrt{xf(x+1)+1}\leq \sqrt{xA^{2^{-n}}(x+2)+1}\\ \hspace*{1.63in}=\sqrt{A^{2^{-n}}(x+1)^{2}+1-A^{2^{-n}}}\leq A^{2^{-(n+1)}}(x+1)$
as $A^{2^{-n}}\geq 1$.

Now $n\to\infty$ gives $f(x)\leq x+1$ for all $x\geq 1$, and the rest is identical to [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=131270]here[/url].
Satisfied?


Extremal combinatorics
combinatorics
IMO Shortlist
graph theory
Chessboard

Let $ n$ and $ k$ be positive integers such that $ \frac{1}{2} n < k \leq \frac{2}{3} n.$ Find the least number $ m$ for which it is possible to place $ m$ pawns on $ m$ squares of an $ n \times n$ chessboard so that no column or row contains a block of $ k$ adjacent unoccupied squares.

We claim that this minimum is $ m \equal{} 4(n \minus{} k)$. This is achievable by placing pawns along a few diagonals which are hard to describe without a proper diagram.

To see that this is the minimum, we divide the board up as follows:
[asy]size(200);
draw((0,0)--(0,1)--(1,1)--(1,0)--cycle);
dot((0.4,1));
dot((0.6,1));
dot((1,0.4));
dot((1,0.6));
draw((0.4,1)--(0.4,0));
draw((0.6,1)--(0.6,0));
draw((1,0.4)--(0,0.4));
draw((1,0.6)--(0,0.6));
label("$2k-n$",(0.5,1),N);
label("$n-k$",(0.2,1),N);
label("$n-k$",(0.8,1),N);
label("$2k-n$",(1,0.5),E);
label("$n-k$",(1,0.2),E);
label("$n-k$",(1,0.8),E);[/asy]

Consider each oblong rectangular subfigure. Each empty row/column in these (whichever is perpendicular to the longer sides) necessitates two pawns; one in each of the adjoining $ n \minus{} k$ x $ n \minus{} k$ squares. We then divide by two because we go over each square block at most twice, and add $ \sum_{cyc} n \minus{} k \minus{}$(number of empty rows/columns in each rectangular box) to account for the pawns which are in the rectangular blocks themselves to get $ 4(n \minus{} k)$ pawns - everything else cancels out.

Motivation: Divide each side into $ k$ and $ n\minus{}k$ blocks from the left and the right (pretty obvious approach), then notice that we can use the rectangular boxes to count corners.

It would seem intuitively that placing pawns along the diagonal of the central box would be less "wasteful" - anyone have a $ 4(n \minus{} k)$ example/solution with that? I actually tried that approach for quite a while, but it seemed fruitless to me.

Actually, I think $m = 3n - 2k$. I have a model for this and a solution. I am still searching my mistake in my proof because the answer should imply that $k \le \dfrac{2}{3}n$. Anyone have other results ?(could there exist a typo and instead of $\dfrac{2}{3}$ should be $\dfrac{3}{2}$ ?)

Now fwolith's proof is correct.Your assertion is not valid for $n=6$(I had found a coloring of 9 squares,while your answer is 10)

My solution:

Number the rows and columns from $0$ to $n-1$.Color every cell by one of the numbers $0$ to $k-1$ such that $i+j+1 \equiv i'+j'+1 \pmod{k}$ means $(i,j)$ and $(i',j')$ have the same number,$(0,0)$ contains $1$,$(0,1)$ contains two,....,$(0,k-1)$ contains $k$,$(0,k)$ contains $0$ again and so on.Now place the pawns on the cells numbered $0$.By the inequality $\frac{n}{2} < k \le \frac{2n}{3}$,$(i,j)$ is coloured $0$ iff and only if $i+j+1=k,2k$ or $3k$.$i+j+1=k$ has $k$ solutions:$(0,k-1),(1,k-2),\cdots,(k-1,0)$.$i+j+1=2k$ has $2n-2k$ solutions:$(2k-n,n-1),(2k-n+1,n-2),\cdots,(n-1,2k-n)$.$i+j+1=3k$ has $2n-3k$ solutions:$(3k-n,n-1),(3k-n+1,n-2),(n-1,3k-n)$.So the total no of cells containing $0$ by this coloring is $k+(2n-2k)+(2n-3k)=4(n-k)$.

Now we show that this is the minimum.Every $k$ consecutive cells should have atleast one pawn.Now suppose that in the coloring mentioned above all pawns are not placed over cells having same number.Then there becomes a gap of $k$ consecutive cells (considering that we place one pawn in each k-length block in order to achieve the minimal coloring).So every pawn should be placed over cells having same number.It is easy to verify that the no of $0$'s are the least,so this is the required minimal coloring.

Here is nice solution. First make coordinate system with lower left square being (1,1) (next on right (2,1),one up from (1,1) is (1,2)   and so on..) Put pawns on all squares with sum of coordinates k+1,2k+1,3k+1 and (1,1) if it is possible.(k+1 and 2k+1 are always possible and 3k+1 is impossible if k=[(2/3)*n]).There are 4(n-k) pawns and it is easily checked that there are no k consecutie empty squares .To prove that you need at least 4(n-k) pawns we can make four  k(n-k) rectangles and fit them inside big square(construction is well known but idk how to describe it without picture).In every rectangle we need n-k pawns so we need at least 4(n-k) pawns.Only problem with first contruction is if k=4L where n=6L , we get 4(n-k)+1 .But we see if we remove pawn from (1,1) it is still satisfying condition and we are done.(this is just a fency way of summing 1,5 hour of work)

I claim that the answer is $m=4(n-k)$. To show that it is sufficient, consider the following diagram (thanks, fwolth):
[asy]size(200);
draw((0,0)--(0,1)--(1,1)--(1,0)--cycle);
dot((0.4,1));
dot((0.6,1));
dot((1,0.4));
dot((1,0.6));
draw((0.4,1)--(0.4,0));
draw((0.6,1)--(0.6,0));
draw((1,0.4)--(0,0.4));
draw((1,0.6)--(0,0.6));
label("$2k-n$",(0.5,1),N);
label("$n-k$",(0.2,1),N);
label("$n-k$",(0.8,1),N);
label("$2k-n$",(1,0.5),E);
label("$n-k$",(1,0.2),E);
label("$n-k$",(1,0.8),E);
label("$A$", (0.15,0.2),E);
label("$B$",(0.5,0.15),N);
[/asy]
Consider only the outer $n-k$ rows/columns on each side. Also, define the weight of a pawn within a row/column to be $1/2$ if it lies within a region like $A$, and $1$ if it lies within a region like $B$. Then, in order to satisfy the condition, each row/column needs to have total weight at least 1, namely either one pawn in its intersection with region $B$, or 1 in each of the intersections with region $A$. Now, each pawn contributes to the total weight by 1, since pawns in region $A$s are counted for both a row and column, and we wish to have net weight $4(n-k)$, for all of the rows/columns. Therefore, we need at least $4(n-k)$ pawns. To show this is sufficient, consider the construction below:
[asy]size(200);
draw((0,0)--(0,1)--(1,1)--(1,0)--cycle);
dot((0.4,1));
dot((0.6,1));
dot((1,0.4));
dot((1,0.6));
draw((0.4,1)--(0.4,0));
draw((0.6,1)--(0.6,0));
draw((1,0.4)--(0,0.4));
draw((1,0.6)--(0,0.6));
draw((0.4,0.8)--(0.6,1),red);
draw((0.8,0.6)--(1,0.8),red);
draw((0.2,0.6)--(0.4,0.8),red);
draw((0,0.4)--(0.2,0.6),red);
draw((0.2,0)--(0.4,0.2),red);
draw((0.4,0.2)--(0.6,0.4),red);
draw((0.8,0)--(1,0.2),red);
draw((0.6,0.4)--(0.8,0.6),red);
label("$2k-n$",(0.5,1),N);
label("$n-k$",(0.2,1),N);
label("$n-k$",(0.8,1),N);
label("$2k-n$",(1,0.5),E);
label("$n-k$",(1,0.2),E);
label("$n-k$",(1,0.8),E);
[/asy]

The answer is $4(n-k)$. Mark each point on the chessboard with a coordinate $(x,y)$ where the bottom left square is $(0,0)$, the square above $(i,j)$ is $(i,j+1)$ and the square to the right of $(i,j)$ is $(i+1,j)$.

Place a pawn each square $(i,j)$ for which $k\mid i+j$. Suppose there exist a line of $k$ consecutive unoccupied squares. Without loss of generality, let them be in the same row and let them be $(i,j+r)$ for $0\le r\le k-1$. However, let $r\equiv -i-j\pmod k$ then $k\mid i+j+r$, contradiction.

To prove that at least $4(n-k)$ is necessary, call a pawn row-strong if its $x$-coordinate is between $n-k$ and $k-1$, inclusive. Define column-strong similarly. If a row does not have a row-strong pawn, then there needs to be at least two pawns in that row. There are $2(n-k)$ columns which make a pawn not row-strong, and each of these columns must have a pawn.

Call a row or column whose fixed coordinate is not between $n-k$ and $k-1$ inclusive strong if it has a row-strong pawn and a column-strong pawn, respectively. Let there be $r_0$ strong rows and $c_0$ strong columns. There are $2n-k$ rows/columns not between $n-k$ and $k-1$. Each non-strong row or non-strong column must have two pawns. Thus, there are at least \[\text{max}(2(2(n-k)-r_0), 2(2(n-k)-c_0))\ge 4(n-k)-r_0-c_0\] pawns in non-strong rows. These pawns are neither row-strong nor column-strong. There are at least $r_0$ row-strong pawns and $c_0$ column-strong pawns. Thus, there are at least $4(n-k)$ pawns in total.

[hide=Solution]
We claim the answer is $4(n-k)$.\\
$ $\\
To construct a configuration for which $m=4(n-k)$, we begin by placing a pawn in the first row in the $n-k^{th}$ column, we then move one square down and right and place another pawn and we keep doind this until we reach the $k-1^{th}$ column. All the previous rows satisfy the given condition, because the pawn always has at least $k-1$ squares to the right and at least $k-1$ to the left. (*)\\
After this, we keep placing pawns diagonally to the previous as before, but we also place a pawn to its left, so that there are exactly $k-1$ squares between them, in order to satisfy the given condition. We do this until we reach the edge of the board, at which point we'll be on the $k^{th}$ column, because we started at row $2k-n$, column $k-1$ and we stopped at row $n$.(**)\\
Since the second diagonal (the one which starts at row $2k-n+1$) is now at the $n-k-1^{th}$ row, by adding another pawn to continue the diagonal, we reach row $n-k$, so we can repeat (*). We then keep alternating between (**) and (*).\\
This also satisfies the condition for the columns, because we can rotate the board $90^\circ$ clockwise and reflect over its horizontal line of symmetry.\\
We now prove that we do (*) exactly twice:\\
Suppose we can't do it twice completly, then:
$$k+2k-n>n$$
$$k>\frac{2n}{3}$$
We reach a contradiction, therefore we can do (*) at least twice.\\
Suppose we can do (*) more than twice, then:
$$2k>n$$
$$k>\frac{n}{2}$$
We reach a contradiction again, so we can do (*) exactly twice.\\
During (*), in each row there is one pawn, while in (**) there are two, hence: $$m=2(2k-n)+2(n-2k+n)=4(n-k)$$
$ $\\
We now prove $4(n-k)$ is the lower bound:\\
We consider some configuration where the minimum number of pawns is used. Since there must be at least one pawn in every row and column, we can take some set $N$ of $n$ pairs which represent the row and column of some pawn in the configuration, such that no row appears in more than one pair.\\
We call a number $x\in [1,n]$ $good$ if $x\in [n-k,k-1]$ and we call it $bad$ otherwise (in other words, if a pawn can be placed in that row/column without needing another pawn in the same row/column to satisfy the condition, it is $good$).\\
Let $n_1$ be the number of pairs which consist of two $bad$ numbers, let $n_2$ be the number of pairs which consist of one $good$ number and one $bad$ number and let $n_3$ be the number of pairs which consist of two $good$ numbers in $N$, then:
$$n_1+n_2+n_3=n$$
Since there are $2k-n$ $good$ numbers, $n_3\leq 2k-n$. Moreover, $n_2=2(2k-n-n_3)$, because a pair of $good$ numbers makes two pairs of one $good$ number and one $bad$ number. We can now compute $n_1$ in terms of $n,k,n_3$:
$$n_1=n-n_2-n_3$$
$$n_1=n-4k+2n+2n_3-n_3=3n-4k+n_3$$
For every pair of two $bad$ numbers, there are two pawns in a row, while for every pair which contains two $good$ numbers, there is one pawn in that row. For every pair which has the row number as the $good$ number (there are $n_2/2$ such pairs), there is another pair on the same column which has the same property (because the column number is $bad$), thus:
$$2n_1+2\frac{n_2}{2}+n_3=m$$
$$4(n-k)+n_3=m$$
Since $n_3\geq 0$, this proves the lower bound and we're done.
[/hide]


eila ci sono anche io!!!!
Salve a tutti sono Vanessa e sono italiana  :D
C'e' qualcuno che studia matematica alluniversita?
Ho postato un paio di cose nel forum algebra problemi irrisolti ma nessuno mi risponde  :(
Eila c'e nessuno
 :whistling:

ciao Vanessa!
da che parte dell'Italia vieni?
sei anche sull'oliForum?

ciao eucla ma ti è arrivato il mio messaggio?
cmq se vupi ripondermi tramite e-mail non lo fare fallo solamente tramite messaggio privato
potresti darmi una mano su alcuni teoremi di teoria dei gruppi?
sto davvero in fifficolta e non so a chi chiedere :(
magari se conosci qualcuno ....
grazie mille

mi sa che se fai l'università ne sai più di te di teoria dei gruppi  :)
io sono in 4°...cmq ora guardo mi hai messo curiosità :lol:


calculus
integration
derivative
real analysis
real analysis unsolved

Let x(t) be a function.

If x'(t) (the time derivative of x) is bounded and converges to zero, does
it implies that x(t) is also bounded. I know that I cannot conclude that
x(t) has a limit (Barbalat lemma), but can I at least say that it is bounded?
Something that may help is the fact that boundedness of x' implies that
x is uniformly continuous.

$ f(x)\equal{}\sqrt{x}$ on $ [1,\infty)$

Hey, thanks... after I submit the question I found a counter example as well. For intance,
x'(t) = 1/(t+1) is bounded and converge to zero as t--> infinity, for all t >=0.
x(t) = log(t+1) which is not bounded.


[quote="M4RI0"]If: $a_{1}=1(n \geq 1), a_{n+1}=2a_{n}+\sqrt{3a_{n}^{2}-3}$

Prove that: $a_{3n+1}=a_{n+1}(a_{n+1}^{2}+1)$.

All I could prove is the relation: $a_{n+1}+a_{n-1}=4a_{n}$. Any idea :|  :?:[/quote]
I think you meant $a_{n+1}=2a_{n}+\sqrt{3a_{n}^{2}-2}$ for which we have $a_{n+1}+a_{n-1}=4a_{n}$ (as you pointed out.)

Here ( http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=pmc+360&t=101057 )is a detailed solution to a problem that contains the same recurrence relation as in the problem above except that you need to prove something else in it.

Hope this helps.


Prove that $ (a^3\plus{}b^3)^2 \le (a^2\plus{}b^2)^3$ for real a,b.

huh?
It is equivalent to proving -
$ 3a^2b^4 \plus{} 3a^4b^2 \ge 2a^3b^3$
or,
$ 3a^2 \plus{} 3b^2 \ge 2ab$
and by AM-GM,
$ a^2 \plus{} b^2 \ge 2|a|\cdot|b|$
so the inequality is true with equality for $ a\equal{}0 ,b\equal{}0 ,a\equal{}b\equal{}0$


conics
ellipse
geometry
analytic geometry
LaTeX

1.   In the complex plane, an ellipse has foci at 1 + 2i and -5 + 2i.  If the ellipse is tangent    to the real axis and its area can be expressed as , find a + b, where b is a square free integer.

A.  7		B.  15		C.  17		D.  9		E.  NOTA

from the VHHS Comprehensive test.

Answer

[hide]B[/hide]

How do you do this problem?

[quote="hurdler"]1.   In the complex plane, an ellipse has foci at 1 + 2i and -5 + 2i.  If the ellipse is tangent    to the real axis and its area [b]can be expressed as [/b], find a + b, where b is a square free integer.
[/quote]

i think its a typo...

[quote="hurdler"]... area can be expressed [b]as , find[/b] a + b,...[/quote]

What can the area be expressed as?

EDIT: After searching, I found the problem to be
"In the complex plane, an ellipse has foci at $ 1 \plus{} 2i$ and $ \minus{} 5 \plus{} 2i$.  If the ellipse is tangent to the real axis and its area can be expressed as $ a\pi\sqrt {b}$, find $ a \plus{} b$, where $ b$ is a square free integer."

This is essentially an ellipse with foci at $ (1,2)$ and $ ( \minus{} 5,2)$ and center $ ( \minus{} 2,2)$.

Since the line containing the foci is parallel to the x axis (same as real axis in complex plane), the major axis is also parallel to the x axis.

Let the length of the major axis be $ 2a$, the length of the minor axis be $ 2b$, such that the radii are $ a,b$,

and the distance from the center to one of the foci be $ c \equal{} 3$, so that we have

$ \frac {(x \plus{} 2)^2}{a^2} \plus{} \frac {(y \minus{} 2)^2}{b^2} \equal{} 1$. (Not important, just felt like posting this.)

Since the ellipse is tangent to the real axis, we have $ b \equal{} 2$.

By definition, an ellipse is the locus of all points $ P$ such that $ PF_1 \plus{} PF_2 \equal{} 2a$.

We can let $ P$ equal $ ( \minus{} 2,0)$ so that we have $ 2\sqrt {( \minus{} 2 \minus{} 1)^2 \plus{} (0 \minus{} 2)^2} \equal{} 2\sqrt {13} \equal{} 2a\implies a \equal{} \sqrt {13}$.

Now we have our radii, so our answer is $ 2\pi\sqrt {13}\implies2 \plus{} 13 \equal{} 15\implies\boxed{B}$.

In general, we have $ b^2 \plus{} c^2 \equal{} a^2$, as long as $ 2a$ is the major axis and $ 2b$ is the minor axis. (Notice how this is easily derived from my substitution for $ P$.)

My guess is something like $ a \plus{} \sqrt{b}$.

No, it's more like $ \pi a\sqrt b$.

[hide="My solution"]
In the coordinate plane, the foci are at $ (1,2)$ and $ (\minus{}5,2)$ respectively. The line adjoining these two foci is horizontal, which means that the major and minor axes are parallel to the coordinate axes.

Since it is tangent to the x-axis, the vertical (minor) axis $ b$ is simply $ 2$. Also, the center of the ellipse is the midpoint of the foci, or $ (\minus{}2,2)$. So the distance $ c$ from the center to a focus is $ 3$.

Now, the major axis $ a$ can be calculated with this handy-dandy formula: \[ a^2 \equal{} b^2 \plus{} c^2\] So plugging in, we have $ a^2 \equal{} 2^2 \plus{} 3^2 \equal{} 13 \implies a \equal{} \sqrt {13}$.

So the area of the ellipse is $ \pi ab \equal{} \pi \cdot 2 \cdot \sqrt {13}$.

Now I just realized that my variables in the problem conflicted with the variables in the answer that I'm about to give. so I will delete my cache of variables now.

$ \pi \cdot 2 \cdot \sqrt {13} \implies a \equal{} 2, b \equal{} 13 \implies a\plus{}b\equal{}\boxed {15}$, as desired[/hide]

Oh sorry, I did not realize that bit of info was in LaTeX, and that it would not go with the copy and paste. My apologies.

Ok, thanks Yongyi.

I could have solved this, but I misread - I thought that those two coordinates were points on the ellipse, and not the foci. :blush: So I had gotten a different answer.....I need to remember to carefully reread when I am stuck on a problem...


inequalities
trigonometry
inequalities solved

If $a,b,c>0$ such that $2abc+ab+bc+ca=1.$ Prove that: $\sqrt{bc}+\sqrt{ca}+\sqrt{ca}\leq\frac{3}{2}$. ;)

[quote="cezar lupu"]It can be easily proved in the same conditions that a+b+c>=3/2 .[/quote]
The second inequality is true, but the sign of the first inequality is reversed. We can simply substitute $\cos z = ab$ and so on with $\sum x=180$

It is just direct substitution $a=\frac{x}{y+z}$ and so on. I was playing around on the forum with such tricky substitutions some very long time ago.


calculus
inequalities
function
trigonometry
inequalities theorems

Just calculus. It is messy, but put in $C=180-A-B$ into the last cotangent and we have an expression in A,B with bounds on A. Use partial differentiaion from there. Straightforward, and not very messy expressions. I can think of no synthetic way of doing this due to the forms of the terms. Seems standard to me.

OK I will try to change C with 180-A-B Actually I am secondary school student abd I am not so good at calculus

Thanks[b] me@home[/b]

[quote="delegat"]Is there any way  to get upper and lower bounds of expression:
$AcotA$+$BcotB$+$CcotC$ where $A$ $B$ and $C$ are angles of acute angled triangle.[/quote]
Maybe the  expression is $a\cot A+b\cot B+c\cot C$ $?$

To [b]arqad[/b]y:

No I wrote correct form of problem Actually This is crux problem and it asks to find upper and lower bounds of this expression. I tried to do something but it seems difficult.
Inequality also involves number [b]Pi[/b]=3.14

[b]Lemma 1.[/b] The function $ f(\nu)=\nu\cot\nu$ is concave on $ [0,\pi/2].$
[b]Proof.[/b] We have that $ \frac{\partial(\nu-\tan\nu)}{\partial\nu}=-\tan^{2}\nu$ and thus $ \nu-\tan\nu$ is decreasing. But this implies
\[ f''(\nu)=\frac{2(\nu-\tan\nu)}{\tan\nu\sin\nu^{2}}\leq \frac{2(0-\tan0)}{\tan\nu\sin^{2}\nu}=0 \]
and thus $ f$ is indeed concave.

As the triangle is acute, we have $ \left(\frac{\pi}{2},\frac{\pi}{2},0\right)\succ \left(\alpha,\beta,\gamma\right)\succ\left(\frac{\pi}{3},\frac{\pi}{3},\frac{\pi}{3}\right)$ and therefore
\[ \frac{\pi}{\sqrt{3}}\geq\alpha\cot\alpha+\beta\cot\beta+\gamma\cot\gamma>1. \]
Equality is never achieved on the right side, but tends to when $ \alpha\rightarrow 0,$ $ \beta,\gamma\rightarrow\frac{\pi}{2}.$


[quote="SirErnest"]I am condused.  I think my time zone changed again.  Thats what i hate about Indiana.  I went to the math jam section and tried to get in the classroom but hte join button wasnt on.  If I live in Terre Haute, on the border between Illinois and Indiana, what time zone am I in?[/quote]

Right now you should be in the Eastern times zone, but from April to October, you're in Central (I think).

This site will tell you: [url]http://www.worldtimeserver.com[/url]  Click on major cities on the left and then select Terre Haute.

A quick solution to your problem would be to just PM one of your friends in EST and ask them what time it is.

If you are in Indiana, I think,..  at least in some cases, ( or so I was told  in  a hotel lobby in Indiana) it depends on which side of the street you are.. :D

(I was too busy to worry about this, as my dog was still amused  and was in a good mood because  she passed through a town called "Gnaw Bone".

[quote="eryaman"]Uhhh...shouldn't you know your own time zone? If you know the time where you live, the time zone should be evident.  Or am I missing something here?[/quote]
Uhhh... well I Know what time it is here but I dont know how it compares to the time on AoPS and its confusing when half of your state changes time and half doesn't.

I'm GMT-5, and you posted at 6:06pm from my viewpoint. Then you can figure out your time zone relative to GMT


Fie triunghiul $ABC$ si $M$ un punct mobil pe $(BC)$. Construim $P, Q$, picioarele perpendicularelor din $M$ pe $AB$, respectiv pe $AC$. Determinati pozitia punctului $M$, astfel incat  $PQ$ sa aiba lungimea minima. Aceeasi cerinta pentru $PQ$ de lungime maxima.

$A,P,M,Q$ sunt conciclice si, dupa ce verificam vreo doua cazuri, rezulta ca $PQ = AM \sin A$. Cum $\sin A > 0$, $PQ$ e minim cand $AM$ e minim ($AM$ = inaltime) si $PQ$ e maxim cand $AM$ e maxim (pentru $M=B$ daca $AB \geq AC$).


Let $(a_n)$ be the sequence defined as follows: $a_1 = 1$, $a_2 = 2$, and for $n \ge 3$, $a_n$ is the smallest positive integer such that:
(i) $a_n$ does not appear in $\{a_1, a_2, \dots, a_{n - 1}\}$, and
(ii) $|a_n - a_{n - 1}|$ does not appear in $\{|a_2 - a_1|, |a_3 - a_2|, \dots, |a_{n - 1} - a_{n - 2}|\}$.
So for example, the first few terms are $a_3 = 4$, $a_4 = 7$, $a_5 = 3$, $a_6 = 8$, $a_7 = 14$.

Prove that every positive integer eventually appears in the sequence.

[url=http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eismum.cgi]Here[/url]'s the link to the article in the Encyclopedia of Integer Sequences. It says there, however, the "the sequence is almost certainly a permutation of the positive integers".

Does this mean that this really is an open problem? How, then, can it be in a problem set from MOP?

I don't think it's an open question, it seems quite easy to me. Suppose it's false, and let $k$ be the smallest number not appearing in the sequence.
Let $t$ be the moment when all numbers les than k appear (i.e. $a_t$ is the last number from $1 \cdots k-1$in the sequence.
Let $u=max\{|a_{i+1}-a_i|\}$ for all $i$ s.t. $a_i<k$
Now there is a moment $m>t+k$ s.t. $a_m$ is greatest than each of
$a_1, \cdots a_{m-1}$. Moreover we can choose $a_m>k+u$
I claim $a_{m+1}$ must be $k$.
Indeed, if $a_m-k=a_{i+1}-a_i$ for some $i$ then we must have $a_i<k$or
$a_{i+1}<k$ Hence $a_m=k+a_{i+1}-a_i \leq k+u$ contradiction.

[quote="iura"]Indeed, if $a_m-k=a_{i+1}-a_i$ for some $i$ then we must have $a_i<k$ or $a_{i+1}<k$ Hence $a_m=k+a_{i+1}-a_i \leq k+u$ contradiction.[/quote]

iura, I've stared at this for hours, and I still don't understand.  Could you explain it more explicitly?  And do you mean $a_m - k = |a_{i + 1} - a_i|$?

Yes I mean $a_m-k=|a_{i+1}-A_i|$ WLOG we suppose
$a_{i+1}>a_i$ the other case being analogous.
Then $a_m-k=a_{i+1}-a_i$ and since $a_{i+1}<a_m$ we deduce $a_i<k$/
Now that means $a_{i+1}-a_i \leq u$ (look above at the definition of $u$).
Hence $a_m \leq k+u$ which is contradiction

iura, I think I see now, thank you.

I have one other question.  I've attached a graph of the first 100 terms of this sequence.  The terms appear to collect into three lines, but I have no idea why.  Can anybody prove this?


Let $ a\ge b\ge c > 0$. Prove that:

$ 3abc\cdot (\dfrac{a^2}{c} \plus{} b \plus{} \dfrac{c^2}{a})\le (a^2 \plus{} b^2 \plus{} c^2)^2$

[quote="great math"]Are you sure your problem given correctly?
I found it quite easy to cope with! :)
Meanwhile, it is not necessary to use the condition : $ a \geq b\geq c$
I think the correct problem is
Let $ a \geq b\geq c$ , then prove that
$ 3abc\cdot (\dfrac{a^2}{c} \plus{} b \plus{} \dfrac{c^2}{a})\le (a^2 \plus{} b^2 \plus{} c^2)^2$
and obviously, it is more challenging!  :roll:[/quote]

I have edited. I just forgot to put $ 3$ in left. :)


algebra
polynomial
number theory proposed
number theory

a) Prove that there are two polynomials in $ \mathbb Z[x]$ with at least one coefficient larger than 1387 such that coefficients of their product is in the set $ \{\minus{}1,0,1\}$.
b) Does there exist a multiple of $ x^2\minus{}3x\plus{}1$ such that all of its coefficient are in the set $ \{\minus{}1,0,1\}$

a)Consider this polynomial :
$ f(x) \equal{} (x \minus{} 1)(x^{2} \minus{} 1)(x^{4} \minus{} 1)\cdots(x^{2^{n}} \minus{} 1)$
coefficient of this polynomial are $ 0,1, \minus{} 1$ . and we can write :
$ f(x) \equal{} g(x)h(x)$ where :
$ g(x) \equal{} (x \minus{} 1)^{n}$ and $ h(x) \equal{} (x \plus{} 1)(x^{3} \plus{} x^{2} \plus{} x \plus{} 1)\cdots(x^{2^{n} \minus{} 1} \plus{} ... \plus{} 1)$
and u can check that if we choose $ n$ sufficient big both of these polynomials have a coefficient bigger than $ 1387$ .
b)No
Assume that we can write $ f(x) \equal{} (x^{2} \minus{} 3x \plus{} 1)g(x)$ where all coefficient of $ f$ are $ 0,1, \minus{} 1$ and $ g(x)\neq 0$
we can write $ f(x) \equal{} a_{n}x^{n} \plus{} a_{n \minus{} 1}x^{n \minus{} 1} \plus{} \cdots \plus{} a_{1}x \plus{} a_{0}$ . and we can assume $ a_{n} > 0$
$ Lemma$ : if $ s\geq 2$ then : $ s^{n} > s^{n \minus{} 1} \plus{} s^{n \minus{} 2} \plus{} \cdots \plus{} s \plus{} 1$ .
now consider $ r \equal{} \frac {3 \plus{} \sqrt5}{2} > 2$
we know that $ r^{2} \minus{} 3r \plus{} 1 \equal{} 0$ so we must have $ f(r) \equal{} 0$ but all of coefficient of $ f$ are $ 0,1, \minus{} 1$ and $ a_{n} > 0$ so we must have $ a_{n} \equal{} 1$ Thus we can easily prove by lemma $ f(r) > 0$ . Contradiction !


limit
trigonometry
real analysis
real analysis unsolved

Find \[ \lim_{m \to \plus{} \infty, n \to \plus{} \infty} \cos^{2m}(n!\pi.x)\] with \[ x \in R\]

That's a well-known limit. See [url=http://www.mathlinks.ro/viewtopic.php?t=244949]here[/url] for example.


Suppose that $ \{X_n\}$ is a stationary process $ f: \mathbb{R}^2 \to\mathbb{R}$ is a Borel function and $ Y_i \equal{} f(X_i, X_{i \plus{} 1})$. Is $ \{Y_i\}$ also stationary?

If "stationary" means "narrow sense stationary", then surely yes.

By stationary, I mean, strict sense stationary.

$ X$ stationary iff for any $ n,k$ and any bounded measurable function $ g$
\[ E[g(X_{n},\dots,X_{n\plus{}k\minus{}1}) \equal{} E[g(X_1,\dots,X_k)].\]
Now choose an appropriate $ g$  :wink:


calculus
integration
trigonometry
function
real analysis
real analysis unsolved

Study the convergence of $\int_{0}^{+\infty}x^{a}|\cos x|^{x^{b}}dx$
where $a,b $ real number

It's easy to prove that for $|a|<1$ this integral converges:
From Cauchy-Maclaurin's criterium:
$ \int_{0}^{\infty} x^a |\cos x|^{x^b} dx < \infty \Longleftrightarrow \sum_{n=0}^{\infty} n^a |\cos n|^{n^b} <\infty$
It's obvious that
$\sum_{n=0}^{\infty} n^a |\cos n|^{n^b} \leq \sum_{n=0}^{\infty} n^a$
so for $|a|<1$ integral converges.

*are you sure your Cauchy-MacLaurin test is correct ? your function has to satisfies some hypotheses I suppose. (Monotony i guess.. but i don't know this test ..)
* must be a mistake at the end because $\sum \sqrt{n}$ does not converges for example ..

LOL sorry I was drunk when I wrote it - I've read $\sum a^n$ instead of $\sum n^a$...
-_-'

Please delete it or sth - no more beer before solving math problems:P
And Cauchy-Maclaurin is:

If $[a,\infty) \ni x \to f(x) \geq 0$ and $f$ converges monotonically to 0 when $x\to \infty$
Then $\sum_{n=a}^{\infty} f(n) < \infty$ iff $\int_a^{\infty} f(x)dx < \infty$.


algebra
polynomial
number theory
prime numbers
algebra unsolved

Prove that for any n there exists a polynomial $ f\in \mathbb Z [x]$ such that all number $ f(1) < f(2) < ..<f(n)$ are
a) prime numbers
b) powers of $ 2$

Work $ f_{n \minus{} 1}(x) \equal{} f(1) \plus{} a_1(x \minus{} 1) \plus{} a_2(x \minus{} 1)(x \minus{} 2) \plus{} ... \plus{} a_{n \minus{} 1}(x \minus{} 1)...(x \plus{} 1 \minus{} n)$.
When we chose $ f_{n \minus{} 1}$, we can chose $ a_n$, suth that $ f_n(n \plus{} 1) \equal{} f_{n \minus{} 1}(n) \plus{} a_n n!$.
In case b) must be $ v_2(f_{n \minus{} 1}(k))\ge v_2(k!)$.


Hello everyone, I am in the process of looking for colleges and am curious if anyone can offer me any good schools to consider. I am interested in PURE mathematics, no physics, no stats, no anything. My particular interests are Real and Harmonic analysis. If anyone can give me a good school I would appreciate it greatly.

Alex

It's hard to know what would  be a good fit if we don't know anything about you.


[quote="spx2"]i would like you people to help me know everything there is to know about the phi function of euler[/quote]
Well, $\varphi(n):=\#\{1\le m\le n:\gcd(m,n)=1\}$ so, if $p$ prime, then $\varphi(p)=p-1$, in general if $\gcd(m,n)=1$ then $\varphi(m,n)=\varphi(m)\varphi(n)$, and in general if $n=p_1^{a_1}\cdots p_r^{a_r}$ with $p_j$ primes, then $\varphi(n)=n\left(1-\frac 1{p_1}\right)\dots \left(1-\frac 1 {p_r}\right)$, in fact there are more interesting ptoperties, if you have interest in any in particular, ask us!.

Best regards.

Robertux the Crazy Math

You forgot the most important one:

$m$ divides $a^{\phi(m)} - 1$ if $(a, m) = 1$

http://mathworld.wolfram.com/TotientFunction.html

That site is good when you have a question like this. "I want to know about so-and-so." That site has just about everything you would need.


probability
combinatorics proposed
combinatorics

What's the probability of getting two numbers [color=red]random[/color] which are [color=red]coprime[/color]?

(We may use  the expanded form :) )

A really well-known result : $\frac 6 {\pi^2}$.

Pierre.


[i]Την άσκηση την είδα χθες τυχαία σε ένα τεύχος του 2000 στο περιοδικό Cardinal (Ρουμανία - Craiova)). Εκεί ζητούσαν διανυσματική λύση.Την βάζω για εμπλουτισμό και στο δικό μας φορουμ.Όσο απλές λύσεις και να βρείτε , γράψτε μια γραμμούλα ως υπόδειξη για τους Junior.[/i]

[b]ΑΣΚΗΣΗ[/b]

Δίνεται τρίγωνο $ ABC$ και μια ευθεία [b](ε)[/b] που περνά από το $ A$. Έστω $ D,E$    οι προβολές των $ B,C$  στην (ε). Οι κάθετες από τα $ D,E$ προς τις $ AC,AB$  αντίστοιχα , τέμνονται στο $ Z$. Nα αποδειχθεί ότι $ ZA \perp BC$.

[quote="silouan"]κ .Μπάμπη σας αφιερώνω τη λύση . (για τα γράμματα δείτε το σχήμα ) (Έστω ότι η ΖΑ τέμνει την BC στο F)

Το SDCE είναι εγγράψιμο άρα $ AS\cdot AC \equal{} AD\cdot AE$ και το DKEB είναι εγγράψιμο άρα $ AD\cdot AE \equal{} AK\cdot AB$

άρα $ AS\cdot AC \equal{} AK\cdot AB$ άρα το SBCK εγγράψιμο άρα $ \angle{AZK} \equal{} \angle{KSA} \equal{} \angle{ABC}$ επομένως τα τρίγωνα ABF, AZK είναι όμοια και τέλος  :)[/quote]

Σιλουανέ,
 ωραία και σύντομη λύση , από το βιβλίο του Θεού (κατά Erdoes)!
 Ευχαριστώ για την αφιέρωση και με τιμά :) .

[u]Μπάμπης[/u]

Edw uparxei kai i genikeusi... :)

http://www.mathlinks.ro/viewtopic.php?t=193474


I've found that, no matter how much I try, I cannot study for more than 3 hours straight. By the end of 3 hours, at least one of the following occurs: I get very bored; I cannot concentrate; my eyes hurt. What do you do when something like this happens?

By studying, I mean reading a book or working on the computer.

Take a 20 minute break at least once every hour and a half.

An optometrist told me I should take a break from the computer or from reading at least once every two hours (and probably much more often than that) since reading and using the computer puts a strain on the eyes by making them do something for which they were not meant to be used.

[quote="mathnovice"]I've found that, no matter how much I try, I cannot study for more than 3 hours straight.[/quote]

That's probably a very general finding. There is an excellent discussion of practice in the famous article on expert performance entitled "[url=http://www.cebiz.org/cds/ericsson_charness.pdf]Expert Performance: Its Structure and Acquisition[/url]" by K. Anders Ericsson and Neil Charness. This article is cited in all the best scientific literature about how people reach high levels of performance (and I think it is cited in Count Down) and has great advice for reaching your highest level. Yes, you must practice. Practice makes the difference between mediocre and expert performance. But it is possible to practice too long, and it is necessary to take breaks at the right intervals. Read all about it in the article.

The tip I learned when I was an undergraduate, long before the Ericsson and Charness article was published, is that mental fatigue tends to be subject-matter specific. If you've done math until you're tired, switch to reading a book of literature. If you are tired of all book work, practice piano. If tired of all that indoor stuff, go outside and play sports. Don't overdo any one thing for too long.

[quote="ComplexZeta"]tokenadult, you seem to have a large collection of interesting articles that you cite regularly on this forum. How did you find out about them?[/quote]

My job description these days is "principal" of our homeschool, and my wife generously does most of the work to pay the bills so that I have time to do research on better education. I do lots of Web browsing, and I do substantial library research, including research at a huge academic library where I first made a photocopy of the Ericsson and Charness article before it was available for free downloading. Maybe I'll turn all this research into a book, and thus income, some day, but meanwhile I try to apply it to my four children and to their friends--the local friends we have in our homeschooling support group and the online friends we have here on AoPS and elsewhere in cyberspace. I had a rather dissatisfactory experience in school as a kid, so I like to find out how young people in the current generation can learn more efficiently and more happily.

I appreciate everything you've helped us out with on this forum. I also take breaks like you were suggesting... sometimes if I am stuck on a computer somewhere, I will get up, log off, and go to the snack machine just to stretch... heehee...

gah, i can't bear studying for a single hour. i'll always end up throwing a bouncy ball against a wall or doing other stuff. although i have found out i can play video games for 3 hours.

If you study more than 3 hrs./day you're not being lazy enough :-).

I have studied for numerous hours of the night before with 10 minute breaks every 2 hours. I would sujest though if you get bored easily have something there to drink or bubble gum or something to keep you entertained as you study. Take a break when you need one and start to get bored or study for awhile take a break and run a little or exersize for about 10-20 minutes and then go back to studying.  I find if you get your brain to working again by exersizing then you can concentrate again.

[quote="joml88"]The only video game that I can play for that long is Halo and maybe a few others like Splinter Cell.  I don't normally play video games that much although I will play them a lot when Halo 2 comes out.  They better not push the date back any more. :-x[/quote]
yeah, same here. i have an x-box, but i almost never play it when i'm by myself. now that summer's here, i think it'll change. my friend has like 15 games, i just have 2. a baseball game and halo. i could definitely play halo for 3 hours straight, especially w/ 3 other friends. and about halo 2, it's ridiculous how they're pushing it back. i remember two years ago they said it'd be ready for the '03 holiday season. now, it's november, '04

Not to derail the thread or anything, but Fable is going to be awesome.  I think it looks better than Halo 2.

Hehe.. I find it funny how all of the console kids are crying about Halo. Im primarily a computer game player and we are always getting delayed. I think it was Diablo 2 if i remember correctly that was delayed a year and a half. I think I remember one time where a game was delayed for 2 years and then wasnt released. O and if you think that halo is good, then youd be amazed by the FPS's (First person shooter) on the computer. Just like consoles are much better for sports games, computers are much much better for fps's and rts's (Real Time Strategy... like Starcraft for the old-timers).


Hey everyone,
On my site http://matharena.net/ I will be hosting a Practice AMC contest in two weeks.  You can read all about it there.  I hope many participate!

i want to inform u that i did the test....  :)

however at the end it said "your result will be sent to your e-mail"
i forgot with which one of my e-mail account i registered  :oops:

if you would look up to my mail and send my e-mail in a private message.

also your site doesn'T have a profile, so i couldn't look up....  :mad:

sorry if i should send this wth private message :?: