I'm not sure if there is any easy way to do this problem. I used bit of algebra (well, not "bit" actually but no hard idea was used) and got answer in terms of $k,$ which is an integer that is resulted from $\sqrt{m-1} + \sqrt{m+15}$. It is $\displaystyle k = \sqrt{2(m+7) \pm 2 \sqrt{(m+7)^2 - 64}}$. One obvious root is 1 but I'm not sure if there are anymore.

[hide]Let $m-1=x^2, m+15=(x+a)^2$
$\Rightarrow 2ax+a^2=16$

We know $0<a<5$ since $x>0$.

$a=1 \Rightarrow 2x+1=16$, no integral solutions.
$a=2 \Rightarrow 4x+4=16, \Rightarrow x=3 \Rightarrow m=10$.
$a=3 \Rightarrow 6x+9=16$, no integral solutions.
$a=4 \Rightarrow 8x+16=16 \Rightarrow x=0 \Rightarrow m=1$.

So there is only $\boxed{2}$ solution.[/hide]

[quote="Ramtin Madani"]How many $m\in N$ is available such that $\sqrt{m-1}+\sqrt{m+15}\in Z$?

\[ \text{(A)}\ 1\ \ \ \ \text{(B)}\ 2\ \ \ \ \text{(C)}\ 3\ \ \ \ \text{(D)}\ 4\ \ \ \ \text{(E)}\ 5  \][/quote]

[hide="another solution..."]it is pretty obvious that each term has to be an integer, so let m=k^2-15, then we get

$k^2-16=q^2$
$k^2=4^2+q^2$ so we have to find all pythagorean triples, for the ones with no common factor, $a^2-b^2,2ab,a^2+b^2$,
first $4=2ab$, $a=2,b=1$, $k=5$, $m=10$
then $4=(a+b)(a-b)$
each RHS factor must be positive, so set them equal to the factors,
$a+b=4$, $a-b=1$, no solution
$a-b=2$, $a+b=2$, $a=2$, $b=0$, $k=4$, $m=1$

so a total of 2 solutions $\implies \boxed{B}$[/hide]

[hide="and another..."]
Note that there must be two perfect squares 16 apart.  Since consecutive squares differ by odds, the square roots must differ by evens (so there is an even amount of odds you're adding... and 16 is even).  Looking for roots 2 apart, 3 and 5 jumps out, giving m = 10.  Looking for roots 4 apart, we find 0 and 4, giving m = 1.  Since, when the roots are farther apart, the difference between the squares is higher, if we were to find roots 6 apart that satisfy, they would have to be smaller than 0 and 4, since the difference between squares is smaller for smaller roots... therefore, there are no solutions past being 4 apart, so we just have 2.
[/hide]


The following has stumped me....

The combined resistance of two resistors R1 and R2 in a
parallel circuit is given by the formula

total r= 1/1/r+1/r

I am sorry but I don't know how to use the symols for making the equasion with fractions. [/u][/quote]

:huh:
EDIT Typed right but without parenthesis.
$\frac{1}{R_{total}}=\frac{1}{r_{1}}+\frac{1}{r_{2}}+...+\frac{1}{r_{n}}$
Or the form you gave
$R_{total}= \frac{1}{\frac{1}{r_{1}}+\frac{1}{r_{2}}+...+\frac{1}{r_{n}}}$
It has to do with how they are in "parallel", it is simply their orientation that causes this relation.
Though in "series" it is simply
$R_{total}= r_{1}+r_{2}+...+r_{n}$

Erm, it's called LaTex, so if you want that, it's just dollar signs between the thing you wanna post, so if there was something like $total r= 1/1/r+1/r$, there you go! It will just be:

[code]$total r= 1/1/r+1/r$[/code]

easy as pie, eh?

the only thing i can't get down on LaTex is the division signs, not the slash signs (/) but the one with a dot then a line then a dot.

toadoncart, do you mean $\div$?
[code]$\div$[/code]

Does anyone have some guidence for my question regarding the total resistance?

I had some help with a series circut but the questions asks according to a parallel.

The total resistance can be found by the formula.
Do you want to know why it is as such?
http://en.wikipedia.org/wiki/Series_and_parallel_circuits#Parallel_circuits

A simple derivation
$I = \frac{V}{R}$
Then the $I_{total}= I_{1}+I_{2}+...+I_{n}\Rightarrow$ $I_{total}= \frac{V}{R_{1}}+\frac{V}{R_{2}}+...+\frac{V}{R_{n}}\Rightarrow$ $\frac{V}{R_{total}}= \frac{V}{R_{1}}+\frac{V}{R_{2}}+...+\frac{V}{R_{n}}\Rightarrow$ $V \cdot \frac{1}{R_{total}}= V \cdot \left( \frac{1}{R_{1}}+\frac{1}{R_{2}}+...+\frac{1}{R_{n}}\right) \Rightarrow$ $\frac{1}{R_{total}}= \frac{1}{R_{1}}+\frac{1}{R_{2}}+...+\frac{1}{R_{n}}$


[quote="tyy"]Does any one know where i can get some challengin geoemtry problems (i already have geoemtry revisited)? ;)  :D[/quote]

Do you mean by book or by online?

[quote="pohoatza"]search by google or somethingh alse the problems that were am olimpiads and others contests :)[/quote]

This is the MathCounts forum, not the Olympiad Forum.

[quote="tyy"]Yeah, online geometry books that contains challenging problems.[/quote]

I don't know where to find em, but certainly amazon.com has some good books.

BTW, your signature should read pi^2/6 not pi/6, okay.

[quote="easyas3.14159..."]Introduction to Geometry on this site has some good problems.(It's not really an introductory geometry book, btw.)[/quote]

It teaches the basic topics of geometry but it gives complex problems within that topic in my opinion.


Is it possible to create a triangle out of three smaller congruent triangles?

Sure -- draw an equilateral triangle $ \triangle ABC$ with center $ O$. Then, $ \triangle OAB , \;  \triangle OBC , \; \triangle OCA$ satisfy your condition.


Prove if $k,n \in N$ and $n \ge 2$ then:
$(n - 1)^2 \left| {n^k - 1} \right. \Leftrightarrow n - 1\left| k \right.$

If part:
[hide]$n^k-1=(n^{k-1}+n^{k-2}+...+1)(n-1)$
Now, the part can be simplified.
$n-1 |(n^{k-1}+n^{k-2}+...+1)$.
Now, notice that the left side has k terms.  If the equation is looked at through mod n-1, every term on the left is a 1 mod n-1.  Since there's k terms, $k\equiv 0 \mod {n-1}$.  This is a restatement of $n-1 |k$.[/hide]

I am a little confused with my attempt at a solution to the "only if" portion of the problem.
I rewrote n^k -1 as (n-1)(n^(k-1) + ... + 1) but then found that this implies that (n-1)|(n^(k-1) + ... +1). This means that if we let f(n) = n^(k-1) + ... + 1, then f(1)=0 but also, by simple substitution, f(1)=k, so that k=0.
This solution is wrong though, since I know that n=3 works for k=4, so k is not limited to 0, and it appears as if the first solution is correct.
What am I doing wrong?

$\Leftarrow \ \text{part}$

[hide]This is equivalent to proving $n^{n-1}\equiv 1\pmod{(n-1)^2}$ because if this were true, then $n^k\equiv (n^{n-1})^\frac{k}{n-1}\equiv 1^\frac{k}{n-1}\equiv 1$. Conversely, if the congruence works for any $k$ which is a multiple of $n-1$, it must work for $n-1$ itself.

Let $m=n-1$. Then $n^{n-1}=(m+1)^m$ and the modulus is $m^2$.

Expand $(m+1)^m$ by the binomial theorem. The terms where $m$ is raised to a power of 2 or greater clearly are multiples of $m^2$, and the coefficient on the $m$ term is $\binom{m}{1}$; these multiply to get another $m^2$. That leaves the constant term, 1. So $n^{n-1}\equiv 1\pmod{(n-1)^2}$.[/hide]


Let N={0, 1, 2, 3,...}. Define F:N×N→N by F(i,j)=j+(1/2)(i+j)(i+j+1).

Show that F is a bijection and that it has something to do with the following picture.

[img]http://myweb.hinet.net/home9/tooeeasy5/000.JPG[/img]

It's easier to see what's going on from the picture than from the formula. The picture clearly defines a bijection (follow the diagonals in the north-west direction), and then you check that the formula agrees with it. The horizontal axis is $ i$, the vertical one is $ j$.

Some of my students who is studying in Paris, France have asked this problem a few days ago. I am wondering  :?:


1.  How many different triangles have whole number-long sides if none of the sides is longer than 4 units?

2.  You have a 2, 4, 6, and a 7 unit long segment.  Using any of these four segments as many time you want, how many different triangles can you make?

Umm... it would be nice if you listed what they were.

Ohh, right. The isosceles triangles should include every combination except those with $(2,2,n)$.

[quote="xenon_nightmare152"]i believe you missed 442, 662, and 772 which would bring the total to 15[/quote]
yup those were the ones


I'm working with

[code]\thispagestyle{fancy}
\lhead{\textbf{text}\hfill \textbf{text N\textsuperscript{o}1}\\[-1ex]\hrulefill\\text\hfill text}
\renewcommand{\headrulewidth}{0pt}
\begin{center}
\textbf{text N\textsuperscript{o}1}
\end{center}[/code]

but I can't delete the page number. I was trying to put \thispagestyle{empty} below \thispagestyle{fancy}, but by doing that it disappears the whole headline.


geometry
integration
calculus
calculus computations

a) A solid is constructed so that it has a circular base of radius r centimeters and every plane section perpendicular to a certain diameter of the base is a square, with a side of the square being a chord os the circle. Find the volume of the solid.

b) If the solid decribed in part (a) expands so that the radius of the base increases at a constant rate of 1/2 centimeters per minute, how fast is the volume changing when the radius is 4 centimeters?

- And, yes, i typed everything word for word. there are no typos, its just confusing. @_@ :?:

A slice through this object along a plane perpendicular to the base circle and passing $ y$ units from the center of the circle shows a square of side length $ 2\sqrt{r^2\minus{}y^2}$ and thus area $ 4(r^2\minus{}y^2).$ Hence the area is

$ A\equal{}\int_{\minus{}r}^r4(r^2\minus{}y^2)\,dy\equal{}8\int_0^r(r^2\minus{}y^2)\equal{}
\left.8\left(r^2y\minus{}\frac13y^3\right)\right|_0^r\equal{}\frac{16r^3}3.$

As for the related rates problem: you now have a formula for $ A$ in terms of $ r.$ Can you compute $ \frac{dA}{dr}\,?$

yes, i think i can do that rest. its just part (a) was so confusing it was driving me crazy lol well thank u very much =)


[quote="Peter"]It is possible to show that $ \csc\frac{3\pi}{29}\minus{}\csc\frac{10\pi}{29}\equal{} 1.999989433...$. Prove that there are no integers $ j$, $ k$, $ n$ with odd $ n$ satisfying $ \csc\frac{j\pi}{n}\minus{}\csc\frac{k\pi}{n}\equal{} 2$.[/quote]

Reference Included. It's from 'AMM, Problem 10630, Richard Strong'


Solve the equation system:

$ \left\{ \begin{array}{l}
 \sqrt[3]{{(x^3  \minus{} 7)}} \plus{} y^2  \minus{} 2y \plus{} 3 \equal{} 0 \\
 x^2  \plus{} x^2 y^2  \minus{} 2y \equal{} 0 \\
 \end{array} \right.$

The solution is very great!  :D  :D  :D

hello, what is your range of definition? The complex numbers?
Sonnhard.

[hide="Integer solution"]
As $ x^2(y^2\plus{}1)\ge 2y$ where equality holds for $ x^2\equal{}1$ and $ y\equal{}1$, it follows that $ x^2\equal{}1$ and $ y\equal{}1$. Checking cases for $ x$, we get $ x\equal{}\minus{}1$ and $ y\equal{}1$.
[/hide]

[hide]this is my solution:

$ {(1) \leftrightarrow (y \minus{} 1)^2 \equal{} \minus{} 2 \minus{} \sqrt [3]{{(x^3 \minus{} 7)}} \ge 0 \leftrightarrow x^3 \le \minus{} 1 \leftrightarrow x \le \minus{} 1}$

$ {(2) \leftrightarrow x^2 \equal{} \frac {{2y}}{{1 \plus{} y^2 }} \le 1 \leftrightarrow \minus{} 1 \le x \le 1}$

So $ x \equal{} \minus{} 1$ anh $ y \equal{} 1$.

Yeah!  :D  :D[/hide]


[quote="rainynightstarz"]do they have AIME after they finish grading all the AMC? if they do should i get my score already? if u didn't get ur score , but u qualified might u get the AIME test anyway?[/quote]

Your coach should have been notified VERY soon after your school's scores were received at the AMC office about who in your school qualified for the AIME. The AIME packets are sent to the schools that have AIME qualifiers, as is detailed in the [url=http://www.unl.edu/amc/d-publication/d1-pubarchive/2003-4pub/04USAMO/04aimeusamoTM.html]AIME coaches handbook[/url]. Have you spoken to your coach (or whoever it was who set up your AMC 10B test) since you took the test? Everyone should know results about AIME qualifying by now, I'm pretty sure.

The packet probably got sent to the wrong place or is sitting in the counselor's office or something. It doesn't matter. Your teacher can call AMC to ask them to send the AMC results by email. The more important question is whether the AIME exam packet is lost. That inlcudes the names of the qualifiers, but more importantly it includes the exams. The qualifiers cannot take the exam if they don't have it. What does you teacher say when you ask about this?

i ask: did u get the AMC 10 B score yet?
she says: no

There should be an emailed student report sent to your school.  We got ours 3/15.  And besides, if you are quite sure that you got over 120, you're in for sure.

Good luck!

i think i got 120.5, but if i got one extra problem wrong , then i won't make it

but as long as the mail thing arrive and i made to AIME i will be able to take AIME on tuesday right?

if your teacher doesn't have the results, then she may not have the EXAMS! That means no one at your school can take it.

if that's the case you could still probablly take the B then, since it's pprobablly too late to do anything about the A exams.

[quote="Syntax Error"]can you take the aime B at another school even if you were supposed to take the aime A but missed it?[/quote]

You'd have to arrange that agreeably to the other school. You can take the later date if you unavoidably miss the first date, as explained in the [url=http://www.unl.edu/amc/d-publication/d1-pubarchive/2003-4pub/04AIME/html/04-04aime-text.html]AIME manual[/url]. But remember that "[b]under no circumstances can a student take both AIME's.[/b]" It costs money to take the later test--and you may have to pay for a suitable proctor as well. You need to be sure to talk to both schools involved as soon as possible.

its weird that this other person that took AMC10 B with me got his score already


Let \[ x\equal{}y^{2^{2009}}\]. Then $ Y\equal{}y\plus{}y^2\plus{}...\plus{}y^{2^{2008}}\equal{}2009y^a, a\equal{}\frac{2008*2^{2009}}{2009^2}$.
Obviosly $ Y\ge 2009*y^b, b\equal{}\frac{1\plus{}2\plus{}...2^{2008}}{2009}\equal{}\frac{2^{2009}\minus{}1}{2009}>a$.
Therefore we had not real solution, when $ y>1$ (equavalent $ x>1$). Only solution $ x\equal{}1$. It had second real  solution  $ 0<x<1$


B[hide]y guess and check, and getting it on my first guess, treething makes 4.5 errors a sentence.

So in 100 sentences, he makes 450 errors.[/hide]

Actually, that's a little off. Keep in mind that the number of errors in 25 sentences isn't rounded.

[hide]Let's see, he makes approximately 36 errors per 8 sentences, so he makes 111 or 113 errors per 25 sentences. Because he averages approximately 5 per sentence, so it has to be 113 per 25 sentences. Multiply by four and get 452.[/hide]

treething makes five grammatical errors per sentence (rounded to the nearest integer) he makes 36 grammatical errors per eight sentences (rounded to the nearest integer) and he makes an Integral odd amount of grammatical errors every 25 sentences. how many grammatical errors does he make per 100 sentences?

[hide]treething makes 4.5 to 5.49 errors per sentence. He makes 4.5-ish
errors per 1. He can't make more than 4.5625 per 8, or that would round to more than 36. So the range is 4.5 to 4.5625. 4.52 works for 113. Therefore, 113*4=452.[/hide]

[hide]25 is 3 * 8 sentences + 1 sentence, which is (36* 3) + 5 which is 108 + 5 = 113 * 4  because 25 * 4 = 100, so it's [b]452[/b][/hide]

I remember seeing this problem before and i'm pretty sure my answer is right.

[hide]36/8 = 4.5, 4.5*25 = 112.5, so the answer should by near 4.5.  i did 4.51*25 (112.75) so i knew that the answer had to be 113, which is an integer.  113*4 = 452, the answer[/hide]

I think Treething posted this problem last year and the answer was
[hide][b]452[/b][/hide]


The test is referred either way, amy, or AIME. I don't think I've heard "usamo" though, just U-S-A-M-O, but some may call it that.

For what it is worth, within the AMC office we say it as "Amy".  Among the committee that creates the contest, we call it the "Amy" also.  But occasionally I hear A-I-M-E, and occasionally use it too.  However, when I am speaking to an "outsider", for instance a newspaper reporter seeking information about the contest or the participants, I refer to it as the "Invitational".  That has lots of positive associations with sports events, conveys the right image, and is easier to say.

Steve Dunbar
MAA Director, American Mathematics Competitions

[quote="not_trig"]I've heard both for both. It doesn't matter that much unless you're talking with people who don't know what it is - then it's better to spell it out :O[/quote]

Ditto.

I've heard most of the pronounciations above, but i pronounce AIME as if the E were silent, as in "My aim is to qualify for the AIME this year", where aim and AIME are pronounced the same. However, many of my friends call it the "amy".

Also, i pronounce USAMO as U-S-A-M-O, but i know people who say you-sammo, and i know others(who don't know what it is) who call it the math olympics because it has "olympiad" in its name :rotfl: .

just my 2 cents...

[quote="AMCDirector"]I refer to it as the "Invitational".  That has lots of positive associations with sports events, conveys the right image, and is easier to say.
[/quote]

Interesting.  I honestly never thought of that before, and it does make a lot of sense.

[quote="pascal12"]Likewise, how do you pronounce "USAMO"?[/quote]
you-sammo

ive heard it both ways for both as well.
guess it just depends who you talk to.

[quote="MrTeacher"]When I'm trying to explain the AMC sequence to my students, do I refer to the AIME as the "Ay-Eye-Emm-Eee" (spell it out), or do I call it the "Amy"? How about the USAMO? Spell it out, or "You-Sammo"?

Once upon a time, back when I worked in the computer industry, our motherboards had what was called an AMI BIOS. In that case, we [i]always[/i] referred to it as an "Ay-Emm-Eye Bios", and [i]never[/i] an "Amy Bios".

Like the title says, a stupid question. Hopefully someone here can give me a stupid answer.  :D

MrTeacher[/quote]

Teachers at my school pronounce it "aim"

I'm not sure how many teachers at my school know what AIME is let alone know how to pronounce it but... :D

I pronounce it "amy"


Are homeschooled ppl smarter than most public schooled ppl?

I doubt there's any correlation. You'd have to do a pretty big study and test people from regular school and home school, and then you'd have to account for the fact that if homeschooled kids test better, it might just be that they're subjected to a better curriculum--not that their IQs are higher. Then you'd have to poll the kids' parents for the reason they pulled the kids out of regular school, and see if that suggests a reason for any discrepancies. But home schooling is more a statement about the parent than the student.

[quote="paganinio"]maybe non-schooled ppl are smartest.[/quote]non-schooled people can never be the smartest - maybe you ment homeschooled? (Reid Barton is a good argument for this theory :) ).

[quote="Valentin Vornicu"]non-schooled people can never be the smartest - maybe you ment homeschooled? (Reid Barton is a good argument for this theory :) ).[/quote]

and Anders Kaseorg, Alison Miller, Matt Ince

Define public-schooled people. Is it people who aren't homeschooled, or people who go to a public school, because where would Private schools fit in that?

I wouldn't say smarter, but in most cases probably better educated. Plus homeschooler don't have to deal with those idiots that have a problem with being smart.

I don't know if you'd find that homeschoolers as a whole have higher IQ.  I've heard that the vast majority of homechoolers (something like 90%) are children of religous fundamentalists (this isn't limited to Christians, though they form the majority of the demographic in this country), the ones who think that evolution is just a "theory" .  Clearly these people won't tend to be the brightest and thus one might surmise that there kids wouldn't be smarter than average.

Err, sorry, but I definitely would NOT generalize Christians calling them not bright.  I'm sure there are very smart Christians out there.

But overall I would say I think that at least in Texas, home schoolers are better.  Most Texas public schools, including the one I attend, aren't exactly the best.  For example, my World Geography teacher has trouble speaking and reading English, as she was born in India.  Also, the only math competition my high school offers is UIL, which a lot of is a waste of time.  So I'm basically stuck either trying to get the teachers to register for something or just self-studying.
However, in the Austin region where I live, AAHS (austin area home schoolers) tend to do pretty well at Mathcounts.  The middle school I went to was the 2nd best public school in central texas, and we were still beated every year by the private school and the homeschoolers most of the time.

Homeschoolers tend to fall into two categories.

The first category is families who homeschool for philosophical or religious reasons.  These parents tend to know they are going to homeschool even before they have kids.  I would imagine that the distribution of smart kids among these families is similar to that in schools.

The second category is families who decide to homeschool because they feel the school will not or is not meeting the needs of their children.  In this category I think there is a disproportionate number of "highly gifted" kids who don't "fit" well in a traditional school setting (public or private).   Some of these kids begin in traditional schools and their parents pull them out when they see the "spark" of love of learning disappearing from their eyes, when they see that the child is not going to learn anything in school but how to wait quietly while others learn, or when the child gets dangerously depressed, etc.

The phenomenon you are seeing with homeschoolers doing well on competitions is partly because of this latter situation, but also because a bright homeschooled kid in many cases gets more time to concentrate on his or her areas of interest/talent, which allows him/her to do better on contests.

So, I would say that there is a disproportionate number of really bright kids who are homeschooled.  And a disproportionate number of contest winners who are homeschooled.  And in general, the individual attention of homeschooling probably provides better opportunities for most homeschooled kids (as long as the parents are smart enough and well enough educated themselves).  But it would be hard to say whether or not in general homeschooled kids are "smarter than" public-schooled kids.

[quote="eryaman"]Err, sorry, but I definitely would NOT generalize Christians calling them not bright.  I'm sure there are very smart Christians out there.[/quote]
He didn't say Christians, he said students that are exposed only to creationist stuff and not to evolution. I wouldn't say those students are stupid, because it's most certainly not their fault they're stuck with a bad curriculum. Many of them will grow up to reject creationism anyway (and retain Christianity, which creationism does not in any way equate to) when they study real evolution in a real biology class instead of the strawmen many creationists set up.

Can anyone elaborate on the tradional path and curriculum of homeschooled people ? Thanks.  :)

It seems like this topic has turned into a discussion about whether among the very smartest kids, home-schoolers are disproportionately represented, not whether the AVERAGE homeschooler is smarter than the average public school child.  I hold that on average, the people who are homeschooled for religous reasons will not be smarter than others.  There are some children who are homeschooled because they are gifted, but then there are others who are homeschooled because they are mentally disabled. So as a whole, I don't think that homeschoolers will be smarter than public school children.

[quote="themonster"] There are some children who are homeschooled because they are gifted, but then there are others who are homeschooled because they are mentally disabled. [/quote]

I think the latter is [i]much [/i]less common.   In large part, this is because public schools do a good job meeting the needs of learning disabled children, or paying to place them in other institutions where their needs can be met.  This is required by law.  Many states and certainly federal law do not require any accomodations for gifted students. :(

I can speak from personal experience about both homeschooling and public schooling...
I attend public school for most subjects but since elementary school I have been pulled out of school for math; one of my parents is a mathematician.  Being extremely interested in the subject, and having my parents' help at my disposal, I'm now way ahead, topic-wise, of anyone in my school district in math (although that's not too hard considering my town lol).  So homeschooling is definately helpful in some cases.
On the other hand, they tried to pull me out of history one year because my school teacher was really bad; I was learning from a tutor, but I had NO INTEREST in history whatsoever.  Without the pressures of deadlines for assignments and stuff, I learned absolutely nothing, and the next year I went back to public school for history.
So I think homeschooled kids learn more if and only if they have the interest and if their parents know what they're talking about.

[quote="hmmm"]So I think homeschooled kids learn more if and only if they have the interest and if their parents know what they're talking about.[/quote]That is a very objective opinion. You must have gifted parents with some amounts of spare time, more than the usual needed to raise a kid, to assure that your homeschool education is more beneficial than the one you'ld get in either a public or a private school.

For the record, I wasn't homeschooled, and I don't homeschool my children.  But I still believe that homeschooling can be an excellent opportunity for many children, even if their parents aren't particularly gifted in every area.

Just because a child is homeschooled doesn't mean that their parents much teach them everything themselves, for starters.  And there are advantages to the individualized attention and curriculum that homeschoolers get that is hard to duplicate in a public or private school setting.  There are lots of resources available to homeschooling families to allow them to offer their children the best education possible, even in areas where the parents themselves are not experts.

For example, a homeschooler could enroll in math classes through AoPS if his/her parents were unable to teach math at an appropriate level ;-)

[quote="mathmom"]For example, a homeschooler could enroll in math classes through AoPS if his/her parents were unable to teach math at an appropriate level ;-)[/quote]Yes, but so would any other regular school student ;)

Home schooling fails in one very important thing: it does not provide interaction between the child and other children of his age. When I was young in my school there were kids that were coming to school only to meet with other colleagues. I think this is very hard to be covered on the homeschooled group.

[quote="Valentin Vornicu"]
Home schooling fails in one very important thing: it does not provide interaction between the child and other children of his age. [/quote]

This is a myth.  In reality, homeschooled kids find a great myriad of ways in which to find positive social interaction opportunities, with kids not only of their exact age, but with others as well.  Once you leave school, you will never again be expected to socialize only with others exactly your age.  In addition, the social interaction in some schools is far from a positive experience for many kids :(

[quote="Valentin Vornicu"]Home schooling fails in one very important thing: it does not provide interaction between the child and other children of his age.[/quote]

Hardly true. I'm a homeschooler but that doesn't mean I have no social life. For starters, I take every one of my courses with a class of some sort, half of them are with the homeschool association in our area and the other half are online. In either case, the class sizes are about 15-25, making it only slightly below my class sizes from public school.

Outside of that, extracurricular activities provide lots of chances for homeschoolers to get together and in many cities homeschoolers even form teams for some competitions to compete against public and private schools. Homeschoolers also host parties; they go to the mall, watch movies with their friends, and enjoy a wide variety of other social activities that public schooled kids don't have the time to do. If anything, I think there's a better social life for many homeschoolers than the average public schooled kid that sets high goals academically.

In public school, I wouldn't even be having a so-called Christmas break and all what with the projects and reports and busy work that teachers leave for the holidays.

[quote="Valentin Vornicu"][quote="mathmom"]For example, a homeschooler could enroll in math classes through AoPS if his/her parents were unable to teach math at an appropriate level ;-)[/quote]Yes, but so would any other regular school student ;) [/quote]

That's also very debatable. In eighth grade, I was doing homework every night til after midnight, getting up at 6:30 to go to the high school for classes, then going through normal school til about 4:00. I could also count on an hour or more of after-school activities everyday to keep me at school til 5:00 or 5:30 or even later. Dinner was a 15-minute deal, and then it was naptime for half an hour and homework for the rest of the night. AoPS wouldn't have been an option.

Of course, that was a year when I had teachers that particularly enjoyed giving work just for the sake of having enough grades for our averages but the idea is similar. A good majority of my friends go to public school and there are maybe 5% of them that would even have time to consider an AoPS class. This is the 5% that don't take all honors classes and attempt to maintain an A average in all of them.

Each school and district is probably different in how much homework the students have, but I think I speak for most academically-gifted students in America when I say that they simply don't have the time or energy (or will) to take uncredited math courses for their enjoyment. However, of course, that's not the case with students on this site since they obviously have all set time apart for AoPS. :-)


Find the greatest k such that the following inequality holds for any non-negative real numbers a,b,c:
$\frac{a+kb}{c+kb}+\frac{b+kc}{a+kc}+\frac{c+ka}{b+ka}\ge 3$

see here some discussions:
http://www.mathlinks.ro/viewtopic.php?p=804475#804475


This is from: [url=http://www.icoool.com/index.asp]http://www.icoool.com/index.asp[/url]

A卷(四级标准,适用于长江以南)题型----单项选择,选出下列词句的正确意思:

1."哎呀吗呀,他对象长得老棵碜了"一句中,"棵碜"指:
A好看;B丑;C长相一般;D美若天仙

2.嘿,那边有人干仗呢."干仗"的意思:
A发生口角;B扭秧歌;C打架;D作战

3.咱俩嘎点儿啥."嘎'的意思:
A打赌;B割;C打扫卫生;D吃

4.这孩子真各应人."各应"是指:
A烦人,讨厌人;B招人喜欢;C天真活泼;D呆若木鸡

5.这间屋子真埋汰."埋汰"是指:
A脏;B整洁;C豪华;D简陋

6.选出"欠儿登"的同义词:
A有债务;B哪儿有事哪儿到;C有债权;D牛

7.选出"吭哧瘪肚"的同义词:
AIQ高;BIQ低;C速度快;D速度慢

B.卷(六级考试,北方人也未必全答对)
题型----单项选择,选出与下列词语最接近的解释:

1.骂血人了:
A赞美对方是一个有血有肉的人;B很漂亮;C脸有血色;D太狠了,极至

2.磨叽:
A拉磨;B一种病名;C罗里罗嗦;D哭了

3.劲儿劲儿的
A力大无比;B皮筋;C执着,有精神头;D没完没了

4.撒漠:
A往四处看;B武术;C拣东西;D荒漠化

5.卖呆儿
A拐卖人口;B个体户;C闲着无事看热闹;D神经病

6.叽咯浪:
A拌嘴;B戏水;C撞船;D食品

C卷(八级标准,足以考蒙东北人)

1.请指出"嘎拉哈"是什么东西?
A一种食品;B一种用动物关节制的儿童玩具;C一种家用器具;D一种农用器具

2."戈能"是指:
A垃圾;B一种能量;C一种粥;D没能耐

3."豁楞"是下面的哪一个?
A摇摆;B豁牙子;C搅和;D房檐

4."无极六兽"是指:
A六种动物;B精神饱满;C性情凶残;D百无聊赖

5."老哇子"是指:
A公鸡;B老母鸡;C乌鸦;D小伙子

A2 在北京也很常用. 经常听到有人说: 我昨天跟某某在厕所干了一仗......
A4 各应=找抽.
B1 太狠了
B2 不利索.
B3 没完没了. 现在似乎还有" 臭牛bi" 的意思.

我只知道这些......

[size=134]应该大部分都对的....如有错，欢迎更正！
A卷：
1.　 B丑
2.    A发生口角
3.    D吃
4.    A烦人,讨厌人
5.    A脏
6.    B哪儿有事哪儿到
7.    D速度慢
B卷：
１.   D太狠了,极至
2.    C罗里罗嗦
3.    D没完没了
4.    A往四处看
5.    C闲着无事看热闹
6.    A拌嘴
C卷:
1.    B一种用动物关节制的儿童玩具
2.    C一种粥
3.    C搅和
4.    C性情凶残
5.    B老母鸡[/size][/b]

干仗是吵架么? 我觉得是打架.

不知廣東話有否類似的考試?
但我也未必會得到很好成績 :P

[quote="ifai"]不知廣東話有否類似的考試?
但我也未必會得到很好成績 :P[/quote]
我们这里会讲广东话的还蛮多的吗,呵呵^^^

[quote="luce86"]呵呵，干仗是吵架[/quote]
是么!? 原来这个词到了北京之后变了味. 那也就是说我隔三岔五就要和人"干一仗"... :rotfl:


$ S \equal{} \binom{20}{2} \plus{} 2 \binom{20}{3} \plus{} 3 \binom{20}{4} \plus{} \cdots \plus{} 19 \binom{20}{20}$

Find the number of divisors that $ S$ has (positive numbers).

Level: AIME

4 doesn't seem to be the right answer.

Let's wait till if another person is willing to try out for this problem. Here's my hint.

Consider $ \sum^{20}_{n \equal{} 1} k \binom{20}{n}$ where $ k$ is a number. Play with this... Can you set $ k$ to any value that'll help you to solve this problem?

Hmm, no one is giving a shot so I'll give the major hint away.

The problem is created by this equation:

$ \sum^{20}_{n\equal{}1} n \binom{20}{n} \minus{} \sum^{20}_{n\equal{}1} \binom{20}{n}$

The latter is a classic identity so it shouldn't be hard to recognize. The first one is what you need to figure out. Any taker?

I'm not sure what you're waiting for.  I think both of the solutions given are correct.

Oh man... I feel so stupid now... I just realized where I have gone wrong.  :oops:

I somehow had this in my paper:

$ 2^{20} \equal{} \binom{20}{1} \plus{} \binom{20}{2} \plus{} \cdots \plus{} \binom{20}{20}$

I forgot to add $ \binom{20}{0}$.... blah. My bad   :|


[quote="2424"]Prove that $ \sum\frac {a \plus{} b}{a^2 \plus{} ab \plus{} b^2}\geq\frac {21}{4(a \plus{} b \plus{} c)}$ with $ a,b,c > 0$[/quote]

"Use" Iran 96  :P
Ineq
$ < \equal{} > (ab \plus{} bc \plus{} ca)(\frac {1}{a^2 \plus{} ab \plus{} b^2} \plus{} \frac {1}{b^2 \plus{} bc \plus{} c^2} \plus{} \frac {1}{c^2 \plus{} ca \plus{} a^2}) \ge \frac {9}{4}$
We have:
$ LHS \ge (ab \plus{} bc \plus{} ca)(\frac {1}{(a \plus{} b)^2} \plus{} \frac {1}{(b \plus{} c)^2} \plus{} \frac {1}{(c \plus{} a)^2}) \ge \frac {9}{4}$


For each integer $n\geq 3$, find the least natural number $f(n)$ having the property

$\star$ For every $A \subset \{1, 2, \ldots, n\}$ with $f(n)$ elements, there exist elements $x, y, z \in A$ that are pairwise coprime.

$[\frac{n}{2}]+[\frac{n}{3}]-[\frac{n}{6}]+1$
where $[ x ]$ means the largest integer that less than or equal to $x$.


A harmonic sequence is a sequence of numbers in which the reciprocals form an arithmetic sequence. In the harmonic sequence $ \frac12, \frac14, \frac16, \ldots$ when the 50th term is divided by the $ n$th term, the quotient is 4. What is the value of $ n$?

The nth term is in the form 1/2n.  The 50th term is 1/100.  1/100 divided by 1/400 is 4.  1/400 is the 200th term.

Instead of finding the values independently, we notice that the quotient $ \frac{\frac{1}{n}}{\frac{1}{4n}}$ is $ 4$. With $ n\equal{}50$, the $ 4n$ term must be $ 50\times4\equal{}\boxed{200}$. This is true for any sequence (try it yourself and see!)


function
trigonometry
algebra
domain
topology
calculus
calculus computations

26.  If $ k$ is a real number, and

\[ f(x) = \left\{ \begin{array}{c}{1}
sin(\frac{1}{x}) \; \mathrm{if}\, x \neq 0 \
k\; \mathrm{if}\, x = 0
\end{array}\right.\]

and if the graph of $ f$ is NOT a connected subset of the plane, then the value of $ k$

(A) could be -1
(B) must be 0
(C) must be 1
(D) could be less than 1 and greater than -1
(E) must be less than -1 or greater than 1

Well if I understand the question correctly, it's asking a value of $ k$ such that the graph of the given piecewise defined function is not continuous.

In order for that to occur, we must have

$ \sin(\frac{1}{0}) \not\equal{} k$

because the graph of a continuous function would have the two parts be equivalent at $ x\equal{}0$.

Now $ \frac{1}{0}$ can be interpreted as $ \infty$ and the value of $ \sin(x)$ as $ x$ approaches infinity is not definite. It can be anywhere from $ \minus{}1$ to $ 1$ because that is in the range of $ \sin(x)$.

Given that, it immediately makes options $ A$-$ D$ false because those values could possibly make the graph continuous. Thus, the answer is $ D$.

*Note: I have never done a problem like this, so I may be wrong. But I am pretty sure that I'm right.

Which part of the y axis contains limit points of the graph of the function?

Wait  minute: how can it be continuous if the limit doesn't exist? ... therefore any choice of k makes it not connected

I'm assuming connected = continuous (i'm only used to see the term "connected" apply to shapes with dimension > 1)

also just checking - the domain of x is the reals, right? As opposed to complex?

No, connected does not mean continuous. It means that we can not draw two open sets in the plane which both intersect the graph, but without sharing any points from the graph, and whose union contains the whole graph (the graph cannot be separated by open sets).

the thing to think about here: if $ k\in [\minus{}1,1]$, what can you say about about the open set containing $ (0,k)$?  must it intersect the graph of $ f$ in some other place?


If $a$,$b$ and $c$ are positive real numbers distinct of $1$, with $a^{b}.b^{a}=c^{b}.b^{c}=c^{a}.a^{c}$ prove that:

$\frac{a(b+c-a)}{log_{10}a}=\frac{b(a+c-b)}{log_{10}b}=\frac{c(a+b-c)}{log_{10}c}$

Let $a^{b}b^{a}=c^{b}b^{c}=c^{a}a^{c}=k\ (k>0)$. By $\log a^{b}b^{a}=log k, \ blog a+alog b=log k\ \cdots [1]$. Similarly we have $b\log c+c\log b=log k\ \cdots [2]$, $alog c+clog a =log k\ \cdots \log k\ \cdots [3]$. $[1]*c+[2]*a$ gives $bc\log a-ab\log c=(c-a)\log k\ \cdots [4]$. Then $[4]+[3]*b$ gives $2bc\log a =(b+c-a)\log k$. Since $a,\ b,\ c$ are positive numbers such that $a,\ b, c\neq 1$, we have $\frac{a(b+c-a)}{\log a}=2abc\log k$. By symmetry of $a,\ b, c$, we obtain $\frac{b(c+a-b)}{\log b}=2bca\log k,\ \frac{c(a+b-c)}{\log c}=2cab\log k$, yielding $\frac{a(b+c-a)}{\log a}=\frac{b(c+a-b)}{\log b}=\frac{c(a+b-c)}{\log c}$, which completes the proof.

$[1]*c+[2]*a$

I'm sorry, but I didn't understand this part. What does the * mean?


[quote="averin"]Hm...so I saw this problem two years ago and couldn't solve it.  Two years later, I still can't solve it.  Maybe one of you smart people can show me how.

"Mr. and Mrs. Lewis go to a party with three other couples, and several handshakes took place.  No one shook hands with himself or herself, no spouses shook hands, and no two people shook hands more than once.  When each other person told Mr. Lewis how many hands he or she shook, the answers were all different.  How many handshakes did Mr. and Mrs. Lewis each make?"[/quote]
Were there only four couples as you mentioned?
Each person told only to Mr.Lewis, was Mrs. Lewis also included in those persons?
{
if answer to the first is yes
and to he second is no, there could be any no. of handshakes attributed, with some exceptions. there will not be a unique answer.
if answer to the first is yes
and to he second is yes, then it is not possible since one of them would have shaken hands six times and Mrs. Lewis would have been included in it.
}
Keep Smiling
Malay

similar problem, maybe you can help me out too

[quote]Quote:
The handshake problem: I invite 10 couples to a party at my house. I ask everyone present, including my spouse, how many people they shook hands with. It turns out that everyone questioned (I didn’t question myself) shook hands with a different number of people. If we assume that no one shook hands with his or her spouse, how many people did my spouse shake hands with?

Please remember: I did not ask myself any questions.
Please also remember: Although it may seem so at first, this problem is NOT intractable.

I kinda understand the question but can't think of any ways to solve this atm, any hints ?

Thanks.

Mon Oct 23, 2006 3:56 am
my attempt
is the answer 9? First i listed the number of handshakes for ppl questioned 0 -> 18. What i found is that the person who shakes hand 18 times is coupled with the person who shakes 0 times, 17 with 1 and so forth. Only a person with 9 handshakes is left while the person who questions everyone is calculated to have made 9 handshakes, thus his wife made 9 handshakes.

However, I'm not 100% sure, still very confused
[/quote]

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=handshake&t=115721[/url]

Total of eight people. No two peeps shake more than once, .sooo Mr. Lewis is still eligible to shake everyone's hands 'cept Mrs. Lewis. That won't violate the rules if two other people attempt to shake only once.

That means Mr. Lewis can shake 6 hands.

Mrs. Lewis can shake four hands only since two has already dropped out from the shakin' game because "no two people can shake more than once." Sooooo Mrs. Lewis shakes 4.

its 3...

If you assume Lewis couple with 1 couple. Easy enough, you can see Mr. and Mrs. Lewis only shake once each.. And the other couple, one person shaked 2 hands(with Mr. and Mrs. Lewis) and the other one didn't shake hands at all.
Therefore, Lewis couple & 1 more couple=1 hand shake for each Lewis.

If you assume the Lewis couple with 2 couple shaking hands. You can see one couple with one person shake 4 hands and the other one shake 0 times. The other couple with one person shake hands 3 times and the other shake hand once. Lewis couple shakes hand twice each.
2 more couple=2hand shakes.

Therefore, you can get the answer is $\boxed{3}$, when Lewis couple and 3 more couple shaking hands.

Of course, you can brute force and draw 8 people out. But I suggest you to think at the easier cases to attack this problem

Well, you can generalize it that if Mr. and Mrs. Lewis shakes hand with x more couples, each Lewis would shake hand x times
 :D


According to the intro to number theory Legendre proved that there are no equations with rational coefficients that always give a prime number for every integer input. Are there any equations with irration coefficients?

well an irrational times a rational is irrational, so a polynomial with irrational coefficients would give irrational values for integer input

[quote="maokid7"]I remember a proof that showed that $f(a+f(a))$ is a multiple of $f(a)$ thus proving that a polynomial can never always output primes.[/quote]

This is only for polynomials with integer coefficients, though. That would clearly trivialize the problem.

the proof i saw was in ACoPS and it used the binomial therom. i dont see how it would be differnt if the coefficients were not integers.

Maybe if $f(a)$ is an integer for all integers $a$... since if $f(a)$ is not an integer "multiples" are not well-defined.


[quote="Thunder33"]This little problem came into my mind a few days ago:

Suppose that you have a natural number N which factorizes into M distinct primes (possibly with multiplicities). Let P be the set of all distinct primes which N factorizes to. So, there exists a subset of P for which the sum of all its members is a number coprime to N. True or not true?

For example, for 14=7*2, the subset {7,2} of the prime set {7,2} has this property (7+2=9 which is coprime to 14). In fact, if m=2, it is easy to prove that the subset aforementioned exists. Furthermore, it is the entire set of the distinct primes of N.

However, the general case (m>2) seems to be more complicated, and I'm not even sure if it's true, but I've yet to find a counterexample.[/quote]

Such a subset always exists if $m>1$

1) case $m=1$ : obviously no subset exists.

2) case $m=2$ : $n=p^{a}q^{b}$ and $p+q$ is obviously coprime with $p^{a}q^{b}$

3)  case $m\geq 3$. Let $p<q$ the $2$ littlest prime divisors of $n$
If $(p+q)$ is not coprime with n, it exist in $P$ a prime $r>q$ which divides $p+q$. Since $p+q<2q$ and $r>q$, we have $r=p+q$ and so $p=2$ (else $p+q$ would be even, so nonprime).
So, the $3$ littlest divisors of $n$ are $2,q,q+2$
Then, if $2+(q+2)=q+4$ is not coprime with $n$, then :
Either $2$ divides $q+4$, which is impossible
Either $q$ divides $q+4$, which is impossible
Either $q+2$ divides $q+4$, which is impossible
Either $s>q+2$ divides $q+4$, and $s=q+4$, which is impossible, since $q$,$q+2$ and $q+4$ can't be together primes ($3$ must divide one of them).

So, either $(p+q)$ is coprime with $n$, either $2$ and $q+2$ are prime divisors of $n$ and $2+(q+2)$ is coprime with $n$.

Q.E.D.


geometry
analytic geometry
combinatorics unsolved
combinatorics

Let $ S$ be a set of all points of a plane whose coordinates are integers. Find the smallest positive integer $ k$ for which there exists a 60-elemen subset of set $ S$ with the folloiwing condition satisfied: for any two elements $ A,B$ of the subset there exists a point $ C$ contained in $ S$ such that the area of triangle $ ABC$ is equal to $ k$.

Is this from IMO Shortlist?
This was at Brazilian IMO Tst.

This was Korean TST too when k is a special number.
I think this is IMO shortlist.


trigonometry
function
limit
calculus
calculus computations

How do you find the limit of $ \frac {x \sin x} {x^{2}\plus{}5}$ as x approaches $ \infty$.

Also, this would probably solve my question:
When you have $ \frac {undefined} {\infty}$, what do you get?

What do you mean by "undefined"? If you mean "doesn't have a limit", dividing by something large can either squeeze it down to zero or leave it as something that doesn't have a limit. Since $ x\sin x$ is unbounded, this is effectively still an indeterminate form.

hello,write your term in the following form:
$ \frac{x\sin(x)}{x^2\plus{}5}\equal{}\frac{\frac{\sin{x}}{x}}{1\plus{}\frac{5}{x^2}}$ and here you can find the limit if $ x$ tends to infinity.
Sonnhard.

[quote="t0rajir0u"]The general strategy here is bounding.  Since $ \minus{} 1 \le \sin x \le 1$, the expression you want is bounded above by $ \frac {x}{x^2 \plus{} 5}$ and below by $ \minus{} \frac {x}{x^2 \plus{} 5}$.  What limits do these expressions have as $ x$ goes to infinity?[/quote]

They both go to 0, right?

I don't really get what you mean by "bounding".
If you could explain that to me, I would greatly appreciate it.

The relevant theorem here is [url=http://en.wikipedia.org/wiki/Squeeze_theorem]the squeeze theorem[/url].  If the expression you want is bounded above by an expression that goes to $ 0$ and bounded below by an expression that goes to $ 0$, then it goes to $ 0$.

The term bounding represents a value at which a specific graph never surpasses (either negative or positive).
For instance, the function $ \arctan(x)$ is bounded by $ \frac{\pi}{2}$ and $ \minus{}\frac{\pi}{2}$.
And $ \tanh(x)$ is bounded by $ \minus{}1$ and $ 1$.
A ''bound'' could, to some, be considered a limit. As t0rajir0u says, $ \frac{x\sin(x)}{x^2\plus{}5}$ is bounded by two specific functions based on the behavior of $ \sin(x)$.

Essentially, he is implying the result: if $ |f(x)| \leq x^2$ prove $ \lim_{x\rightarrow 0} f(x) \equal{} 0$. Thus $ x^2$ is "bounding" $ f(x)$ near $ x\equal{}0$.

[quote="iamhe"]When you have $ \frac {undefined} {\infty}$, what do you get?[/quote]
It's probably best not to think about that in formulas, as if it were algebra. We don't have a literal $ \infty;$ rather, we have something that becomes arbitrarily large as we approach the limit in question. So what we really have is a numerator of "potentially large but wiggly" divided by a denominator of "large". And it we phrase it that way, the question immediately becomes [b]how[/b] large? Once we start thinking that way, we see that that "large but wiggly" numerator is no larger than $ x,$ while the denominator looks like $ x^2.$ As $ x\to\infty,$ $ x^2$ is "much larger" than $ x^2.$ (I really should say "arbitrary much larger".) And now we realize that since the denominator is much larger than the numerator, the whole thing goes to zero.

That's very informal reasoning - but everyone who is quoting the squeeze theorem is really saying the same thing in more formal language.

$ \lim_{x\to\infty}\frac{x^2\sin x}{x^2\plus{}5}$ does not exist. This expression remains bounded but does not "settle down" to any limit. And $ \lim_{x\to\infty}\frac{x\sin x}{\sqrt{x^2\plus{}5}}$ also does not exist, in the same way. Both of these differ from your original example by giving different answers to my question of "How large?"


vector
linear algebra
linear algebra unsolved

Let $e_1,...,e_m$ be vectors in $n$-dimensional vector space $V$ with $m\geq n+2$. Prove that there exist scalars $\alpha _1,...,\alpha _m$ not all of them equal  to $0$ such that:

$\sum _{i=1}^m \alpha _i e_i=0$    and    $\sum _{i=1}^m \alpha _i=0$

Consider a relation of the form $\sum_{i=1}^{n+1}a_ie_i=0$, not all $a_i$'s being zero, and one of the form $\sum_{j>1}b_je_j=0$, again, not all the $b_j$'s being zero. If either $\sum a_i=0$ or $\sum b_j=0$, we take $a_{i+2}=0$ or $b_1=0$ respectively, and we're done. If, on the other hand, both sums are non-zero, replace each $a_i$ with $\frac{a_i}{\sum a_i}$, each $b_j$ with $\frac{b_j}{\sum b_j}$, and subtract the relations from one another.


vector
trigonometry
real analysis
real analysis theorems

I have a interested problem in normed space $R^{n}$
.It is ....
Suppose that $e_{1},e_{2},...,e_{m}$
are linearly independent vectors in normed space $R^{n}$
.And suppose that $\left\{{x_{i}^{k}}\right\}_{k = 1}^\infty$
;i=1;2,...;m are m  real-sequences so that $x_{1}^{k}.e_{1}+x_{2}^{k}.e_{2}+...+x_{m}^{k}.e_{m}\to 0$ when $k \to \infty$.Prove that
\[x_{i}^{k}\to 0;when\;k \to \infty \left({i \in \overline{1,m}}\right) \]
 :idea:  :idea:
 :!: I am waiting a solution  :lol:  :oops:  :lol:

[quote="mlok"]Let $P_{j}$ be the linear span of $\{e_{i}\colon i\ne j\}$, and let $P_{j}^{\perp}$ be the orthogonal complement of $P_{j}$. The orthogonal projection of $\sum_{i=1}^{m}x_{i}^{k}e_{i}$ onto $P_{j}^{\perp}$ is $cx_{j}^{k}e_{j}$, where $c$ is $\pm$ the cosine of the angle between $e_{j}$ and $P_{j}^{\perp}$. Hence $x_{j}^{k}\to 0$ as $k\to\infty$.[/quote]

I don't understand your solution.Is  $P_{j}^\bot$  subspace of $\left\langle{e_{1},e_{2},...,e_{m}}\right\rangle$ ?.Why  orthogonal projection of $\sum_{i=1}^{m}x_{i}^{k}e_{i}$ onto $P_{j}^{\perp}$ is $cx_{j}^{k}e_{j}$ ?
You can explain clear more...?


Let  a,b,c,d be distinct  real  numbers and  a and b are roots of   x^2-2cx-5d=0.
If  c  and  d are roots  of  x^2-2ax-5b=0   then  find  a+b+c+d=?

We have :$a+b=2c$ and $ab = -5d$ from the first equation.
$c+d = 2a$ and $cd = -5b$ from the second.
Summing the first and the third relations leads to $b+d=a+c$.
Substituing onto the second and the fourth, leads to $2ac-a^2 = -10a+5c$ and $2ac-c^2 = -10c+5a$.
Thus, by difference, since $a\neq c$ (otherwise we would have $a=b=c=d$), we have $a+c=15$.
Therefore, $a+b+c+d=30$.

Pierre.


A sequence of numbers is formed according to the following rule:

$ a_1\equal{}19, a_2\equal{}77$ and

$ a_n\equal{}\frac{1\minus{}a_{n\minus{}1}}{a_{n\minus{}2}}\text{ for }n\ge3.$

Prove that the sequence repeats. What is the 2007th member of the sequence?

[quote="exoeragonius"]A sequence of numbers is formed according to the following rule:

$ a_1 \equal{} 19, a_2 \equal{} 77$ and

$ a_n \equal{} \frac {1 \minus{} a_{n \minus{} 1}}{a_{n \minus{} 2}}\text{ for }n\ge3.$

Prove that the sequence repeats. What is the 2007th member of the sequence?[/quote]

idk if this should be in pre-olympiad or not

[hide]
we've got that $ a_1 \equal{} a_1$ and that $ a_2 \equal{} a_2$

well $ a_3 \equal{} \frac {1 \minus{} a_2}{a_1}$

so $ a_4 \equal{} \frac {1 \minus{} \frac {1 \minus{} a_2}{a_1}}{a_2} \equal{} \frac {a_1 \plus{} a_2 \minus{} 1}{a_1a_2}$

and $ a_5 \equal{} \frac {1 \minus{} \frac {a_1 \plus{} a_2 \minus{} 1}{a_1a_2}}{\frac {1 \minus{} a_2}{a_1}} \equal{} \frac {a_1a_2 \minus{} a_1 \minus{} a_2 \plus{} 1}{a_1a_2}\cdot \frac {a_1}{1 \minus{} a_2} \equal{} \frac {1 \minus{} a_1}{a_2}$

then $ a_6 \equal{} \frac {1 \minus{} \frac {1 \minus{} a_1}{a_2}}{\frac {a_1 \plus{} a_2 \minus{} 1}{a_1a_2}} \equal{} \frac {a_1 \plus{} a_2 \minus{} 1}{a_2}\cdot \frac {a_1a_2}{a_1 \plus{} a_2 \minus{} 1} \equal{} a_1$

finally $ a_7 \equal{} \frac {1 \minus{} a_1}{\frac {1 \minus{} a_1}{a_2}} \equal{} a_2$, which is pretty cool

so it repeats like in a loop of 5 for any given initial $ a_1$ and $ a_2$

moreover, since $ 2007 \equiv 2\mod5$, the 2007th term of that sequence would be $ 77$
[/hide]


One computer in a lab is programmed to back up data every five minutes. Another computer is programmed to back up data every two minutes. Find the number of times in twenty-four hours that the two computers back up data at the same time.

We find the LCM of $ 5$ minutes and $ 2$ minutes, which is $ 10$ minutes.  In $ 1$ hour, the computers will back up data at the same time $ 6$ times.  Therefore, in $ 24$ hours, the computers will back up data at the same time $ 6 \cdot 24$ times, which is also $ \boxed{144}$.

Someone check my answer--I normally make dumb mistakes  :(

Yes but if it backed it up once at the beginning and once at the end, wouldn't the answer be 144+1=145 for one extra backup?

It's sorta vague...but I think it means 144...


functional equation
Functional inequality
algebra
IMO 1977
IMO Shortlist
IMO

Find all functions $f : \mathbb{N}\rightarrow \mathbb{N}$ satisfying following condition:
\[f(n+1)>f(f(n)), \quad \forall n \in \mathbb{N}.\]

[quote="Rust"]There are more solution. For example
$1) \ f(n)=n,$
$2) \ f(n)=n-(-1)^n,$
and combinations 1), 2).[/quote]

The second one doesn't seem to work because the inequality is strict. For instance, take any $n$ odd:

$f(n+1) = (n+1)-1 = n$
$f(f(n)) = f(n+1) = n$

so strict inequality does not hold.

I have a solution please tell me if I am wrong .We suppose that
$f$ is decreasing in an interval $G$ . Then for $k\in G$ we have from ypothesis
$f(k+1)>f(f(k))$ so $f(k)>k+1$  so $f(p)>p+1$ for any $p$ in $G$ . So $f(p)=p+m$ where $m$ is a positive integer $m\geq 2$

By hypothesis  we find that $p+m+1>p+2m$ or $1>m$ contradiction .
So $f$ is an increasing function .So by hypothesis again $f(n+1)>f(f(n))$ implies that $n+1>f(n)$ . But from the fact that $f$ is an increasing function we find that $f(n)\geq n$ so $f(n)=n$

[quote="silouan"]$f$ is decreasing in an interval $G$ . Then for $k\in G$ we have from ypothesis
$f(k+1)>f(f(k))$ so $f(k)>k+1$  so $f(p)>p+1$ for any $p$ in $G$ . [/quote]
Exactly for any $p$ in $G$, suth that $f(p)\in G$ and $p+1\in G$.

A fairly simple solution is as follows. It assumes that the mapping $ f$ is bijective, and increasing. For the sake of brevity (I'm lazy) those lemmata have been left out.

$ f(n+1)>f(f(n))$
We apply the inverse of $ f$ to both sides
$ n+1>f(n)$

Similarly, we can say that

$ f(n)>f(f(n-1))$
$ \therefore n>f(n-1)$

Putting these together, we get that
$ f(n+1)>n>f(n-1)$

This means that $ n+1>f^{-1}(n)>n-1$, but since $ f$ is bijective, $ f^{-1}$ must be an integer between $ n-1$ and $ n+1$ exclusive.

Thus $ f^{-1}(n)=n$
$ f(n)=n$
QED

I thought this was a new problem and so I posted it elsewhere to share it with others(It was locked and I was referred to this thread).Anyway,I shall post my solution now:

Let $r=\text{min}\{f(n):n\in \mathbb{N}\}$ and $m\in \mathbb{N}$ so that $r=f(m)$.For the sake of contradiction, assume that $m>1$.Then $r=f(m)>f(f(m-1))$, which contradicts the minimality of $r$.So, $m=1$ and $f(1)<f(2)$.In the set $\{f(n):n\ge 2\}$,$f(2)$ is the minimum(using the same argument).If $f(1)>1$ then $f(1)\ge 2$.Clearly,$f(f(1))\ge 2$, a contradiction to the initial condition.So, $f(1)=1$.Consider $g(n)=f(n+1)-1$. So, $g(g(n))=g(f(n+1)-1)<f(n+2)-1=g(n+1)$.$g$ satisfies the same condition as $f$.Therefore,$g(1)=1$ and hence $f(2)=2$. By induction on $n$, $f(n)=n$

[b]Lemma.[/b] If $n\geq m$, then $f(n)\geq m$.
[i]Proof.[/i] We proceed by induction on $m$. For $m=1$ this is trivial. Suppose $f(n)\geq m$ for all $n\geq m$. If $n\geq m+1$, using the induction hypothesis twice,
$$ n-1\geq m\hspace{4mm}\Rightarrow\hspace{4mm}f(n-1)\geq m\hspace{4mm}\Rightarrow\hspace{4mm}f\left(f(n-1)\right)\geq m, $$
and it follows that $f(n) > f\left(f(n-1)\right)\geq m$. $\square$

Now, going back to the original question, the lemma implies that $f\left(f(n)\right)\geq f(n)$, so $f(n+1) > f(n)$, i.e., $f$ is strictly increasing. Since $f$ is strictly increasing, $f(n+1) > f\left(f(n)\right)$ implies $n+1 > f(n)$, i.e., $f(n)\leq n$. Combining this with the lemma, which asserts $f(n)\geq n$, we get $f(n) = n$.

We claim that the only solution is $f(x)=x$. It is easy to check that this works. We now show that this is the only solution, so suppose $f$ is a solution.

[u]Lemma:[/u] $f(x)\ge x$.

[u]Proof of Lemma:[/u] We will prove that $f(x)=n\implies x\le n$. We proceed by strong induction on $n$. Suppose $f(x)=m\implies x\le m$ for all $m<n$. Now suppose $f(x)=n$. We have that $f(x)>f(f(x-1))$, so $f(f(x-1))\le n-1$, so $f(x-1)\le n-1$ by the induction hypothesis. Thus, $x-1\le n-1$ by the induction hypothesis, so $x\le n$. Therefore, the claim is proved. $\blacksquare$

We see $f(n+1)>f(f(n))\ge f(n)$, so $f$ is strictly increasing, so if $a\ge b$, then $f(a)\ge f(b)$. Suppose there was some $n$ with $f(n)\ge n+1$. Then, $f(n+1)>f(f(n))\ge f(n+1)$, a contradiction. Thus, there is no $n$ with $f(n)\ge n+1$, so $f(n)=n$ for all $n\in\mathbb{Z}_{>0}$.

It's IMO 1977 P6. See [url=http://artofproblemsolving.com/community/c6h1145343p5403334]here[/url].

The only solution is $f(n) = n$, which we prove by strong induction.
Base check: let $a$ be the input for which $f$ is minimized. Suppose for contradiction that $a > 1$. Then substitute $n = a - 1$ into the given inequality to find that an input of $f(a - 1)$ gives a lower output, a contradiction, thus $a = 1$, and $a$ is the only input for which $f$ gives its minimum output (i.e $f$ is injective for this input).
We have 3 assumptions: (1) Assume that $f(n) = n \forall n \le k - 1$, (2) that $f$ is injective $\forall$ inputs $n \le k$, and (3) that $f$ takes its $k$th least output at $f(k)$.
Let $f(a)$ be the $(k + 1)$th least output of $f$. First, $a > k$ as $f(a) > f(n) \forall n \le k$ and $f$ is injective for these values. Also, by the given inequality, we have $f(f(a - 1)) < f(a)$. But $f(a)$ is the $(k + 1)$th least output of $f$, so $f(f(a - 1))$ is among the $k$ least outputs. For these values, $f$ is injective and $f(n) = n \forall n \le k - 1$. Either $f(f(a - 1))$ gives the $k$th least output or it is among the $k-1$ least outputs. If it is in the $k-1$ least outputs then $1 \le f(a - 1) \le k-1$ which in turn implies that $2 \le a \le k$ which is absurd as $a > k$. Thus $f(f(a - 1))$ gives the $k$th least output of $f$. But we assumed $f(k)$ gives the $k$th lowest output and $f$ is injective here. Thus $f(a - 1) = k$. $\therefore f(a - 1) > k - 1$, so $f(a - 1)$ is at least the $k$th least output of $f$. Also $k > f(n) \forall n < k$ and the $k$th least output of $f$ is $f(k)$. Hence $f(k) = k$, which shows (1) hold for the induction. Also, $f$ is injective here, so $f(a - 1) = k = f(k)$ implies that $a = k + 1$, so the $(k + 1)$th least output of $f$ is $f(k + 1)$, so (3) holds. Also, $f$ is injective at this point (as $a = k + 1$ was the only value of $a$), so (2) holds. This completes the induction, so the only solution is $f(n) = n$

We claim that the only solution to this functional equation is $f \equiv n$.

Lemma: For all $n \in \mathbb{Z}_{>0}$, $f(n) \ge n$.

We will show that for all $k \ge 0$, $n > k$ implies $f(n) > k$, which implies the desired result for $k = n-1$. To show this, we will induct on $k$.

Base Case: For $k=0$, this is trivial.

Induction Step: Given that $n > k$ implies $f(n) > k$, we will show that $n > k+1$ implies $f(n) > k + 1$. Suppose that $n > k+1$. Since $n-1 > k$, we see that $f(n-1) > k$, and so $f(f(n-1)) > k$. By the given equation, we have that $f(n) > f(f(n-1)) > k$. Since $f$ is a function on the integers and $k$ is an integer, we see that $f(n) \ge f(f(n-1) + 1 \ge k + 2$. Equivalently, $f(n) > k+1$, which is the desired result. $\blacksquare$

Now, we will show that $f$ is monotonically increasing. Suppose this is not the case, that is, suppose that $f(k) \ge f(k+1) > f(f(k))$ for some $k$. By our lemma, we then have that $f(k) > f(f(k)) \ge f(k)$, which is absurd. Thus, we must have that $f(k) < f(k+1)$ for all $k$, so $f$ is monotonically increasing.

Since $f$ is monotonically increasing, the given equation implies that $n+1 > f(n)$ for all $n$. However, $f(n) \ge n$ for all $n$ by our lemma. Thus, $f(n)$ must be $n$ for all $n$, as $f$ is defined on the integers.

for reference

[hide = solutoin]We will first show that $f(n) \geq n$.

$\textbf{Claim:}$ For all defined $m \geq n$, we have $f(m) \geq n$.
$\textbf{Proof:}$ We will do this with induction on $n$. The base case of $n = 1$ is trivial, as $f(m) \geq 1$ for all defined $m$ since the range of $f$ is $\mathbb{Z}^{+}$. Now, suppose that for some positive integer $k \geq 2$, we have that for all $m \geq k$, $f(m) \geq k$. If $m \geq k+1$, then $m-1 \geq k$, so thus $f(f(m-1)) \geq f(m-1) \geq m-1 \geq k$. Then, by the problem statement, $f(m) > f(f(m-1)) \geq k$, so $f(m) \geq k+1$, completing our inductive step and therefore our induction. $\square$

By our claim, we note that $n-1 \geq n-1 \implies f(f(n-1)) \geq n-1$ and by the problem, $f(n) \geq f(f(n-1))$ so $f(n) > n - 1 \implies f(n) \geq n$, as desired.

Now we show that $f(n) < n+1$. First, note that $f(n+1) > f(f(n)) \geq f(n)$ by our claim, so $f$ is strictly increasing. That is, for all $x \geq y$, $f(x) \geq f(y)$. FTSoC, assume $f(n) \geq n+1$ for some $n$. Then, we get$$f(n+1) > f(f(n)) \geq f(n+1)$$by $f$ strictly increasing, a contradiction, completing our proof.

Since $n+1 > f(n) \geq n$ for all defined $n$, $f(n)$ must be $n$ for all defined $n$, and we are done. $\blacksquare$[/hide]

Clearly $f(n)\ge 1$ for all $n\ge 1$. Thus $f(n+1)>f(f(n))\ge 1$ for all $n\ge 1$, so $f(n)\ge 2$ for all $n\ge 2$. So for all $n\ge 2$, $f(n+1)>f(f(n))\ge 2$ (since $f(n)\ge 2$), meaning $f(n)\ge 3$ for all $n\ge 3$. We can keep on going and use induction to show $f(k)\ge k$ for all $n\ge k$. We also have $f(n)\ge n$ for all $n$. Hence, $f(n+1)>f(f(n))\ge f(n)$ for all $n$, meaning $f$ is increasing. Thus, we can "unwrap" $f(n+1)>f(f(n))$ to get $f(n)<n+1$, meaning $f(n)=n$ for all $n$, as desired.

[url=https://dgrozev.wordpress.com/2020/03/22/a-functional-equation-imo-1977-problem-6/]IMO 1977, problem 6[/url]

The solution idea is a weird induct, where we want to prove $f(n) \geq n$. Evidently this holds for $f(1)$. Now for $f(2)$, we can write the inequality $f(2) > f(f(1))$. Since $f$ maps to positive integers, $f(2) > 1$ or equivalently $f(2) \geq 2$. Now for the general case:

Assume that $i = 1, 2, \cdots, k - 1$ all satisfy $f(i) \geq i$. Write the inequality $f(k) > f(f(k - 1))$, but by plugging in $n = f(k - 1) - 1$ which is obviously $\geq k - 2$, we get $f(f(k - 1)) > f(f(f(k - 1) - 1))$. The idea is now clear; we continue like this, with the lower bound decreasing by $1$ every single time, and we are done.

Now, use $f(x) \geq x$ on the original statement in order to get $f(n + 1) > f(f(n)) \geq f(n)$, or $f(n + 1) > f(n)$, hence $f$ is strictly increasing.

Next, assume that $f(k) > k$ for some value of $k$. Then, plugging in $n = k$, we find that $f(k + 1) > f(f(k)) = f(\text{something geq k + 1})$, which is obviously bad. Hence we are done.

[I'm not sure if I did an IMO problem, check it please!]

Claim 1: $f(n) \ge n$.
Proof 1: $f(n)>f(f(n-1))=a_{n-1}>f(f(f(n-1)-1)))=a_{n-2}>...>f(f(f(f(...f(1)-n+1))))=a_1$.
By Induction on $n$, $f(n)\ge n$.

Claim 2: $f(n)>f(n-1)>...>f(2)>f(1)$.
Proof 2: $$f(n)>f(f(n-1))\ge f(n-1)$$. Hence the claim is true.

Claim 3: $f(n)=n \forall n$.
Proof 3:
If $f(n)=n+k$ for some $k\ge 0$. Then, $$2k+n+1\ge f(f(n-1))$$.
Since $2k+n+1$ is a possible value it is always less than $f(n)=n+k \implies k<1 \implies k=0 \implies f(n)=n \forall n$.

Let $P(n)$ be the given assertion.
We first prove by induction that $f(x)\ge x$. For $1$ it is obvious, and the same is true of $2$ since $P(1)\Rightarrow f(2)>f(f(1))\ge1$.
Assume that $f(n)\ge n$ for some $n\in\mathbb N$. The induction step folows.
Let $a_1=n$ and $a_{i+1}=f(a_i)-1$ for all $1\le i\le n$.

First we prove that $\bigcup_{i=1}^n\{a_i\}\subseteq\mathbb N$.
Since $a_i\ge0$, it suffices to consider the case where some $j$ satisfies $a_j=0$. Simple induction gives $f(a_{j-t})=t$ for $t<j$, so $f(n)=j-1$. This implies $j\ge n+1$, hence the result.

Then:
$$P(a_1),P(a_2),\ldots,P(a_n)\Rightarrow f(a_1+1)>f(a_2+1)>\ldots>f(a_n+1).$$
Since there are $n-1$ "$>$" signs, we obtain $f(a_1+1)>n$, or $f(n+1)\ge n+1$.

This proves that $f(x)\ge x$.
$P(n)\Rightarrow f(n+1)>f(f(n))\ge f(n)$, so $f$ is strictly increasing. Then the original assertion implies $f(n)<n+1$, so $\boxed{f(n)=n}$ for each $n$.

Here's a wacky solution that I think is really cool.
Suppose $f$ works, and consider the function $g(x)=f(x+1)-1$. Observe that $g: \mathbb{N} \to \mathbb{N}$, since if $f(x+1)=1$ for $x \in \mathbb{N}$, we have $1>f(f(x))$ which is absurd. Substituting $g$ into our inequality, we obtain $g(x)+1>g(g(x-1))+1 \implies g(x)>g(g(x-1))$ for $x\geq 2$, which is equivalent to $g(x+1)>g(g(x))$ for $x \in \mathbb{N}$. Hence $g$ also satisfies the inequality.
From this, induction gives that $g_k(x)=f(x+k)-k$ is a solution to the inequality for all positive integers $k$. Since $g_k$ is $\mathbb{N} \to \mathbb{N}$, it follows that $f(1+k)-k\geq 1 \implies f(x) \geq x$ for $x\geq 2$. Since it's obvious that $f(1) \geq 1$, it follows that $f(x)\geq x$ for all $x \in \mathbb{N}$. Hence $f(n+1)>f(f(n)) \implies f(n+1)>f(n)$, so $f$ is strictly increasing. Thus $f(n+1)>f(f(n))$ also implies $n+1>f(n) \implies f(x) \leq x$ for all $x \in \mathbb{N}$, so $f(x)=x$ for all $x$, which indeed works. $\blacksquare$

[s]I finally solved a P6[/s]
[b]Claim:[/b] $f$ is strictly increasing
[i]Proof.[/i] We will start by proving that $f(1)$ is the least value $f$ can take, and that there does not exist $x$ such that $f(x)=f(1)$. Suppose not. Let $f(k+1)$ be the minimum value $f$ can take for some $k \geq 1$. Then note that $f(k+1)>f(f(k))$. This is a contradiction. Now I will prove that $f(2)$ is the minimum value $f$ can take for $\mathbb{N}-\{1\}$. Suppose not. Let $f(k+1)$ be the minimum for $k \geq 2$. Then note that $f(k+1)>f(f(k))$. But $f(k) >1$ since $k \geq 2$. So this is a contradiction. Continuing henceforth, we see that $f$ is increasing $\blacksquare$.
--------
Now we clearly have $n+1 > f(n)$. We also have $f(n)>f(n-1)> \dots > f(1)$. Therefore $f(n+1) \geq f(1)+n-1=1+n-1=n$ since $2>f(1) \implies f(1)=1$. $n+1 > f(n) \geq n$ clearly implies $f(n)=n$ which works too $\blacksquare$

[hide=Solution]
We will prove by induction that $f(k)$ is the $k$th smallest number in the range of $f$.

Base Case: $k=1$
Suppose that $f(a)$ is the smallest number in the range of $f$. If $a\geq2$, then we have $f(a)>f(f(a-1))$, which is a contradiction since $f(a)$ is the smallest number in the range of $f$. Therefore, $a=1$, so $f(1)$ is the smallest number.

Inductive Step:
Suppose that $f(a)$ is the $k$th smallest number in the range of $f$. Then, $f(a)>f(f(a-1))$ since $a\neq1$. By the inductive hypothesis, $f(a-1)$ must be at most $k-1$, which also implies that $a-1\leq k-1$, or $a\leq k$. Since $f(x)$ is the $x$th smallest number for $x\leq k-1$, we must have $a=k$.

Therefore, $f$ is strictly increasing, so $f(n+1)>f(f(n))$ implies $n+1>f(n)$, or $n\geq f(n)$. However, since $n\geq f(n)\geq f(1)+n-1$, we must have $n\geq f(1)+n-1$. The only possibility is $f(1)=1$. Then, $n\geq f(n)\geq n$, so $f(n)=n$ for all $n$.

I have my solution to this problem [url=https://artofproblemsolving.com/community/c6h2831507p25056981]here[/url].

Let $m_1,m_2, \ldots$ be a permutation of $1,2, \ldots$ such that $f(m_1) \leq f(m_2) \leq \cdots$. We will prove that $f(k)=k$ for every positive integer $k$, we will prove this by induction.

[b]Claim 1:[/b] $f(1)=1$

Recall that $f(m_1)$ is minimal. First, we will show that $m_1=1$. Assume for the sake of contradiction that $m_1>1$. We have $f(m_1) > f(f(m_1-1))$ which contradicts minimality. So, $f(1)$ is minimal. Recall $f(m_2)$ is second to minimal. So, \[ f(m_2) > f(f(m_2-1)) \implies f(m_2-1) = 1 \] so, $1$ is in the image of $f$ which must be minimal. Implying that $f(1)=1$ $\square$

Notice that this also implies that $m_2-1=1 \implies m_2 = 2$

Now assume that $f(i)=i$ for all $i \leq k$. We will show that $f(k+1)=k+1$. Notice that $f(m_{k+1}) > f(f(m_{k+1}-1))$ so it must be the case that $f(m_{k+1}-1)  \in \{1,2, \ldots, k \}$. This implies that $f(f(m_{k+1}-1)) = f(m_{k+1}-1)$. So, $f(m_{k+1})>f(m_{k+1}-1)$. Thus, it must be the case that $m_{k+1}-1 = m_\ell$ for some $\ell \leq k$. Assume $\ell < k$. Then $m_{k+1}=m_\ell + 1 = m_t$ for some $t \leq k$. Thus, $f(m_{k+1}) = f(m_t) = t$ which contradicts the fact that $f(m_{k+1})$ is the $k+1$'st smallest element in the image of $f$. So, $m_{k+1}-1=m_k$. This pretty easily implies $f(k+1)=k+1$. Completing induction. $\blacksquare$

Nice problem. :)

We are given $f:\mathbb{N}\rightarrow\mathbb{N}$ such that $f\left(n+1\right)>f\left(f\left(n\right)\right)$. Clearly, $f\left(n\right)\neq n+1$.

[color=#f00]Claim 1:[/color] if 1 has a pre-image, then it must be 1.
[color=#00f]Proof.[/color] If there exists $n>1$ such that $f\left(n\right)=1$, then $1=f\left(n\right)>f\left(f\left(n-1\right)\right)$, which is a contradiction.
So, for $n>1$ then $f\left(n\right)>1$.

[color=#f00]Claim 2:[/color] if 2 has a pre-image, then it must be 2.
[color=#00f]Proof.[/color] Note that $f\left(1\right)\neq2$. Suppose $n>2$ exists, such that $f\left(n\right)=2\Rightarrow2=f\left(n\right)>f\left(f\left(n-1\right)\right)$, forcing $f\left(f\left(n-1\right)\right)=1\Rightarrow f\left(n-1\right)=1\Rightarrow n-1=1\Rightarrow n=2$, a contradiction.

[color=#f00]Claim 3:[/color] $f\left(n\right)\geq n$ for all positive integers $n$.
[color=#00f]Proof.[/color] Let $k=f\left(n\right)$. Using the previous claims as base case/s, the inductive step is as follows: suppose there exists a pre-image of $k$ which is $m>k$. So, $f\left(m\right)=k>f\left(f\left(m-1\right)\right)\Rightarrow f\left(f\left(m-1\right)\right)\le k-1\Rightarrow m=1+f\left(f\left(m-1\right)\right)\le k$, a contradiction. So, pre-image/s of every number is/are less than that number. Hence, $f\left(n\right)\geq n$.

[color=#f00]Claim 4:[/color] if $n$ has a pre-image, then it must be $n$.
[color=#00f]Proof.[/color] Suppose there exists $k\neq n$ such that $f\left(k\right)=n$. Note that $n=f\left(k\right)\geq k$, and $k=n-1$ is not possible since $f\left(n\right)=f\left(k+1\right)>f\left(f\left(k\right)\right)=f\left(n\right)$, a contradiction. So, $k\le n-2$. Suppose $d=n-k\geq2$. Now,
\[f\left(n\right)=f\left(f\left(k\right)\right)<f\left(k+1\right)\Rightarrow f\left(k+2\right)>f\left(f\left(k+1\right)\right)\geq f\left(k+1\right)>f\left(n\right)\]
\[\Rightarrow f\left(k+3\right)>f\left(f\left(k+2\right)\right)\geq f\left(k+2\right)>f\left(n\right)\]
\[\cdots\]
\[f\left(n\right)=f\left(k+d\right)>f\left(f\left(k+d-1\right)\right)\geq f\left(k+d-1\right)>f\left(n\right),\]
which is a contradiction. So, the pre-image of $n$ must be $n$.

As a result, the function must be injective. Moreover, if $n$ does not have a pre-image, then the unique pre-image of $f\left(n\right)$ must be $f\left(n\right)\neq n$, a contradiction. So, the function is surjective as well. Hence, the function is bijective. So, it is the identity function $\boxed{f(n)=n}$.

First by induction, $f(k) \geq n$ for all $k\geq n$. The base case $n = 1$ is trivial. For the inductive step,
\[ f(n+1) > f(f(n)) \geq f(k) \geq k \]
So $f(n+1) \geq k+1$ when $n+1 \geq k+1$.

Now, $f(n)$ is increasing. Assume FTSOC that $f(n) \geq f(n+1)$. Then
\[ f(n) \geq f(n+1) > f(f(n)) \]
which is a contradiction.

Finally, note that if $f(n) > n$, then
\[ f(n+1) > f(f(n)) \geq f(n+1) \]
a contradiction. Thus, it must be that $f(n) = n$. $\blacksquare$

[hide="Nice problem!"]
We will show that for all $k \ge 0$, $n > k$ implies $f(n) > k$, which implies the desired result for $k = n-1$. To show this, we will induct on $k$.

Base Case: For $k=0$, this is trivial.

Induction Step: Given that $n > k$ implies $f(n) > k$, we will show that $n > k+1$ implies $f(n) > k + 1$. Suppose that $n > k+1$. Since $n-1 > k$, we see that $f(n-1) > k$, and so $f(f(n-1)) > k$. By the given equation, we have that $f(n) > f(f(n-1)) > k$. Since $f$ is a function on the integers and $k$ is an integer, we see that $f(n) \ge f(f(n-1) + 1 \ge k + 2$. Equivalently, $f(n) > k+1$, which is the desired result. $\blacksquare$

Now, we will show that $f$ is monotonically increasing. Suppose this is not the case, that is, suppose that $f(k) \ge f(k+1) > f(f(k))$ for some $k$. By our lemma, we then have that $f(k) > f(f(k)) \ge f(k)$, which is absurd. Thus, we must have that $f(k) < f(k+1)$ for all $k$, so $f$ is monotonically increasing.

Since $f$ is monotonically increasing, the given equation implies that $n+1 > f(n)$ for all $n$. However, $f(n) \ge n$ for all thus, $f(n)$ must be $n$ for all $n$, as $f$ is defined in $\mathbb{N}$.

$$\mathbb{Q.E.D.}$$
[/hide]

Define the sequences $\{a_n\}$ and $\{b_n\}$ such that $a_1=t$, $a_n = f(a_{n-1}-1) $, and $b_n = f(a_n)$. Then
[list]
[*] $\{b_n\}$ is strictly decreasing, so it must terminate.
[*] Thus $\{a_n\}$ must also terminate, so 1 must be in this sequence, implying 1 is in the range of $f$.
[*] Note that if $r>1$ and $f(r)$ is the lowest value in the range of $f$, $f(r) > f(f(r-1))$ gives a contradiction. Hence $r=1$, and is the only value such that $f(r)=1$.
[/list]

We continue with induction. Suppose for all $j \leq k$, we have $f(x)=j \iff x=j$. Then there must exist some integer $m$ such that
\[a_m=1, a_{m-1}=2, \ldots, a_{m-k+1}=k.\]

The value of $a_{m-k+1}$ tells us $a_{m-k}=k+1$, so $k+1$ is in the range of $f$. Assume again by contradiction the lowest value in the range of $f$ greater than $k$ is $f(r)$, where $r>k+1$. Then
\[f(r-1) > k, f(r) > f(f(r-1)) \implies f(r) > k+1,\]

contradiction. Thus $f(x)=k+1 \iff x=k+1$, and we're done. $\blacksquare$

We uploaded our solution [url]https://calimath.org/pdf/IMO1977-6.pdf[/url] on youtube [url]https://youtu.be/4IQ4c92USgo[/url].

Years later this problem still makes me laugh...

I will show that $f(n) \geq k$ for all $n \geq k$ by induction on $k$. In particular, the result for $k=1$ is vacuous.

Now, observe that for any $n \geq k$, we have $f(n+1) > f(f(n)) = f(\geq k) \geq k$ hence $f(n+1) \geq k+1$, which completes the induction.

In particular, this implies that $f(n) \geq n$ for each $n$. Then, let $r$ be the minimum of $f(n)-n$ across all $n$, and suppose $f(k) - k = r$ for some $k$. It follows that $$r+k = f(k) > f(f(k-1)) = f(\geq k-1+r) \geq k-1+2r.$$ This forces $r = 0$, so $f \equiv n$.


Prove that $ (a_n)_{n\ge 1} , a_n > 0$ is arithmetic progression $ \Longleftrightarrow (\forall)n\ge 2\ ,\ \sum_{k \equal{} 2}^n\frac {1}{\sqrt {a_{k \minus{} 1}} \plus{} \sqrt {a_k}} \equal{} \frac {n \minus{} 1}{\sqrt {a_1} \minus{} \sqrt {a_n}}$

Should this be in high school basics??

-rts2007


inequalities
trigonometry
Euler
inequalities proposed

$ \odot\ \blacktriangleright\ \frac pr\ \le\ \tan A \plus{} \tan B \plus{} \tan C\ \le\ \frac pr\cdot \frac {1}{8\cos A\cos B\cos C}\ .$

$ \odot\ \blacktriangleright\ \tan B\tan C \plus{} \tan C\tan A \plus{} \tan A\tan B\ \ge\ \left(\frac {R \plus{} r}{r}\right)^2\ .$

[quote="Virgil Nicula"]$ \odot\ \blacktriangleright\ \frac pr\ \le\ \tan A \plus{} \tan B \plus{} \tan C\ \le\ \frac pr\cdot \frac {1}{8\cos A\cos B\cos C}\ .$ [/quote]
WE have: $ \frac{p}{r}\equal{}cotg\frac{A}{2}cotg\frac{B}{2}cotg\frac{C}{2}  \equal{}\sum \ cotg\frac{A}{2}$
But $ \sum \ cotg\frac{A}{2} \le \sum \ tan\frac{A}{2}$ (since $ tanA\plus{}tanB \ge  2 cotg\frac{C}{2}$)
 $ tanA\plus{}tanB\plus{}tanC\equal{}tanAtanBtanC \equal{}\frac{sinA.sinB.sinC}{cosAcosBcosC} \le 3\sqrt{3} \frac{1}{8cosAcosBcosC} \le \frac{p}{r} \frac{1}{8cosAcosBcosC}$
Then,we have Q,.E.D :)

[b][quote="Virgil Nicula"]$ \odot\ \blacktriangleright\ \frac pr\ \le\ \tan A \plus{} \tan B \plus{} \tan C\ \le\ \frac pr\cdot \frac {1}{8\cos A\cos B\cos C}\ .$

[/quote]

We have :

$ \tan A \plus{} \tan B \plus{} \tan C \equal{} \frac{2pr}{p^2\minus{}(2R\plus{}r)^2}$

And

$ cosA.cosB.cosC\equal{}\frac{p^2\minus{}(2R\plus{}r)^2}{
4R^2}$

 *$ \frac pr\ \le\ \tan A \plus{} \tan B \plus{} \tan C \Leftrightarrow 4R^2\plus{}4Rr\plus{}3r^2 \geq p^2$

It's Gerretsen 's inequality ..... :)

* $ tan A \plus{} \tan B \plus{} \tan C\ \le\ \frac pr\cdot \frac {1}{8\cos A\cos B\cos C}\Leftrightarrow 2r \leq R$

It's Chapple-Euler 's inequality :)

[quote="Virgil Nicula"]

$ \odot\ \blacktriangleright\ \tan B\tan C \plus{} \tan C\tan A \plus{} \tan A\tan B\ \ge\ \left(\frac {R \plus{} r}{r}\right)^2\ .$[/quote]

I think it's wrong  :maybe: ...... Try $ a\equal{}0,95 ; b\equal{}0,95;c\equal{}1$

Am i right ???? :D[/b]

[quote="Virgil Nicula"] $ \odot\ \blacktriangleright\ \tan B\tan C \plus{} \tan C\tan A \plus{} \tan A\tan B\ \ge\ \left(\frac {R \plus{} r}{r}\right)^2\ .$[/quote]

[b][u]Proof[/u].[/b] $ \sum \tan B\tan C \equal{} \prod\tan A\cdot\sum \cot A \equal{}$ $ \sum\tan A\cdot\sum\cot A\ge\frac pr\cdot\sum\frac {\cos A}{\sin A} \equal{}$ $ \frac {2p}{r}\cdot\sum\frac {\cos^2 A}{\sin 2A}\stackrel{C.B.S.}{\ \ \ge\ \ }$

$ \frac {2p}{r}\cdot\frac {\left(\sum\cos A\right)^2}{\sum \sin 2A} \equal{}$ $ \frac {2p}{r}\cdot\frac {\left(\frac {R \plus{} r}{R}\right)^2}{4\prod\sin A} \equal{}$ $ \frac {p\cdot(R \plus{} r)^2}{r\cdot 2R^2\prod \sin A} \equal{}$ $ \frac {p\cdot (R \plus{} r)^2}{r\cdot S} \equal{} \frac {p(R \plus{} r)^2}{pr^2} \equal{} \left(\frac {R \plus{} r}{r}\right)^2$ .


function
inequalities
algebra
polynomial
algebra unsolved

[quote="aodeath"]Find all functions

[tex]\displaystyle f:\mathbb{R}\rightarrow \mathbb{R}[/tex]

such that

[tex]\displaystyle f(x+f(x))+y-f(x)=f(f(x+y))[/tex][/quote]
put \[x=y=0 \rightarrow f(0+f(0))+0=f(0)+f(f(0)) \rightarrow
f(0)=0\]
then put \[x=0 \rightarrow f(0+f(0))+y=f(0)+f(f(0+y)) \rightarrow
f(0)+y=0+f(f(y)) \rightarrow f(f(y))=y  (1)\]
\[ \rightarrow f(x+f(x))+y=f(x)+f(f(x+y)) \rightarrow f(x+f(x))+y=f(x)+x+y \rightarrow f(x+f(x))=x+f(x) \rightarrow f(x)=x\]
and we done!!..
 :D

You surely need the continuity of f.

First note that f(0)=0 (i) (by taking x=0 in your equation).

Also, taking y = 0:

f(x+f(x)) - f(x) = f(f(x)) (ii)

Now taking y = - x,

f(x+f(x)) -x - f(x) = f(f(0)) = 0 (iii)

(ii) and (iii) give f(f(x)) = x (iv)

Now you can show that your original equation is actually equivalent to (ii) and (iv) (trivial).

Assuming the continuity of f we have its monotonicity.

(1) Let's assume f is increasing: you can then pick a neighborhood V of 0 such that g(V) - where g is the continuous function g(x) = x+f(x) - contains an open interval ]-e,e[ where e>0 . But then for any x in R, you can find an y in ]-e,e[ such that x/2^n = y = x'+f(x'), with n a positive integer. But then f(y) = y and f(2^n*y) = 2^n*y by (ii), meaning f(x) = x.

(2) Now remains the case where f is decreasing. You may start by stating the same than (1) except that your interval may be of the form [0,e[ or ]-e,0] for some e>0 ( because you don't know the sign of g anymore when f is decreasing ). But still in that case you have f(x) = x for some x in R ( be it x>0 or x<0 ) and f(0)=0, which is absurd since f is decreasing, whence (2) never happens.

Conclusion: f(x) = x.

If you remove the continuity hypothesis, you have many other ugly solutions, for instance f(2) = 3, f(3) = 2, and f(x) = x elsewhere. Such solutions can be characterized but you'll never get anything nice.

--
Julien Santini

I can't understand you because i think my solution is clear and
$f(x)=x$ is the answer of this question but i don't know what is your problem(because i think in you message you get my answer and it is absolutly correct)

Jensen, your solution is not quite allright because you haven't proved that $x+f(x)$ takes all real values.

we prove that $f(f(x))=x$
suppose that
$f(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_0$
$ \rightarrow$ $f(a_nx^n+...+a_0)=x$
$ \rightarrow $ $a_n(a_nx^n+...+a_0)^n+...+a_1(a_nx^n+...+a_0)+a_0=x$
$ \rightarrow$ we supposed that $a_n$ is not $0$
 $\rightarrow n=1$(?)
$\rightarrow  a_1(a_1x+a_0)+a_0=x$
$ \rightarrow  (a_1)^2x+a_1 \times a_0=x$
$ \rightarrow a_1\times a_0=0$
$ \rightarrow a_0=0$(because $a_1$ can not be $0$)
and $ \rightarrow (a_1)^2x=x \rightarrow a_1=1 \rightarrow$
$ f(x)=x$ and now we r done. :D

But now you're making an even bigger assumption - i.e. that the solution is a polynomial.

i am sorry i saw another $f(x)$ problem that $f(x)$ was polinomial and ithink this one is too.ok i will proove it when it is not polinomilad.

[quote="aodeath"]
[tex]\displaystyle f(x+f(x))+y-f(x)=f(f(x+y))[/tex][/quote]
This equation can be also satisfied wit f(x)=-x:
f(x-x)+y+x=f(-x-y)
x+y=x+y.

first what did you coclude with that and then are you sure you can put $f(x)=-x$ and after it are you sure you will get that?! :?:

Well, i was trying to find mistakes in my calculations but please try the function f(x)=-x yourself.
Jensen: In your first proof u said f(x)=x because f(x+f(x))=x+f(x), but f(x)=-x satisfies this too, coz f(0)=0.  And in the second proof u said
$a _1^2=1$ implies that  $a _1=1$ but it can also be $a _1=-1$. I think f(x)=-x satisfies all conditions of the task.


let a,b,c>0  prove that:
a^4.c^4+b^8+c^8>_a^3.b^2.c^3+b^6.c^2+c^7.a                                                                                MY LIFE IS MYFRIENDS

[quote="myfriendly"]Let $ a,b,c > 0$  prove that:
$ a^4c^4 \plus{} b^8 \plus{} c^8 \geq a^3b^2c^3 \plus{} b^6c^2 \plus{} c^7a$                                                                                MY LIFE IS MYFRIENDS[/quote]
I've edited.

[hide=" A solution"]This problem is nice indeed.

Let $ \sqrt{ac} \equal{} u$

It turns out to prove that

$ u^8 \plus{} b^8 \plus{} c^8 \ge b^2u^6\plus{} b^6c^2\plus{} c^6u^2$

We are done![/hide]


The three roots of the cubic $ 30 x^3 \minus{} 50x^2 \plus{} 22x \minus{} 1$ are distinct real numbers between $ 0$ and $ 1$.  For every nonnegative integer $ n$, let $ s_n$ be the sum of the $ n$th powers of these three roots.  What is the value of the infinite series
\[ s_0 \plus{} s_1 \plus{} s_2 \plus{} s_3 \plus{} \dots \, ?\]

[hide]If the roots are $ a,b,c$, then the infinite series is equal to $ \frac{1}{1\minus{}a}\plus{}\frac{1}{1\minus{}b}\plus{}\frac{1}{1\minus{}c}$. If we let $ P(x)\equal{}30x^3\minus{}50x^2\plus{}22x\minus{}1$, then the polynomial $ Q(x)\equal{}P(x\plus{}1)\equal{}30(x\plus{}1)^3\minus{}50(x\plus{}1)^2\plus{}22(x\plus{}1)\minus{}1$ has roots $ a\minus{}1,b\minus{}1,c\minus{}1$. We expand and combine like terms to obtain $ Q(x)\equal{}30x^3\plus{}40x^2\plus{}12x\plus{}1$. The sum of the reciprocals of the roots of this polynomial is $ \minus{}\frac{12}{1}$, so $ \frac{1}{a\minus{}1}\plus{}\frac{1}{b\minus{}1}\plus{}\frac{1}{c\minus{}1}\equal{}\minus{}12$. Our answer is then this multiplied by -1, or $ \boxed{12}$.[/hide]

[hide=sad solution]Notice that if we add up all these equations up to infinity, we have

$$\sum_{k=4}^{\infty} s_k = 29s_3-21s_2+s_1$$

Therefore adding on $s_0,s_1,s_2,s_3$ and finding them via newton sums we get $12$ as the answer[/hide]

Very nice.

[hide=Solution]Let the three roots be $a,b,c$. Then we want $(a+b+c)+(a^2+b^2)+(a^3+b^3+c^3)+\ldots$, which is equivalent to $\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}$. This is $\frac{(1-a)(1-b)+(1-a)(1-c)+(1-b)(1-c)}{(1-a)(1-b)(1-c)}$, which by Vieta evaluates to $\boxed{12}$.[/hide]


Let a set $S$ of 2004 points in the plane be given, no three of which are collinear. Let ${\cal L}$ denote the set of all lines (extended indefinitely in both directions) determined by pairs of points from the set. Show that it is possible to colour the points of $S$ with at most two colours, such that for any points $p,q$ of $S$, the number of lines in ${\cal L}$ which separate $p$ from $q$ is odd if and only if $p$ and $q$ have the same colour.

Note: A line $\ell$ separates two points $p$ and $q$ if $p$ and $q$ lie on opposite sides of $\ell$ with neither point on $\ell$.

Neat! :) It's true for every even $n$ instead of $2004$.

Consider the graph $G$ on vertex set $S$ in which two vertices are connected iff they are separated by an even number of lines. We want to prove that $G$ is bipartite, which is the same as showing that it has no odd cycles.

Assume now that $x_1,\ldots,x_{2k+1},x_1$ is an odd cycle in $G$. This means that for all $i,\ x_i,x_{i+1}$ are separated by an even number of lines (here $x_{2k+2}=x_1$). Let's check the parity of the number of pairs $(e,\ell)$, where $e$ is an edge $(x_i,x_{i+1})$, and $\ell\in\mathcal L$ separates $x_i,x_{i+1}$. On the one hand, this number is even because by assumption, for each $e=(x_i,x_{i+1})$ there is an even number of $\ell$ which separate $x_i,x_{i+1}$. On the other hand, each line which does not pass through any of the $x_j$ cuts an even number of edges $(x_i,x_{i+1})$, as does each line $x_u,x_v$, and each line which passes through exactly one of the $x_j$ cuts exactly one edge. There's an odd number of such lines, since there's an odd number of $x_j$'s, and for each $x_j$ there is exactly one such line for every point in $S\setminus\{x_i\}$, which has odd cardinality. Adding all of these up, we see that our second count gives an odd number of pairs $(e,\ell)$ as above, contradicting the first count, which said that this number had to be even.

grobber always comes up with so much nice stuff :)

[quote="grobber"]Neat! :) It's true for every even $n$ instead of $2004$.

Consider the graph $G$ on vertex set $S$ in which two vertices are connected iff they are separated by an even number of lines. We want to prove that $G$ is bipartite.[/quote]

And so if $p$ and $q$ have the same colour, then the number of lines which separate $p$ from $q$ is odd.
But how about "only if"?

Call a pair of vertices [i]good[/i] if they have the same color with odd number of lines separating them OR they have different colors with even number of lines separating them.

First, color $P_1$ red. Then, color each other $P_i$ red or blue such that $(P_1,P_i)$ is good.

[b]Lemma:[/b] If $(P,Q)$ and $(P,R)$ are good, then $(Q,R)$ is good.
[b]Proof:[/b] Let $d(PQ)$ be the number of lines separating $P$ and $Q$, i.e. cutting through segment $\overline{PQ}$.

[b]Claim 1:[/b] $d(PQ)+d(PR)+d(QR)$ is odd.
[b]Proof of Claim 1:[/b] Consider two types of lines.
[u]Type 1:[/u] Line joining two points that are not $P,Q,R$.
If it cut through $\triangle PQR$, it must cut at exactly two points. So, in total they cut through $\triangle PQR$ for even times.

[u]Type 2:[/u] Three lines joining an other point $X$ with $P,Q$, and $R$.
If $X$ lies outside $\triangle PQR$, then exactly one of these three lines cut $\triangle PQR$ at exactly one point.
If $X$ lies inside $\triangle PQR$, then each of these three lines cut $\triangle PQR$ at exactly one point, so in total at three points.
Since there are 2001 other points $X$, in total they cut through $\triangle PQR$ for odd times.

Therefore, $d(PQ)+d(PR)+d(QR)$ is odd.

Back to the lemma, WLOG $P$ is red.
[u]Case 1:[/u] $Q$ and $R$ are red --- Then, $d(PQ)$ and $d(PR)$ are odd, so $d(QR)$ is odd, so $(Q,R)$ is good.
[u]Case 2:[/u] $Q$ and $R$ are blue --- Then, $d(PQ)$ and $d(PR)$ are even, so $d(QR)$ is odd, so $(Q,R)$ is good.
[u]Case 3:[/u] $Q$ is red, $R$ is blue --- Then, $d(PQ)$ is odd, $d(PR)$ is even, so $d(QR)$ is even, so $(Q,R)$ is good.
[u]Case 4:[/u] $Q$ is blue, $R$ is red --- Similar to Case 3.

From the lemma, it follows that every pair of vertices is good.

Is @grobber's solution really correct? It is not really clear, if the cycle is not geometrically convex, why a line through no points of the cycle $C$ cuts an even number of edges in $C$. And even if the cycle is geometrically convex, it is very possible that a line through a vertex of $C$ does not pass through the geometric interior of $C$ and hence does not separate any edges (contrary to "separating exactly one", as @grobber claims).

Nevertheless, an odd cycle solution can be built from @carefully's claim 1 - proving it shows that the graph has no triangle and then if one considers a smallest odd cycle (supposing such exists), by taking an edge $v_1v_2$ and a third vertex $v_3$ not adjacent to $v_1$ or $v_2$ the total number of cutting lines between $v_1$ and $v_2$, $v_2$ and $v_3$, and $v_3$ and $v_1$, must be even by the graph relation, whilst it is odd by @carefully's claim. This gives a contradiction and we are done.


let $p ,n$ be integers $>0$ , and $f$ be function defined by :$f(x)=x^{3}-3x+3$ and $(x_{n})$ be sequence such that :$x_{n+1}=x_{n}-\frac{f(x_{n})}{f'(x_{n})}, x_{0}=a$
prove that for every $p$ there exist $a$ such that $(x_{n})$ is a periodic sequence.


Hey,
Here's a riddle I'm having trouble solving:
[i]There are N points on a plane. Find the two points that are closest, in time better than O(N^2).[/i]

Any idea?
Thanks :)

Hm, yes, this is a well-known problem. We were given this problem once and I didn't like the solution one bit. I still don't  :) . So just put it into google and you may find something like this:

http://www.cs.brown.edu/courses/cs016/book/slides/ClosestPoints2x2.pdf

However, if you don't want to go straight ahead and look at the answer, you can try to find a solution yourself:

[hide="Hint 1"]It can be done in N x log N. So how about quicksort?[/hide]
[hide="Hint 2"]Divide and conquer.[/hide]
[hide="Hint 3"]If you've already found 2 points that are $d$ from each other, you now only need to check pairs that are closer.[/hide]


two identical jars are filled with a solution of alcohol and water. the ratio of alcohol to water in the first jar is p:1, in the second, q:1. the contents of the two jars are combined. what is the ratio of alcohol to water in this mixture?

first of all, this is in ch 29 of aops1. second, the solution says that the amount of alcohol in the first jar is pV/(p+1), and amount of water is V/(p+1), if V is the volume, and likewise for the second jar. how did they get that?

Well since the ratio is p:1, that means that there is p amount of alcohol for every p+1 amount of solution, hence the fraction p/(p+1). Since the volume is V, we just multiply to find pV/(p+1). Same deal with the water.


[quote="spx2"]$a,b,c \in R$ with $a,b >0$
proove that:
$\sqrt{a^{2}+1}+\sqrt{b^{2}+4}+\sqrt{c^{2}+9}\geq \frac{a+b+c+6}{\sqrt{2}}$[/quote]

$\Longleftrightarrow \sqrt{\frac{a^{2}+1}{2}}+\sqrt{\frac{b^{2}+4}{2}}+\sqrt{\frac{c^{2}+9}{2}}\geq \frac{a+b+c+6}{2}$

but by power mean:
$\sqrt{\frac{a^{2}+1}{2}}\geq \frac{a+1}{2}$
$\sqrt{\frac{b^{2}+4}{2}}\geq \frac{b+2}{2}$
$\sqrt{\frac{c^{2}+9}{2}}\geq \frac{c+3}{2}$

adding them up, we have our ineq.

thank you pohoatza
i guess you used
http://www.artofproblemsolving.com/Wiki/index.php/RMS-AM-GM-HM
RMS-AM inequality
thank you
it was pretty lame of me to ask this inequality but i could only get
close to it and not solve it entirely becuase i forgot the mean inequality
thank you again


Prove that , for all $ a, b, c\in\mathbb{R}^{\plus{}}$ such that $ a \plus{} b \plus{} c \equal{} 3$, we have:
$ \sqrt {\left( \frac {a^2 \plus{} 2c^2}{b \plus{} c} \plus{} \frac {b^2 \plus{} 2a^2}{c \plus{} a} \plus{} \frac {c^2 \plus{} 2b^2}{a \plus{} b} \right) \left( \frac {a}{(b \plus{} c)^2} \plus{} \frac {b}{(c \plus{} a)^2} \plus{} \frac {c}{(a \plus{} b)^2} \right)}\ge\frac {3\sqrt {3}}{2\sqrt {2}}$

 :D
Sorry if it's too big......  :maybe:

[quote="Potla"]Prove that , for all $ a, b, c\in\mathbb{R}^{ \plus{} }$ such that $ a \plus{} b \plus{} c \equal{} 3$, we have:
$ \sqrt {\left( \frac {a^2 \plus{} 2c^2}{b \plus{} c} \plus{} \frac {b^2 \plus{} 2a^2}{c \plus{} a} \plus{} \frac {c^2 \plus{} 2b^2}{a \plus{} b} \right) \left( \frac {a}{(b \plus{} c)^2} \plus{} \frac {b}{(c \plus{} a)^2} \plus{} \frac {c}{(a \plus{} b)^2} \right)}\ge\frac {3\sqrt {3}}{2\sqrt {2}}$

 :D
Sorry if it's too big......  :maybe:[/quote]
Nice inequality.Thanks,potla :lol:
we  have:
$ \sum\ {\frac{a^2}{b\plus{}c}} \ge \frac{a\plus{}b\plus{}c}{2} \equal{}\frac{3}{2}$
$ 2\sum\ {\frac{c^2}{b\plus{}c}} \ge a\plus{}b\plus{}c \equal{}3$
$ \sum\ \frac{a}{(b\plus{}c)^2} \ge \frac{9}{4(a\plus{}b\plus{}c)} \equal{}\frac{3}{4}$
Then,we have:Q.E.D :)

[quote="Potla"]Prove that , for all $ a, b, c\in\mathbb{R}^{ \plus{} }$ such that $ a \plus{} b \plus{} c \equal{} 3$, we have:
$ \sqrt {\left( \frac {a^2 \plus{} 2c^2}{b \plus{} c} \plus{} \frac {b^2 \plus{} 2a^2}{c \plus{} a} \plus{} \frac {c^2 \plus{} 2b^2}{a \plus{} b} \right) \left( \frac {a}{(b \plus{} c)^2} \plus{} \frac {b}{(c \plus{} a)^2} \plus{} \frac {c}{(a \plus{} b)^2} \right)}\ge\frac {3\sqrt {3}}{2\sqrt {2}}$

 :D
Sorry if it's too big......  :maybe:[/quote]
Prove this inequality with condition $ a^2\plus{}b^2\plus{}c^2\equal{}3$,Potla  :) .
It than stronger inequality of you.
I hope you will like it,my friend :lol:

[quote="quykhtn-qa1"][quote="Potla"]Prove that , for all $ a, b, c\in\mathbb{R}^{ \plus{} }$ such that $ a \plus{} b \plus{} c \equal{} 3$, we have:
$ \sqrt {\left( \frac {a^2 \plus{} 2c^2}{b \plus{} c} \plus{} \frac {b^2 \plus{} 2a^2}{c \plus{} a} \plus{} \frac {c^2 \plus{} 2b^2}{a \plus{} b} \right) \left( \frac {a}{(b \plus{} c)^2} \plus{} \frac {b}{(c \plus{} a)^2} \plus{} \frac {c}{(a \plus{} b)^2} \right)}\ge\frac {3\sqrt {3}}{2\sqrt {2}}$

 :D
Sorry if it's too big......  :maybe:[/quote]
Prove this inequality with condition $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$,Potla  :) .
It than stronger inequality of you.
I hope you will like it,my friend :lol:[/quote]
Potla, do you like this inequality? :maybe: .I hope you will like it

[quote="quykhtn-qa1"][quote="FelixD"]I have a very nice proof for $ a^{10} \plus{} b^{10} \plus{} c^{10} \equal{} 3$  :)[/quote]
Please post solution[/quote]
The inequality is homogenous, so the condition doesn't matter..

[quote="Potla"][quote="nguoivn"][quote="quykhtn-qa1"]Prove this inequality with condition $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$,Potla  :) .
It than stronger inequality of you.
I hope you will like it,my friend :lol:[/quote]
[quote="FelixD"]I have a very nice proof for $ a^{10} \plus{} b^{10} \plus{} c^{10} \equal{} 3$  :)[/quote]
It's obviously trues with all $ a, b, c > 0$
[quote="Mathias_DK"]The inequality is homogenous, so the condition doesn't matter..[/quote]
Yes, I agree with you. That's funny  :P[/quote]

That's why I made this inequality! :roll:[/quote]
I am impressed by your inequality! :lol:

[quote="enndb0x"]hmm ...What are homogenous inequalities   :blush:[/quote]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=887826#887826]See here[/url]. :)
An inequality is called homogenous if it remains unchanged after switching the variables from $ a, b, c\cdots$ to $ ka, kb, kc\cdots$ $ (k>0)$ so that we can assume $ a\plus{}b\plus{}c\equal{}1$ or $ abc\equal{}1$ . The concept is just that if I make the constant $ k>0$ arbitarily big or arbitarily small then the inequality does not change, leading to the conclusions.
 :blush:


quadratics
inequalities
number theory unsolved
number theory

hi!!

find all pairs of integers $ (a,b)$;$ (c,d)$ such that:

$ a\plus{}b\equal{}c.d$

$ c\plus{}d\equal{}a.b$

thanx.

Consider the quadratic equation with roots p,q. $ Then (x\minus{}p)(x\minus{}q) \equal{} x^2 \minus{} (p\plus{}q)x \plus{} pq$

or use inequality as we know if $ x,y \ge 2$ then $ x\plus{}y \le xy$
$ (c\plus{}d)(a\plus{}b)\equal{}abcd$
for example we can suppose that WLOG $ a\plus{}b \le ab$ which implies $ c\plus{}d \ge cd$


limit
trigonometry
calculus
calculus computations

Find:
\[ \begin{array}{l}
 a)\mathop {\lim }\limits_{x \to 0} \frac{{1 \minus{} \cos 2x.\cos 3x.\cos 5x}}{{x^2 }} \\
 b)\mathop {\lim }\limits_{x \to 0} \frac{{1 \minus{} \cos 4x.\cos 3x.\cos 5x}}{{\sin ^2 x}} \\
 c)\mathop {\lim }\limits_{x \to 0} \frac{{1 \minus{} \sqrt {1 \minus{} x^2 } \cos x.\cos 3x}}{{x^2 }} \\
 d)\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[3]{{8 \plus{} 6x}} \minus{} \sqrt {4 \plus{} x^2 } }}{{x^2 }} \\
 \end{array}
\]

Homework?
Well, in that case you will first show what you have done so far.

If you have already covered Taylor's Series, then replace cos(x) with $ (1 \minus{} \frac {x^2}{2})$ and cos(5x) with $ (1 \minus{} \frac {(5x)^2}{2})$ and so on.

For the third problem you should try to undo rationalization first (i.e multiply the top and the bottom by $ (1 \plus{} \sqrt {1 \minus{} x^2 } \cos x.\cos 3x)$ and then use the first two terms in the expansion of cosine.

I would approach the third problem by using the binomial series.

${ \sqrt[3]{8+6x}=2(1+\frac{3x}4)^{\frac13}= 2(1+\frac13\cdot\frac{3x}4+\frac{\frac13\cdot\frac{-2}3}2 (\frac{3x}4)^2+O(x^3}))$

$ =2+\frac{x}2-\frac{x^2}{16}+O(x^3)$

and

$ \sqrt{4+x^2}=2(1+\frac{x^2}{4})^{\frac12}= 2(1+\frac12\cdot\frac{x^2}4+O(x^4))$

$ =2+\frac{x^2}4+O(x^4)$

Subtract, discover that the $ \frac{x}2$ term dominates, divide by $ x^2,$ see that the limit doesn't exist, wonder if the problem is stated as intended.


If k, n are natural non-zero numbers, n>2, prove that the equation
x^n - y^n = 2^k , has no solutions if x,y are natural non-zero numbers.

"Fie k i n numere naturale nenule, n > 2. S se demonstreze c ecuaia:
			x^n - y^n = 2^k
	nu are soluii n mulimea numerelor naturale nenule"

Suppose that x,y,k,n satisfiy the statements of the problem.

Then, x,y have the same parity. If they are even, then, we may divides the equation by 2 n until they become both odd or the rhs becomes 1. In the latter, the equation is x n  - y n  = 1, which is impossible since x,y are positive integers so that their n-th power differ by at least n (from the binomial expansion).
Thus, wlog we may assume that x,y are odd and we still note k the power of 2 in the rhs).

Let p be any prime divisor of n, and let a = x n/p  and b = y n/p . Thus a p  - b p  = 2 k = (a-b)(a p-1  + a p-2 b +...+ b p-1 )
Thus a-b = 2 t  for some positive integer t < k.
Then :
2 k = a p  - b p
= (b+2 t ) p  - b p
= pb  p-1 2 t mod[2 t+1 ]
Thus p is even, that is p = 2. It follows that n = 2 q for some q  \geq  2 (since n > 2)

Let X = x n/2  and Y = y n/2 .
Then X, Y are odd and X 2  - Y 2 2 = 2 k , from which we deduce that X - Y and X+Y both are perfect power of 2, and that X = 2 b + 2 a  and Y = 2 b  - 2 a  for some nonegative integers a < b. Since they are odd, we have a = 0, that leads to X = Y + 2, that is x n/2  - y n/2  = 2.
But, we know that n/4 is a positive integer, so let X' = x n/4  and Y' = y n/4 . Then (X') 2  - (Y') 2  = 2 where X' and Y' are positive inetegers. Then, X' + Y' = 2 and X' - Y' = 1, from which we deduce that X' = 3/2, a contradiction.

Pierre.

Could someone explain me why (b+2^t )^p - b^p = pb^(p-1)2^t mod[2 ^(t+1) ] " ??

It follows from binomial expansion and Newton formula.


linear algebra
matrix
abstract algebra
superior algebra
superior algebra solved

Prove that we have a exact sequence ( groups): $ 1\to Z_2\to SU(2)\to SO(3)\to 1$, where $ SU(2)$ is a unita matrix group order 2, $ SO(3)$ is special orthogonal group.
Help me!

work with the field $ \mathbb{H}$ of quaternions: $ \mathbb{R}^3 \cong \langle i,j,k \rangle$ and $ SU(2) \cong S^3 \cong \{z \in \mathbb{H} : \parallel{}z\parallel{} \equal{} 1\}$. then conjugation yields a homomorphism $ SU(2) \to GL_3(\mathbb{R})$, whose image lies in $ SO(3)$. you may check that this yields an epimorphism $ SU(2) \to SO(3)$ with kernel $ \{\pm 1\}$.